Calhoun: The NPS Institutional Archive
Theses and Dissertations
Thesis and Dissertation Collection
1988-09
Parametric analysis of airland combat model
in high resolution.
Lee, Jae Yeong
http://hdl.handle.net/10945/23342
sons
•^'
^
:
NAVAL POSTGRADUATE SCHOOL
Monterey, California
—
I
PARAMETRIC ANALYSIS OF AIRLAND COMBAT
MODEL IN HIGH RESOLUTION
by
Jae Yeong Lee
J
i
1
September 1988
Thesis Advisor
Samuel H. Parry
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HIGH RESOL-
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Lee.
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September 19S8
Terms
i
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necessary
and
identify by block
number)
Lanchester and Helmbold Equations, Utility and Game Theory,
Weiss parameter. Lethality of Firing Theory
continue on reverse if necessary and identify by block number)
high resolution deterministic combat model is analyzed in this thesis. Actual Republic of Korea (ROK) terrain data
is employed in the model.
The goal of the thesis is to anahze key parameters which are routineh used in high resolution
combat models. These parameters are attrition rate coefficients, force size, courses of action, and the Weiss parameter (in the
equation for Helmbold type combat). The model's scenario divides the battlefield mto three regions: indirect fire, minefields,
and direct fu-e. Lethality of Firing Theor>' and Lanchester type differential equations are used to compute unit casualties and
unit speed in a discrete time increment. The models output (unit casualties and survivors, duration of battle, loss exchange
19
Abstract
(
A
termed of Measures of Effectiveness (MOEs), winch are analyzed by Utility Theor\- and Game Theor\ methodolis applied to each battle option to determine how changes to one or more input parameters aflect
the model's output. Additionally, the model operates in an interactive mode using network attribute data. The model can
easily be expanded or modified to satisfy a user's requirements by adding submodels or changing input data.
rate) are
ogies.
Sensitivity analysis
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Combat Model
in
High Resolution
Parametric Analysis of Airland
by
Yeong
Major, Republic Of Korea Army
B.S., Korea Military Academy, 1980
Lee,
Submitted
Jae
in partial fulfillment
of the
requirements for the degree of
MASTER OF SCIENCE
IN
OPERATIONS RESEARCH
from the
NAVAL POSTGRADUATE SCHOOL
September 1988
ABSTRACT
A
high resolution deterministic combat model
Republic of Korea
is
(ROK)
terrain data
to analyze key parameters
These parameters are
Weiss parameter
divides
the
employed
analyzed in this
which are routinely used
(in the
in high resolution
Lethality of Firing Theor\'
three
regions;
indirect
and Lanchester type
pute unit casualties and unit speed
in a discrete
and survivors, duration of
fire,
exchange
Measures of EiTectiveness (MOEs), which are analyzed by
Theor>' methodologies.
how
the
Sensitivity analysis
is
and
direct
fire.
equations are used to com-
time increment.
loss
combat models.
of action, and the
minefields,
differential
battle,
Actual
The model's scenario
equation for Helmbold type combat).
into
thesis.
model. The goal of the thesis
in the
attrition rate coefTicients, force size, courses
battlefield
(unit casualties
is
is
Utility
The model's output
rate) are
termed of
Theory and
Game
applied to each battle option to determine
changes to one or more input parameters affect the models output. Additionally,
model operates
easily be
in
an interactive mode using network attribute data. The model can
expanded or modified
to satisfy a user's requirements
changing input data.
in
by adding submodels or
TABLE OF CONTENTS
INTRODUCTION
I.
A.
B.
C.
1
BACKGROUND
PURPOSE AND GOAL
METHODOLOGY
1
1
2
MODEL DEVELOPMENT
IL
A.
B.
3
SCENARIO
3
L
General
3
2.
Assumptions
3
ALGORITHM
7
1.
Basic model
2.
Units casualties
3.
4.
7
10
a.
Indirect fire
10
b.
Minefield packages
13
c.
Direct
13
fire
Unit speed
19
a.
Indirect
b.
Direct
fire
and Minefields
19
20
fire
Advanced model
21
MODEL OUTPUT ANALYSIS
HI.
A.
B.
C.
D.
E.
25
OVERVIEW
MEASURES OF EFFECTIVENESS (MOES)
COMPARISON BETWEEN BATTLE AND BATTLE2
UTILITY THEORY
1
25
28
31
35
1.
General
35
2.
Application
35
a.
Linear Utility
35
b.
Squared
36
c.
Square root
Utility
37
Utility
GAME THEORY
38
IV
I.^.
1.
General
38
2.
Application
39
a.
b.
c.
F.
casualties)
39
MOE-2 (Blue survivors)
MOE-3 and MOE-4
41
USEOFUTILITYVS GAME THEORY FOR THE ANALYST
SENSITIVITY ANALYSIS
IV.
41
43
44
A.
GENER.AL
44
B.
FORCE
44
C.
ATTRITION RATE COEFFICIENTS
D.
E.
F.
V.
MOE-1 (Red
SIZE
1.
Sensitivity for attrition rate coelTicients
2.
ANalysis OfVAriance
(ANOVA)
46
49
WEISS PAR^^METER
54
COUNTERMEASURES
MODEL ENRICHMENT
59
SUMMARY AND FUTURE DIRECTIONS
A.
SUMMARY
B.
46
FUTURE DIRECTIONS
64
68
68
69
APPENDIX
A.
MODELS INPUT DATA
APPENDIX
B.
BASIC
APPENDIX
C.
EXEC FOR BASIC MODEL
86
APPENDIX
D.
ADVANCED MODEL
87
APPENDIX
E.
EXEC FOR ADVANCED MODEL
LIST
MODEL
OF REFERENCES
BIBLIOGR^^PHY
70
71
114
115
116
INITIAL DISTRIBUTION LIST
117
VI
LIST OF TABLES
Table
1.
Table
2.
Table
3.
Table
4.
Table
5.
Table
Table
Table
FORCE ELEMENTS
THE MATRIX OF RED AND BLUE OPTIONS
THE RELATIVE LETHAL AREAS FOR A 155MM HOWITZER
COMPARISON OF MOES BETWEEN BATTLE! AND BATTLE2
SENSITIVITIES OF MOES FOR ATTRITION RATE COEFFIINITIAL
3
9
.
.
.
.
12
31
48
7.
CIENTS
RESULTS OF ONE-WAY ANOVA TEST WHEN THE MAXIMUM
ATTRITION R.ATE COEFFICIENTS ARE CHANGED
RESULTS OF ONE-WAY ANOVA TEST WHEN THE WEISS PA-
57
8.
RAMETER BLUE ENGAGING RED ARE CHANGED
PAYOFF MATRIX FOR RED AND BLUE COURSE OF ACTIONS
(MOE-I MOE-2)
61
6.
vu
54
LIST
OF FIGURES
Figure
1.
Map
Figure
2.
Network representation of the
Figure
3.
Flow chart of computer program
Figure
4.
The output of
Figure
5.
Relation between the defender's casualty rate and the attacker defender
of area used
in the
the Basic
4
scenario
battle area
5
for the basic
model
8
model (BATTLE2 case)
10
14
force ratio
Figure
6.
Relation between the attacker's casualty rate and the attacker defender
force ratio
15
17
Figure
7.
Relationship between range and attrition coefficient
Figure
8.
Battle case
Fiszure
9.
Speed of attacking force versus
when
a Blue
threatcd from two directions
is
IS
20
their attrition coefficient
Figure
Flow chart of the advanced model
22
Figure
The output of the advanced model
24
F'igurc
Red and Blue
26
Figure
The Red
Figure
Distance between Red and Blue forces o^er time
28
Figure
Speed of Red force on each avenue over time
29
Figure
The
Figure
Ratio of
Figure
Figure
MOEs
MOEs
Figure 20
Three different
Figure 21
\\'eighted utility \alues for each function
Figure 22
Initial
Figure 23
Solution for Value of the
Ficure 24
Red and Blue
force sizes over time
by different types of Blue weapons
casualties
MOE
values for each options of
MOHs
in
B.XTTLEl
27
Red and Blue (B.ATTLE2)
30
BATTLE2
32
types based on
type
for each options of
Red and Blue (BATTLEl with Avenuc-21)
for each options of
Red and Blue (B.ATTLEl with Avenue-22)
utilit>
41
42
(.\10E-1)
when
the Blue
ambush
point
is
rein-
45
forced
Figure 25. Comparison of
Figure 26. Change of
ficients, Oq
34
38
matrix and reduced matrix for .\10E-1
force sizes over time
.
33
36
distribution functions
Game
.
MOEs when
.MOE
and
46
the Blue forces are reinforced
values based on different
maximum
attrition rate coef-
47
bo
Fieure 27. Ratio of change for
MOE
values based on different
vm
a^
and
bo
47
Figure 28. Fitting the Red casualties
Normal
Figure 30.
MOE
to
50
Distribution
Figure 29. Fitting the Blue survivors
Normal
(MOE-1) of the Advanced model output
(MOE-2) of the Advanced model output
to
Distribution
51
outputs for both battle cases; original battle, and both
Oq
and
bo
increased by 10 percent
52
Figure 31. Change of Basic model output
(MOE-1 and MOE-2) based on Weiss
parameter for Blue engaging Red
55
Figure 32. Change of Basic model output (Red to Blue casualty ratio and force ratio)
based on Weiss parameter for Blue engaging Red
Figure 33. Change of the Basic model output
55
(MOE-1 and MOE-2) when Weiss
parameters for both Red and Blue approach the same value
Figure 34.
Change of Red Blue casuahy and
force ratio in Basic
56
model when Weiss
parameters for both Red and Blue approach the same value
Figure 35. 3-Dimensional view of payoff matrix for
Red
56
casualties
62
Figure 36. 3-Dimensional view of payoff matrix for Blue survivors
62
Figure 37. 3-Dimensional view of payolT matrix for the
Red
to Blue casualty ratio.
Figure 3S. 3-Diniensional view of payoff matrix for the
Red
to Blue force ratio.
Figure 39.
Two
different functions for
Figure 40. Comparison of
RED BLUE
minimum
.
speed
casualty ratio
63
.
63
65
{MOE-3) between
Basic and
Enriched model
66
Figure 41. Comparison of duration of battle
model
{MOE-4) between
Basic and Enriched
66
IX
ACKNOWLEDGEMENTS
For
their significant contributions
lowing people
•
toward
this thesis,
I
would
like to
thank the
fol-
:
Professor Samuel H. Parr\' for his tolerance, professional assistance, and thorough
re\iew.
•
•
U.S.
Army LTC. Bard Mansager
Most
especially,
tience,
my
for his
wife. In-Sook,
and earnest support.
and
encouragement and guidance.
my
daughter, Jin-Ho for their prayers, pa-
INTRODUCTION
I.
BACKGROUND
A.
Since the Korean
significant
War
(1950-1953), the Republic of Korea
amount of money
to defend against the threat
developed,
it
has become necessar\- for the
has invested a
posed by North Korea and to
As
maintain a peaceful atmosphere on the Korean Peninsula.
omy have
(ROK)
ROK
and the econ-
industr>'
Department of Defense to
apply more efficient methods to operate and control the huge defense system.
Army, one major source of defense expenditure has been
simulation models which can be used in
Atlantic Treaty Organization
geographically having
it
many
(NATO)
Therefore,
cisely
and
concept,
when
the
ROK
hoped that
terrain, the
Unlike the North
Korean peninsula
hills,
and
comple.x
ver\'
is
reservoirs.
This means
enemy's avenues of approach when they attack.
develops an airland combat simulation model,
realistically describes the
it is
developing different combat
terrain conditions.
small streams, mountains,
relatively easy to channalize the
is
ROK's
in
For the
Korean geography than other models.
this thesis will
be an useful addition
it
more
Based on
prethis
develop better
in helping to
combat simulation models.
PURPOSE AND GOAL
B.
The purpose of
this thesis
is
to analyze the parameters
As
high resolution airland combat models.
common
scenario
is
which are heavily used
the deterministic
used to approach this goal.
model
In this scenario.
is
of different routing options.
It is
developed, a
Red (enemy
attacks Blue (friendly force) through an arbitran." terrain network giving
in
Red
a
force)
number
assumed that Red can take limited avenues of ap-
proach since the maneuverability of Red
is
restricted to certain arcs (routes)
due to de-
stroyed bridges or a strong probabihty of ambush.
Specifically, the goal of this thesis
Blue
commander based upon
strate several decision
to suggest the best courses of action for the
various types of airland battle situations and to
methods from an analytic viewpoint.
can be easily modified and applied to
terrain data.
is
test
demon-
The model developed here
other battle situations
if
one has more precise
METHODOLOGY
C.
To
(DMZ) was
near the Demilitarized Zone
se-
and a network was developed having several possible paths for Red forces based
lected
upon
ROK
build a scenario, an area in
the road conditions of that actual terrain area.
Generally.
Red attempts
to attack
Blue's fortified positions through the shortest distance path or shortest time path.
There are several methodologies available to find these paths.
rate methodologies previously explored
1]
and McLaughlin [Ref
from a high resolution
tributes
terrain data base
Regard each intersection of the roads as
an
arc.
round
battle.
is
was
de-
The determination of network
at-
to
model
terrain
was developed by Choi [Ref
3
:
p. 27].
node and the road between the nodes as
to be used as inputs for the models.
lethality of Firing Theor>'
Lanchester's
Law
Linear
equations are used for direct
also
a
2].
at the
and add the speed, distance, and width to estimate the trafficabihty of each
This information
single
(ALARM)
by the AirLand Rearch Model
Xaval Postgraduate School. The use of network methodology
veloped by Craig [Ref
This effort will incorpo-
fire
is
is
adopted for
In this deterministic model,
artillen." fire in
rate coefficient
In other words,
Reds
speed
This
will
is
a function
to derive the dilTcrent outputs.
If the
the output can be desciibed by an
compared and analyzed.
M
Red
by
M
Chapter
options and Blue has
N
2.
changed
options,
These matrices could be Red casualties, Blue survivors, time
Because these values are
significant factors of the battle, they are defined as the values of
very-
Measures Of Efiec-
(MOEs). After generating the required matrices of MOEs, they were analyzed
using the techniques of Utility Theor>' and
analysis
in
matrix of each specific value which can be
of the battle, or the ratio of Red and Blue casualties.
liveness
decrease as they
side, the battle situations are
force has
N
of Blue's attrition
will
be explained in more detail
Based on the several options available to each
It is
their attrition rate coefficient
on Red. since the niauvcrability of the Red force
get closer to the Blue positions.
Hembold
dependency and ambush attack.
assumed that the speed of the Red force depends on
caused by the Blue force.
the beginning of the
used for casualties in the minefields.
as well as range
arc.
was
Game
also conducted for selected parameters.
Theor>'.
Additionally, sensitivity
MODEL DEVELOPMENT
II.
SCENARIO
A.
General
1.
The Red regiment has
three infantr\' battalions
and one reserve company and
the Blue battalion has three infantr>' companies and one reconnaissance platoon.
commander can employ
its
reconnaissance platoon for a special mission.
the reconnaissance platoon of Blue will
approach and Blue
Figure
2).
The
Only
nodes.
will
battle will terminate if
Red
unit
The
is
1.
:
1500
150
1650
3
1
companies
recon. pltn.
510
40
550
:
:
battle area in this scenario
This area
is
is
taken from one of the
relatively flat except for
ROK Army
one large mountain.
located in the west central part of the battle area.
One
Cheon. which flows from the northwest to the southeast, poses no
to the
Red
force
when
this terrain area is
arcs
and nodes
2.
Initial
infantr\-man.
assigned sectors.
San, which
Table
of the model.
Blue force
reserve co.:
=
in
force
Total
•
shown
in the initial runs
FORCE ELEMENTS
3 battalions
1
the most important avenue of
one of the Red forces reaches either of these two
were considered
infantn.- forces
INITIAL
1.
In the model,
defend from two well fortified positions (Node-27, Node-28 in
force elements of both sides are
Table
ambush Red on
Blue
is
attacking with either
shown
shown
in Figure
in Figure
1
mounted
divisions'
Geumhag
small river,
Daegyo
significant
problem
or dismounted units.
and a representation of the network
The map of
in
terms of
2.
Assumptions
The assumptions and
description of the scenario are as follows:
1.
There are three major avenues of approach for the Red force.
2.
The Red
force on the second avenue (center) will be divided at node- 14, with two
equal forces and attack to nodes 27 and 28, respectively (This assumption could
^
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ft -.^^-^'5
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Fiavi^D^
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Figure
1.
Map
•^
wi V<J-
'?i\fiCN:i 5;
CokW
0«n g
•--fe'-.-- '-TV'
~^,
of area used in the scenario
a«KDi
STTTN
N
AVENUE-3
30
49
72
(/3.
JS)
16]
C24)
2a>-(«)
25)
23
20)
42
Figure
2.
Network representation
of
tiie
battle area
—
Seleaed Arcs
—
Unselected Arcs
be changed so that the Red force on the second avenue will attack only one of two
Blue positions as described in Chapter 3). Therefore, denote Avenue-21 used by
Red attacking to node-28 and Avenue-22 used by Red attacking to node-27.
3.
Red
force initiates
maneuver from
their
boundarv
at the
same time (from nodes
1.2,3).
4.
Red force has four options to allocate their units (battalions) for each avenue of
approach, where attacker (Red) actions are treated as states of nature, as follows:
Option
1
3.
A.
5.
Fort-l(node 27)
Co.
2 Co.
1.
1
2.
1
1
Bn.
Bn.
2Bn.
OBn.
Fort-2(node 28)
2 Co.
Co.
1
Blue force has four options for how to allocate the three scatterable mine packages
they currently possess. Those options are as follows:
Optioii
1.
Avenue-
Avenue-2
Pk2.
Pke.
2 Pke.
1
Pke.
3 Pke.
1
J.
OPks.
OPkc.
4.
OPkiZ.
2.
7.
1
Blue force has two options to allocate their units (companies) for each fortified
defense positions as follows:
Option
6.
Bn.
2 Bn.
1 Bn.
3 Bn.
Bn.
OBn.
OBn.
OBn.
2.
A venue-
Avenue-2
Avenue-
1.
A venue-3
Pke.
Pke.
2 Pke.
1
1
1
OPkc.
units (battalions) affect the speed of attacking Red forces on
This implies that Red moves with reduced speed when there are
multiple units on one of the avenues of approach. The Red forces also become
more dense in a given area when multiple units are on one avenue.
The number of Red
each avenue.
8.
Whenever the Red force reaches a minefield, it takes more time to bypass than to
break through. Therefore, the Red force will choose to break through.
9.
The only artillery weapon system that supports the Blue force
howitzer (towed).
is
the 155 millimeter
10.
Red
11.
Blue has two types of minefield packages: preinstalled and scatterable.
12.
The maximum range
13.
indirect fire
weapons
are not represented.
for infantry direct-fire
There are no Close Air Support (CAS)
this scenario.
weapons
aircraft or
is 1.1
km (M60 machine
gun).
chemical weapons employed in
As
Red has
described above.
four options and Blue has eight options.
Therefore, there
are 32 battle cases in this scenario.
ALGORITHM
B.
Basic model
1.
The
rule for representing the battle
clock will begin
nodes
and
2,
1,
when
Red
the
is
force attacks
based on discrete time
from
their original locations,
network representation of Figure
3 in the
The
steps.
2.
The time
battle
which are
step interval
is
one minute.
The
basic
model
is
only one of the possible battle cases in this scenario, that
is,
one option for Red and one option for Blue which represent one element out of the 4
by
These options may be entered
8 matrix.
interactively.
The
basic
model
steps are
described below:
•
Input Characteristics of the network terrain model which include source node (tail
node), sink node (head node), arc number, distance and speed on each arc. location
of minefield, and amount of time for indirect fire. An example of input format is
included in Appendix A.
•
Output Characteristics of battle results which include casualties and survivors for
both sides, the length of battle, proportion of casualties based on weapon types
(indirect fire, minefield, direct fire), avenues of approach, and positions (nodes) of
:
:
the the Blue forces.
:
Read
Step 2
:
Set the battle clock to the next time step
Step 3
:
Detect the
Step 4
:
Compute
Step 5
:
Step
I
the input data
:
initialize the forces, time, location etc..
type such as indirect
the casualties
and survivors
fire,
minefield and direct
for each
fire.
weapon type based on Step
3.
Determine the speed of the attacking force and compute the distance it moved
during that time step. It is assumed that the speed of Red forces is afiected
by indirect
Step 6
weapon
and
fire,
minefields,
and
direct
fire.
Consider battle termination conditions and decide whether the battle will
continue or not. If the battle does not terminate, go to Step 2. If the battle
terminates, print the output.
All these steps occur independently
The flowchart of
the basic
and simultaneously on each avenue.
model
nitions apply:
IF = Indirect Fire battle
MF
=
Minefield acquisition
is
given in Figure
3,
and the following
defi-
read data
set initial conditions
±*
time = time + dt
detectlF. MF. DF
yes
call
IDFIRE
yes
callMFIELD
<.
yes
callDRFIRE
adjust speed based on S =
F( IF,
MF, DF
update location of RED units
battle terminate
!
end
Figure
3.
Flo\> chart of
computer program
for the basic
model
)
DF
=
Direct Fire battle
IDFIRE =
M FIELD
subroutine for computing the casualties by indirect
= subroutine
DRFIRE =
The
Table
for
computing the casualties by minefield.
subroutine for computing the casualties by direct
table for
fire.
Red and Blue options can be represented
fire.
as a 4 by 8 matrix (see
2).
Table
2.
THE MATRIX OF RED AND BLUE OPTIONS
Red's unit allocation on each avenue
Options
1
Blues
Scatterable
mine
unit al-
cation on
each ave-
location
(1 2)
(»
nue
mine
unit al-
allo-
1
(2 1)
in
Table
On
o„
2
1
0:,
0:2
Or.
0:.
1
2
Om
0,2
0,y
0,.
0.:
0.2
0.,
0^
1
0.,
Or.
0,y
O.M
1
Ch,
0,2
0,3
0..
2
0-,
0-2
0-y
0-a
0,2
0.3
o«
1
Ov.
2.
the
Red
force has four options based
and the Blue force has eight options which represent
all
Based on the Blue commander's
geographical situations, the O,, (option-2 for
tactical
on
its
unit allo-
possible combinations
of unit allocation with scatterable mine allocation (see assumptions
ously discussed).
3
0.2
3
As presented
2
On
1
nue
1
1
02
cation on
each ave-
location
1
1
3
Scatterable
Blue's
cation,
1
allo-
02
I
1
4, 5.
and 6 previ-
viewpoint and considering the
Red and option-2
for Blue) battle case
could be expected to occur in a real battle situation with higher probability than the
other cases.
Thus, O^j case was executed as an example output of the basic model and
the results are
shown
in Figure 4.
In Figure 4, the numeric values corresponding to the
the
number of troop
units or minefield packages
avenue of approach or position.
interactively.
(i.e.,
are
Bn, Co, Pkg) each allocates to each
These values remind the user of what he has entered
In fact, this output shows precisely
the results are for each Blue position.
vided in Appendix B.
Red and Blue options
A
how
the battle terminated and
computer program of the basic model
is
what
pro-
RED OPTION ==>
2
1
BLUE OPTION
'>
SCATERABLE MINEFIELD
2
1
(LOCATED ON ARCS) ===> 30 34 42
'^
=>
FORCE ALLOCATION
1
2
=>
RED FORCE IN AVE,-'^22 TOOK BLUE POSITION7i2
225 MINUTES)
(BATTLE TIME
=>
<«
RED CASUALTIES BY EACH TYPE OF BLUE FORCE
INDIRECT-FIRE
AVENUE//
AVENUE// 21
AVENUE/i22
AVENUE/i3
MINE-FIELD
0.
<«
<«
2.
588. 6
19.4
SURVIVORS
55.7
134. 3
173. 3
TOTAL CASUALTIES AND SURVIVORS
751. 7
550.
242.4
The output
>»
CASUALTIES
1650.0
of the Basic model
>»
SURVIVORS
898.3
307.6
(BATTLE2
case)
Units casualties
a.
Indirect fire
Casualties caused by indirect
lethalty
116. 3
186. 7
RED
BLUE
4.
0.
163.
406. 2
CASUALTIES
INITIAL FORCE
>»
47. 1
47.
22. 2
BLUE FORCE CASUALTIES FOR EACH POSITION
POSITION-1
POSITION-2
!
DIRECT-FIRE
0.0
0.
23.4
23.4
46.8
Figure
BATTLE END
:
and weapon
firing rates.
fire
weapons
Target density
10
is
are
computed using
single
round
a major factor in computing casualties
and
decreases as
Red
allocates
more than one
unit (battalion) to one avenue.
words, the number of Red infantrymen in a given area
the targets are represented as
is
could consist of
tillery'
such consideration
computed
many
will increase.
A
air.
In other
Also, in this model,
two dimensional, which means that an
represented as detonating on the ground, not in the
Ai
The dispersion between troops
defined as the distance between adjacent soldiers.
is
artiller}' projectile
target under attack by ar-
personnel performing various functions of combat, and
taken into acount in the computation of lethality. Lethal areas are
is
for single conditions such as
The
prone personnel.
definition of lethal area
given by equation (2.1).
is
+OC /•+00
H) dxdy
J— oo
J
where p(x,y|H)
is
(2.1)
— oo
the conditional probability that a hit
on a
target element located at the
point (x.y) on the ground results in an incapacitation for given height of burst
in this
model) [Ref 4
:
p.
H
(H =
14-9].
Using the Gaussian Lethality Function p(x,ylH) =
e"*"^*^^''^""
forming into polar coordinates, equation (2.1) becomes equation
,
and trans-
(2.2).
^+oo ^+oo
-(x* +
Al =
— oo
v-)/2a'
dxdy
»/— oo
^2- ^+oo
'^
e
rdrdd
= 2na
Red infantr\man
Thus, the probabihty of
kill is
is
assumed
to be killed
The
=
P{hit)
Prop(lethal)
lethal
=
if
he
I
\J
2
x +y
i
(2.2)
is
is
within the lethal area.
the probability of hitting the lethal area multiplied by the
proportion of lethal area of the entire target area, given
P{kill)
2
r
I
where
J
Jo
A
'^^
in
equation
(2.3).
X Prop{leihal)
calculated
(2.3)
by dividing the
total area
of the Red force by the total
area per unit time from equation (2.4), and P(hit)
dimensional target by equation
(2.5).
11
is
computed
for a two-
R,-xA,
=
Propileihaf)
(2.4)
Row^^ X Col^^ X RF^
where
Rf
rate of fire per unit time
:
Ai_
lethal area
:
Row,.
Col,^
RF,
:
:
row spacing of each
unit (infantr\Tnen)
column spacing of each
current
:
of each round
Red
unit (infantr>'men)
force size
2
2
P{hii)
=
exp
+
[[a^
a'^)
{a^
+
2m
a;)]
1/2
''
-i«
)^
2
,
2
+
(2.5)
(
a
-\-
o.
where
a
lethahty constant for each round type (where
:
/J,
:
^1,
:
(7,
:
a^
:
mean
distance of deflection error
mean
distance of range error
Table
by
indirect fire per unit time (/f"„,) are
Weapon and
Caliber
Ml 09 A
155-mm
The source
in
Table
computed by
(2.5).
P{kllf)xRF,
THE RELATl\
3.
2na-)
standard deviation of range error
equation (2.6) based on equations (2.3) through
=
=
standard deviation of dellcction error
Finally, the casualties
IF,,,
Total Ai
(2.6)
E
LETHAL AREAS FOR A 155MM HOWITZER
Area
for Personnel
:
m^)
M107
M4S3
M549
Standing
Prone
Foxhole
103 42.5
70.8 22.3
6.67 2.0
981 580
508; 308
14.2'9.5
100 44.8
82.2 26.0
1.500.83
for 155 milimeter howitzer lethaity data
3 for a
(Open Woods
Projectile
range of 14km.
is
[Ref 4
For example, the value of a
12
:
p. 43-9]
single
which
round
is
shown
lethal area,
70.8m-,
is
used when the Red force
is
prone position and
in the
subjected to Blue in-
is
direct fire attack in the open.
Minefield packages.
b.
It
assumed
is
that the Blue force has five preinstalled minefield packages;
one on Avenue- 1, three on Avenue-2, and one on Avenue-3.
packages are also employed by the Blue commander.
The area
"Lanchester Linear Law" are generally appHed to indirect
Scatterable minefield
fire
fire
assumptions of the
patterns of artillery.
Minefields also exhibit these same patterns, and therefore the Lanchester Linear
Law
equations are appropriate for assuming minefield attrition. The equation for preinstalled
minefield casualties
PMF,,,
=
given in equation (2.7).
X a;,, X .V?,,„, X Tp
P,,,f
The equation
is
(2.7)
for scatterable minefield casualties
is
shown
in
equation
(2.8).
where
PMF,,;
:
SMF,,,
:
number of casualties by
preinstalled minefield
number of casualties by
scatterable minefield
:
attrition coefficient
S„,,
:
attrition coefTicient of scatterable minefield
A'P„,„
A'S„„,
F^
F,
:
:
A'„^
:
:
number of mines
in
one preinstalled minefield
number of mines
in
one scatterable minefield
time to traverse a preinstalled minefield
time to traverse a scatterable minefield
:
number of infantrymen
The time
moves on an
the
of preinstalled minefield
P„,,
traversing the minefield
to traverse each minefield
is
a constant value.
arc that has a minefield, Red's traversing time
amount of time needed
decreases based
upon
to traverse the minefield.
When
on that arc
is
the
Red
force
increased by
However, Red's speed on that arc
the time to traverse the minefield, which will be discussed in detail
in section 3.
c.
Direct fire
Based on consideration of historical combat data, Helmbold has proposed
a modification of Lanchester's equation for
"modern warfare"
13
to account for inefficien-
UJ
/
*"*
^r
l^s
0.03
•"-1-
K
2
3
UJ
—
/
X DEFENDS AND Y ATTACKS
y/^
y^
^r\.
'
'
^-'
y
^
N.
0.02
bd
^1
<
3
S
C
Ll.
UJ
Q
wTo
^^i:^
(THE
NAL
1
1
\^. V^
\
^
(LOGARITHMIC LAW)
^^'<^^
y'^^y'''^/
o
K X
o _
< £
O
q:
u.
^\ZJi>^
^^
-V
X X/
/ //
// y^
1/
1
u.
V
'
If
\
\
\
2.0
3.0
4.0
\
1.0
J
FORCE RATIO. A/0
1
Figure
Relation bet>^een the defender's casualty rate and the attacker/defender
5.
force ratio.
level and
cies
to
D
[NOTE:
denotes that of
of scale for the larger force
modify the
when
fire
rates of
^=
and
A
denotes the attacker
"\'
Helmbold considered the
power function.
a
is
s
force
defender].
relative force attrition (or fire efiectiveness) capability
elTectiveness modification factor
X
tlie
figure.
force sizes are grossly unequal.
factor depending on only the force ratio.
the
above
In the
by
His basic idea
is
a multiplicative
special case in
which
In this case, the casualty
are
)'-">•
with
x(0)
=
.ro
(2.9)
7-=-h{i){^)'-'^x
with
y(0)
=>o
(2.10)
-a(0(
J
dy_
di
where w
is
called the ^yciss parameter,
and equations
equations for Helmbold type combat.
14
(2.9)
and
(2.10) are called the
UJ
.1.
>-
''"
n4
W
(LOGARITHMIC LAW)
•
0.03
VOv^^^ W
Q.
y
\ ^Sw W
a:
5
\
^
0.02
-J
——
1/2
•
^
^X^ILINEAR
\
LESS
LAW)
•
USE
/of
W » 1
^^sfSQUARE LAW)
>
/
'
/ DErENDER'S
liJ
X
EFFICIENT
-
-.-„.___^
^"^^^^
N.
"
i
1/4
•
—
0.01
"
^.^,
FIREPOWER
/
^^---^
o
X ATTACKS AND Y DEFENDS
< ^
1
1
t
\
1.0
2.0
3.0
4 .0
A/0
FORCE RATIO,
Figure
Relation bet>\een the attacker's casualty rate and the attacker/defender
6.
force ratio.
These equations are particularly significant because a simple generalization
of them gives a
much
better
fit
m
to casualty rate curves used
temporary large scale combat models than does Lanchester's
warfare.
As
for the case of constant attrition rate coefTicients
several important con-
classic
(i.e.,
model of modern
"a(t)"
and
"b(t)"
from
equations (2.9) and (2.10) become "a" and "b"), the equations for Helmbold type combat
yield the
Law when w =
Square
Law when w =
1,
the Linear
Law when w = 12, and
the Logarithmic
0.
Figure 5 shows the relation between the defender's casualty rate (expressed
as a fraction of his current force level, x(t))
model dx/dt =
— a(a:/j')'""
y with
X
and the
defending.
attacker, defender force ratio for the
Figure 5
is
X's fractional casualties
plotted per unit time versus the force ratio y;x (denoted in the figure as
in
which
Y
attacks and
X's casuahy rate
is
X
defends.
In Figure
proportional only to the
15
5,
w=
1
AD) for the case
corresponds to the case
number of enemy
firers,
and
in
which
(in the
sym-
metric case in which Y's casualty rate has the same functional form) consequently the
corresponding attrition model
serve that in this case
w=
(i.e..
we
for force ratios
fire
w=
see that
AD
vi\
=
1)
corresponds to a more
y x
>
1
than does
effectiveness modification factor for
w= Wj
when yx >
1
[Ref
5
:
We
equation.
ob-
X's fractional casualties per unit time are directly
AD when V attacks and X defends.
proportional to the force ratio
(2.9),
Law
given by Lanchester's Square
is
p. 299].
w-
w=vv,
of the attacker's firepower
efficient use
Wj v.hen
1
>
The following
is
>
Wi
(jt/j)'""!
(i.e.,
Refering to equation
)
Wj, since
is
the attacker's
greater than that for
a numerical example for Figure
5.
<EXAMPLE>
Let a =
0.
010
b =
0.
020
X = 10 (defender force size)
y = 30 (attacker force size)
=3.0
hence, attacker to defender force ratio A/D = y/x
Using equation (2.9), fractional casualties per unit time
for X (the defender) are as follows:
w=l
w=l/4
w=l/2
w=0
dx/dt
0.
300
0.
173
0.
(l/x)(dx/dt)
0.030
0.
0173
0.0132
Figure 6 shows the same type of plot
defender.
when X
is
0.
100
0.
010
Note, the smaller the value
is
made when
w
(0
<w<
less efTicient
attacker to defender force ratio
However, the reverse phenomenon occurs when
The strength of
incorporates either area
1),
fire
the equations for
or aimed
fire
16
AD
is
less
Y
the attacker and
In this case, the casualty-rate curve corresponding to the Square
hyperbola.
firepower
132
(AD)
than
is
is
Law
the
is
a
use of defender's
greater than
1.0.
1.0.
Helmbold type combat
is
that
with the Weiss parameter, "w".
it
easily
These
Figure
Relationship bet>>een range and attrition coefficient
7.
equations are based on the perception that limitations
target
in
space, terrain
engagement opportunities prevent a large force from using
deterministic
fortified
firepower.
This
for direct fire
from
its full
model uses these Helmbold type combat equations
Blue positions (node 27 and 2S). and the Lanchester's Mixed
special case of
Helmbold type combat where
ambushes on
arc- 19 of the second
is
utilized for
is
determined by the Blue commander.
forces reach the Blue
w=
ambush
point,
I
for
one
side
and
masking and
w=
Law (which
1
2 for
is
the
the other)
avenue of approach. The ambush point
The ambushes
and Blue survivors
will
terminate
when
the
will retreat to join their
Red
main
forces at node-27 or node-28.
The range dependency
the
ambush
the
Red
point).
and
considered for both cases (at nodes 27 and 28, and
This implies that attrition rate coefficients for each side increase as
force approaches the Blue position.
how
function of range shows
weapon
is
Plotting attrition rate coefficients as a
different values
types) affect battle outcomes.
of ^ (power factor based on different
Figure 7 shows the relationship between range
attrition coefficients.
A common
tactic that
proaching a Blue position
is
is
employed to counter multiple Red forces ap-
discussed next.
17
Each Blue position has two possible ave-
Boundary of
Direct-fire
Blue position
Figure
Battle case >\hen a Blue
8.
nucs of approach: Avenue-
1
If a Blue position
proach within
its
forces
threated from t^o directions
and Avenuc-21
for node-27.
Red
is
is
for nodc-2S.
and Avenue-22 and Avenue-3
threaied simultaneously from both avenues of ap-
direct fire region. Blue allocates firepower
on that avenue of approach.
are given by equation
based on the proportion of
The number of Blue
forces firing
on Avenue-a
(2. 11
RF.
''^=''^^ RF^RF.
The number of Blue
BF^
=
(2.11)
forces firing on Avcnue-b are given by equation (2.12)
RF,
BFq X
RF,
-r
(2.12)
RF,
where
BF,
BFi,
RF^
RF^
:
:
:
:
Blue forces firing into Avenue-a
Blue forces firing into Avenue-b
Red
forces attacking
from Avenue-a
Red
forces attacking
from Avenue-b
18
BFq
Once
Blue forces at the
:
the Blue force allocates
its
moment two Red
firepower,
it
forces engage a Blue position.
continues at a steady rate until battle
Figure 8 depicts a Blue position being attacked by
mination.
Red
units
ter-
on two avenues
of approach.
The following equations used
model represent both
in this deterministic
range dependency and the Weiss parameter.
^ = -a,{\--^r>ifr-y
(2.13)
= -bo{^-T^r'iir)'~'^'^
4al
'max
(2-14)
y
'max
at
-^
where
X
y
Red
:
Blue force
:
Oo
:
bo
:
r,
force size
size
maximum
attrition rate coefficient to
maximum
attrition rate
current range between
:
^max
•
maximum
Red from Blue
coefUciem to Blue from Red
Red
force
and Blue force
range between the Red force and the Blue force
(l.I
Km by
assumption).
l^y
H,
:
:
ca,
:
CO,
:
Equation (2.13)
ties
power
factor,
based on Blue weapon types
power
factor,
based on Red weapon types
measure of
efficiency
which the Blue force engages the Red
force.
measure of
efficiency
which the Red force engages the Blue
force.
is
used for Red casualties while equation (2.14)
during a specific time step.
Red and
3.
Note
that different values of
fx
is
used for Blue casual-
and
w
are used for the
the Blue force.
Unit speed
a.
Indirect fire
When Red
and Minefields
forces
move over
they are delayed for a specified
updated speed
is
the arcs which have indirect
amount of time.
computed by equation
(2.15).
19
If the delay time
is
fire
or minefields,
denoted by DT, the
Figure
9.
Speed of attacking force versus their
attrition coefficient [S vs a(r)].
/)/.<^Tiarc)
I
=
a. b, c
(2.15)
where
DIST(arc,)
TIME(arc
DT,
b.
:
)
is
:
amount of time
to transit arc,
delay time on arc,
arc^
:
arcs
which are being subjected
arcj
:
arcs
which have preinstallcd minefields
arc,
:
arcs
which have scatterable minefields
to indirect fire
Direct fire
To update
(2.16)
distance of arc,
:
Red
the location of the
force in the direct
fire
region, equation
used to calculate the Red force speed dunng each time step.
shows that
unit speed decreases as the range
no smaller than a
specified
minimum
speed
The equation
between opposing forces decreases, but gets
(S„,„).
20
where
V
power
:
S„^
S^ij
S^,n
a,
:
Oq
:
Equation (2.16)
in
Chapter
factor,
based on difTerent weapon types
updated speed of attacking force
:
speed of attacking force for past time step
:
:
minimum
speed for the attacking force
current attrition coefficient
maximum
is
attrition coefficient
portrayed graphically in Figure
9.
Actual model results are discussed
3.
Advanced model
4.
To
obtain the required data for analysis, the basic model
is
expanded.
The ad-
vanced model uses the basic model as a submodel and produces two output matrices (see
Figure
The upper portion of the
11).
values for use in Table
MOE. MOE-1
battle,
and Red
2.
which
is
a 4
figure
by
S
is
the output matrix of four selected
matrix for the Red and Blue options for each
through .MOE-4 represent Red casualties. Blue survivors, duration of
to Blue casualty ratio, respectively.
The lower
part of the figure con-
tains utiUty values for three different utility functions (Linear, Squared,
Root) versus Blue options
for
MOE-2
values (Blue survivors).
The
based upon the way the Blue commander weighs the Red options.
of each
utility
The
one of the Blue positions
vanced model
10,
is
listed in
and the following
BATTLE =
1
BATTLE2 =
at
node- 14
(BATTLE!
(BATTLE2
case).
values are
characteristics
function and weighted utility values are discussed in Chapter
through Avenue-2 are equally divided
and Square
utility
In the advanced model, the user can decide whether the attacking
to only
MOE
3.
Red
forces
case), or continue to attack
The computer program of the ad-
Appendix D. The How chart of the advanced model
is
in
Figure
definitions apply:
basic model which has three avenues of approach (Red force on
Avenue-2 will choose only one of two objectives to attack).
model which
with three avenues of approach but divides
force on Avenue-2 will be divided equally at
node- 14 and attacks toward each objective, respectively.
basic
starts
into four because the
Red
21
start
read data
Do
lOi = 1,4
Do 20
i
=1,8:
:
RED options
select
select
BLUE options
callBATTLEI
callBATTLE2
20 continue
i
10 continue
print
MOE.
UTILITY matrix
i
end
Figure 10.
Flou chart of the advanced model
22
WAVE2
=
numeric value
user
:
to
but
Red
\VAVE2=
will
all
12.
it
or 12) which
I, it
Avenue-2 attack
case,
These outputs
(1. 2,
irWAVE2=
will
be
will
node-2S
;
lorces
be a
is
shown
in
Fisure
is
1
1.
23
supposed to be given by the model
case, and all Red forces on
BATTLLl
il\\AVE2=
on
m
2.
Avenue-2
BATTLE2
be analyzed in more detail
of the advanced model
a
it
will also
attack
to
be a
BATTLE
node-27
;
1
if
case.
the next chapter.
The example output
«
MOE VALUES BASED ON THE RED AND BLUE OPTIONS
RED
3
4
949.2
219.9
828.5
225.0
2.95
290.
229.0
2.59
751. 7
307. 6
951. 3
783.
220.0
290.0
268.9
229.0
88
2. 79
817. 1
1
2
1
759.6
802.3
1
349. 9
278.
1
215.0
3.80
795.2
»
BLUE
1
2
2
2
2
3
3
3
3
321.
4
4
4
4
1
225.0
786. 9
321. 7
795.4
278.4
223.0
3.45
771.0
345.5
225.
241.6
283.0
2.93
752.0
923.0
768.6
336. 3
215.
7
3.60
813.9
316.5
224.0
3.49
803.5
7
305.
7
223.0
7
29
773. 9
6
6
7
10
2.
3.
14
280.0
2.
747. 8
298. 7
225.
1
297.5
290.0
2.98
639. 9
346. 3
21
808. 9
297. 9
225.0
290.0
14
21
847. 3
307. 1
283.
3.
1O^
299.
1
215.0
225.
3.05
79
229.0
2.57
3.49
251.4
229.0
2.42
1
44. 56
2
3
48.47
4
44. 05
48. 10
5
6
47.63
7
47.51
62.09
8
Figure
61. 69
u.
The ouipui
631. 3
308.0
229.0
2.61
3.
16
S'G
MOE -2 >>
SQ uARE ROOT U(X)
69.87
73.32
69.90
72.67
24
19
14
14
17
72. 70
37.05
28.00
77.92
72.62
77.86
36. 87
of ihe
1
7
SQUARED U(X)
24.
29.
24.
28.
28.
722.
300.4
280.0
<< WEIGHTED UTILITY VALUES US
LINEAR U(X)
305.
3.
787.
MOE-1
MOE -2
MOE -3
<—== MOE -4
<=
<
229.0
2.45
630.5
3.
/
<==
229.
2. 82
733. 7
251. 1
83
810.
1
230.4
229.0
2.56
787.0
271.0
224. 4
98
3-^5. 5
3.
2.
967. 3
297. 3
225.
225.0
2. 94
632.4
3A2.3
3.
8
8
8
8
3.
230.
2.88
224.0
3.47
215.
3. 77
5
5
5
5
6
6
1
advanced model
24
MODEL OUTPUT ANALYSIS
III.
A.
OVERVIEW
As mentioned
in
Chapter
of air-land combat models
in
1,
the purpose of this thesis
is
to analyze the key variables
high resolution and provide some analytical methodologies.
This chapter focuses on several methods of analysis of model output while the following
chapter explores various sensitivity analyses.
Since this model
is
sequenced by time steps, the user can easily visuahze the battle
For example,
situation as the battle progresses.
the Blue force places a certain size
if
minefield on a certain arc of the network terrain model, the output will
minefield affects the casualties and speed of the
12-15) provide
more
Red
force.
displays of battle situations which occur
12-15 are created based on sample output of the basic
Red
In this battle case, the
force
is
assumed
is
assumed
to
figures (Figure
on each avenue.
model
no
Figures
in the previous chapter.
to take option-2 (allocation of
ions on Avenue-2. one batthon on Avenue-3, and
Blue force
The next four
show how the
unit allocation
two
battal-
The
on Avenue- 1).
employ option-2, which locates one company on position-
(node-28) and two companies on position-2 (node-27),
etc. (see 0^2
of Table 2
in
Chapter
2).
Figure 12 shows
Note
Red
that the
how
the
Red and Blue
forces attacking on
approximately one third of the
serve
Red
company (150
initial
infantrymen).
force sizes are decreasing on each avenue.
Avenue-22 are severely
forces even
The
attrited
and end up with
though they were reinforced by
unit reinforcement
is
supposed to occur
a reif
the
forces are less than the Blue forces after they are engaged in the direct fire battle.
On Avenue-22
of Figure
12,
Red on
that avenue
appears to be the beginning of the direct
of the
initial
caused by the indirect
that the
Red
of approach.
combat on
forces in
fire
casualties
is
fire battle.
position-2.
reinforced at battle time 160 which
The Blue
force lost
more than half
The approximate number of
and ambush attack can also be observed. Note,
casualties
in Figure 12,
on Avenue-22 are much more than those on the other avenues
In other words, the intensity of the battle between
Avenue-22 and Blue force on position-2 was
ver>' high.
Red
forces
on
There are no forces on Avenue-
because the Red force did not allocate any forces there.
Figure 13 depicts the size or proportion of
employed by the Blue
Red
casualties in relation to each
force, such as indirect fire, mines,
25
and
direct fire.
weapon
For example.
RED
rORCCREON 004 MOfUC
uxratKEazEQN DCM rvmrm
1
>
MIOME.-3
.
i:
-^-
-
,,
•
<i
B
B
a
IX
ito
n
•
m
•
IB
IB
BB
nctMMnQ
Figure
Red and Blue
12.
on Avenue-22.
item depicted
a high
is
force sizes over time
proportion of the Red casualties were due to direct
that the time
slope of the curve
(i.e..
when
the
Red
force engaged in direct
fire.
fire is
the point that the slope of curve changes abruptly).
Another
based on the
An ambush
attack occurred from approximately 55 to 90 on the battle clock (for 35 minutes) on
Avenue-2 (actually
Avenue-22
at
this
happened before the Red was divided onto Avenue-21 and
The Red
node- 14).
about 205 battle clock time, and
Avenue-3.
directions
force
at
on Avenue-21 entered the
direct fire region at
about 160 on Avenue-22, and
at
about 210 on
So, after 210 minutes, the Blue position-2 started to be threatened from two
which were from Avenue-22 and Avenue-3.
Avenue-21 (9.45Km
equaled 0) was
:
much
initial
distance between
Red and Blue
force
when
shorter than that of Avenue-3 (13.55Km), the
Avenue-21 and Avenue-3 were engaged
attrited
the battle clock
Red
forces
on
in direct fire with the Blue force for a shorter
period of time (between 205 and 210 minutes).
Avenue-21 were
Note, even though the range of
more by Blue
forces
Avenue-3.
26
That was because the Red forces on
and
their
obstacles than those
on
RED CASUALTIES ON EACH AVENUES
MeA£-22
/MENUE-21
B
TOTAL
S
h
RECT-nUE
I
I
IDIRECT-FIRE
ttlWRECT-FFE
1M
100
10O
la
TiKEOflNurq
AVD4UE-3
100
lao
71M£<UINUrp
Figure
13.
The Red
casualties by different types of Blue
The Blue commander may want
to
know where
and the distance from which they are attacking.
battle terminates but also
who reached one of the
from Avenue-22 took the Blue position
first in this
the
that time,
ambush
to 130 battle clock time
Red
forces are at certain times
Figure 14 shows not only
Blue positions
battle case.
of each curve represents the speed of the Red forces.
constant speed from
neapons
first.
when
The Red
the
force
In Figure 14, the slope
For example, Avenue-3 has a
which means there
is
no
attrition during
and Avenue-21 and Avenue-22 have very small slopes (low speed) during an
attack by Blue forces (from 55 to 90 battle clock time).
The speed of Red
responses.
force
is
another important factor which can cause different Blue
Sometimes, the Blue forces need to
27
know when
the
Red
forces will reach their
RANCE BErmEEN RED AND BLUE FORCE
13.55
1S0
100
nME(MlNUTE3
Figure
Distance betneen Red and Blue forces over time.
14.
positions in order to prepare and decide on an effective course of action.
speed of the attacker can he calculated from Figure
Figure 15 also shows
is
reduced by indirect
nue.
In the direct
fire
fire,
how
fast the
Red
by exponential shaped curve. Note
forje
and the
minefields,
region, the
Red
14.
moves and how
their maneuverability
direct fire of the Blue force
forces are forced to slow their
how
The average
on each ave-
movement
the curve changes during the
ambush
depicted
battle (at
55-90 battle dock).
MEASURES OF EFFECTIVENESS (MOES)
B.
After considering the battle situations, four
how
MOE
they changed in each of the time steps. The four
•
MOE-1
•
MOE-2 number
•
MOE-3
•
MOE-4
:
number of Red
:
:
:
values were selected to estimate
MOEs
are as follows:
casualties
of Blue survivors
duration of battle
ratio of
Red and Blue
casualties (Loss
28
exchange
rates)
SPEED OF RED FORCE ON EACH AVENUE
lAj
^eo
100
lao
TIMQlflNUn)
ISB
thiecmmuti)
AVENUE-3
K
100
leo
TMECkllNUTp
Figure 15.
Larger
the
MOE
MOEs'
and Blue,
Speed of Red force on each avenue over time.
values are better for the Blue forces in
for this scenario are
Table 2 in Chapter
refer to
For MOE-1,
shown on Figure
if
the
Red chose
all
cases.
The
characteristics of
For the particular option of Red
16.
2.
option-2, then the Blue force should select option-
because the Blue force wants to maximize Red casualties (MOE-1).
difierences
among
its
options for Red option- 1.
approximately the same tendency and
there
is
pend on
some
how
relation
between
long the battle
all
this also
Note
happens
MOE-1 and MOE-3.
lasts.
29
that
for
Blue has no large
Red options
MOE- 3.
In other words,
1,
2
and 4 show
This implies that
Red
casualties de-
U.O.E-URQ CASUALTISQ
MA£-2(BLUE SURVMXe)
•
RO
OPTION-I
a RED OPnON-2
V RED OPTlON-3
4 RED
14*
BLUEOPnONS
M.0.E-3(a«,TTU
i
mjJC
*
:
LUEOPixna
BLueoPTote
Figure
16.
MOE
The
OTUNS
MA.E-4<CASUM.T1E5 RATICQ
TIklE)
t
values for each options of
Red and Blue (BATTLE2)
For MOE-2. the Blue forces are affected hy the Red options
through
5 but they are not severely alTected in Blue option 6. 7,
option 2 and 4 of
implies that the
MOE-2 show
Red
artm-*
and
in
Blue option
8.
Note
that
1
Red
MOE-l.
This
casualties have a negative correlation with the Blue survivors
which
almost reverse trends against those of
agrees with the theoretical battle concept.
For MOE-3, Red option-3 has a longer
reason this occurs
is
that
over the Avenue-3 which
the
Red
Red
is
s
main
battle time than
any other options.
The
forces (two battahons) attack to the Blue positions
the longest
forces choose option 2 or 4,
among
the three major avenues of approach.
If
MOE-3
has constant values of 225 and 229,
re-
30
In other words,
spectively.
ions)
when
Red
the
on Avenue-2, the duration of battle
MOE-4
force allocates
more than two
units (battal-
be same regardless of Blue's options.
will
seems to have some relationship with .VlOE-2, but the relationship
entirely clear except for the
relationship between
To summarize
Red
them such
the
MOEs,
benefitial for all
MOEs
Blue's decision
maker wants
option-
would require more
It
1.
is
not
analysis to verify the
as a coefficient of correlation (p).
it
is
pick the best option which
difficult to
for the Blue force.
The course of action
is
is
most
the
strongly based
on how
to drive the battle according to his tactical courses of
action.
COMPARISON BETWEEN BATTLE AND BATTLE2
C.
1
This
section
will
BATTLE2. BATTLE
1
discuss
between
differences
same
battle
Table
refer to the last part of
which was discussed
The
option).
For
in
Chapter
the
AND BATTLE2
to
first?
Red
BATTLE2
with Avenue-22
763.3
609.3
751.7
15S.6
242.S
307.6
264.0
228.0
225.0
1.05
1.9S
3.10
Red on Avenue-22
to Node 27
Red on Avenue-22
Red on Avenue-3
reaches
the
B.4TTLE1
BATTLEl
MOE-1
MOE-2
MOE-3
MOE-4
From
and
differences are given in Table 4.
with Avenue-2
Where
1
Chapter 2 (Red's second option with Blue's second
value names
Who
BATTLE
a description of
The output of BATTLEl came from
2.
COMPARISON OF MOES BETWEEN
4.
BATTLE! and
of
results
has two approaches which the Red force can take on Avenue-2.
So, in total, three battle types are compared.
BATTLE2.
the
Node
27
force's viewpoint,
it
is
better to choose
to
Node
27
Avenue-22 over Avenue-2
based on Red casualties (MOE-1) and duration of battle (MOE-3) since 609.3 < 763.3
and 228 < 264 [Note, the smaller
with
BATTLE2,
casualties
and
minutes. The
kill
if
the
Red
more Blue
Red Blue
MOE
values
force chooses
forces,
is
better for the
Avenue-22
in
Red
force].
BATTLEl,
it
Compared
can reduce
its
even though the duration of battle increases by three
casualty ratio
(MOE-4)
31
is
indifferent as to
whether the Red force
chooses Avenue-21 or Avenue-::
is
much
in
BATTLE
and
(1.95
1
BATTLE:, MOE-4
In
1.98).
larger.
Red
\\'hen the
creases to
its
force chooses
Avenue-:i
greatest value, because the
Red
if
the duration of battle in-
forces attacking through the longest path
(Avenue-3) reached the Blue position (node-:?)
Blue's viewpoint,
BATTLEI.
in
first
and
battle
From
was terminated.
the Blue force attempts to prolong the battle,
attempt to
will
it
on Avenue-: into Avenue-21 (because the duration of battle
channelize
Red
BATTLEI
with Avenue-21
forces
the highest: 264 minutes in Table 4).
is
Installing
for
more
mines and obstacles or conducting ambushes on Avenue-2 should be considered by the
Blue force.
in
As shown
BATTLEI
based on
amonc
(i.e.,
Figure 17, Blue
in
way
the best
BATTLE2
for Red).
battle type,
the three battle types.
is
at a
if
Red chooses Avenue-22
Since Figure 17 was drawn with
shows the
clearly
it
disadvantage
MOE
ratios
proportion of changes
relative
Figure 17 also shows that the attacking force usually has
the advantage in concentrating
its
forces
on one
objective,
which
is
a
well-known
tactical
concept of war.
A21
2.0
A22
A2
D
^-^
:
:
BATTL£1 WrTH AVENUE-21
BAHLEI WITH AVENU£-22
BAHI F?
-
g
A21
A21
o
UJ
:
1-0
k22A2
A2
A2
1
A2
1
1
!
k22
A22!
A21A2J
H
/k21
ics
-
MOE-1
Figure
17.
Ratio of
Figure 18 describes
are on
alties
MOEs
how
MOE-3
MOE-2
all
in
BATTLEI
types based on
the battle case
ratio of
Red
to Blue casualties
32
BATTLE2
MOE's change when
Avenue-2 choose Avenue-21. Compared with
(.MOE-l) and the
MOE-4
BATTLE2
(MOE-4)
type.
the
Red
(Figure 16),
forces that
Red
casu-
are relatively small,
and
M.OX-URED
y.0£-2(BUJE SUirVNDRS)
CASUALT1ES0
RED OPnON-1
•
RED OFnON-2
? RED OFTION-3
L RED 0PT1ON-4
1
•
*
mEOPTOfS
LUEOPnONS
M^J-4<CASUALT1ES
BLUE oimoie
MOEs
Figure 18.
for
BLUE OPTX)»G
each options of Red and Blue
the variance of Blue survivors
(MOE-2) based on
(MOE-3)
ualties decrease
rate of killing per time
number of
is
1
is
battle types.
v>ith
Avenue-2I)
large.
is
Generally,
decreased).
cas-
This fact imphes that the
BATTLE
on Avenue-21
1
BATTLE2.
In other words, the troop density in
how
MOEs
BATTLE
1
is
on
small.
Figure 19 shows
Avenue-22.
1
on Avenue-21, though Red
units (infantr\men) engaging in direct fire in
smaller than that of
Avenue-21
increases in
(BATTLE
Red options
the
BATTLE
the duration of battle
(i.e.,
IUT10)
all
In this battle.
battle
Red
For MOE-1, there
case has a reverse shape from
change when Red forces on Avenue-2 choose
casualties
is
(MOE-1)
a interesting
all
other
are the smallest
phenomena which
Red option
33
is
among
that
all
three
Red option-
cases with respect to the Blue
W.0X-1(RED CASUALTIESQ
•
y.0£-2(BUUE SUIMM9RS}
RED OPTTON-I
RO
OPTlON-2
V RED OPnON-3
A RED OmON-4
BLUE
OJEOPmNS
aiTOS
M-CE-MCASUALTIES
U.0.E-3(B»,TTLE TIME)
RATIO)
f!
pi
i
Bli
BLUE
Figure
19.
options.
cause
Red
for
all
All
battle
\IOE-3 values
32 battle cases (four
forces
aux opvots
each options of Red and Blue
For the duration of
options.
Red
MOEs
CnKtS
(MOE-3).
on Avenue-22. Therefore,
(MOE-2) and
constant for each
In
ratio of
Red
large difTerences
its
among
the
battle time, be-
by the
potions for more than 4 hours.
(MOE-4), values
Blue
are relatively
options.
will
choose
to consider other courses of action against Red.
34
Avenue-22)
model shows that the Blue force
summary, the assumption that the Red force
Avenue-22 forces Blue
>vith
eight Blue options) are terminated
to Blue casualties
Red option based on Blue
no
1
and 235 minutes of
this deterministic
needs other courses of actions in order to hold
survivors
there are
are between 205
Red options with
(BATTLE
BATTLE
1
on
THEORY
UTILITY
D.
1.
General
among
Individuals (or Decision Makers) frequently must choose
that difTer.
subjected.
ual
among
The
who buys
sum
other things, in the degree of risk to which the individual will be
clearest
premium)
(the insurance
much
larger loss (the value of the house)
and a
combination of a small chance of a
large chance of
An
individual
subjecting himself to a large chance of losing a small
both
risks.
He
is
choosing uncertainty
In order to develop a
means
in
sub-MOE's. subjective and judgemental
aspects of outcomes
(i.e.,
the hierarchy in which each decision
2.
an overall
risk factors) into
maker must
MOE
function.
is
a lotter>' ticket
in preference to avoiding
[Ref 6
which
figures of merits or
he
is,
(the price of lottery
to rank probability distributions
way
That
loss.
who buys
in preference to certainty.
This provides a consistent
the theor>- of "Utility".
no
amount
chance of winning a large amount (a prize)
ticket) plus a small
individ-
acceping the certain loss of a small
is
in preference to the
choosing certainty in preference to uncertainty.
is
An
examples are provided by insurance and gambling.
insurance on a house he owns
fire
alternatives
to
:
p. 234].
we must
resort to
combine
different
factors,
and probabihstic
appropriate for the level in
[Ref 7
:
p. 9].
Application
Three diflerent
utility
by using MOE-2. which
is
the
functions are proposed to
number of Blue
survivors.
compare each
These
utility
battle situation
functions are re-
presented as linear, squared, and square root, and are depicted in Figure 20.
The
utility
curves reflect the decision maker's judgement.
a.
Linear Utility
Using the
fact that the utility functions are arbitrar>'
transformation, for convenience
U(550
L(
As
survivors,
survivors,
a Blue decision maker,
than 340, which
is
i.e.,
we
it is
to a positive linear
define:
no
i.e., all
up
losses)
losses)
=
100
=
assumed that
the size of two companies
if
the
(i.e., if
number of survivors decrease
they lost one
company out of three),
the utility value of the Blue force drops with a 10 percent step change.
function for linear utiUty
is
given by equation (3.1).
35
to less
The appropriate
UTILITY
rUNCTI0NS(3 TYPES)
80UME UOO
UNEAR UpO
Br
Figure 20.
Three
U(x)
=
different utility distribution functions.
^r-x
,
-7T-X+10,
x<
340
;c>340
5:>
,
b.
Squared
This
sion.
is
otherwise
(3.1)
Utility
concave upward function which
The appropriate function
for squared utility
36
is
will result in a risk prefering deci-
given by equation (3.2).
^'W=
-3025"
-"^^
•"'
'
oiherwise
,
Square root
c.
This
averse case;
most
utility
(3.2)
Utility
curve
concave downward.
is
It
most individuals and organizations generally
The appropriate function
decisions.
generally represent the risk
are risk averse with respect to
for square root utility
is
given by equation
(3.3).
U{x)=
4.264 ^'F
,
,
If
is
it
.v>0
oiherwise
(3.3)
assumed that Red's options are weighted
as w,, w^, Wj. U4 for each
option, then weighted utility values can be calculated by equation (3.4).
L'(.r)
=
W] L\{x)
+
u-2 L'jix)
+
+
u-3 ^'3(^1:)
W4 L\{x)
(3.4)
where
U(x)
w,
:
:
weighted
utility
value
proportion of weight for
L',{x)
:
utility
value of i-th
Red
As an example, with given
utility
values for each function are
acteristics
of
MOE-2
large weight (0.70)
for the
greater than 340
shown
(i.e.,
8 are
option.
option,
i
well,
=
<
w,
<
1
i
=
1,2,3,4
1,2,3,4
w, (0.05, 0.70, 0.05, 0.20),
in Figure 21.
Red option-2
and
Red
values of
on the Red option-2
Utility case, Blue options 6
i-th
the weighted
This figure represent the char-
because the Blue decision maker puts
(see Figure 16 for
MOE-2). Note,
emphasized because
their
in the
MOE-2
Linear
values are
Blue survives with more than two companies) while the others are
not.
37
DIFFERESrr VALUE
OF U(X)
Sn
SQUARE ROOT U(X)
8UNEAR U(X)
g
5-
SQUARE UCX)
8-
BLUE OPTIONS
Figure 21.
E.
Weighted
utility
values for each function.
GAME THEORY
1.
General
In
game
complete rule
theory, the
for decision
values consist of a 4 by
word
"strategy" has a very definite meaning, namely, "a
making".
[Ref. S
:
By the assumptions,
matrix and the values for each
S
output of the advanced model (Figure
Chapter
II in
option paired with a particular Red option.
ties
p. 2].
2).
MOE
A
all initial
MOE
matrix come from the
strategy
is
a particular Blue
In order to study the characteristic proper-
of games of strategy and find an optimal strategy for the two players (Red and Blue),
the 4 by 8 "Rectangular
value of the game.
Theorem
Game"
[Ref 9
1.
:
:
p.
is
utilized.
34 and
Two theorems
39J.
Let
Mn
A =
^m\
38
are also used to
compute the
m rows
be any matrix, with
X=
l|.v,
....
x„
II
Y=
and
n columns, and
|Lvi
....
let
the expectation function E(X,Y). for any
members of S„ and
vj| that are
S„, respectively,
be de-
fined as follows:
m
E{X,Y)
n
= Y,Y.''^m
(^-^^
i=\ j=\
Then
the expressions
max min E{X,Y)
(3.6)
min max
(3.7)
and
exist
S^
:
:
close subset of
whole space (£„) which have elements,
close subset of
whole space (£J which have elements.
Theorem
let V
Where;
and are equal.
S„
£(A',}')
2.
:
Let E be the expectation function for an
be a real number, and
let
A" and }" be
necessary and sufficient condition that
optimal strategies for
P^
and
v
members of S„ and
m
jc,,
....,
x„.
,y^.
}\
x w rectangular game,
S„, respectively.
Then
be the value of the game and that A" and
P^, respectively, is that, for
< < m and
i
]
V
a
be
<J< n.
1
E{iy)<v<E{XJ).
2.
Application
a.
MOE-1 (Red casualties)
The
initial
4 by 8 matrix
Mixed Strategyi solution
Column
Row
3
3
is
reduced to a
(see Figure 22).
dominate column 4
dominate row 4
The
by 5 matrix which results
relations of
= = = >
= = = >
3
dominance are as
eliminate
eliminate
column
in a
follows:
3
row 4
1
Rectangular game whose matrix has no saddle point, that is, Red could determine optimal
way with probabilities P^P-^..., P^ for each column of matrix, and Blue could determine optimal
wav with probabilities P, P^ ..., P, for each row of matrix. For more explanation in detail, refer [Ref
9:
p. 21).
39
By Theorem
and
V
which
Row
3
dominate row 5
= = = >
eliminate
row
5
Row
7
dominate row
= = = >
eliminate
row
8
2,
S
the reduced matrix suffices to find
numbers
+
JC^
+
j:4
=
J-.
1
0<;r„>;,<l
i=
and ]=
1,2.4
1,2,3,6,7
759.6;f,
+
802.3^:2
+
S28.5;c,
<
v
195.2X,
+
ISl.lx,
+
783.0X,
<
v
786.9.V,
+
795.AX,
+
817.1.V,
<
v
S13.9.r,
+
639. 9a%
+
630.5.V,
<
v
803. 5.V,
+
737.4.V,
+
722. Lv,
<
v
759. 6r,
+
795. 2v,
+
7S6.9v,
+
813.9v,
+
803. 5v-
>
v
802. 3>-,
+
751.7i%
-f 795.41-3
+
639.9j,
+
737.4j--
>
v
S2S.5v,
+
7S3.0r,
+
+
630.5v,
+
722. Ir-
>
v
From Theorem
1.
we know
SH.lvj
Linear Program to find the
and
results).
V.
If the objective
optimum
equation
if
there
is
is
set to find a
(LINDO) [Ref
10]
which
solution (see Figure 23 for the equations
In this solution, the objective equation
Note, that
These conditions
that there exists a solution to the system.
were solved by using Linear. INteractive, Discrete Optimizer
same.
Vi, jj, y'y j^.
satisfy the following conditions:
ATj
utilizes a
x^X2, x^,
is
minimum
set to find a
value of
maximum
value of
the result will be the
v.
a negative element in the matrix of .Mixed Strategy, the
Linear Program method can not be used.
Based on the solution
game
is
to choose options
1, 2.
and
Figure 23, an optimal
and 4 with respective
an optimal way for Blue to play
probabilities 0, 0, 0.89, 0.
in
is
0.11.
value of the
for
Red
probabilities 0, 0.78,
to choose options
The
way
1.
game
2, 3. 6,
is
and
788.8;
i.e..
to play this
and
0.22;
and
7 with respective
Red can play
in
such a way as to make sure of not losing more than 788.8 Red casualties, and Blue can
play in such a
way
as to
make
certain that
Red
40
will lose at least 788.8.
<< Initial matrix (4
2
1
759.6
2
3
795. 2
786. 9
802. 3
751. 7
4
7
771.0
768.6
813.9
803.5
8
773. 9
5
6
«
X
1
8)
>>
795.4
752.
747.8
639. 9
737.4
632.4
Reduced matrix (3
4
3
949.
951.
967.
923.
810.
808.
847.
787.
X
5)
2
828.5
3
783.
817.
3
1
733.
9
630.5
722.
7
631.3
4
2
3
6
759.6
795.2
786.9
802. 3
751. 7
813. 9
639. 9
7
803.5
737.4
828. 5
783.
817. 1
630. 5
722. 1
795.4
matrix and reduced matrix for
Initial
MOE-2
b.
A
1
»
2
Figure 22.
7
3
1
1
1
787.0
MOE-1
(Blue survivors)
similar
method
as the
.MOE-1 case was used
for
VIOE-2.
After elimi-
nating the redundant rows and columns by relations of dominance, the following 2 by
2 .Mixed Strategy matrix
Red
Blue
4
3
7
307.1
251.4
8
300.4
30S.0
This implies that the
8.
is left.
Red
Since a 2 by 2 matrix
favors
is
its
option
3
and
4,
and the Blue favors
An
0.8811;
and the value of the game
optimal startegy for Red
is
is
||0.89, 0.1
1||,
an optimal strategy for Blue
301.2 Blue survivors.
companies (340 infantry men) may survive during the
So, not
is
i|0.12,
more than two Blue
battle.
MOE-S and MOE-4
.MOE-3 and
MOE-4
do not have a mixed strategy, but have a unique
strategy which can be easily found
by inspection.
Red option- 1
option
option- 1.
option 7 and
simple to solve, familiar methods of elementary algebra were
used.
c.
its
is
dominated by
its
Thus, an optimal strategy
is
2,3,
that
41
For
MOE-3
(duration of battle), the
and 4 so that the Red
Red choose
its
will
choose
its
option- 1 with probability
1
«
»
EQUATIONS FOR LINEAR PROGRAM
MAX
V
SUBJECT TO
X2
X4
Y2
Y3
759.6 XI
795. 2 XI
786. 9 XI
813. 9 XI
803. 5 XI
759.6 Yl
802. 3 Yl
828.5 Yl
XI
Yl
2)
3)
4)
5)
6)
V
V
V
V
V
V
V
V
7)
8)
9)
10)
11)
1
Y6 + Y7 =
802. 3 X2
751. 7 X2
795.4 X2
639.9 X2
737.4 X2
795.
Y2
751.
Y2
783.
Y2
1
<=
<=
<=
<=
<=
+ 813.
+ 639.
+ 630.
828.5 X4
783.
X4
817.
X4
630.
X4
722.
X4
786.
Y3
795.4 Y3
817. 1 Y3
Y6
Y6
Y6
Y7 >=
Y7 >=
Y7 >=
803
737
722
END
«
OUTPUT OF LINEAR PROGRAM
LP OPTIMUM FOUND
»
AT STEP
12
OBJECTIVE FUNCTION VALUE
788. 791260
1)
VARIABLE
VALUE
REDUCED COST
788. 791260
V
XI
X2
0.
777479
0. 222521
0.0
0.0
0.0
0.0
19.689911
3.271210
0.
X4
Yl
Y2
Y3
Y6
0.
0.
0.886059
0.
610260
0.0
13.
0.
Y7
0.
113941
Solution for ^alue of the
Figure 23.
and Blue choose
its
Game (MOE-1)
option-6 with probabiUty
which means that the Blue forces can hold
utes.
From
if
For
nated by
its
The
value of the
game
their positions at least 3
a tactical viewpoint, the Blue forces
of battle time
1.
is
224 minutes
hours and 44 min-
have to be reinforced before 224 minutes
Blue does not want to lose their positions.
MOE-4
option
Red/Blue casualty
1, 2,
ratio.
(ratio of
and
3
Red and Blue
so that the
Red
casualties), the
will
Thus, an optimal strategy
42
choose
is
that
its
Red option-4
is
domi-
option-4 to minimize the
Red choose
its
option-4 with
probability
2.
Red
82
Red
1
and Blue choose
its
option-4 with probability
casualties per Blue casualty
1.
The value of the game
which means that one Blue force unit can
kill
2.82
units in the battle.
USE OF UTILITY VS GAME THEORY FOR THE ANALYST
F.
The choice of technique depends on how
For the Utihty approach, the
assigned by the analyst.
states
For
Game
theor\-,
of nature
the analyst desires to view the
,
enemy
we
(i.e.,
unit
force.
with appropriate probabilities, would be
If inteUigence reports are available, these could be
termine the most likely Red option
used to de-
mix on various avenues).
are assuming that both
Red and Blue
forces are using the
same payoff matrix, and that Red and Blue are making strategy decisions according
the
as
is
max(min) and min(max)
an
intelligent player as
criteria, respectively.
opposed
In the analyses,
to states of nature.
Red
is
being treated
Results from this analysis can be
used to determine the frequency with which Red will choose his various options.
43
to
SENSITIVITY ANALYSIS
IV.
GENERAL
A.
The previous chapters have considered, using
analysis based
size
on a given scenario.
and a given
tivity
of
set
In
all
the deterministic
cases, the analysis
model and
was based on
weapon parameters. This chapter examines
its
output,
a fixed force
the numerical sensi-
of the methodologies and the efleci on the model output interpretation as
tains to
Red and Blue
options.
The Analysis of Variance
(ANOVA)
Some
t-Test are applied to support the sensitivity analysis.
and
it
per-
Two Sample
introductor>' analytical
techniques to evaluate countermeasures are also covered, and an example of model
enrichment
The
may
considered.
deterministic
model developed
in this thesis
has used
possibly be changed based on real battle situations.
coefficients
strict
If
is
may
be affected
if
Red
uses
the firepower of opposing forces.
Red
uses a "Mine-plow"
when
it
"Smoke" while they
in the minefield or increasing the
amount of indirect
force.
results
showing large
may
smoke
attack, since
the Blue force's viewpoint, installing
fire
These key parameters are varied
may
more mines
in turn to
determine their effects
httle or
no
be treated as constants and the model simplified accordingly.
sensitivity
will re-
be used as countermeasures
on outputs of the model. Those parameters which are shown to have
on the
attrition rate
traverses the minefield, attrition rate coefficients for
From
Red
For example,
This would also occur in the night battle case.
mines should also be decreased.
against the
many parameters which
should be examined in
detail, since these will affect
effect
Those
model
per-
formance to the greatest degree.
B.
FORCE SIZE
The
the
force size
Blue
forces
infantr>'men).
is
one of the major factor to
are
considered
Two methods
to
be
affect the battle output.
reinforced
In this section,
by one additional platoon (40
of reinforcement are compared with the basic battle case
which has no reinforcement, and the basic battle also has Blue option-2 and Red
option-2 (refered to as O^^ of Table 2 in Chapter
discussed in detail in Figure 12-15 in Chapter
2).
3 as
These battle cases'
when
more platoon. Compared with Figure
44
have been
an example of the basic model output.
Figure 24 shows Red and Blue force sizes over time
reinforced with one
results
the Blue
ambush point
12 in Chapter
3,
is
the rate of
RES FORCE sat ON EACH
»KMUt
BLUE FORCE BOE ON EACH PO«TX>N
B
5
I
POSmON—
AWENUE-21
SS
posmoN-1
AVENUE'
'
IK
100
B
aoo
IB
TIWE (MIMUTQ
Red and Blue
Figure 24.
la
TIME (kONUIQ
force sizes over time >vhen the Blue
ambush
point
is
rein-
forced.
decreasing
Red
minutes).
Since the
(about time 70),
forces
is
ver>' sleep at the
Red
force
posed to occur
on Avenue-22
this implies that the
as that of the Blue force based
if
the
Red
ambush
Red
is
battle region (for battle time 55-90
reinforced during the
force size
on Avenue-22
compared with Figure
of the curve during the ambush battle region
12 in
The shape of
Chapter
is
somewhat
if
the Blue
3
battle
reduced to the same
on the model assumption (the unit reinforcement
forces are less than the Blue forces).
force size curve looks similar
is
ambush
is
sup-
the Blue
except that the slope
steeper than those in Figure
12.
Figure 25 provides significant information
difTerence depending
at the
when
main
the
on whether he reinforces a given platoon
forces areas
ambush
Blue survivors
commander wants
point
on node 27 and
is
28.
The Red
reinforced while there
(MOE-2) between
is
no
casualties
at the
ambush
(MOE-1)
know
the
point or
are the highest
large difference with respect to the
the two reinforcement methods (Rl and
45
to
R2
in
Figure
2.0
p
RO:
NO REINFORCED^ENT
R1:
REINFORCEMENT TO MAIN FORCES(NODE 27 AND 28)
R2:
REINFORCEMENT TO AMBUSH POINT
R2
1.5
R2
R1
R2
R1
R1
RO
RO
1.0
RO
0.5
MOE-2
MOE-1
25).
in
MOEs
Comparison of
Figure 25.
The Red
to Blue casualiiy ratio
Based on Figure
Figure 25.
MOE-4
>vhen the Blue forces are reinforced.
{MOE-4)
linearly increases
25. reinforcing a
platoon
at the
from the RO
ambush
to
R2
case
point appears to
be a better course of action for the Blue conunander. However, the following conditions
should be considered before the fmal decision of whether a Blue platoon
at the
•
main forces areas or ambush
is
reinforced
point.
Availability of well-fortified battle space for one
more platoon
to fight at the
am-
increased
(i.e.,
bush point.
•
Probability of being detected by
the
C.
more
force size
is
ATTRITION RATE COEFFICIENTS
Sensitivity for attrition rate coefficients.
1.
The
model. The
(2.14) in
attrition rate coefficients are
maximum
Chapter
2,
"bQ
or the
in
attriled
equation (2.14)
number of Blue
another important factor
attrition rate coefficients,
can be varied.
which Red forces are
The
Red when Blue ambush
forces there are, the easier they are to be detected).
is
The
"a^' in
which are given by equations (2.13) and
equation (2.13)
is
by Blue, or the number of Red forces
the
maximum
rate at
forces lost per unit of time.
46
in the deterministic
the
maximum
lost per unit
which Blue forces are
attrited
rate at
of time.
by Red,
LEFT BAR
RED CASUALTIES (MOE-1)
RIGHT BAR
BLUE SURVIVORS (MOE-2)
MAX. AHRmON RATE COEFFICIENTS
C
:
:
:
805.5
BOO
^7^^
751.7
—
722.6
B91.8
UJ
Kl
t/i
uj600
^
s
u
m400
33 6.5
32 1.4
307.6
5
B
294.6
283.2
^200
0.9XC
O.BxC
Figure 26.
Change
of
ficients, Oo
MOE
and
I.OxC
1,1
xC
1.2xC
values based on different
maximum
attrition rate coef-
Z>o
•
V
D
r
2
:
:
RED CASUALTIES (MOE-1)
BLUE SURVIVORS (MOE-2)
:
RED TO BLUE CASUALTY RATIO (MOE-4)
Nv
^^^^^
^Cn.
B
o
h
^^^^^^^^5s<ci!r^^
"^ "^^^^
"
^^^^^
i
-»
^^^^-^
s
a*
OS
V/iWAnON OF ATTRfnON
Figure 27.
Ratio of change for
1.0
W!Z
u
1/4
COET. ("MAX. OOEFJ
MOE values based on different a^ and
47
/><,
Figure 26
cussed in a previous section,
are
changed to between
ualties
and
(MOE-l)
MOE
created with the
is
80*^/0
outputs of the basic model, which were
when both maximum
and 120°o of
attrition rate coefficients
increase linearly while Blue survivors
(MOE-2)
and
(Cf,
Red
original values in the basic model.
disb^)
cas-
decrease linearly as
Oo
bo increase.
Figure 27
is
MOE values with decreased or increased attriton
MOE outputs with no changes of attrition rate
the plotted ratio of
rate coefficients to the basic model's
In this figure, unlike Figure 26, attrition rate coefficients (oq and
coefficients.
tered between
50% and
150°o of original values.
are al-
^o)
Figure 27 shows that the ratio of
change for Red casualties (MOE-l) increases with a weak concave downward shape
curve, which
means
The
rate coefficients get larger.
and
MOE-4
change
decreases with a concave
for Blue survivors
change for
upward shape, which implies
(MOE-2) and Red
to Blue casualty ratio
In other words, as
a^
that the rates of
(MOE-4)
and
MOE-2
b^
decrease
increase, fire-
effectiveness (or loss rate exchange) of the Blue force will be gradually reduced.
At
is
slightly decreases as the attrition
figure also depicts that the ratio of
as the attrition rate coefficients increases.
power
MOE-l
that the rate of change for
this time,
maxmum
only one of the
changed and compared.
The
attrition rate coefficients, either Gq or
differences are given in Table
in parentheses represent the percentage of
MOE
5.
value change
^o.
The numerical values
when compared with
those of the basic case (no changes of attrition rate coefficient';)
Table
5.
SENSITIVITIES OF
CIENTS.
U.9 X a,
fin
MOE-l
MOE-:
Based on Table
When
Oo
(a^)
5,
o,
(100%)
X
c7p
3()2.4 (98. 3" o)
307.6 (100°o)
313.0(101.8%)
hn
0.9 X b,
X
b.
790.S
(105.2'-''o)
740.9 (98.6%)
751.7(100%)
762.6
(
101. S'^/o)
289.2(94.0%)
307.6(100%)
326.4
(
106. l°o)
when
decreases by
increases 10"
1.1
a.
751.'
1.1
MOE-l
MOE-2
COEFFI-
712.4 (Q4.S%)
b.
Blue forces
MOES FOR ATTRITION RATE
the constant rate at
which Red forces are
attrited
by
10%, MOE-l decreases 5.2% and .MOE-2 decreases 1.7%.
MOE-l
increases
5.2% and MOE-2
constant rate at which Blue forces are attrited by Red forces
48
increases 1.8%.
(^o),
when
b^
For the
increases by
MOE-1
10°/o,
MOE-1
MOE-2 decreases 6.0%, and when b^
MOE-2 increases 6.1%. Ifthe same amount
decreases \.4% and
increases
1.5% and
are required to either increase
creasing
On
more
is
Of)
efficient for
the contrary, decreasing b^
than increasing
2.
10%
by
fl^
more
ofefTort or cost
by 10° o for the Blue
force, in-
Oq.
(ANOVA).
considered.
known
The use of ANOVA
as the analysis of variance, abbreviated
some
requires
basic assumptions, such as;
the observations that are obtained are independent and normally distributed;
mean of each
observations have the same variance; and the
unknown
ented as a linear combination of certain
One-way Layout (One-way
It is
trix
ANOVA)
will
eight observations (based
on
casualties
is
With
(MOE-1) and Blue
figure, all four
Chapter
2) is
survivors
normal probability
also verified that
Red
(MOE-2)
plot,
To apply an
1.1,
creased by 10%).
AVOVA
the
test,
respectively
The problem
and
result,
Red opAd-
how Red
the normal distribution.
fit
[Note:
is
maximum
both
(i.e.,
Red options between
option has eight independent
2
ma-
In each
all
third
dis-
verify that
32 observations are within the
graph
(MOE-4)
fits
in Figures
28 and
29)].
the normal distribution
MOE-1 and MOE-2.
attrition rate coefficients, Aq
maximum
and
b^,
are
attrition rate coefficients are in-
MOE
output for
the two battle cases; original outputs of the
Advanced
compare the mean values of each
to
model without any changes, and when both
results for
an independent battle
and normal likelihood surface)
to Blue casualty ratio
by using the same methods for
the different
(a cell of the output
Figure 28 and 29 show
distribution.
K-S bound- on norma! probability plot (the
multiphed by
The
p. 644].
be discussed here.
these observations are normally distributed.
It is
:
graphs (histogram and normal density function, normal cumulative
tribution function,
95*^0
11
these assumptions, the actual output of the
normal
fitted to the
[Ref
parameters.
Blue options) on each column (based on each
8
have the same variances.
vanced model
in
the
all
observation can be repres-
assumed that the output of each Basic model run
produced by the Advanced model
tion)
b^.
improving Blue survivors (MOE-2)
efficient for
In this section, an analysis tool
is
bo
10°/'o,
improving Red casuahies (MOE-1) than decreasing
is
ANalysis Of VAriance
ANOVA,
or decrease
decreases
MOE
and
bg
are multiphed
by
1.1.
Each Red
outputs based on eight different Blue options.
both battle cases are given
Kolmogorov-SmirnoY bound [Ref
Oq
1 1
in Figure 30.
:
p. 552).
49
The
NORMAL DENSmr FUNCTION. M-32
NOIBML CUMULATIVE DSTnBUTKM FUNCTION. »^32
:S|-
^-Ti
I
i
WO
TOO
yoE-1
NOmML LKEUHOOO
SJIVACt (mO*')
N-92
Figure 28.
Red
Fitting the
Normal
casualties
(MOE-1)
of the
Advanced model output
to
Distribution.
ROW
In Figure 30,
represents the eight difTerent Blue options, columns
1
through 12 (CI -CI 2) represent the original Advanced model outputs without any modifications,
and columns
outputs with increased
Red
for
option- 1,
Red
maximum
C2 and C22
option-4, etc.
through 32 (C21-C32) represent modified Advanced model
21
for
The
the value of F- Ratio for the
MINITAB
output for the
difference between
attrition rate coefficients.
Red
option-2,
MINITAB
One-Way
One-Way
C3 and C23
Also, CI and C21 represent
for
Red
computer software [Ref
AVOVA test.
ANOVA test.
CI and C21.
50
12]
option-3,
was used
C4 and C24
to
compute
The following example shows the
The example compares the mean
NOIOML DENSmr FUNCTION. M-32
NORMAL CUMULATVE OtSTKIBUnON FUNCTION. N-32
yx-2
NOniAL UKOJHOOD SURFACE (olO^
N-32
NORMAL PROaABIUTY PLOT, N-32
i
•
^^
p^
:|:;:::::|;:;::::{::::::l::::::y:
J
{
.\
•'
i
k^
Figure 29.
,k::::I^'
;
i
S±;t
i
j
*
i
:::::
Fitting the Blue survivors
Normal
(MOE-2)
Distribution.
51
of the
Advanced model output
to
A.
ROW
1
2
3
RED CASUALTIES (MOE-1)
CI
C2
C3
C4
759. 6 802. 3 949. 2 828.5
795. 2 751. 7 951.3 783.0
786.9 795.4 967.3 817. 1
752.0
747.8
639.9
737.4
632.4
923.0 787.0
8
771.0
768.6
813.9
803.5
773.9
B.
BLUE SURVIVORS (MOE-2)
4
5
6
7
ROW
1
2
3
C5
C6
810. 1 733. 7
808. 9 630.5
847. 3 722. 1
787. 7 631.3
C7
C8
349. 9 278. 1 219. 9 230. 1
321. 1 307.6 220.
268.9
321. 7 278.4 241.6 230.4
C21
769.3 832. 8
1 779.3
798.2 825.8
782.3 779.9
791.7 775. 7
839.8 660. 8
828.4 765.4
798.2 652.5
807.
C25
8
B.
RED TO BLUE CASUALTIES (MOE-4)
5
6
7
ROW
1
2
3
4
5
6
7
8
C9
3.80
3.47
3.45
3.77
3.60
3.49
3.29
3.79
Figure 30.
CIO
2.95
3.10
2.93
2.98
2.98
3.14
2.94
3.05
MOE
Cll
C12
C29
88
88
2.
59
3.
2.
2.
79
3.
3.
14
2.56
3.
82
2.45
2.57
2.42
3.
2.
2.83
3.
3.
21
21
3.49
3.
16
2.
2.
61
C26
C23
69
38
36
67
3.51
3.42
3.
3.
22
71
C30
959.9
977.4
936.0
814.2
810.6
846.3
814.4
756.0
814. 7 646.4
845.6 744.4
791.9 646.8
C27
C31
91
2.
3.05
2.
89
93
3.
2.92
3.06
3.
2.
2.
2.
2.
88
2.97
C24
961. 9 857.7
341.5 263.5 206.9
311.4 294.8 207. 1
312. 1 263. 8 228.3
336.6 283.6 211.2
324. 7 284.0 293.0
304. 8 334.3 293. 3
292. 7 284.4 300. 7
335.
330.2 295. 3
345.5 297.3 224.4 271.0
336. 3 298. 7 297.5 251. 1
316.5 346.3 297.9 305.0
305. 7 299. 1 307. 1 251.4
345.5 342. 3 300.4 308.
4
C22
2.
80
80
04
76
17
3. 17
3. 39
3. 11
C28
212.
1
253.3
212.4
255.6
234. 2
291.5
234.5
294.
7
C32
2.54
2.
73
2.51
2.
77
2.39
2.50
2.
36
2.53
outputs for both battle cases; original battle, and both a^ and b^
increased by 10 percent.
52
< EXAMPLE >
ANOVA- ONEWAY on CI C21
ANALYSIS OF VARIANCE
SOURCE
DF
SS
MS
F
FACTOR
1
1267
1267
2.84
ERROR
14
6256
447
TOTAL
15
7523
INDIVIDUAL 95 PCT Cl'S FOR MEAN
BASED ON POOLED STDEV
LEVEL
N
MEAN
STDEV
--+
+
CI
8
784.07
18.94
(
'V
C21
8
801. 87
23.13
POOLED STDEV =
The example compares column
casualties
this
(MOE-l) of Red
example,
of freedom
1
is
14).
and column
table value for the
If the significant level, a,
Consequently, because the value of F-Ratio
2.84
<
4.60), the difference for
tween above two battle cases
All results for the
given in Table
6.
is
)
*
784
21 (CI
option- 1 for both battle cases.
compared with the
and
1
mean of Red
is
is
Fi_,4
and C21). which are Red
The
value of F- Ratio 2.84, in
(F-Distribution with degrees
0.05, the table value for f^^^
less
than the table value for
casualties
ANOVA
test
)
812
798
(MOE-l) on Red
not significant at the 0.05 significance
One-Way
+
(
770
21. 14
+
is
4.60.
F,
,j
(
option- 1 be-
level.
between CI -CI 2 and C21-C32 are
Like the example, no results are significant with a significance level
of 0.05. This implies that increasing both
a^
and
b^
by 10% do not make any differences
at 0.05 significance level for all possible battle cases in this given scenario.
53
RESULTS OF ONE-WAY ANOVA TEST WHEN THE MAXIMUM
ATTRITION RATE COEFFICIENTS ARE CHANGED.
Red options
MOE-1
MOE-2
MOE-4
Table
6.
Red option-
2.84
1.55
0.92
Red option-2
0.66
1.07
2.40
Red option-3
0.03
0.18
0.39
Red option-4
0.35
1.06
0.70
WEISS PARAMETER
D.
In Chapter
2,
the Weiss
how
parameter was discussed by showing
the attacker's (or
defender's) casualty rate per time changes by attacker to defender force ratio (see Figures
and 6
5
in
Chapter
At
2).
that time,
it
was
casualty rate per time changes difierent
In this section,
1.0.
two approaches
also
shown how
the attacker's (or defender's)
when w (Weiss parameter)
for sensitivity analysis of
vv
changed from
is
will
to
be applied.
only w^ (Weiss parameter Blue engaging Red) varies from 1.0 to 0.6. which
First,
implies that the type of fire for the Blue force changes to area-fire from aimed-fire.
Red
Figure 31 shows that both
as
vv^
decreases.
casualties
(MOE-1) and Blue
survivors
(MOE-2)
Red
In other words, fractional casualties per unit time for
increase
(attacker)
increase as w^ decreases, and fractional casualties per unit time for Blue (defender) in-
This fact can be verified from Figures 5 and 6 in Chapter
crease as u; decreases.
[NOTE
Red
:
ure 32 while
(when
vv
=
0,
6 in Chapter
that of
to Blue force size ratio
1
4, 1,2,
2].
MOE-2.
Figure 32
from
varies
\v^
and
w^,
1.0) consistently varies
In other words.
for
Red
Red
varies.
This
w, (Weiss parameter
the type of
the
Red
fire
between 2.92 and 2.66
between 2.92 and 2.66
casualties are
is
more
to Blue casualty ratio
from
in Fig-
for each
and
in Figures 5
MOE-1
is
vv
greater than
sensitive to the parameter, w^.
and Red to Blue force
1.0 to 0.6, while
Red
ratio.
Red
to Blue force ratio
a reasonable tendency in theoretical battle situations.
Second, outcomes are analyzed as
and
is
and fractional casualties per unit time
to Blue casualty ratio increases as w^ varies
decreases as
Blue)
Figure 31 also depicts that the increasing rate of
compared
is
1.0 to 0.6,
(A,D= Red
2.
if
the w^ (Weiss parameter Blue engaging Red)
Red engaging Blue) approach
for the Blue force turns to area-fire
force turns to aimed-fire from area-fire.
54
the
same
value.
In other words,
from aimed-fire, and type of
Figure 33 shows that
Red
fire
for
casualties
1.2
W
r
:
WEISS PARAMETER BLUE ENGAGES TO RED
LEFT BAR
:
RIGHT BAR
:
RED CASUALTIES (WOE-1)
BLUE SURVrvORS (MOE-2)
1.1
B16.8
800.3
784.0
7B7.B
751.7
312.9
311.6
310.3
309.0
307.6
1.0
O.B
W-1.0
Figure 31.
Change
of Basic
parameter
for
W-O.B
W-0.9
W-D.7
W-0.5
(MOE-1 and MOE-2)
model output
based on Weiss
Blue engaging Red.
W
4.0
r
:
WEISS PARAMETER BLUE ENGAGES TO RED
LEFT BAR
RED TO BLUE CASUALTY RATIO
:
RIGHT BAR
:
RED TO BLUE FORCE RATIO
3.5 -
3.45
3.36
3.27
|—
3.19
|—
3.10
3.0
2.92
2.85
2.80
2.73
2.66
2.5 -
2.0
W=1.0
Figure 32.
Change
W=0.9
W=0.8
W=0.7
W=0.6
of Basic model output (Red to Blue casualty ratio and force ratio)
based on Weiss parameter for Blue engaging Red.
55
WB
WR
1.2
r
:
:
WEISS PARAMETER BLUE ENGAGES TO RED
WEISS PARAMETER RED ENGAGES TO BLUE
RED CASUALTIES (MOE-1)
LEFT BAR
RIGHT BAR
1.1
:
:
BLUE SURVIVORS (MOE-2)
-
1.0
7B4.4
777.6
Q
77n
7'°^
764.4
(—
L751. 7^7.6 ^r^«3^3^gn
299.6
295.4
291.1
286.6
0.9
C.B
W5-'.
WR = .5
Figure 33.
Change
W3=.95 W3=.9
WR=.55 WR=.5
WB = .3
WR=.7
WB-.85
WR=.65
of the Basic niodel output
W3-.75
WR=.75
(MOE-1 and MOE-2)
>>hen Weiss
parameters for both Red and Blue approach the same value.
Wa
WR
WEISS PARAMETER BLUE ENGAGES TO RED
;
:
WE!SS PARAMETER RED ENGAGES TO BLUE
RED TO BLUE CASUALTY RATIO
LEH- BAR
:
R.GHT BAR
3*h
:
RED TO BLUE FORCE RATiO
3.2 -
3.1
^^
3.05
3.0
.94
2.92
2.96
3.03
3.0
2.98
3.02
2.9B
2.1197
2.B -
2.6
W3=-.
WR=.5
Figure 34.
Change
WB=.95
WR=.55
WB=.9
WR=.6
W3=.S5
WR=.65
WB=.B
WR=.7
WB=.75
WR=.75
of Red/Blue casualty and force ratio in Basic model
parameters for both Red and Blue approach the same value.
56
nhen Weiss
(\10E-I) increase linearly as
The Red
ratio.
other, while
Red
when
is
\v^
comes greater than the casualty
case above.
first
One-Way
The
test
0.8
and
ANOVA
Table
to Blue force ratio be-
of the
model output and the output when
original basic
changed to
is
0.9
from
All value of F- Ratio
1.0.
7.
4.
'9
0.12
0.31
Red option-2
0.36
0.01
5.60
Red option-3
0.6S
0.05
0.29
Red option-4
0.6^
0.02
3.SS
PA-
are
compared values (MOE-1. MOE-2, and MOE-4) and the number of observasame
as
those
Table 7 (4.79 and
option-2.
shown
,4
5.60).
is
in
4.60.
1
the
previous
and
14)
There
which are
This implies that the
eight observations) with
is
e.xist
MOE-1
section (see
utilized.
Red
in
Red
To confirm
these results.
can be employed because
it
Two Sample
t-
Test^
was assumed and
option-1 and
Red
verified that
57
(F,
is
MOE-4
in
Red
if w^
{Weiss parameter
for both cases.
comparing two
:
MOE
8
The
values)
(Second Edition)
t-Test
observations
in the previous section.
Staiisiics
mean of
option-2.
have same variances and each observation (each battle output
3 Page 506. Morris H. DeGroot, Probability and
Wesley Publishing Company, inc. 19S6.
,4
larger than 4.60
result occurs for the
was applied
sample from a normal distribution as explained
,
casualties of eight Blue options
The same
to Blue casualty ratio of eight Blue options with
f
If the significant level, a
option-1 changes significantly
0.9.
Figure 30).
two values which are
mean number of Red
Blue engaging Red) decreases from 1.0 to
dom
and force
Red option-
0.05. the table value for F,
Red
to Blue casualty
and the Red
each
w, get closer to
RESULTS OF ONE-WAY ANOVA TEST WHEN THE WEISS
RAMETER BLUE ENGAGLNG RED ARE CHANGED.
Red option^;
MGE-l
MOE-2
MOE-4
Distribution with degree of freedom
(i.e..
0.7,
is
Red
and
7.
Since the
in
w,
that
to Blue casualty ratio
test is also applied in the sensitivity analysis
compares the
results for the test are given in
tions
Note
(MOE-2)
ratio thereafter.
Weiss parameter for Blue engaging Red
Table
Red
to Blue casualty ratio decreases as n\
to Blue force ratio increases.
ratio are almost equal
Finally, the
n\ get closer in value, while Blue survivors
Figure 34 depicts the diflerences between
decrease linearly.
and force
and
w^
,
is
a ran-
The above
by Addison-
two
The following examples show
significant cases are tested as examples.
of the two sample
the results
with null hypothesis and alternative hypothesis in case of the
t-test
two-tail test.
Example (Case
Let
A'
1
mean number of Red
be the
casualties of eight Blue options with
Red
casualties of eight Blue options with
Red
option- 1 in original Basic model output.
}'
Let
mean number of Red
be the
option- 1 after changing
v.\
from LO.
to 0.9
Test the following hypothesis with a significant level a
H,
Let
SI-
:
A'
=
= 0.05.
}•
and S\ be the sums of squares as
n
n:
Sl = >
follows:
(A-,
-
Xj
= )
Sl
(
}',
- yS'
(4.1)
;=U
Also
let
the statistic
U
be defined by the following relation:
Since the alternative hypothesis
U<
c,
or
l'>
c.
Pr(L'<c,)
,
is
two-sided, the test procedure would reject
where the constants
+ ?t{U>C:) =
is
2.18.
degrees of freedom (m+n-2),
of the
statistic
U
is
c,
and
c.
are chosen so that
when Hq
is
either
true,
a.
In this example, the sample sizes are
defined by equation (4.2)
Ci
//q if
m=8
and n=8, and the value of the
Also, by use of a table of the
and
c^
are -2.145
greater than 2.145
given specified level of significance, a
=
(U >
0.05.
58
c,).
and 2.145,
t
statistic
U
distribution with 14
respectively.
the null hypothesis
is
Since the value
rejected for the
Example (Case 2)
Let
.V
be the
mean Red
Red
to Blue casualty ratio of eight Blue options with
option-2 in original Basic model output.
Y be
Let
the
mean Red
option-2 after changing
w-\
Blue casualty ratio of eight Blue options with Red
to
to 0.9
from
In this example, the sample sizes are
defined by equation (4.2)
V
statistic
is
specified level of significance, a
These examples
in
Table
8.
=
{U >
and n=8, and the value of the
statistic
U
C2),
the null hypothesis
is
rejected for the given
0.05.
verify the results
from Table
Recall,
m=8
Also, as in the previous example, since the value of the
2.37.
greater than 2.145
is
1.0.
from the One-Way
there were only
8.
which were greater than 4.60 (table value
two
ANOVA
test
which
significant cases (4.79
is
shown
and 5.60)
for F-distribution with degree of freedom
1
and
14).
E.
COUNTERMEASURES
The
practicing
eral principles
weapon systems
analyst must be well aquainted with the
of countermeasures. for his task
of proposed countermeasures.
there
is
While
it is
often that of evaluating the potential
Also, countermeasures are
Indeed, the battle
survivability, especially for defenders.
advantages.
is
usually impossible to anticipate
a rather satisfactory or useful
mode
more gen-
is
often likely to improve
often based on "see-sawing"
enemy
reactions to an initiative,
of approach to this problem. The tank was
the countermeasure to the machinegun. while antitank mines
and shaped charge war-
heads are countermeasures to the tank threat.
In this model, two Blue courses of action are considered against two
of action.
firing
It is
assumed that the Blue force can increase
time of the Blue
minefield, while the
artillery
Red
and may also use smoke
also
force
weapons and can
may
mine density
for restricting the firepower of the Blue force's direct
The quantitative amounts of
is
also increase
by expanding the
in a given
use a mine-plow to break through a Blue minefield
assumed that Red smoke does not
which course of action
indirect fire
Red courses
affect its
own
It is
direct fire.
effects for the different parameters,
employed, are defined as follows:
59
fire.
depending on
If the
•
Red
forces use mine-plows
decrease by
:
70*^/0
the attrition rate coefficient for
minefields.
•
If the
force
Red forces use smoke
from Blue force.
:
decrease by 10°
•
If the Blue forces increase indirect
minutes from lu minutes.
•
If the Blue forces increase mine density
minefield bv lOO^/o.
Table
fire
o
expand
:
time of Blue
firing
increase the
;
Red
the attrition rate coefficient to
to 20
artillen."
number of mines
in a given
provides payoff matrices for Red and Blue courses of action, which shows
S
both Red casualties and Blue survivors {MOE-1 and MOE-2). Based on Table
8, if
Blue force can use only one of two courses of action, increasing
is
(MOE-l
than increasing mine density
Red
forces use the mine-plows.
Blue force.
:
316.4 > 313.1
).
number of mines
Blue
will
From
in a
Red and Blue
for the Blue force
lire
the above facts, the Blue
minefield do not have
may need another
that if
:
course of action to
employ
forces
much
:
822.6
commander can
effect
>
774.4,
better
is
MOE-2
on Red's mine-plows, so that
Red mine-plows. Table
attrite
the
perceive that the
S also
then available courses of action, the
all
how Red
casualties
change based on each course of action.
selects Its
when
S.
MOE-1
:
shows
Red
force
71S.S< 751.7,
305.0 < 307.6).
Figures 35 and 36 depict
ference in
)
better
values are better for the
(MOE-1
ha\e larger advantages than the Blue force (see Table
MOE-2
305. 9> 303.1
:
MOE
Recall thai the larger
indirect fire
However, when the Red forces use smoke, increasing mine density
than increasing indirect
for Blue
722. S> 705.1. .\10n-2
:
its
the
Red
For example,
casualties for Blue courses of action 2
course of action
difference of
(MOE-1) and Blue
MOE
3
or 4
Figure 35, there
in
and
(R3 or R4). Fhese two
survivors
3
(B2 and B3)
figures also
if
show
is
(MOE-2)
a small dif-
the
Red
force
a quantitative
values for each course of action, so the reader can easily
tell
the dif-
ferences by the height of each pillar.
Figures 37 and 38
show how Red
to Blue casualty ratio
on Red and Blue's respective course of
the trend of the
change of Red
Blue courses of action.
action.
From
to Blue casualty ratio
and force
ratio
change based
these figures, the reader can see
and force
ratio for
each Red and
Figure 38 has almost the reverse shape of that of Figure 37,
which implies that the Red
to Blue casualty ratio
nciiative correlation.
60
and the Red
to Blue force ratio
have
Table
8.
PAYOFF MATRIX FOR RED AND BLUE COURSE OF ACTIONS
(MOE-l/MOE-2).
RED
Force
use Mine-plow
Courses
c
use
Smoke
increase
Indirect
increase
fire
density
Force
do not
increase
Indirect
Mine
increase
Mine
density
use
NO
YES
BLUI
do not use Mine-plow
r action
Smoke
NO
YES
YES
718.8 305.0
758.1 310.3
888.3330.0
927.5 336.6
NO
683. 5 300.7
722.8;305.9
774.4 313.1
813.7 318.
YES
655.9 298.3
705.1 303.1
S22.6 316.4
862.0 322.2
NO
631.6,296.3
670.2/300.4
712.4/302.4
751.7 307.6
fire
In Figures 35 and 36, the following abbreviations apply:
Rl = Red force uses both .Mine-plow and Smoke.
R2 = Red
force uses only Mine-plow.
R3 = Red
force uses only
R4 = Red
force does
Smoke.
NOT
use either Mine-plow or Smoke.
Bl = Blue force increases both Indirect-fire and Mine density.
B2 = Blue force increases only
Indirect-fire.
B3 = Blue force increases only Mine density.
B4 = Blue force does
NOT
increase either Indirect-fire or
In Figures 37 and 38, the following abbreviations apply:
1
on Red courses of action = Rl
in Figures 35
and 36
Red courses of action = R2
in Figures 35
and 36
on Red courses of action = R3
in Figures 35
and 36
2 on
3
4 on Red courses of action =
R4
in Figures 35
and 36
1
on Blue courses of action
Bl
in Figures 35
and 36
2
on Blue courses of action
B2
in Figures 35
and 36
3
on Blue courses of action
B3
in
4 on Blue courses of action
= B4
Figures 35 and 36
in Figures 35
61
and 36
Mine
density,
RE3 CASJALTIES (M0E-1J
rc:^
Figure 35.
3-Dimensional view of pa\off matrix for Red casualties.
BLL'E
SURV\aDRS (MOE-2)
fP^
.tc'^HS^^^"*^
Figure 36.
3-Dimensional vie^ of payoff matrix for Blue survivors.
62
RED TO BLUE CASUALTY RATIO (MOE-3)
Bourses OP
Figure 37.
^0^
3-Dimensional vien of payoff matrix for the Red to Blue casualty
ratio.
RED TO BLUE FORCE RATIO
3.2
^loO^SES
Figure 38.
3-Dimensional
vie>v of
OF «n>OS
payoff matrix for the Red to Blue force ratio.
63
F.
MODEL ENRICHMENT
The important question answered
performing
chapter
To support
reasonable fashion?"
in a
in this
is
better for the
it
Red
force to
large versus
when
the
"yes" for this question then
"How
are they related ?"
Blue force ratio
is
Is
(i.e.,
ration of battle
is
(MOE-4)
?"
Red
arise
size ratio affect the at-
move
to Blue force ratio
These questions
provide
this question, this section will
answers to the following questions, "Can a Red and Blue force
tacking force's speed?"
model
"Is this deterministic
is
fast
when
If the
small?).
and "How^ does
Red
the
it
to
answer
affect the du-
because the duration of battle in the
Basic model did not van.- by changing any o[ the parameters during the sensitivity analyses in previous sections.
In order to solve these problems, equation (2.16) in
by equation
Sr.e^.
=
Chapter
was modified
2
as given
(4.3).
^old
^
"
^^
~
^
)
^^-/'^^''^^
-^mm
-^min
=
0.3 X
(
-j^
(4.3)
)
where
V
power
:
S,,,,
a^
flo
speed of attacking force for previous time step
:
S,,,,„
based on different \\'capon types
updated speed o[ attacking force
:
S-^^
factor,
:
minimum
speed for the attacking force
current attrition coelllcient
:
maximum
:
RF^
BF,
:
:
attrition coelTicient
Red
current
force size
current Blue force size
In equation (4.3),
it
is
assumed
of Red (attacker) force.
to the
ratio
is
Red
value.
to Blue force ratio affects the
In other words, the
maneuver
The
at least
The hnear function
The function shown
minimum
Red
Red
to Blue force ratio {S^,^ocRFJBF,).
greater than one.
force can
that
speed
is
0.3
It is
also
minimum
speed
is
assumed the Red
speed
proportional
to Blue force
a constant value under the assumption that the
0.3km per hour
for the
in right
is
force's
minimum
minimum
regardless of the
speed
is
shown
Red
not a linear function of Red Blue force
64
to Blue force ratio
in left portion
portion of Figure 39 represents the case
ratio.
Red
when
of Figure 39.
the attacker's
In this chapter, the
MMAMI •tZD-OiMr/IV)
taatiM sPQi>>ax{ir/W)
m-
«
^
.
a
^
^
i^
is
-^^"'^^
•
•^,,.''-^'''''^
a
3
'
•
9
^^.^-^^^^
ga
is
.
^^.^
•
LO
14
j>
Two
Figure 39.
U
1
u
different functions for
1
a
1
1
1
minimum
^^^^
1
xt
IB
\»
XD
speed.
linear function case, given in left portion of Figure 39,
is
applied to the basic model and
analyzed in Figures 40 and 41.
(MOE-3) between
the
speed) and the Enriched model (which
re-
Figure 40 shows differences of the Red to Blue casualty ratio
Basic model (which has constant
minimum
gards the attacker's
each parameter
is
minimum
speed as a linear function of Red Blue force ratio)
when
altered as follows:
NO CHANGE no parameters are changed
t MD Blue increases the Mine density.
:
:
f IF Blue
increases the Indirect
:
USE MP Red
:
USE SM
:
fire.
uses Mine-plow.
Red uses Smoke.
Figure 40 illustrates that as the mine density and indirect
so do
Red Blue
casualty ratios.
Even though there
increase for both models
When mine-plow and smoke
models, Red Blue casualty ratios decrease.
standpoint.
fire
exist
These
somewhat
65
are
employed
results are reasonable
for
both
from a mihtary
differences in increasing or decreasing
LEFT BAR
5
4
MODEL OUTPUT
BASIC
:
RIGHT BAR
:
ENRICHED MODEL OUTPUT
—
o
-
1
1
Z
3
2
-
—
-n
-
1
-
NO CHANGE
Figure 40.
Comparison of
t
M3
t
RED/BLUE
USE MP
IF
USE SM
casualty ratio
(MOE-3) between
Basic and
Enridied model.
LEFT 3AR
:
RIGHT BAR
232
BASIC MODEL OUTPUT
:
ENRICHED MODEL OLnPUT
-
W
•
5,
^228
%
o
z
o
-
1
-
1
—
I
1
-
-
NO CHANGE
Figure 41.
Comparison of duration
1
MD
of battle
model.
66
t IF
USE MP
(MOE-4) between
JSE
I
SM
Basic and Enriched
Red Blue casualty
ratio
a similar tendency based
between Basic and Advanced model, both models change with
on alteration of each parameter.
Figure 41 shows difTerences of duration of battle
and the Enriched model.
density or indirect
fire,
As
and
expected, battle time
battle time
is
shorter
is
(MOE-4) between
longer
when Red
when Blue
uses
its
the Basic model
increases
its
mine
mine-plow or smoke.
Based on the above analysis with both Figures 40 and 41, the Enriched model appears
to produce
more reasonable
results than the Basic model.
67
SUMMARY AND FUTURE DIRECTIONS
V.
SUMMARY
A.
This thesis has developed a deterministic air-land combat model using network attribute terrain data.
regiments,
it
may
Even though the model
utilized terrain data for battalions
(personnel, minefields, indirect
fire,
military planner with information
The priman' focus of the
hi this respect, the
m
Chapters
and
3
volatile to change.
model
Re-
for the at-
forth), the
model outputs (MOEs) provide the
on preparing better defensive plans.
thesis
parameters produced reasonable
were most
and so
in the
in the
Based on the disposition of the forces
tacking force to reach the defending force.
was performed
an assigned battle area
Three avenues of approach were employed
public of Korea.
model outputs,
The model was run under
accept terrain data for larged sized forces.
a given scenario of specific battle conditions using
and
was
to
demonstrate various methods
model provided
4.
for analyzing
sufficient data for analysis,
which
In Chapter 4, sensitivity analysis varying specific
results,
and
also determined those parameters
The important
results
from
which
sensitivity analyses are as
follows:
1.
If Blue
has proper conditions (discussed
ing an additional platoon at the
in
Chapter 4) on ambush points, reinforcis better than at the main force
ambush point
areas.
2.
As both maximum
attrition rate coefficients {a^
ti\eness (or loss rate exchange) of Blue force
3.
is
and
b^
)
increase, firepower efTcc-
gradually reduced.
Both Red casualties and Blue survivors hicrease as iv^ (Weiss parameter for Blue
engaging Redt decrease, but Red casualties are more sensitive to w^ than Blue survi\ors.
4.
When Weiss parameters for both Red and Blue approach the same value, both Red
and Blue casualties decrease and the Red Blue casualty ratio also decreases, which
implies that firepower effectiveness of R.ed force increases.
5.
The number of mines
in
Blue minefields do not have
much
effect
on Red's mine-
plows.
6.
If
Red
uses smoke, increasing the
increasing Blue indirect
number of mines
in
Blue minefields
is
better than
fire.
In addition to the above results, the model pointed out that Blue needs to consider
courses of action for attrition to
Red on Avenue-22
68
in the
BATTLE
1
new
case which was
C
discussed in section
of Chapter
The model was enriched by using
3.
modified
a
equation for computing the attacking units' speed.
B.
FUTURE DIRECTIONS
The
model developed
deterministic
The author focused
model.
is
methodologies, not covered in this
model
cient for each
is
how
weapon
neither a complete nor perfect
needed for improving the model. Other analytical
thesis,
to specify
is
on analyzing the model outputs with various
his efibrts
methodologies. Additional research
deterministic
in this thesis
and determine more
limitation of this
realistic attrition rate coeffi-
system, to be used in Lanchester's Linear and Square law.
However, simulation methods may be a possible approach to
tion rate coefficients for
A
should be explored.
weapon
find
more accurate
systems.
There are numerous factors which affect a battalion or regimental
Therefore, the model developed in this thesis can be expanded by adding
types such as Close Air Support
inclusion of these factors
outputs.
istic
As
If the battle area
and so
to be changed.
required.
used to
model more complicated, yet allows
the
different
has
weapon
more combat
systems.
The
more
real-
for
many
types in Figure 7 of Chapter 2) could be used.
terrain features (such as small
some
hills,
streams, bridges,
attrition rate coefficients
Simulations
may
may need
help to obtain data outputs for comparison to real world
and evaluation with
combat plans or
recommanded
historical data
is
necessary before this model
model be
that the deterministic nature of the
outcomes traceable to
is
scenarios.
specific input parameters.
gested that stochastic variables be treated by a
is
to generate
In other w^ords,
it
is
Monte Carlo
rameters are varied for different runs to maintain audit
retained, as op-
model
posed to applying stochastic processes, because the purpose of
battle
battle.
In order to deal with these factors, different analytical approaches are
test actual
It is
makes
forth) in small battle areas,
Verification
data.
(CAS) and other newly developed weapon
size
one way of dealing with new weapon systems, different value of m
for
{power factor based on
reseviors.
attri-
trail
this
process.
not sug-
Rather, the pa-
control for the purposes of
analysis.
Finally, because only
model
as
an
initial
run,
dismounted units
mounted
(infantry^ forces)
units (vehicles such as tanks,
should be considered for future runs.
were considered
APCs, and
in this
trucks) also
This requires some modifications to the model to
incorporate different parameters based on vehicle types.
69
APPENDIX
This
a
IS
formation of the input data for both models (Basic and Advanced) for
dismounted (Infantrv) unit
Arc
No.
1
1
1
5
6
7
2
2
3
4
4
31. 5
2. 1
0.006
4.
67.5
16.5
4.5
0.
4,
1. 1
0.
356
506
4
15. 33
0.358
9
41.
32.
28.
40.
60.
15
2. 75
1.9
0.
5
11
5
7
9
5
6
6
11
7
12
13
10
19
11
15
15
16
17
18
19
8
9
9
10
10
10
11
11
12
12
13
14
14
15
15
16
16
17
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
20
20
21
22
22
23
24
24
42
43
44
25
26
27
Speed
2
6
3
8
10
11
12
13
14
cases.
Passing
Distance
Nodes
Width
(tail -head) time(min)
2
3
4
MODELS INPUT DATA
A.
7
8
20
14
15
13
14
25
16
17
16
18
17
18
23
24
21
20
21
28
22
23
27
27
25
26
26
25
57
67
29
1.
2.
15
2.
35
0.
0.558
0. 004
0.
3.
21.43
31.5
1.
Oil
254
112
0.554
25
0.
2. 1
106
1.45
004
006
0. 002
0. 026
0. 002
0. 002
0. 002
0. 098
0. 054
0. 154
0. 049
0.406
0. 066
0. 258
1. 3
0.
0. 9
0.
4.
4.5
4.
3.5
4. 5
3. 5
3.
3.5
4.
15. 71
33. 75
25.
15
0.
3.5
2.25
0.
4.
54.0
3. 6
65.
75.
3.25
1.
1.
25
75
5. 35
3.
107.0
18. 67
34. 29
19. 71
1.4
2.0
1.
15
27.43
52.5
1.
6
11. 25
19. 33
26.
15. 43
17. 14
14.25
8.
67
86
10.5
51. 43
12.
21. 75
3. 5
0.
75
75
0. 7
3.0
1.45
13.5
18. 86
9. 43
0. 9
12.
0. 7
8.
67
27
8.57
12. 86
8.0
8.0
28
33. 75
204
0. 084
0.076
0. 328
0. 354
0. 256
0. 049
0. 176
0.506
0. 224
0. 704
0.454
0. 678
0.574
0.274
0. 328
0.408
0. 066
1.0
0. 95
0. 65
0.
1. 1
0.55
0.65
0.5
0.
75
0.6
0.6
2.
152
25
70
3.
4.
3.
3.
3.
4.5
3.5
3. 5
3.5
4.
4.
4. 5
3.
3. 5
3.5
4.
4.5
3.5
4.
3.5
4.
4.
3.5
3.5
3.5
4.5
3.5
3.5
4.5
4.5
4.
APPENDIX
This
is
a
BASIC
B.
computer program using Fortran-77
--< BASIC MODEL >---
MODEL
for the basic model.
LEE, JAE YEONG (1988.2.6) ---
MADE BY
INTEGER
NA, NN, NOA, NOP, RBN, RRESER, BOO, PLTN
PARAMETER (NA=44, NN=28, N0A=3, N0P=2 PLTN=40)
PARAMETER (RBN=500, RRESER=150, BCO=170)
REDAVE(NOA) ,SMAVE(NOA) ,PMAVE(NOA)
INTEGER
ARC(NA) ,TAIL(NA) ,HEAD(NA) ,TAVE(NOA)
INTEGER
INTEGER
PMARC(5),SMARC(3),IDARC(4,0: 1001)
INTEGER
IFA(4, 1000), MFA(4, 1000), DFA(4, 1000)
P0IF,BLUEP1,BLUEP2,ANS1,ANS2,ANS3,ANS4
INTEGER
INTEGER
INPUT, OUTPUT, TP1,TP2,DIF,DPMF,DSMF,INCRE,WID
IFANS,MFANS,DFANS,TEND
INTEGER
REAL
DIST(NA) ,WIDTH(NA) ,TIME(NA)
REAL
SPEED(NA) ,OSPEED(NA) ,ADSP(NA)
BREAKP(4),BREAKT(4)
REAL
REAL
RF(4,0: 1001 ) ,BF( 4,0: 1001) ,WHERE( 4,0: 1001) ,SP( 4,0: 1001)
REAL
IFC(4,0: 1001) ,MFC(4,0: 1001) ,RDFC(4,0: 1001),
,
IMDC(4,0: 1001), BDFC(4,0: 1001)
RBP(4) ,IFSUM(4) ,MFSUM(4) ,RDFSUM(4) ,BDFSUM(4)
IFCAS MFCAS DFCAS BLEFTl BLEFT2 WHOLD
CLOCK, RANGE, BRDI ST, AR,RB
1
REAL
REAL
REAL
,
DATA
DATA
DATA
DATA
DATA
DIF,DPMF,DSMF
PMAVE
PMARC
RANGE
DATA
DATA
DATA
DATA
REDAVE
POIF
SMAVE
BLUE PI, BLUER
RB
,
,
,
,
/10,20,15/
71,3,1/
715,19,23,25,32/
/l.l/
/O.O/
70,2,1/
72/
70,2,1/
/1, 2/
INPUT = 12
OUTPUT = 13
•OPEN OUTPUT FILES
•FILE 'TSTEP' HAS RED AND BLUE FORCE SIZES ON EACH AVENUE
•FILE 'REDCAS' HAS RED CASUALTIES ON EACH AVENUE
OPEN ( UNIT = OUTPUT, FILE = TSTEP )
OPEN ( UNIT =21,
FILE = REDCAS )
!
'
'
'
'
WRITE(6,9)
DO 10 I = 1, NA
READ( INPUT, 11) ARC(I),TAIL(I),HEADCI),TIME(I),DIST(I),WIDTH(I),
1
SPEED(I)
WRITE( 6,11) ARC( I ) ,TAIL( I ) ,HEAD( I ) ,TIME( I ) ,DIST( I ) ,WIDTH( I )
71
SPEED(I)
CONTINUE
FORMAT(lX,50('-'),/2X,'ARC TAIL-HEAD TIME
DIST
1
SPEED' ,/lX,50('-'))
F0RMAT(3X,I2,2(3X,I2) ,2X,F6. 2,6X,F4. 2,4X,F5. 3,3X,F5. 3)
1
10
9
WIDTH
'
11
OPTION (NA,N0A,REDAVE,P0IF,SMAVE,BLUEP1,BLUEP2, SPEED,
DIST, TIME, PMARC,SMARC, ANSI, ANS2,ANS3)
IF(ANS1 .EQ. 0) GO TO 999
IF(ANS2 .EQ. 0) GO TO 999
IF(ANS3 .EQ. 0) GO TO 999
CALL
1
DO 19 I = 1,NA
CALL ASPEED ( I ,NA, ANSI SPEED, ADSP)
SPEED(I) = ADSP(I)
OSPEED(I) = SPEED(I)
,
19
WRITEC 13 25 )REDAVE( 1 ) ,REDAVE( 2) ,REDAVE( 3) SMAVE( 1) SMAVE( 2)
SMAVE(3) ,SMARC( 1) ,SMARC(2) ,SMARC(3) BLUER 1 ,BLUEP2
WRITEC 6 25 ) REDAVEC 1 ) REDAVEf 2 ) REDAVE( 3 ) SMAVE( 1 ) SMAVE( 2 )
1
SMAVEC 3 ) SMARCC 1 j S.MARC( 2 ) SMARC( 3 ) BLUEPl BLUEP2
F0RMAT(//1X,'RED OPTION ===> \3I3 ,/lX, BLUE OPTION',
SCATERABLE MI.NEFIELD ===>', 313,
1/^X,
(LOCATED ARCS)
==>',3I3,
1/4X,'
1/4X,'^^ FORCE ALLOCATION =====>' ,213,//)
,
,
,
,
,
25
,
.
,
'
NNN =
I USED =
NEQ12 =
NEQ34 =
I = 1,4
IFSUM(i) = 0.
MFSUM(I) = 0.
RDFSUM(I) = 0.
BDFSUM(I) =0.0
CONTINUE
DO
5
C-- -DEFINE INITIAL FORCE LEVEL AND LOCATION
RF(1,0) = REDAVE(l)-"-RBN
RF(2,0) = (REDAVE(2)-RBN)/2
RF(3,0) = (REDAVEC 2 )^'-RBN)/2
RF(4,0) = REDAVEC 3 )-"-RBN
DO 28 I = 1,4
WHERE(I,0) = 0.
IFCd.O) = 0.
MFC(I,0) = 0.
RDFCd.O) = 0.
IMDC(I,0) = 0.
BDFC(I,0) = 0.
RBP(I) = RB^'--RF(I,0)
28
BF(1,0) = ELU-EPr-'-'BCO + PLTN/2
BF(2,0) = BLUEPl--'-BCO + PLTN/2
BF(3,0) = BLUEP2^--BC0 -f PLTN/2
BF(4,0) = BLUEP2--'-BC0 + PLTN/2
NOC =
72
,
,
,
'•••-
5
,
,
1
!
,
INCRE = 1
DO 9999 IT = 1,500, INCRE
CLOCK = IT
DO 30 lA = 1,4
BREAKP(IA) = 0.
BREAKT(IA) = 0.
IFC(IA,IT) = IFC(IA,IT-1)
MFC(IA,IT) = MFC(IA,IT-1)
RDFC(IA,IT) = RDFC(IA,IT-1)
IMDC(IA,IT) = IMDC(IA,IT-1)
BDFC(IA,IT) = BDFC(IA,IT-1)
RF(IA,IT) = RF(IA,IT-1)
BF(IA,IT) = BF(IA,IT-1)
WHOLD = VHERE(IA,IT-1)
CALL DETECT ( I A, IT, INCRE, WHOLD, WID)
IF(WID .EQ. 0) GO TO 999
IDARC(IA,IT) = WID
BRDIST = 9999.
SP(IA,IT) = SPEED(WID)
TS = (SPEED(WID)/60)^'-INCRE
WHERE(IA,IT) = WHOLD + TS
C---IS THERE INDIRECT FIRE ? OR MINEFIELD ? OR DIRECT FIRE
CALL IDFIND ( lA CLOCK, POIF, IFANS)
CALL MFFIND ( PMARC SMARC ,WID ,MFANS)
CALL DRFIND ( I A, WHOLD, RANGE, BRDIST, DFANS)
?
,
,
IFA(IA,IT) = IFANS
MFA(IA,IT) = MFANS
DFA(IA,IT) = DFANS
IF( IFANS. EQ.O .AND.
MFANS. EQ.
.AND.
DFANS. EQ.O)
GOTO
C-- -COMPUTE THE CASUALTY OF INDIRECT-FIRE
IF( IFANS .EQ. 1) THEN
CALL IDFIRE ( lA, IT,RF,REDAVE,POIF, IFCAS)
RF(IA,IT) = RF(IA,IT) - IFCAS
IFSUM(IA) = IFCAS + IFSUM(IA)
IFC(IA,IT) = IFC(IA,IT) + IFCAS
!
END IF
C---COMPUTE THE CASUALTY BY MINE-FIELD
IFCMFANS.EQ. 1
.OR.
MFANS. EQ. 2) THEN
CALL MFIELD ( lA, IT,MFANS INCRE,NA,WID,TIME,RF,MFCAS)
RF(IA,IT) = RF(IA,IT) - MFCAS
MFSUM(IA) = MFCAS + MFSUM(IA)
MFC(IA,IT) = MFC(IA,IT) + MFCAS
END IF
!
,
C-- -COMPUTE THE CASUALTY BY DIRECT-FIRE
1
1
IF(DFANS .EQ. 1 .OR.
IF( RF(IA,IT) .LT.
lUSED EQ.
RF(IA,0) .GT.
!
DFANS EQ.
BF(IA,IT)
.
.
2) THEN
.AND.
AND.
.
0.
)
73
THEN
30
RF(IA,IT) = RF(IA,IT) + RRESER
I USED = 1
END IF
CALL DRFIRE (DFANS ,DFA,BRDIST, lA, IT,RF,BF, AR,RDFCAS BDFCAS
PLTN, RANGE, NNN, AGO, AMO)
,
1
RF(IA,IT) = RF(IA,IT) - RDFCAS
BF(IA,IT) = BF(IA,IT) - BDFCAS
RDFSUM(IA) = RDFCAS + RDFSUM(IA)
BDFSUM(IA) = BDFCAS + BDFSUM(IA)
RDFC(IA,IT) = RDFC(IA,IT) + RDFCAS
BDFC(IA,IT) = BDFC(IA,IT) + BDFCAS
ENT)
IF
.AND.
IF(RF(IA,0) .GT. 0.
RF(IA,IT) LE. RBP(IA)) THEN
CALL BREAK ( lA, IT,NOC,WHOLD,RF,BF,RRESER,BREAKP,BREAKT)
EN^ IF
.
C---COMPLTE THE REDUCED SPEED DUE TO THE INDIRECT FIRE OR MINEFIELD OR
UNDER THE DIRECT FIRE BASED ON THE RANGE OF TVs'O FORCES
C
!
CALL
RSPEED
(
1
lA IT, NA, SPEED, TIME ,DIST,WID, RANGE ,AR,RF,
IFANS,MFANS, DFANS, OSPEED,ACO,AMO)
.
SP(IA,IT) = SPEEDfWID)
TS = (SPEED(WID)/60)--'--INCRE
WHERE(IA,IT) = WHOLD + TS
IMDC(IA,IT) = IFC(IA,IT) + MFC(IA,IT) + RDFC(IA,IT)
30
36
37
CONTINUE
IF(NEQ12 .NE. 0) GO TO 36
BF(1,IT) = MIN( BF(1,IT) ,EF(2,IT)
EF(2,IT) = MIN( BF(1,IT) ,BFC2,IT)
IF(NEQ34 .NE. 0) GO TO 37
BF(3,IT) = MIN( BF(3,IT) ,BF(4,IT)
BF(4,IT) = MIN( BF(3,IT) ,BF(4,IT)
IF(NEQ12 .NE. 0) GO TO 38
IF(DFA(1,IT) EQ. 2
AND.
DFAf2,IT) .EQ. 2 .AND.
RF(2,IT) .GT. 0. ) THEN
RF(1,IT) .GT. 0.
.AND.
1
BF(1,IT) = (RF(1,IT) / (RF(1,IT)+RF(2,IT))) • BF(1,IT)
BF(2,IT) = (RF(2,IT) / (RF(1,IT)+RF(2,IT))) * BF(2,IT)
NEQ12 = IT
END IF
IF(NEQ34 .NE 0) GO TO 39
IF(DFA(3,IT) .EQ. 2
DFA(4,IT) .EQ. 2 .AND.
AND.
THEN
1
RF(3,IT) GT. 0.
AND.
RFC 4, IT) .GT. 0. )
(RF(3,IT)+RF(4,IT))) ^ BF(3,IT)
BF(3,IT) = (RF(3,IT)
(RF(3,IT)+RF(4,IT))) * BF(4,IT)
BF(4,IT) = (RF(4,IT)
NEQ34 = IT
END IF
WRITE (OUTPUT, 40) IT,RF(1,IT),RF(2,IT),RF(3,IT),RF(4,IT)
WRITE( OUTPUT, 40) IT BF( 1 0) -( BDFC( 1 IT)+BDFC( 2 IT) )
BF( 2 0) -( BDFC( 1 IT)+BDFC( 2 IT) )
1
) -( BDFC( 3
IT)+BDFC( 4 IT) )
BF( 3
1
BF(4,0)-(BDFC(3,IT)+BDFC(4,IT))
1
.
38
.
.
39
,
,
,
,
,
,
,
,
,
,
74
VRITE(OUTPUT, 42) IT, VHERE( 1 IT) ,WHERE( 2 IT) ,WHERE( 3 IT) VHERE(4 IT)
WRITE( OUTPUT, 42) IT, SP( 1 IT) ,SP( 2 IT) ,SP( 3 IT) SP(4, IT)
,
,
,
,
,
,
,
,
,
40
41
42
WRITE(21,40) IT,IFC(1,IT),IFC(2,IT),IFC(3,IT),IFC(4,IT)
WRITE(21,40) IT,MFC(1,IT),MFC(2,IT),MFC(3,IT),MFC(4,IT)
WRITE(21,40) IT,RDFC(1,IT),RDFC(2,IT),RDFC(3,IT),RDFC(4,IT)
WRITE(21,40) IT,IMDC(1,IT),IMDC(2,IT),IMDC(3,IT),IMDC(4,IT)
VRITE(21,40) IT,BDFC(1,IT),BDFC(2,IT),BDFC(3,IT),BDFC(4,IT)
F0RMAT(1X,I6,4(3X,F6. D)
F0RMAT(1X,I6,4(3X,I6))
F0RMAT(1X,I6,4(3X,F6.3))
9999
CONTINUE
CALL
999
1
C
C
C
PRINT (RBN,BC0,BLUEP1 ,BLUEP2,RRESER,PLTN,NA, OUTPUT,
IFSUM,MFSUM,RDFSUM,BDFSUM)
PRINT FOR CHECKING THE INPUT DATA
VRITE(6,9)
DO 998 I = 1, NA
WRITE(6,11) ARC( I) ,TAIL( I) ,HEAD( I) ,TIME( I) ,DIST( I) ,WIDTH(I)
1
SPEED(I)
WRITE(6 '(//)')
STOP
THANK YOU
'this MODEL IS COMPLETE
END
998
'
!
*
!
SUBROUTINE FOR SELECTING THE RED AND BLUE OPTIONS
1.
L JU ^. ^. .*« ^. ^. ^. J. >t.
SUBROUTINE OPTION (NA,NOA,REDAVE ,POIF,SMAVE ,BLUEP1 ,BLUEP2 SPEED,
DIST, TIME, PMARC,SMARC, ANSI, ANS2,ANS3)
,
1
INTEGER
INTEGER
REAL
1002
REDAVEC NOA) POIF SMAVE( NOA) BLUEPl BLUEP2
,
,
,
,
PMARC(5),SMARC(3),ANS1,ANS2,ANS3,ANS4
SPEED(NA) ,DIST(NA) ,TIME(NA)
WRITE(6, '(///)')
PRINT
HOW DO RED FORCE ALLOCATE THEIR UNIT ON EACH AVENUE
PRINT
(CHOOSE THE NUMBER OF OPTION 1,2,3 OR 4)'
PRINT ^S
*
PRINT ^'S
OPTION
AVE-//1
AVE-|'/2
AVE-//3
*
*
PRINT *,
1.
1 BN
1 BN
1 BN
"
*
PRINT" ^S
2.
BN
2 BN
1 BN
^
*
PRINT ''%
3.
BN
1 BN
2 BN
*
*
PRINT *,
4.
BN
3 BN
BN
*
PRINT *,
0.
EXIT THE RPOGRAM
PRINT ^^
READ(5,*) ANSI
GO TO (1011,1012,1013,1014) ANSI
REDAVE(l) = 1
REDAVE(2) = 1
REDAVE(3) = 1
GO TO 1010
REDAVE(l) =
••-,
-•'•-,
''•-
!
1011
1012
75
?
1013
1014
REDAVE(2) =
REDAVEO) =
GO TO 1010
REDAVE(l) =
REDAVE(2) =
2
REDAVEO) =
2
1
1
GO TO 1010
REDAVE(l) =
REDAVE(2) = 3
REDAVE(3) =
GO TO 1010
IF(ANS1 EQ. 0) THEN
GO TO 1999
ELSE
=> TRY AGAIN
PRINT^>,
ERROR >>
ENTER THE NUMBER 1-4 OR
GO TO 1002
END IF
WRITE (6 (///))
PRINT
'HOW DO BLUE FORCE ALLOCATE THE SCATTERABLE MINES ?
PRINT
(CHOOSE THE NUMBER OF OPTION 1,2,3 OR 4)'
PRINT
PRINT *
OPTION
AVE-fil
AVE -,-'^2
AVE-,^r'3
1.
PRINT
PKG
1 PKG
1 PKG
PRINT
2.
PKG
2 PKG
1 PKG
3.
PRINT
PKG
1 PKG
2 PKG
4.
PRINT 'PKG
3 PKG
PKG
0.
PRINT '>
EXIT THE PROGRAM
VfVrVrV.-VrvVVrVrVf'!Vyr'!ViVVriVVrVrVrVfVriVVfVriVVrVfVfVrVr'!V-'-VrVr';VVr'5'rVfVr'!V'5V'A'')V'>V'5V^V
PRINT
READ(5,->)
ANS2
GO TO (1031,1032,1033 1034) ANS2
SMAVE(l) = 1
SMAVE(2) = 1
SMAVE(3) = 1
GO TO 1025
SMAVE(l) =
S:1AVE(2) = 2
SMAVE(3) = 1
GO TO 1025
SMAVE(l) =
SMAVE(2) = 1
SMAVE(3) = 2
GO TO 1025
SMAVE(l) =
SMAVE(2) = 3
SMAVE(3) =
GO TO 1025
IF(ANS3 EQ. 0) THEN
GO TO 1999
ELSE
=> TRY AGAIN
PRINT'^
ERROR >>
ENTER THE NUMBER 1-4 OR
GO TO 1010
END IF
«
1010
!
^•>-
-••-
'
''-
'"'
^••-
'"'
-,v
!
''
1031
1032
1033
1034
«
1025
WRITE(6, '(///)')
PRINT *, 'WHICH ARC WILL HAVE SCATTERABLE MINES ?
PRINT ^'^
ENTER THE THREE INDEX NUMBER OF ARCS FROM
(
!
'
'
76
1
TO 44)'
(SMARC(I), 1=1,3)
READ(5,*)
IF(SMARC(1).GT. 44 .OR. SMARC( 2) GT. 44 OR. SMARC( 3) GT. 44) THEN
'« ERROR
INPUTS MUST BE LESS THAN 44=>TRY AGAIN
PRINT--'-,
GO TO 1025
ELSE
WRITE(6,1026) (SMARC(I), 1=1,3)
SCATTERABLE MINES WILL BE ON ',
1026 F0RMAT(1X,'« NOTE
112,' ,' ,12,' ,' ,12,' ARCS')
END IF
.
»
»
1020
.
.
:
'
!
:
WRITE(6, '(///)')
PRINT*, 'HOW DO BLUE FORCE ALLOCATE THEIR UNIT ON EACH POSITION ?'
(CHOOSE THE NUMBER OF OPTION 1 OR 2)'
PRINT *,
PRINT *,
* OPTION P0SITI0Wn(N0DE28)
PRINT *,
POSITION#2(NODE27) *'
*•
*
1.
2 CO
1 CO
PRINT *,
*
*'
2.
1 CO
2 CO
PRINT *,
*'
'V
0.
EXIT THE RPOGRAM
PRINT *,
PRINT '%
READ(5,'"^)
ANS3
IF(ANS3 .EQ. 1) THEN
BLUER 1 = 2
BLUEP2 = 1
GO TO 1999
ELSE IF(ANS3 EQ. 2) THEN
BLUEPl = 1
BLUEP2 = 2
GO TO 1999
ELSE IF(ANS3 EQ. 0) THEN
GO TO 1999
ELSE
PRINT'^ '« ERROR
ENTER THE NUMBER 1,2 OR
TRY AGAIN
GO TO 1020
END IF
RETURN
!
.
.
»
1999
=>
:
'
!
E.ND
>''
2.
SUBROUTINE FOR DETECTION WHICH ARC THE RED FORCES ARE
SUBROUTINE
INTEGER
REAL
DATA
DATA
DATA
DATA
3011
DETECT
(
lA, IT, INCRE,WHOLD,WID)
INCRE,WID
AU4),A21(8),A22(6),A3(6),WH0LD
Al
A21
A22
A3
72.75,6.35,9.35,10.25/
/I. 15,3. 3,4. 7,5.45,6.4,7. 1,8.55,9.45/
/I. 15 ,3. 3,4. 7 ,6. 15 ,6. 8, 7. 45/
/4. 5,6. 6,8. 85,12. 35,12. 95,13. 55/
GO TO (3011,3012,3013,3014) lA
IF(WHOLD .LE. Al(l)) THEN
WID = 5
GO TO 1390
ELSE IF(WHOLD. GT. Al(l)
.AND.
WID = 15
GO TO 1390
77
WHOLD. LE. Al(2)) THEN
?
ELSE IFCWHOLD. GT. Al(2)
.AND. WHOLD. LE. Al( 3) ) THEN
WID = 33
GO TO 1390
ELSE IFCWHOLD. GT.AK 3)
.AND. WHOLD. LE. Al(4) ) THEN
WID = 35
GO TO 1390
ELSE IFCWHOLD GT. A1C4)) THEN
WRITEC13,1380)
IT-1
F0RMATC1X,'RED FORCE IN AVE#1 TOOK BLL^ POSITIOW/1
BATTLE END
',14,' MINUTE)")
1/5X,'C BATTLE TIME
PRINT -'s 'RED FORCE IN AVE^.'l TOOK BLUE P0SITI0N7/1
BATTLE ENT)
PRINT '-, 'BATTLE TIME =',IT-1,' MINUTE'
GO TO 1399
END IF
IFCWHOLD LE. A21C1)) THEN
WID = 4
GO TO 1390
ELSE IFCWHOLD. GT.A21C1) .AND. WHOLD. LE. A21C 2)) THEN
WID = 7
GO TO 1390
ELSE IFCWHOLD. GT.A21C2) .AND. WHOLD. LE. A21C 3) ) THEN
WID = 19
GO TO 1390
ELSE IFCWHOLD. GT.A21C 3) .AND. WHOLD. LE. A21C4) ) THEN
WID = 1^
GO TO 1390
ELSE IFCWHOLD. GT.A21C4) .AND. WHOLD. LE. A21C5) ) THEN
WID = 29
GO TO 1390
ELSE IFCWHOLD. GT.A21C 5) .AND. WHOLD. LE. A21C 6) ) THEN
WID = 32
GO TO 1390
ELSE IFCWHOLD. GT.A21C 6) .AND. WHOLD. LE. A21C 7 ) ) THEN
WID = 34
GO TO 1390
ELSE IFCWHOLD. GT.A21C 7) .AND. WHOLD. LE. A21C 8) ) THEN
WID = 35
GO TO 1390
ELSE IFCWHOLD GT. A21C8)) THEN
WRITEC 13,1381)
IT-1
BATTLE ',
F0RMATC1X,'RED FORCE IN AVE,^i21 TOOK BLUE P0SITI0N//1
',14,'
I'END
MLNUTE)')
,/5X,'CBAlTLE TIME
BATTLE END
PRINT -^ 'RED FORCE IN AVE.>'21 TOOK BLUE POSITIONy/1
PRINT -^ 'BATTLE TIME =',IT-1,' MINUTE'
GO TO 1399
END IF
IFCWHOLD LE. A22C1)) THEN
WID = 4
GO TO 1390
ELSE IFCWHOLD. GT.A22C1) .AND. WHOLD. LE. A22(2) ) THEN
WID = 7
GO TO 1390
ELSE IFCWHOLD. GT.A22C2) .AND. WHOLD. LE. A22(3)) THEN
WID = 19
GO TO 1390
ELSE IFCWHOLD. GT.A22C 3) .AND. WHOLD. LE. A22C4) ) THEN
.
1380
=>
:
3012
'
!
:
'
!
.
.
1381
=>
'
!
,
3013
:
:
.
78
'
!
WID = 25
GO TO 1390
ELSE IFCWHOLD. GT. A22(4) .AND. VHOLD. LE. A22(5) ) THEN
WID = 30
GO TO 1390
ELSE IFCWHOLD. GT. A22(5) .AND. WHOLD. LE. A22(6) ) THEN
WID = 39
GO TO 1390
ELSE IFCWHOLD GT. A22(6)) THEN
IT-1
WRITE(13,1382)
F0RMAT(1X,'RED FORCE IN AVE^^22 TOOK BLUE P0SITI0N?',f2
BATTLE
1 END !' ,/5X,' (BATTLE TIME
',14,' MINUTE)')
BATTLE END
PRINT *, 'RED FORCE IN AVE//22 TOOK BLUE P0SITI0N#2
PRINT *, 'BATTLE TIME =',IT-1,' MINUTE'
GO TO 1399
END IF
IFCWHOLD LE. A3(l)) THEN
WID = 2
GO TO 1390
AND. WHOLD. LE. A3C2)) THEN
ELSE IFCWHOLD. GT.A3(1)
WID = 11
GO TO 1390
AND. WHOLD. LE. A3(3)) THEN
ELSE IFCWHOLD. GT.A3C 2)
WID = 13
GO TO 1390
ELSE IFCWHOLD. GT.A3( 3)
AND. WHOLD. LE. A3C4)) THEN
WID = 23
GO TO 1390
ELSE IFCWHOLD. GT.A3C4)
AND. WHOLD. LE.A3C5)) THEN
WID = 42
GO TO 1390
ELSE IFCWHOLD. GT.A3C 5)
AND. WHOLD. LE. A3(6)) THEN
WID = 43
GO TO 1390
ELSE IFCWHOLD GT. A3C6)) THEN
IT-1
WRITEC13,1383)
F0RMATC1X,'RED FORCE IN AVE#3 TOOK BLUE P0SITI0N#2
BATTLE
1' END !' ,/5X,'C BATTLE TIME ==> ',14,' MINUTE)')
PRINT ^', 'RED FORCE IN AVE#3 TOOK BLUE P0SITI0N//2
BATTLE END
PRINT '"s 'BATTLE TIME =',IT-1,' MINUTE'
GO TO 1399
END IF
RETURN
WID =
RETURN
END
.
1382
'
3014
'
=>
,
'
!
.
.
1383
1390
1399
*
3.
'
SUBROUTINE FOR IDENTIFYING THE INDIRECT-FIRE
!
SUBROUTINE IDFIND ( IA,CLOCK,POIF, IFANS)
INTEGER POIF, IFANS
REAL
CLOCK, STIF.FTIF
DATA
STIF,FTIF /5. 0,15.0/
IFCCLOCK .GE. STIF .AND.
IFCPOIF .EQ. 2) THEN
CLOCK
79
.
LT.
FTIF) THEN
,
'
!
IFCIA.EQ. 2 .OR.
I FANS = 1
GO TO 1444
THEN
ELSE
IFANS =
GO TO 1444
END IF
ELSE IFCPOIF .EQ. 3) THEN
IF(IA .EQ. 4) THEN
IFANS = 1
GO TO 1444
ELSE
IFANS =
GO TO 1444
END IF
END IF
ELSE
IFANS =
END IF
RETURN
END
1444
'^
lA. EQ. 3)
SUBROUTINE FOR IDENTIFYI.N'G THE MINE -FIELD
4.
!
SUBROUTINE MFFIND ( PMARC SMARCWID, MEANS)
INTEGER MFANS,WID,PMARC(5) ,SMARC(3)
.
DO 1510 I = 1,5
IF(WID .EQ. PMARC(I)) THEN
MFANS = 1
GO TO 1599
END IF
CONTINUE
DO 1520 J = 1,3
IF(WID .EQ. SMARC(J)) THEN
MFANS = 2
GO TO 1599
END IF
CONTINUE
MFANS =
RETURN
END
510
1511
3
1520
1521
1599
*
SUBROUTINE FOR IDENTIFYING THE DIRECT-FIRE
5.
SUBROUTINE
1
1
DRFIND
(
IA,WHOLD,RANGE,BRDIST,DFANS)
INTEGER
REAL
IC,IA,DFANS
WH0LD,PATH(4) ,RAxNGE,BRDIST,AMBUSH
DATA
DATA
PATH
AMBUSH
/ 10.
25 ,9. 45 ,7, 45
,
13.
55/
/4. 0/
IF(WHOLD.GE. AMBUSH-RANGE/2.
WHOLD LT. AMBUSH
.
(IA.EQ.2.
.OR.
!
.AND.
AND.
.
THEN
lA. EQ. 3))
80
DFANS = 1
BRDIST = AMBUSH - WHOLD
GO TO 1699
ELSE IF( WHOLD GE. PATH( lA) -RANGE .AND.
THEN
WHOLD .LT. PATH(IA)
1
)
DFANS = 2
BRDIST = PATH(IA) - WHOLD
GO TO 1699
END IF
1600 CONTINUE
DFANS =
1699 RETURN
END
.
*
SUBROUTINE FOR COMPUTING CASUALTIES BY INDIRECT-FIRE
6.
SUBROUTINE IDFIRE(IA,IT,RF,REDAVE,POIF,IFCAS)
INTEGER NR0UND,IA,IT,REDAVE(3),P0IF
LETHAL, R0WSP,C0LSP, MUX, MUY,RHOX,RHOY, RATIO, PKILL
REAL
REAL
RF(4,0: 1001),IFCAS
DATA
DATA
DATA
NROUND LETHAL
MUX,MUY
RHOX,RHOY
18, 70. 8/
/15. ,10./
/15. ,10./
/
,
IF(REDAVE(POIF) EQ.
ROWSP =12.
COLSP = 12.
GO TO 1669
ELSE IF(REDAVE(POIF)
ROWSP =10.
COLSP = 10.
GO TO 1669
ELSE IF(REDAVE(POIF)
ROWSP = 8.
COLSP = 8.
GO TO 1669
END IF
.
1
EQ.
2)
THEN
EQ.
3)
THEN
RATIO = (NROUND'> LETHAL)
PHI = 3. 141592654
1669
THEN
)
/
(ROWSP'<COLSP*RF( lA, IT)
A = ((LETHAL-"-NR0UND)/(2''^PHI))*'V. 5
PKILL = RATIO
(A''^A/((A-"-A+RH0X''">2)*(A'>A+RH0Y**2))**. 5) *
EXP(-.5*((MUX*'V2)/(A'^A+RH0X''f*2) + (MUY**2)/( A'^A+RH0Y**2)
1
IFCAS = PKILL * RF(IA,IT) / 2.
--'^
RETURN
END
*
SUBROUTINE FOR COMPUTING CASUALTIES BY MINEFIELDS
7.
SUBROUTINE MFIELD
'
(
lA, IT,MFANS INCRE,NA,WID,TIME,RF,MFCAS)
,
INTEGER
REAL
MFANS,INCRE,NA,NOPM,NOSM,WID
TIME(NA),RF(4,0: 1001) ,PCOEF,SCOEF,MFCAS,TPM,TSM
DATA
NORM ,NOSM
/300,200/
81
) )
DATA
DATA
TPM
,TSM
PCOEF.SCOEF
/20. 0,15.0/
00020, 00015/
/.
.
IFCMFANS .EQ. 1) THEN
MFCAS = (PCOEF-->RF(IA,IT)*NOPM'KTPM/60.
ELSE IFCMFANS EQ. 2) THEN
MFCAS = (SCOEF'>RF(IA,IT)'^NOSM'KTSM/60.
END IF
RETURN
END
))
/
(TIME(WID)/INCRE)
))
/
(TIME(WID)/INCRE)
.
^'eirit•>T•>\^)\•^!•;r•:r-;trk•>xh•>'r•:f•}eiririr>'r>^>^irir•>^^(•^c•>'rir•hir•>^>^iTic•^^^
>''
SUBROUTINE FOR COMPUTING CASUALTIES BY DIRECT FIRE
8.
SUBROUTINE DRF IRE ( DFANS DFA BRD I ST I A IT RF BF AR RDFCAS BDFCAS
PLTN, RANGE, NNN,ACO,AMO)
INTEGER DFANS,DFA(4,1000),FLTN
RF(4,0: 1001) BF(4, 0: 1001) ,PLBP,FBF(0: 1001)
REAL
BRDIST,RANGE,RED,BLU,RDFCAS,BDFCAS,AR,BR,MU,PL
REAL
,
1
,
,
,
,
,
,
,
,
,
DATA
DATA
DATA
DATA
DATA
PLBP
AOMEGA,BOMEGA
AC,BC
AM,BM
/
MU
/l.O/
0.
0/
/O. 50,1. 00/
/O. 0380 ,0. 0160/
/O. 23 0. 00010/
,
ACO = AC
AMD = AM
IF(NNN .GT. 0) GO TO 1810
PL = PLTN
1810
FBF(IT) = PL
C---FIND ENGAGED FORCES OF BOTH SIDES ON DIRECT FIRE BATTLE
RED = RF(IA,IT)
BLU = BF(IA,IT)
.'
C---CASE-,^il (MIXED LAW)
BLUE FORCE AMBUSHES THE RED FORCE
IF( DFANS .EQ. 1) THEN
AR = AM '^ (1. 0- BRD I ST/RANGE )--'"''^MU
BR = BM
(1. 0-BRDIST/RANGE)^'^'--MU
:
--^
ELU = FBFCIT)
IF(BLU .LT. PLBP) THEN
RDFCAS = 0.0
BLU = 0.0
ELSE
RDFCAS = AR
BLU
END IF
BDFCAS = BR * BLU ^^ red
FBF(IT) = BLU - BDFCAS
PL = FBF(IT)
NNN = NNN + 1
GO TO 1900
''•
.
C---CASE-#2 (HEMBOLT EQUTION)
ELSE IFCDFANS EQ. 2) THEN
IF(RED .LE. 0.
.OR.
BLU
BDFCAS = 0.
RDFCAS = 0.
!
.
.
LE.
82
0.)
THEN
!
!
GO TO 1900
END IF
AR = AC '^ (1.0-BRD I ST/RANGE )''"^MU
BR = BC
''
(1. 0-BRDIST/RANGE)''"'«-MU
RDFCAS = AR-->((RED/BLU)-'"K1. 0-BOMEGA))*BLU
BDFCAS = BR--K(BLU/RED)''"''(1.0-AOMEGA))''^RED
1900
END IF
RETURN
END
*
SUBROUTINE FOR COMPUTING THE ADJUSTED SPEED
9.
SUBROUTINE
ASPEED
(
I
!
,NA, ANSI SPEED, ADSP)
,
INTEGER
REAL
ANS1,A0A1(4),A0A2(3),A0A3(6)
SPEED(NA),ADSP(NA)
DATA
DATA
DATA
AOAl
A0A2
A0A3
75,15,33,35/
/4,7,19/
/2 11 13,23,42,43/
,
,
ANSI
GO TO (201,202,203,204)
ADSP(I) = SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2011 J = 1,3
IF(I .EQ. A0A2(J))
ADSP(I) = 0.9*SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2012 J = 1,6
IF(I .EQ. A0A3(J))
ADSP(I) = 0.9^SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2013 J = 1,3
IF(I .EQ. A0A2(J))
ADSP(I) = 0. S-SPEEDC I)
201
202
2011
203
2012
204
2013
2010
RETURN
END
* 10.
SUBROUTINE FOR COMPUTING THE REDUCED SPEED
SUBROUTINE
RSPEED
(
!
lA, IT, NA, SPEED, TIME ,DIST,WID, RANGE ,AR,RF,
IFANS,MFANS,DFANS,OSPEED,ACO,AMO)
1
INTEGER
REAL
REAL
NA,WID,IFANS,MFANS,DFANS
AR,DIST(NA),SPEED(NA),TIME(NA),MINSPD,RF(4,0: 1001)
OSPEED(NA)
DATA
MINSPD
/O.
500/
IFdFANS
.EQ. 1) THEN
SPEED(WID) = DIST(WID)'''60./(TIME(WID)+10.
END IF
IF( MEANS .EQ. 1)
THEN
SPEED(WID) = DIST(WID)->-60. /(TIME(WID)+20.
END IF
83
)
)
IF(MFAKS .EQ. 2) THEN
SPEED(WID) = DIST(WID)*60. /(TIME(WID)+15. )
END IF
IF(DFANS .EQ. 1) THEN
SPEED(WID) = SPEED(WID) * ( 1. 0-AR/AMO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(WID) = OSPEED(WID)
END IF
IF(DFANS .EQ. 2) THEN
IF(RF(IA,IT) .EQ. 0. ) THEN
SPEED(WID) = 0.
ELSE
SPEED(VID) = S?EED(WID) * (1.0-AR/ACO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(WID) = OSPEED(WID)
ENT)
ENT)
IF
IF
RETURN
END
* 11.
TO FIND THE TIME AND LOCATION WHEN RED FORCE REACH BREAK -POINT
vVAVfVciVVcVrV.-VfVrVrrVVrVfVrVrVf'sVVrVf^ViVVrVrVr^V-VyfVrVrV.-V.-V.-VrV-VrVr^VVTyrV.-Vr-.ViVV.-VriVV^^
SUBROUTINE
BREAK
(
lA IT,NOC ,\s'HOLD,RF,BF,RRESER,BREAKP,BREAKT)
,
INTEGER RRESER
BREAKP(4),BREAKT(4),WH0LD,RF(4,0: 1001) ,BF(4 ,0: 1001)
REAL
RF(IA,IT) = 0.
BF(IA,IT) = BF(IA,IT-1")
IF(NOC .EQ. 0) THEN
RFfIA,IT) = RF(.IA.IT-1) + RRESER
BREAKP(IA) = WHOLD
BREAKT(IA) = IT
END IF
NOC = NOC + 1
RETURN
END
C
1111
VfVfVrVf'5VVrVrV.-V.-Vr^VVrVrVrVf^VVriV'5VVrVrVfVr:VV,-^VVr;VVrVfVfVrV,-VfVfVrV,-VTVrVfiVVfVf:fciV'^
'•'
TO PRINT THE CASUALTIES AND SURVIVORS FOR BOTH SIDES
12.
SUBROUTINE
PRINT
1
(
!
RBN BCO BLUER 1 BLUEP2 RRESER PLTN NA OUTPUT
IFSUM MFSUM RDFSUM BDFSUM)
,
,
,
,
,
,
,
,
,
,
RBN BCO BLUEPl BLUEP2 NA RRESER PLTN OUTPUT
IFSUM(4) ,MFSUM(4) ,RDFSUM(4) ,BDFSUM(4)
TIFSUM,TMFSUM,TRDF,TBDF
CHARACTER--^9
AVENUE ( 4
3
1
AVENUE//22
AVENUE//
DATA
AVENUE / AVENUE?^ 1
AVENUE,-'/2
/4*0. 0/
DATA
TIFSUM,TMFSUM,TRDF,TBDF
INTEGER
REAL
REAL
,
,
,
,
'
'
100
,
'
,
,
'
,
'
,
'
,
'
'
/
WRITE(OUTPLT, '(//)')
WRITE (OUTPUT, 100)
F0RMAT(5X,'«< RED CASUALTIES BY EACH TYPE OF BLUE FORCES >»',
l/60('-'),
1/15X, INDIRECT-FIRE' ,4X, 'MINE -FIELD' ,4X, 'DIRECT-FIRE' ,/60( -' ))
DO 101 lA = 1,4
TIFSUM = IFSUM(IA) + TIFSUM
TMFSUM = MFSUMCIA) + TMFSUM
'
'
84
!
TRDF = RDFSUM(IA) + TRDF
TBDF = BDFSUM(IA) + TBDF
WRITE( OUTPUT, 102) AVENUE( lA) IFSUM( lA) ,MFSUM( lA) ,RDFSUM( lA)
CONTINUE
VRITE( OUTPUT, 103) TIFSUM,TMFSUM,TRDF
FORMAT(3X,A9,5X,F6. 1,10X,F6. 1,9X,F6. 1)
F0RMAT(60( '-'),/ 17X,F6. 1,10X,F6. 1,9X,F6. 1)
+ RRESER
TRF = RBN'>3.
+ PLTN
TBF = BC0*3.
TRCAS = TIFSUM + TMFSUM + TRDF
TRSUR = TRF - TRCAS
TBCAS = TBDF
TBSUR = TBF - TBCAS
BFCASl = BDFSUM(l) + BDFSUM(2)
BFCAS2 = BDFSUM(3) + BDFSUM(4)
BFSURl = BLUEPl'^BCO + PLTN/2 - BFCASl
BFSUR2 = BLUEP2''^BC0 + PLTN/2 - BFCAS2
,
101
102
103
WRITE(0LTPUT, '(//)')
WRITE( OUTPUT 105 ) BFCAS 1 BFSURl BFCAS2 BFSUR2
F0RMAT(5X,'«< BLUE FORCE CASUALTIES FOR EACH FORT >»'
l/60( '-') ,/15X, 'CASUALTIES' ,5X, SURVIVORS' ,/60( -' ) ,/3X, 'F0RT//1'
,
105
,
,
,
'
'
15X,F6. 1,10X,F6. l,/3X, 'F0RT,^/2' ,5X,F6. 1,10X,F6. 1,/60(
WRITE(OUTPUT, '(//)')
WRITE( OUTPUT, 106) TRF, TRCAS TRSUR, TBF, TBCAS TBSUR
,
106
'
-'
))
,
F0RMAT(5X,'«< TOTAL CASUALTIES AND SURVIVORS >»'
l/60('-'),/13X,' INITIAL FORCE' ,5X, CASUALTIES ,6X,
I'SURVIVORS' ,/60( -• ) ,/5X, 'RED' ,9X,F6. 1,10X,F6. 1,10X,F6.
1/5X,'BLUE' ,8X,F6. 1,10X,F6. 1,10X,F6. l,/60('-'))
'
'
'
RETURN
END
85
1,
,
APPENDIX
This
is
C.
EXEC FOR BASIC MODEL
a execution code of Foriran-77 for the basic
FILEDEF 12 DISK THESIS DATA
EXEC WF77 BASIC (NOEXT NOWAR
86
model
in
Appendix
B.
APPENDIX
This
is
a
ADVANCED MODEL
D.
computer program using Fortran-77
--< ADVANCED MODEL >--- MADE BY
for the
advanced model.
LEE, JAE YEONG (1988.2.6)
INTEGER MR, MB
PARAMETER (MR=4, MB=8)
REAL
RED(MR,MB) ,BLUE(MR,MB) ,BATIME(MR,MB) ,ROB(MR,MB)
INTEGER ANSWER, REDAVE(3),P0IF,SMAVE(3),BLUEP1,BLUEP2,SMARC(3)
INTEGER OUTPUT, TEND, WAVE2
INTEGER ARC(44) ,TAIL(44) ,HEAD(44)
REAL
DIST(44) ,WIDTH(44) ,SPEED(44) ,TIME(44) ,0SPEED(44) ,ADSP( 44)
REAL
TRCAS,TBSUR,TBCAS,VALUE(4)
REAL
LINE ( MB ) SQUA( MB ) ROOT( MB ) LIN SQU ROO
WRITE(6, '(///)•)
PRINT -"-, 'HOW DO RED FORCES ON AVENUE -2 ATTACK ?'
,
777
,
.
,
.'. .'. .*. .•. ^*. iJ. .*. ^. m>^ .J. ij. .y i-*. .«.
,
,
y . .*. .u y. y. i^t. .1. y . y , .'. y . y. ^. y.
^ y* j^ y* J. j^ y^ y
.
y^ ^^
.J.
< EXPLANATION >
PRINT ^S
<OPTION>
PRINT ^'S
1
ALL FORCES ATTACK TO THE NODE 28
PRINT ^S
2
ALL FORCES ATTACK TO THE NODE 27
PRINT ''-,
3
DIVIDE EQUALLY AND ATTACK TO NODE 27 AND
PRINT ^^
NODE 28, RESPECTIVELY
PRINT '•,
EXIT THE PROGRAM
—
PRINT
READ(5,'0 ANSWER
IF (ANSWER .EQ. 0) GO TO 888
IF (ANSWER. NE.l .AND. ANSWER. NE. 2 .AND.
ANSWER. NE. 3 .AND. ANSWER. NE.
GO TO 777
)
!
•--'----».-—*--•--•—»
•••-,
L
55
56
57
58
j^ y .
GO TO (55,56,57)
WAVE 2 = 1
GO TO
WAVE 2 = 2
GO TO
WAVE 2 = 12
GO TO
•.-,j-»'-j.-'.,,.t,j.^',j.^t,^,j-jt.»t..
ANSWER
INPUT = 12
OUTPUT = 24
OUTPUT FILES
'M0ES2' HAS ALL MOE VALUES FOR EACH RED AND BLUE OPTIONS.
'UTIL2' HAS UTILITY VALUES FOR EACH UTILITY FUNCTION.
•FILE 'WH0W2' REPRESENTS HOW THE BATTLE TERMINATES!
OPEN ( UNIT = OUTPUT, FILE = 'M0ES2' )
OPEN ( UNIT = 25
FILE = •UTIL2' )
OPEN ( UNIT = 26
FILE = 'WH0W2' )
C-- •OPEN
C-- •FILE
C-- FILE
!
,
,
DO
1
IB = 1,MB
CALL
BDATA
(
IB,WAVE2 ,P0IF,SMAVE,BLUEP1 ,BLUEP2 ,SMARC)
87
,
DO
2
IR = 1,MR
CALL
RDATA (IR,REDAVE)
DO 10 I = 1, 44
READ( INPUT, 11) ARC( I) ,TAIL(
I) ,HEAD( I) ,TIME( I) ,DIST( I)
WIDTH(I),SPEED(I)
1
CALL ASPEDl (WAVE2 I IR,SPEED, ADSP)
IF(WAVE2 .EQ. 12) CALL ASPED2 ( I ,IR, SPEED, ADSP)
SPEED(I) = ADSP(I)
OSPEED(I) = SPEED(I)
CONTINUE
F0RMAT(3X,I2,2(3X,I2),2X,F6.2,6X,F4. 2,4X,F5.3,3X,F3.
,
,
10
11
C--- READ THE ORIGINAL DATA AGAIN
REWIND INPUT
1)
!
C
IF(WAVE2 .EQ. 1 .OR.
WAVE2 EQ. 2) THEN
CALL BATLEl (WAVE2 ,TRCAS ,TBSUR ,TEND ,TBCAS
IB IR REDAVE POIF SMAVE BLUEPl BLUEP2 SMARC
ARC, TAIL, HEAD, TIME, DIST, WIDTH, SPEED, OSPEED)
ELSE IF(WAVE2 EQ. 12^ THEN
CALL BATLE2 (TRCAS ,TBSUR,TEND,TBCAS,
IB IR REDAVE POI F SMAVE BLUEPl BLUEP2 SMARC
ARC, TAIL, HEAD, TIME, DIST, WIDTH, SPEED, OSPEED)
END IF
.
1
1
,
,
,
,
,
,
,
,
,
,
,
,
.
1
1
,
,
j^
RED(IR,IB) = TRCAS
BLUE(IR,IB) = TBSUR
BATIME(IR,IB) = TEND
ROB(IR,IB) = TRCAS/TBCAS
DO 4 I = 1,44
SPEED(I) = OSPEED(I)
4
CONTINUE
2
IB ,RED( 1 IB) ,RED( 2 IB) ,RED( 3 IB) ,RED(4, IB)
WRITEC OUTPUT, 900)
IB BLUE( 1 IB ) BLU£( 2 IB ) BLUE( 3 IB ) BLUE( 4 IB )
IB,BATIME(1,IB),BATIME(2,IB),BATIME(3,IB),BATIME(4,IB),
1
1
IB,R0B(1,IB),R0B(2,IB),R0B(3,IB),R0B(4,IB)
F0RMAT( 3X, 13 ,4( 3X,F6. 1) ,/3X,I3,4( 3X,F6. 1) ,/3X, I3,4(3X,F6. 1),
/3X,I3,4(3X,F6. 2) )
1
,
1
900
905
,
,
,
,
,
,
,
,
,
,
DO 905 J = 1,4
VALUE(J) = BLUE(J,IB)
C-- -COMPUTE DESIRED VALUES FOR DIFERRENT UTILITY FUT^CTIONS
CALL
UTIL (MR,MB,IB,VALUE,LIN,SQU,ROO)
LINE(IB) = LIN
SQUA(IB) = SQU
ROOT(IB) = ROO
!
910
IB ,LINE( IB) ,SQUA( IB) ,ROOT( IB)
WRITEC 25 ,910)
FORMATC 3X, I3,3(3X,F6. 2) )
1
CONTINUE
888
STOP
END
*
I.
'THIS MODEL IS COMPLETE
'
THANK YOU
!
!
SUBROUTINE FOR DEFINING RED OPTIONS.
SUBROUTINE RDATA (IR,REDAVE)
INTEGER IR,REDAVE(3)
1151
1152
1153
1154
1155
GO TO (1151,1152,1153,1154)
REDAVE(l) = 1
REDAVE(2) = 1
REDAVE(3) = 1
GO TO 1155
REDAVE(l) =
REDAVE(2) = 2
REDAVE(3) = 1
GO TO 1155
REDAVE(l) =
REDAVE(2) = 1
REDAVE(3) = 2
GO TO 1155
REDAVE(l) =
REDAVE(2) = 3
REDAVE(3) =
GO TO 1155
IR
RETURN
END
ji. ..u^.. w*. j^ .•.. .1* .*. .'. .'« .*. .*. ^. .*. >•< .t« ^^ .*.. jm ^^ .*. j^ -"^ ^. ^. ^. .*« ji. ^. m^^ mf. >'. ^. ^^ ^« ^*. mf* j^ JL. >*. .*. k*« ^. •). y. »*^ ^^ .J'
'•-
II.
^ ^ "V*^' *^' "V rV Vt^ Vf Vc V' Vt V' V? Vc
SUBROUTINE FOR DEFINING BLUE OPTIONS.
SUBROUTINE BDATA
(
IB ,WAVE2 POIF SMAVE BLUEPl BLUEP2 SMARC)
,
,
,
,
,
INTEGER IB,WAVE2,P0IF,SMAVE(3),BLUEP1,BLUEP2,SMARC(3)
IIB = IB
IF(IB .LE. 4) THEN
BLUEPl = 1
BLUEP2 = 2
GO TO 1250
ELSE
BLUEPl = 2
BLUEP2 = 1
IB = IB - 4
END IF
1250
1251
^ Vt *V V* V'
GO TO (1251,1252,1253,1254)
POIF = 2
SMAVE(l) = 1
SMAVE(2) = 1
SMAVE(3) = 1
SMARC(l) = 30
IB
89
'^' V' V'
1252
SMARC(2) = 35
SMARC(3) = 42
IF(WAVE2 .EQ.
SMARC(l) =
SMARC(2) =
SMARC(3) =
ELSE IF(WAVE2
SMARC(l) =
SMARC(2) =
SMARC(3) =
END IF
GO TO 1267
POIF = 2
SMAVE(l) =
SMAVE(2) = 2
SMAVE(3) = 1
SMARC(l) = 30
SMARC(2) = 34
THEN
1)
30
34
42
.EQ
.
THEN
2)
35
39
42
SMARCO) = 42
1253
IF(WAVE2 .EQ.
SMARCCl) =
SMARC(2) =
SMARCC3) =
ELSE IF(WAVE2
SMARCCl ) =
SMARC(2) =
SMARC(3) =
END IF
GO TO 1267
POIF = 2
1)
THEN
34
35
42
.EQ
30
.
THEN
2)
39
42
SMAVECn =
SMAVE(2) = 1
SMAVE(3) = 2
SMARCCl ) =30
SMARCC2) = 42
SMARCC3) = 43
IF C WAVE 2 .EQ.
SMARCCl) =
SMARCC2) =
SMARCC3") =
ELSE IFCWAVE2
SMARCCl) =
SMARCC2) =
SMARCC3) =
1)
THEN
35
42
43
.EQL
42
43
IF
GO TO 1267
POIF = 2
SMAVECl) =
ENT)
1254
THEN
2)
39
-
-
"""
SMAVEC2) = 3
SMAVEC3) =
SMARCCl) = 30
SMARCC2) = 34
SMARCC3) = 35
IFCWAVE2 .EQ. 1) THEN
SMARCCl) = 32
SMARCC2) = 34
90
SMARCO) =
35
ELSE IF(WAVE2 .EQ
SMARC(l) = 25
SMARC(2) = 30
SMARCO) =
2)
THEN
39
END IF
GO TO 1267
IB = IIB
1267
RETURN
END
****************************************************************
* III. SUBROUTINE TO EXECUTE THE "BATTLE l" TYPE BATTLE
***********************'>'{***********************************************
!
THIS PROGRAM DEAL WITH THE BATTLE CASE THAT HAS 3 AVENUES OF
APPROACH FOR RED FORCES CONTINUEOUSLY.
(I.E.
RED FORCES ON AVENUE-2
ATTACK TO JUST ONE OF TWO BLUE POSITIONS).
,
BATLEl (WAVE2 ,TRCAS,TBSUR,TEND,TBCAS
I B IR REDAVE POIF SMAVE BLUER 1 BLUEP2 SMARC
1
ARC, TAIL, HEAD, TIME, DIST, WIDTH, SPEED, OSPEED)
INTEGER
NA, NN, NOA, NOP, RBN, RRESER, BCO, PLTN
PARAMETER (NA=44, NN=28, N0A=3, N0P=2 PLTN=40)
PARAMETER (RBN=500, RRESER=150, BCO=170)
INTEGER
REDAVE(NOA) ,SMAVE(NOA) ,PMAVE(NOA)
ARC(NA) ,TAIL(NA) ,HEAD(NA)
INTEGER
PMARC(5),SMARC(3),IDARC(3,0: 1001)
INTEGER
INTEGER
IFA(3, 1000), MFA(3, 1000), DFA(3, 1000)
INTEGER
P0IF,BLUEP1,BLUEP2,ANS1,ANS2,ANS3,ANS4
INPUT, OUTPUT, TP1,TP2,DIF,DPMF,DSMF,INCRE,WID
INTEGER
INTEGER
IFANS MEANS ,DFANS TEND, WAVE2
REAL
DIST(NA) ,WIDTH(NA) ,TIME(NA) ,SPEED(NA) ,OSPEED(NA)
REAL
BREAKP(3),BREAKT(3),RBP(3)
REAL
RF(3,0: 1001) ,EF( 3 0: 1001) ,WHERE( 3 0: 1001) SP( 3 ,0: 1001)
IFC(3,0: 1001) ,MFC( 3 0: 1001) ,RDFC( 3 ,0: 1001),
REAL
IMDC(3,0: 1001), BDFC(3,0: 1001)
1
REAL
IFSUM(3),MFSUM(3),RDFSUM(3),BDFSUM(3)
REAL
IFCAS ,MFCAS DFCAS BLEFTl BLEFT2 WHOLD
REAL
CLOCK, RANGE, BRDIST,AR,RB
SUBROUTINE
1
,
,
,
,
,
,
,
,
,
,
,
,
,
DATA
DATA
DATA
DATA
DATA
DIF,DPMF,DSMF
PMAVE
PMARC
RANGE
,
,
,
710,20,15/
71,3,1/
/15, 19, 23, 25, 32/
11.11
KB
ANSI = IR
NNN =
lUSED =
NEQ12 =
NEQ23 =
C---DEFIN1: INITIAL FORCE LEVEL AND LOCATION
27
DO 28 I = 1,N0A
RF(I,0) = REDAVE(I)"-RBN
91
!
,
,
28
WHERE(I,0) = 0.
IFC(I,0) = 0.
MFC(I,0) = 0.
RDFC(I,0) = 0.
IMDC(I,0) = 0.
BDFC(I,0) = 0.
IFSUM(I) = 0.
MFSUM(I) = 0.
RDFSUM(I) =0.0
BDFSUMd^ = 0.
RBP(I) = RB'''RF(I,0)
IF(WAVE2 .EQ. 1) THEN
BF(1,0) = BLUER l-'^BCO + PLTN
BF(2,0) = BLUEPl^'^BCO + PLTN
BF(3,0) = BLUEP2'-BC0
ELSE IFCWAVE2 EQ. 2) THEN
BF(1,0) = BLUER r'^BCO
BF(2,0) = BLUEP2---BC0 + PLTN
BF(3,0) = BLUEP2---BC0 + PLTN
END IF
.
NOC =
INCRE = 1
DO 9999 IT = 1,500, INCRE
CLOCK = IT
DO 30 I A = 1,N0A
BREAKP(IA) = 0.
BREAKT(IA) = 0.
IFC(IA,IT) = IFC(IA,IT-1)
MFC(IA,IT) = MFC(IA,IT-1)
RDFC(IA,IT) = RDFC(IA,IT-1)
IMDC(IA,IT) = IMDC(IA,IT-1)
BDFC(IA,IT) = BDFC(IA,IT-1)
RF(IA,IT) = RF(IA,IT-1)
BF(IA,IT) = BF(IA,IT-1)
WHOLD = WHERE (I A, IT- 1)
CALL DETETl (WAYE2 lA, IT, INCRE, WHOLD, VID)
IF(WID .EQ. 0) GO TO 999
IDARC(IA,IT) = WID
BRDIST = 9999.
SP(IA,IT) = SPEED(WID^
TS = (SPEED(WID)/60)-'aNCRE
WHERE(IA,TT) = WHOLD + TS
,
C---IS THERE INDIRECT FIRE ? OR MINEFIELD ? OR DIRECT FIRE ?
CALL IDFNDl ( I A, CLOCK, PO IF, I FANS)
CALL MFFNDl ( PMARC ,SMARC WID, MEANS)
CALL DRFNDl (WAVE2 I A, WHOLD, RANGE BRDIST, DFANS)
,
,
,
IFA(IA,IT) = IFANS
^^FA(IA,IT) = MEANS
DFA(IA,IT) = DFANS
92
IF(IFANS.EQ.
.AND.
MFANS. EQ.
.AND.
C-- -COMPUTE THE CASUALTY OF INDIRECT-FIRE
IFdFANS
DFANS. EQ. 0)
GO TO 30
!
THEN
CALL IDFIRl (IA,IT,RF,REDAVE,POIF,IFCAS)
RF(IA,IT) = RF(IA,IT) - IFCAS
IFSUM(IA) = IFCAS + IFSUM(IA)
IFC(IA,IT) = IFC(IA,IT) + IFCAS
.EQ.
1)
END IF
C-- -COMPUTE THE CASUALTY BY MINE -FIELD
MFANS. EQ. 2) THEN
.OR.
IFCMFANS.EQ. 1
!
CALL MFILDl ( lA, IT, MFANS INCRE,NA,WID, TIME, RF.MFCAS)
RF(IA,IT) = RF(IA,IT) - MFCAS
MFSUM(IA) = MFCAS + MFSUM(IA)
MFC(IA,IT) = MFC(IA,IT) + MFCAS
,
END IF
C-.. COMPUTE THE CASUALTY BY DIRECT-FIRE
IF(DFANS .EQ. 1 .OR.
DFANS EQ. 2) THEN
IF( RF(IA,IT) .LT. BF(IA,IT)
.AND.
lUSED .EQ.
1
.AND.
RF(IA,0) .GT. 0.0
1
THEN
)
RF(IA,IT) = RF(IA,IT) + RRESER
lUSED = 1
END IF
CALL DRFIRl (DFANS ,DFA,BRDIST, lA, IT, RF,BF,AR,RDFCAS ,BDFCAS,
PLTN, RANGE, NNN)
1
!
.
RF(IA,IT) = RF(IA,IT) - RDFCAS
BF(IA,IT) = BF(IA,IT) - BDFCAS
RDFSUM(IA) = RDFCAS + RDFSUM(IA)
BDFSUM(IA) = BDFCAS + BDFSUM(IA)
RDFCCIA,IT) = RDFC(IA,IT) + RDFCAS
BDFC(IA,IT) = BDFC(IA,IT) + BDFCAS
END IF
IF(RF(IA,0) .GT.
CALL BREAKl
END IF
.AND.
RF(IA,IT) LE. RBP(IA)) THEN
lA, IT,NOC ,WHOLD,RF,BF,RRESER,BREAKP,BREAKT)
0.
(
.
C---CO.MPUTE THE REDUCED SPEED DUE TO THE INDIRECT FIRE OR MINEFIELD OR
UNDER THE DIRECT FIRE BASED ON THE RANGE OF TWO FORCES
C
!
CALL
RSPEDl
(
lA, IT, NA, SPEED, TIME, DIST,WID, RANGE, AR,RF,
I FANS,
1
MFANS, DFANS, OSPEED)
SP(IA,IT) = SPEED(WID)
TS = (SPEED(WID)/60)^-INCRE
WHERE(IA,IT) = WHOLD + TS
IMDC(IA,IT) = IFC(IA,IT) + MFC(IA,IT) + RDFC(IA,IT)
30
CONTINUE
IF(WAVE2 .EQ. 1) THEN
IF(NEQ12 .NE. 0) GO TO 9999
93
BF(1,IT) = MIN( BF(1,IT),BF(2,IT) )
BF(2,IT) = MIN( BF(1,IT),BF(2,IT) )
BF(3,IT) = BF(3,IT)
IF(DFA(1,IT) .EQ. 2 .AND.
DFA(2,IT) EQ. 2 .AND.
RF(1,IT) .GT. 0.
.AND.
RF(2,IT) GT. 0.) THEN
BF(1,IT) = (RF(1,IT) / (RF(1,IT)+RF(2,IT))) * BF(1,IT)
BF(2,IT) = (RF(1,IT) / (RF( 1 ITj+RF( 2, IT) ) )
BF(2,IT)
BF(3,IT) = BF(3,IT)
KEQ12 = IT
.
1
.
•>''
,
E.ND
IF
ELSE IF(WAVE2 EQ. 2) THEN
IF(NTQ23 .NE. 0) GO TO 9999
BF(1,IT) = BF(1,IT)
BF(2,IT) = MIN( BF(2,IT),BF(3,IT) )
BF(3,IT) = MIN( BF(2,IT),BF(3,IT) )
IF(DFA(1,IT) .EQ. 2 .AND.
DFA(2,IT) EQ. 2 .AND.
RF(1,IT) .GT. 0.
.AND.
1
RF(2,IT) GT. 0.) THEN
BF(1,IT) = BFCl.IT)
BF(2,IT) = (RF(2,IT) / (RF(2 IT)+RF(3 IT) ) ) * BF(2,IT)
BF(3,IT) = (RF(3,IT) / (RF( 2 IT)+RF( 3, IT) ) ) '^ BF(3,IT)
NEQ23 = IT
END IF
END IF
.
.
.
,
,
,
CONTINUE
9999
CALL
999
PRINTl
1
( RBN.BCO ,BLUEP1 ,BLUEP2 ,RRESER,PLTN,NA,
WAVE2 IFSUM MFSU.M RDFSUM BDFSUM
TRCAS,TBCAS,TRSUR,TBSUR)
,
,
1
TEND = IT
RETURN
END
-
,
,
1
^.
-'•-
III~1.
.ju j:. ^. J. ^V -*' -^ ^' -I- •^' >>' «'' -^' -J' -J' ^' •J' >*' -A--*--^-
SUBROUTINE FOR LOCATING THE ARC THE RED FORCES ARE ATTACKING.
SUBROUTINE
INTEGER
REAL
DATA
DATA
DATA
DATA
DETETl (WAVE2 lA, IT, INCRE ,WHOLD,WID)
,
INCRE,WID,WAVE2
A1(4),A21(8),A22(8),A3(6),WH0LD
Al
A21
A22
A3
/2. 75,6. 35,9. 35,10. 25/
/I. 15,3.3,4. 7,5.45,6.4,7. 1,8.55,9.45/
/I. 15 , 3. 3 ,4. 7 , 6. 15 6. 8 7. 45 7. 45 , 7. 45/
/4. 5,6. 6,8. 85,12.35,12.95,13.55/
,
,
,
IF(IA .EQ. 1) THEN
IFCWHOLD .LE. Al(l)) THEN
WID = 5
GO TO 1390
ELSE IFCWHOLD. GT.Al(l)
.AND. WHOLD. LE. Al( 2)
WID = 15
GO TO 1390
ELSE IFCWHOLD. GT.AK 2)
.AND. WHOLD. LE. A1C3)
WID = 33
GO TO 1390
94
)
THEN
)
THEN
.AND. WHOLD. LE. Al(4) ) THEN
ELSE IF(WHOLD. GT. Al(3)
WID = 35
GO TO 1390
ELSE IF(WHOLD GT. Al(4)) THEN
IT-1
WRITE(26,1380)
F0R.MAT(1X,'RED FORCE IN AVE/Zl TOOK BLUE P0SITI0N//1
BATTLE END
1/5X,' (BATTLE TIME ==> ',14,' MINUTE)')
GO TO 1399
END IF
ELSE IF(IA .EQ. 2) THEN
GO TO (1311,1312) WAVE2
IF(WHOLD .LE. A21(l)) THEN
WID = 4
GO TO 1390
ELSE IF(WHOLD.GT. A21(l) .AND. WHOLD. LE. A21(2) ) THEN
WID = 7
GO TO 1390
ELSE IF(WHOLD.GT. A21(2) .AND. WHOLD. LE. A21(3)) THEN
WID = 19
GO TO 1390
ELSE IF(WHOLD. GT. A21(3) .AND. WHOLD. LE. A21(4) ) THEN
WID = 24
GO TO 1390
ELSE IF(WHOLD.GT. A21(4) .AND. WHOLD. LE. A21( 5) ) THEN
WID = 29
GO TO 1390
ELSE IF(WHOLD.GT. A21(5) .AND. WHOLD. LE. A21(6) ) THEN
WID = 32
GO TO 1390
ELSE IF(WHOLD. GT. A21(6) .AND. WHOLD. LE. A21( 7) ) THEN
WID = 34
GO TO 1390
ELSE IF(WHOLD.GT. A21(7) .AND. WHOLD. LE. A21( 8) ) THEN
WID = 35
GO TO 1390
ELSE IF(WHOLD GT. A21(8)) THEN
WRITE(26,1381)
IT-1
F0RMAT(1X,'RED FORCE IN AVE//21 TOOK BLUE P0SITI0N#1
BATTLE ',
I'END !' ,/5X,'( BATTLE TIME ===> ',14,' MINUTE)')
GO TO 1399
END IF
IF( WHOLD .I.E. A22(l)) THEN
WID = 4
GO TO 1390
ELSE IF(WHOLD. GT. A22(l) .AND. WHOLD. LE. A22(2) ) THEN
WID = 7
GO TO 1390
ELSE IF(WHOLD. GT. A22(2) .AND. WHOLD. LE. A22(3) ) THEN
WID = 19
GO TO 1390
ELSE IF(WH0LD.GT.A22(3) .AND. WHOLD. LE. A22(4)) THEN
WID = 25
GO TO 1390
ELSE IF(WHOLD.GT. A22(4) .AND. WHOLD. LE. A22(5) ) THEN
WID = 30
GO TO 1390
.
1380
1311
:
.
1381
1312
:
95
!'
ELSE IF(WHOLD. GT. A22(5) .AND. WHOLD. LE. A22(6) ) THEN
WID = 39
GO TO 1390
ELSE IF(WHOLD GT. A22(6)) THEN
WRITE(26,1382)
IT-1
F0RMAT(1X,'RED FORCE IN AVE//22 TOOK BWE POSITIOW/2
BATTLE ',
',14,' MINUTE)')
I'END !' ,/5X,'( BATTLE TIME
GO TO 1399
END IF
ELSE IF(IA .EQ. 3) THEN
IF(WHOLD .LE. A3(l)) THEN
WID = 2
GO TO 1390
.AND. WHOLD. LE. A3(2) ) THEN
ELSE IFCWHOLD. GT. A3(l)
WID = 11
GO TO 1390
ELSE IFCWHOLD. GT.A3( 2)
.AND. WHOLD. LE. A3(3)) THEN
WID = 13
GO TO 1390
.AND. WHOLD. LE. A3C4) ) THEN
ELSE IFCWHOLD. GT.A3( 3)
WID = 23
GO TO 1390
ELSE IFCWHOLD. GT.A3C4)
.AND. WHOLD. LE. A3C5) ) THEN
WID = 42
GO TO 1390
ELSE IFCWHOLD. GT.A3C5)
.AND. WHOLD. LE. A3C6) ) THEN
WID = 43
GO TO 1390
ELSE IFCWHOLD GT. A3C6)) THEN
WRITEC26,1383)
IT-1
BATTLE',
F0RMATC1X,'RED FORCE IN AVE.^f3 TOOK BLUE POSITIOW/2
1' END
,/5X, 'CBATTLE TIME ===> ',14,' MINUTE)')
GO TO 1399
END IF
END IF
RETURN
WID = C
RETURN
END
.
1382
=>
:
.
1383
:
!
'
1390
1399
^-
III -2.
SUEROITINE FOR IDENTIFYING THE INDIRECT-FIRE
SUBROUTINE IDFNDl C lA, CLOCK, POIF, IFANS)
INTEGER POIF, IFANS
CLOCK, STIF,FTIF
REAL
DATA
STIF,FTIF /5. 0,15.0/
IF(CLOCK .GE. STIF .AND.
IFCIA .EQ. 2) THEN
IFANS = 1
GO TO 1444
ELSE
IFANS =
GO TO 1444
END IF
ELSE
CLOCK
96
.
LT.
FTIF) THEN
!
I FANS =
END IF
RETURN
END
1444
* III -3.
SUBROUTINE FOR IDENTIFYING THE MINE-FIELD
!
SUBROUTINE MFFNDl (PMARC,SMARC,WID, MEANS)
INTEGER MFANS,WID,PMARC(5),SMARC(3)
DO 1510 I = 1,5
IF(WID .EQ. PMARC(I)) THEN
MEANS = 1
GO TO 1599
END IF
CONTINUE
DO 1520 J = 1,3
IF(WID .EQ. SMARC(J)) THEN
MFANS = 2
GO TO 1599
END IF
CONTINUE
MFANS =
RETURN
END
1510
1511
1520
1521
1599
*
I
II -4.
SUBROUTINE FOR IDENTIFYING THE DIRECT-FIRE
tj^ *u .'- .'- -•- J- J- .*- ».'- y- .'- »- y- .'- .'- .'- J- .'- .'- y- y- *• ••- y- y- y- -•- ^•- y- y- y- y- y- y . y- y- y- y- y-
SUBROUTINE
1
!
j- y- y- y- y . y- y- y * y- y , y - y- y- y- y- y- y-y ^ VrV'
*'r
Vr ^r Vr V' *V
DRFNDl (VAVE2 IA,WHOLD,RANGE ,BRDIST,DFANS)
,
INTEGER
REAL
IA,DFANS,WAVE2
WHOLD,PATH( 3), RANGE, BRDI ST, AMBUSH
DATA
AMBUSH
/4. 0/
PATH(l) = 10. 25
PATH(2) = 9.45
IF(VAVE2 .EQ.
PATH(3) = 13. 55
IF(WHOLD .GE.
WHOLD LT.
.
2)
PATH(2) = 7.45
AMBUSH-RANGE/2.
AMBUSH
.AND.
AND.
.
THEN
DFANS = 1
BRDIST = AMBUSH - WHOLD
GO TO 1699
ELSE IF( WHOLD GE. PATH( lA) -RANGE .AND.
1
WHOLD .LT. PATH(IA)
THEN
)
DFANS = 2
BRDIST = PATH(IA) - WHOLD
GO TO 1699
END IF
DFANS =
1699 RETURN
END
1
lA
.EQ.
2
)
.
y-y- y- y- y- y- - •; y- y • y- y- y- y- y- y- y- jl. y- y- y- -•- y- y- y- y- y- y- y- y- y * y- y- y- y- y- y^ y- y- y- y- y- y^. y^ y^. y- y- y^ y- y^ y.
97
^ y^ y- y
.
y^ y . y^ y . y^ y- y^ y- y. y^ y^
* III-5.
SUBROUTINE FOR COMPUTING CASUALTIES BY INDIRECT-FIRE
SUBROUTINE IDFIR1( lA IT,RF ,REDAVE ,POIF, IFCAS)
INTEGER NROUND I A IT REDAVE( 3 ) POIF
REAL
LETHAL ROWSP COLSP MUX MUY RHOX RHOY RATIO PKILL
REAL
RF(3,0: 1001), IFCAS
,
,
,
,
NROUND LETHAL
MUX, MUY
RHOX, RHOY
DATA
DATA
DATA
,
,
,
/ 18
,
,
,
,
7
,
,
,
,
8
.
/15. ,10.
/15. ,10.
/
IF(REDAVE(POIF) EQ. 1) THEN
ROWSP =12.
COLSP = 12.
GO TO 1669
ELSE IF(REDAVE(POIF) EQ. 2) THEN
ROWSP = 10.
COLSP = 10.
GO TO 1669
ELSE IF(REDAVE(POIF) EQ. 3) THEN
ROWSP = 8.
COLSP = 8.
GO TO 1669
END IF
PRINT '•% IT,IA,RF(IA,IT)
RATIO = (NROUND--- LETHAL) / (ROWSP*COLSP*RF( lA, IT)
PHI = 3. 141592654
.
.
.
1669
A =
)
LETHAL-'^NROUND ) / ( 2---PH 1 1 ) ''•^ 5
PKILL = RATIO
(A'--A/((A-'>A+RH0X--'^^'-2)'>(A'''A+RH0Y*'''2))*'^5)
1
EXP( - 5--n ( MUX^'"-"-2 ) / ( A----A+RH0X'--^>2 ) + ( MUY''^'"'2 ) / ( A'^A+RH0Y'''->2 )
IFCAS = PKILL '• RF(IA,IT)
( (
.
'"^
•'•^
.
RETURN
END
Vr;VVrVfVfV^V.-^>^V-,VVrVrVrVr-/r•)VVrVrV?VrVrV.-VrVrVrVr^V:VVrVrVrV-VrVcV>-V^Vr^VVrVr^VVrVfVfV^
'"^
-,V -V
III-6.
!': -.V
-.',-
-,V -.V
-,'.-
SUBROUTINE FOR COMPUTING CASUALTIES BY MINEFIELDS
Vf -.'r Vr Vr
-.',- -.',-
V.- -.'r -,V -,',-
•/,-
-A- -.V -,V
',-
Vr Vf -V V,-
-,',-
•,': •>'.-
Vr -,V -.VVr V: -,'r -,V -,':
-,',-
Vf Vr ic Vr ;> ?V •;V >'.- -.V
-.',-
Vr Vr -,V
-.'.-
-A- --'r
VrVr ;> iV iV iV ^V iV Vr -V
SUBROUTINE MFILD1( lA, IT,MFANS INCRE ,NA,WID,TIME ,RF,MFCAS)
,
INTEGER
REAL
MFANS INCRE NA NORM NOSM WID
T1ME(NA),RF(3,0: 1001) ,PCOEF,SCOEF,MFCAS,TPM,TSM
DATA
DATA
DATA
NOPM ,NOSM
TPM
,TSM
PCOEF,SCOEF
,
,
,
,
,
/300,200/
/20. 0,15.0/
/.
00020
,.
00015/
IF( MFANS .EQ. 1) THEN
MFCAS = (PCOEF''--RF(IA,IT)'VNOPM'>(TPM/60. ))
ELSE IF(MFANS EQ. 2) THEN
/
(TIME(WID)/INCRE)
/
(TIME(WID)/INCRE)
.
MFCAS = (SCOEF*RF(IA,IT)*NOSM^''(TSM/60.
END IF
RETURN
END
*
I
II -7.
))
SUBROUTINE FOR COMPUTING CASUALTIES BY DIRECT FIRE
SUBROUTINE DRFIR1(DFANS ,DFA,BRDIST, lA, IT,RF,BF,AR,RDFCAS,BDFCAS
98
)
PLTN, RANGE, NNN)
1
INTEGER
REAL
REAL
DFANS,DFA(3,1000),PLTN
DATA
DATA
DATA
DATA
DATA
PLBP
AOMEGA,BOMEGA
AC,BC
AM,BM
MU
RF(3,0: 1001) ,BF( 3 0: 1001) ,PLBP ,FBF( 0: 1001)
BRDI ST RANGE RED BLU RDFCAS BDFCAS AR BR MU PL
,
,
,
,
,
,
,
,
,
,
/lO.O/
/O. 50,1. 00/
70.0490,0.0250/
/O. 23,0.00010/
/l.O/
IF(NNN GT. 0) GO TO 1810
PL = PLTN
1810 FBF(IT) = PL
C---FIND ENGAGED FORCES OF BOTH SIDES ON DIRECT FIRE BATTLE
RED = RF(IA,IT)
BLU = BF(IA,IT)
.
BLUE FORCE AMBUSHES THE RED FORCE
IF(DFANS .EQ. 1) THEN
AR = AM '> (1 -BRDI ST/RANGE )'"">MU
(1 -BRDI ST/RANGE )''^''--MU
BR = BM
BLU = FBF(IT)
IF(BLU LT. PLBP) THEN
RDFCAS = 0.
BLU = 0.0
ELSE
RDFCAS = AR '^ BLU
END IF
BDFCAS = BR '^ BLU
RED
FBF(IT^ = BLU - BDFCAS
PL = FBF(IT)
NNN = NNN + 1
GO TO 1900
C---CASE-^/l (MIXED LAW)
:
!
!
'"^
.
''^
C---CASE-#2 CHEMBOLT EQUTION)
ELSE IF(DFANS EQ. 2) THEN
IF(RED .LE. 0.
.OR.
BLU LE. 0.) THEN
BDFCAS = 0.
RDFCAS = 0.
GO TO 1900
END IF
AR = AC * (1 -BRDI ST/RANGE )**MU
BR = BC * (1-BRDIST/RANGE)*--'^MU
RDFCAS = AR'H(RED/BLU)*'K1.0-B0MEGA))*BLU
BDFCAS = BR'V((BLU/RED)**(1.0-A0MEGA))*RED
END IF
1900 RETURN
END
!
.
.
* III-8.
*
SUBROUTINE FOR COMPUTING THE ADJUSTED SPEED BASED ON
THE NUMBER OF RED UNITS (BATTALIONS) ON EACH AVENUE.
SUBROUTINE
INTEGER
ASPED1(WAVE2,
I
,
IR,SPEED, ADSP)
WAVE2,IR,A0A1(4) ,A0A2(8) ,A0A3(6) ,A0A21(8) ,A0A22(8)
99
REAL
SPEED(44),ADSP(44)
DATA
DATA
DATA
DATA
AOAl
A0A21
A0A22
A0A3
75,15,33,35/
74,7,19,24,29,32,34,35/
/4, 7, 19, 25, 30, 39, 39, 39/
/2, 11, 13,23,42,43/
DO 2009 K = 1,8
A0A2(K) = A0A21(K)
IF(WAVE2 EQ. 2)
A0A2(K) ^ A0A22(K)
2009
.
GO TO (201,202,203,204)
IR
ADSP(I) = SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2011 J = 1,8
IF(I .EQ. A0A2(J))
ADSP(I) = 0.9*SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2012 J = 1,6
IF(I .EQ. ACA3(J))
ADSP(I) = 0.9^-SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2013 J = 1,3
ADSP(I) = 0. 8''^SPEED( I)
IF(I .EQ. A0A2(J))
201
202
2011
203
2012
204
2013
RETURN
END
2010
)VVrVf';ViVVrVrV.-V.-VrVrVrVf:VVrVrVrVriVVrVrVrVriViVVfVrVrVrVr^ViVVrVf^VVfVrVrV.-'>VV.-VfVfvVVrVrV.^
•>
I
II -9.
SUBROUTINE FOR COMPUTING THE REDUCED SPEED BASED ON
EACH WEAPON TYPE.
SUBROUTINE
1
RSPED1( lA, IT, NA, SPEED, TIME ,DIST,WID, RANGE, AR,RF,
IFANS,MFANS,DFANS,OSPEED)
INTEGER
REAL
REAL
NA ,VID IFANS MEANS ,DFANS
DATA
DATA
MINSPD
ACO,AMO
,
,
AR,DIST(NA),SPEED(NA),TIME(NA),MINSPD,RF(3,0: 1001)
OSPEED(NA)
/O. 5/
/O. 06,0. 20/
IFdFANS
.EQ. 1) THEN
SPEED(WID) = DIST(WID)*60. /(TIME(WID)+10.
END IF
IF(MFANS .EQ. 1) THEN
SPEED(WID) = DIST(WID)^'60./(TIME(VID)+20. )
END IF
IF( MEANS .EQ. 2)
THEN
SPEED(VID) = DIST(WID)*60./(TIME(WID) + 15. )
END IF
IF(DFANS .EQ. 1) THEN
SPEED(VID) = SPEED(WID) * (1-AR/AMO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(WID) = OSPEED(WID)
END IF
IF(DFANS .EQ. 2) THEN
100
IF(RF(IA,IT) .EQ. 0. ) THEN
SPEED(WID) = 0.
ELSE
SPEED(WID) = SPEED(WID) * (1-AR/ACO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(WID) = OSPEED(VvID)
END IF
END IF
RETURN
END
* III -10.
*
SUBROUTINE TO FIND THE TIME AND LOCATION
WHEN RED FORCE REACH BREAK-POINT
!
SUBROUTINE
BREAKl
(
lA, IT,NOC,WHOLD,RF,BF,RRESER,BREAKP,BREAKT)
INTEGER RRESER
REAL
BREAKP(3),BREAKT(3),WH0LD,RF(3,0: 1001) ,BF(3,0: 1001)
RF(IA,IT) = 0.
BF(IA,IT) = BF(IA,IT-1)
IF(NOC .EQ. 0) THEN
RF(IA,IT) = RF(IA,IT-1) + RRESER
BREAKP(IA) = VHOLD
BREAKT(IA) = IT
END IF
NOC = NOC + 1
RETURN
END
C
nil
* III -11.
•^
SUBROUTINE TO PRINT THE CASUALTIES AND SURVIVORS
FOR BOTH SIDES IN BATTLE 1 TYPE.
PRINTl (RBN,BC0,BLUEP1 ,BLUEP2 RRESER, PLTN.NA,
WAVE2 IFSUM MFSUM RDFSUM BDFSUM
TRCAS,TBCAS,TRSUR,TBSUR)
SUBROUTINE
,
1
,
1
INTEGER
REAL
REAL
REAL
,
,
RBN BCO BLUEP 1 BLUEP2 NA RRESER PLTN WAVE2
IFSUM(3),MFSUM(3),RDFSUM(3),BDFSUM(3)
TIFSUM,TMFSUM,TRDF,TBDF
TRCAS,TRSUR,TBCAS,TBSUR
,
,
,
,
TIFSUM = 0.
TMFSUM = 0.0
TRDF = 0.0
TBDF = 0.
TRCAS
TBCAS
TRSUR
TBSUR
,
= 0.
= 0.0
= 0.
= 0.
DO 101 lA
TIFSUM
TMFSUM
TRDF =
= 1,3
= IFSUM(IA) + TIFSUM
= MFSUM(IA) + TMFSUM
RDFSUM(IA) + TRDF
101
,
,
,
TBDF = BDFSUM(IA) + TBDF
CONTINUE
+ RRESER
TRF = RBN^'^S.
+ PLTN
TBF = BC0-'-3.
TRCAS = TIFSUM + TMFSUM + TRDF
TRSUR = TRF - TRCAS
TBCAS = TBDF
TBSUR = TBF - TBCAS
IF(WAVE2 .EQ. 1) THEN
BFCASl = BDFSUM(l) 4 BDFSUM(2)
BFCAS2 = BDFSUMO)
BFSURl = BLUER I'^BCO + PLTN - BFCASl
BFSUR2 = BLUEP2--^BC0 - BFCAS2
ELSE IF(WAVE2 EQ. 2) THEN
BFCASl = BDFSUM(l)
BFCAS2 = BDFSUM(2) + BDFSUMO)
BFSURl = BLUEFl'-^BCO - BFCASl
BFSUR2 = BLUEP2-'--BC0 + PLTN - BFCAS2
END IF
101
.
•
TRCAS
TRSUR
TBCAS
TESUR
=
=
=
=
TIFSUM -^ TMFSUM + TRDF
TRF - TRCAS
TBDF
TBF - TBCAS
[
RETURN
END
VrVfVr'iVVe'iVVrTVVrVrVriVVrVrVfVriViVVrVrVrVT^VVrVrVriVVr'sVVrVrVrVTiViVVrVrVrVr-.ViV'A'VriVVfVr'j'r'jViV
SUBROUTINE TO EXECUTE THE "BATTLE2" TYPE BATTLE.
IV.
^V^V^•r^V^lr-r^'
rVrVfjViViV
V.-Vf/'rV.-s'r-V'A'^V'j'r'jV'jV-VVr-V'jVVr'jVV.-VriViV'jV:
THIS PROGRAM DEAL WITH THE BATTLE CASE THAT HAS 3 AVENUES OF
APPROACH FOR RED FORCES INITIALLY, BUT ATTACK TO BLUE WITH 4 AVENUES
OF APPROACH FINALLY.
RED FORCES ON AVENUE-2 DIVIDE EQUALLY
(I.E.
AND ATTACK TO BOTH BLUE POSITIONS
NODE 27 AND NODE 28).
,
:
BATLE2 (TRCAS TBSUR, TEND, TBCAS
IB IR REDAVE POIF SMAVE BLUEPl BLUEP2 SMARC
1
ARC, TAIL, HEAD, TIME, DIST, WIDTH, SPEED, OSPEED)
INTEGER
NA, NN, NOA, NOP, RBN, RRESER, BCO, PLTN
PARAMETER (.\A=44, NN=28, N0A=3 N0P=2 PLTN=40)
PARAMETER (RBN=500, RRESER=150, BCO=170)
INTEGER
REDAVE(NOA) ,SMAVE(NOA) ,PMAVE(NOA)
INTEGER
ARC(NA) ,TAIL(NA) ,HEAD(NA)
INTEGER
PMARC(5),SMARC(3),IDARC(4,0: 1001)
INTEGER
IFA(4, 1000), MFA(4, 1000), DFA(4, 1000)
INTEGER
POIF, BLUER 1,BLUEP2,ANS1,ANS2,ANS3,ANS4
INTEGER
INPUT, OUTPUT, TP1,TP2,DIF,DPMF,DSMF,INCRE,WID
INTEGER
IFANS ,MFANS ,DFANS ,TEND
REAL
DIST(NA),WIDTH(NA),TIME(NA),SPEED(NA),OSPEED(NA)
REAL
BREAKP(4),BREAKT(4),RBP(4)
REAL
RF(4,0: 1001) ,BF( 4 ,0: 1001) ,WHERE(4,0: 1001) ,SP(4 ,0: 1001)
REAL
IFC(4,0: 1001) ,MFC( 4,0: 1001) ,RDFC(4 0: 1001),
IMDC(4,0: 1001), BDFC(4,0: 1001)
1
REAL
IFSUM(4) ,MFSUM(4) ,RDFSUM(4) ,BDFSUM(4)
REAL
IFCAS MFCAS DFCAS BLEFTl BLEFT2 WHOLD
SUBROUTINE
,
1
,
,
,
,
,
,
,
,
,
,
,
,
102
,
,
,
CLOCK RANGE BRDI ST AR RB
REAL
DATA
DATA
DATA
DATA
DATA
,
,
,
DIF,DPMF,DSMF
PMAVE
PMARC
RANGE
,
/10,20,15/
/1, 3,1/
/15 19 23 ,25 , 32/
,
,
/I. 1/
/O. 0/
RB
ANSI = IR
NNN =
lUSED =
NEQ12 =
NEQ34 =
DO 5 I = 1,4
IFSUM(I) = 0.
MFSUM(I) =0.0
RDFSUM(I) = 0.
BDFSUM(I) = 0.
CONTINUE
5
C-- -DEFINE INITIAL FORCE LEVEL AND LOCATION
RF(1,0) = REDAVE(1)''^RBN
RF(2,0) = (REDAVE(2)-'>RBN)/2
RF(3,0) = (REDAVE(2)''^RBN)/2
RF(4,0) = REDAVE(3)*RBN
DO 28 I = 1,4
WHERE(I,0) = 0.
IFC(I,0) = 0.
MFC(I,0) = 0.
RDFC(I,0) = 0.
IMDC(I,0) = 0.
BDFC(I,0) = 0.
28
RBP(I) = RB'>RF(I,0)
BF(1,0) = BLUER l'"^BCO + PLTN/2
BF(2,0) = BLUEPl-'^BCO + PLTN/2
BF(3,0) = BLUEP2--'^BC0 + PLTN/2
BF(4,0) = BLUEP2--'-BC0 + PLTN/2
NOC =
INCRE = 1
DO 9999 IT = 1,500, INCRE
CLOCK = IT
DO 30 lA = 1,4
BREAKP(IA) =0.0
BREAKT(IA) = 0.0
IFC(IA,IT) = IFC(IA,IT-1)
MFC(IA,IT) = MFC(IA,IT-1)
RDFC(IA,IT) = RDFC(IA,IT-1)
IMDC(IA,IT) = IMDC(IA,IT-1)
BDFC(IA,IT) = BDFC(IA,IT-1)
RF(IA,IT) = RF(IA,IT-1)
BF(IA,IT) = BF(IA,IT-1)
raOLD = V,TIERE(IA,IT-1)
CALL DETET2( lA IT INCRE ,VTiOLD ,WID)
IF(WID .EQ. 0) GO TO 999
,
,
103
!
IDARC(IA,IT) = WID
BRDIST = 9999.
SP(IA,IT) = SPEED(WID)
TS = (SPEED(WID)/60V'-INCRE
WHERE(IA,IT) = WHOLD + TS
C---IS THERE INDIRECT FIRE ? OR MINEFIELD ? OR DIRECT FIRE
CALL IDFND2 ( IA,CLOCK,POIF, IFANS)
CALL MFFND2 ( PMARC SMARC ,WID,MFANS)
CALL DRFND2 ( I A, WHOLD, RANGE, BRDIST, DFANS)
?
,
IFA(IA,IT) = IFANS
MFA(IA,IT) = MFANS
DFA(IA,IT) = DFANS
IF( IFANS. EQ.O .AND.
MFANS. EQ.
.AND.
DFANS. EQ.O)
GOTO
30
C-- -COMPUTE THE CASUALTY OF I.NDIRECT-FIRE
IF( IFANS .EQ. 1) THEN
CALL IDFIR2 ( lA, IT,RF,REDAVE ,POIF, IFCAS)
RF(IA,IT) = RF(IA,IT) - IFCAS
IFSUM(IA) = IFCAS + IFSUM(IA)
IFC(IA,IT) = IFC(IA,IT) + IFCAS
END IF
!
C-. -COMPUTE THE CASUALTY BY MINE -FIELD
MFANS. EQ. 2) THEN
IF(MFANS.EQ. 1
.OR.
CALL MFILD2 ( lA, IT, MFANS INCRE,NA, WID, TIME, RF,MFCAS)
RF(IA,IT) = RF(IA,IT) - MFCAS
MFSUM(IA) = MFCAS + MFSUM( lA)
MFC(IA,IT) = MFC(IA,IT) + MFCAS
!
,
END IF
C-- -COMPUTE THE CASUALTY BY DIRECT-FIRE
!
IF(DFANS .EQ. 1 .OR.
DFANS EQ. 2) THEN
.AND.
IF( RF(IA,IT) .LT. EF(IA,IT)
.A.ND.
lUSED .EQ.
THEN
RF(IA,0) .GT. 0.0
)
RF(IA,IT) = RF(IA,IT) + RRESER
lUSLD = 1
END IF
CALL DRFIR2 (DFANS ,DFA, BRDIST, lA, IT, RF ,BF,AR,RDFCAS,BDFCAS,
PLTN, RANGE, NNN,ACO,AMO)
.
1
1
1
RF(IA,IT) = RF(IA,IT) - RDFCAS
BF(IA,IT) = BF(IA,IT) - BDFCAS
RDFSUM(IA) = RDFCAS + RDFSUM(IA)
BDFSUM(IA) = BDFCAS + BDFSUM(IA)
RDFC(IA,IT) = RDFC(IA,IT) + RDFCAS
BDFC(IA,IT) = BDFCaA,IT) + BDFCAS
END IF
IF(RF(IA,0) .GT.
CALL BREAK2
END IF
RF(IA,IT) LE. RBP(IA)) THEN
.AND.
lA, IT,NOC ,VvT10LD,RF,BF,RRESER,BREAKP,BREAKT)
0.
(
.
104
»
C-- -COMPUTE THE REDUCED SPEED DUE TO THE INDIRECT FIRE OR MINEFIELD OR
C
UNDER THE DIRECT FIRE BASED ON THE RANGE OF TWO FORCES
!
CALL
RSPED2
(
1
lA, IT, NA, SPEED, TIME ,DIST,WID, RANGE, AR,RF,
I
FANS, MEANS, DFANS,OSPEED,ACO,AMO)
SP(IA,IT) = SPEED(WID)
TS = (SPEED(WID)/60)*INCRE
WHERE(IA,IT) = WHOLD + TS
IMDC(IA,IT) = IFC(IA,IT) + MFC(IA,IT) + RDFC(IA,IT)
CONTINUE
30
36
37
IF(NEQ12 .KE. 0) GO TO 36
BF(1,IT) = MIN( BF(1,IT),BF(2,IT) )
BF(2,IT) = MIN( BF(1,IT),BF(2,IT) )
IF(NEQ34 NE. 0) GO TO 37
BF(3,IT) = MIN( BF(3,IT),BF(4,IT) )
BF(4,IT) = MIN( BF(3,IT),BF(4,IT) )
IF(NEQ12 NE. 0) GO TO 38
DFA(2,IT) EQ. 2 .AND.
IF(DFA(1,IT) .EQ. 2 .AND.
.AND.
RF(2,IT) GT. 0.) THEN
1
RF(1,IT) .GT. 0.
BF(1,IT) = (RF(1,IT) / (RF(1,IT)+RF(2,IT))) *
BF(2,IT) = (RF(2,IT) / (RF(1 IT)+RF( 2 IT) ) ) *
NEQ12 = IT
END IF
IF(NEQ34 NE. 0) GO TO 9999
IF(DFA(3,IT) .EQ. 2 .AND.
DFA(4,IT) EQ. 2 .AND.
1
RF(3,IT) .GT. 0.
.AND.
RF(4,IT) GT. 0.) THEN
BF(3,IT) = (RF(3,IT) / (RF( 3 IT)+RF(4, IT) ) ) *
EF(4,IT) = (RF(4,IT) / (RF( 3 IT)+RF( 4 IT) ) ) '>
NEQ34 = IT
END IF
.
.
.
.
,
38
,
BF(1,IT)
BF(2,IT)
,
BF(3,IT)
BF(4,IT)
.
.
.
,
,
CONTINUE
9999
CALL
999
PRINT2
1
1
TEND = IT
RETURN
END
* IV- 1.
-
RBN,BCO, BLUER 1 ,BLUEP2 ,RRESER,PLTN,NA,
IFSUM,MFSUM,RDFSUM,BDFSUM,
TRCAS,TBCAS,TRSUR,TBSUR)
(
1
SUBROUTINE FOR LOCATING THE ARC THE RED FORCES ARE ATTACKING.
SUBROUTINE
INTEGER
REAL
DATA
DATA
DATA
DATA
DETET2
(
lA, IT, INCRE, WHOLD, WID)
INCRE, WID
Al(4),A21(8),A22(6),A3(6),WHOLD
Al
A21
A22
A3
/2. 75,6. 35,9.35,10. 25/
/I. 15,3.3,4. 7,5.45,6.4,7. 1,8.55,9.45/
/I. 15 , 3. 3,4. 7 ,6. 15 ,6. 8, 7. 45/
/4. 5,6. 6,8. 85,12. 35,12. 95,13.55/
105
3011
GO TO (3011,3012,3013,3014) lA
IF(WHOLD .LE. Al(l)) THEN
WID = 5
GO TO 1390
ELSE IF(WHOLD. GT. Al(l)
.AND. WHOLD. LE. Al(2) ) THEN
WID = 15
GO TO 1390
ELSE IF(WHOLD. GT. Al(2)
.AND. WHOLD. LE. Al( 3) ) THEN
WID = 33
GO TO 1390
.AND. WHOLD. LE. Al(4) ) THEN
ELSE IFCWHOLD.GT. Al(3)
WID = 35
GO TO 1390
ELSE IF( WHOLD GT. Al(4)) THEN
IT-1
WRITE(26,1380)
F0RMAT(1X,'RED FORCE IN AVEZ/l TOOK BLL^ POSITIOW/l
BATTLE END
',14,' MINUTE)')
1/5X,' (BATTLE TIME
GO TO 1399
END IF
IF( WHOLD
LE. A21(l)) THEN
WID = 4
GO TO 1390
ELSE IF(WHOLD. GT. A21(l) .AND. WHOLD. LE. A21( 2) ) THEN
WID = 7
GO TO 1390
ELSE IF(WHOLD. GT. A21(2) .AND. WHOLD. LE. A21( 3) ) THEN
WID = 19
GO TO 1390
ELSE IF(WHOLD. GT. A:1(3) .AND. WHOLD. LE. A21(4) ) THEN
WID = 24
GO TO 1390
ELSE IFCWHOLD.GT. A21(4) .AND. WHOLD. LE. A21( 5) ) THEN
WID = 29
GO TO 1390
ELSE IF(WHOLD. GT. A21(5) .AND. WHOLD. LE. A21(6) ) THEN
WID = 32
GO TO 1390
ELSE IF(WHOLD. GT. A.?l(6) .AND. WHOLD. LE. A21( 7) ) THEN
WID = 34
GO TO 1390
ELSE IFCWHOLD.GT. A21( 7) .AND. WHOLD. LE. A21( 8) ) THEN
WID = 35
GO TO 1390
ELSE IF(WHOLD GT. A21(8)) THEN
WRITE(26,1381)
IT-1
BATTLE ',
F0R>L^T(1X,'RED FORCE IN AVE^^21 TOOK BLUE P0SITI0W;1
',14,' MINUTE)')
I'END !' ,/5X,'( BATTLE TIME
GO TO 1399
END IF
IF(WHCLD LE. A22(l)) THEN
WID = 4
GO TO 1390
ELSE IF(WHOLD. GT. A22(l) .AND. WHOLD. LE. A22(2) ) THEN
WID = 7
GO TO 1390
ELSE IF(WHOLD. GT. A22(2) .AND. WHOLD. LE. A22( 3) ) THEN
.
1380
3012
=>
:
.
.
1381
3013
=>
.
106
:
!'
WID = 19
GO TO 1390
ELSE IF(WHOLD.GT. A22(3) .AND. WHOLD. LE. A22(4) ) THEN
WID = 25
GO TO 1390
ELSE IFCWHOLD. GT. A22(4) .AND. WHOLD. LE. A22(5) ) THEN
WID = 30
GO TO 1390
ELSE IFCWHOLD. GT.A22( 5) .AND. WHOLD. LE. A22(6) ) THEN
WID = 39
GO TO 1390
ELSE IFCraOLD GT. A22(6)) THEN
IT-1
WRITE(26,1382)
BATTLE
1382 F0RMAT(1X,'RED FORCE IN AVE#22 TOOK BLUE POSITIOW/2
1 'END !' ,/5X,' (BATTLE TIME
',14,' MINUTE)')
GO TO 1399
END IF
3014
IFCWHOLD LE. A3(l)) THEN
WID = 2
GO TO 1390
.AND. WHOLD. LE. A3(2) ) THEN
ELSE IFCWHOLD. GT.A3(1)
WID = 11
GO TO 1390
.AND. WHOLD. LE. A3C 3) ) THEN
ELSE IFCWHOLD. GT.A3C 2)
WID = 13
.
=>
:
.
GO TO 139^
ELSE IFCWHOLD. GT.A3C 3)
.AND. WHOLD. LE. A3(4) ) THEN
WID = 23
GO TO 1390
ELSE IFCWHOLD. GT.A3C 4)
.AND. WHOLD. LE. A3C5) ) THEN
WID = 42
GO TO 1390
ELSE IFCWHOLD. GT.A3C 5)
.AND. WHOLD. LE. A3C 6) ) THEN
WID = 43
GO TO 1390
ELSE IFCWHOLD GT. A3C6)) THEN
IT-1
WRITEC26,1383)
BATTLE',
1383 F0RMATC1X,'RED FORCE IN AVE,-'/3 TOOK BLUE POSITIOW/2
1' END
,/5X,'CBATTLE TIME ==> ',14,' MINUTE)')
GO TO 1399
END IF
1390 RETURN
1399 WID =
RETURN
END
.
:
'
!
* IV-2.
SUBROUTINE FOR IDENTIFYING THE INDIRECT-FIRE
!
SUBROUTINE IDFND2 C IA,CLOCK,POIF,IFANS)
INTEGER POIF.IFANS
REAL
CLOCK, STIF,FTIF
DATA
STIF,FTIF /5. 0,15.0/
IFCCLOCK .GE. STIF .AND.
IFCPOIF .EQ. 2) THEN
IFCIA. EQ. 2 .OR.
CLOCK
.
LT.
lA.EQ. 3) THEN
107
FTIF) THEN
'
IFANS = 1
GO TO 1444
ELSE
IFANS =
GO TO 1444
END IF
ELSE IFCPOIF .EQ. 3) THEN
IF(IA .EQ. 4) THEN
IFANS = 1
GO TO 1444
ELSE
IFANS =
GO TO 1444
END IF
END IF
ELSE
IFANS =
END IF
RETURN
END
1444
•-^
IV-3.
SUBROUTINE FOR IDENTIFYING THE MINE-FIELD
!
SUBROUTINE MFFND2 ( PMARC SMARC ,WID,MFANS)
INTEGER MFANS WID PMARC( 5 ) SMARC( 3
,
,
,
,
DO 1510 I = 1,5
IF(WID .EQ. PMARC(I)) THEN
MFANS = 1
GO TO 1599
END IF
j
CONTINUE
DO 1520 J = 1,3
IF(WID .EQ. SMARC(J)) THEN
MFANS = 2
GO TO 1599
END IF
CONTINUE
MFANS =
RETURN
END
1510
1511
1520
1521
1599
* IV-4.
SUBROUTINE FOR IDENTIFYING THE DIRECT-FIRE
SUBROUTINE
1
1
DRFND2
(
lA ,WHOLD,RANGE ,BRDIST,DFANS)
INTEGER
REAL
IC,IA,DFANS
WHOLD,PATH( 4), RANGE, BRDI ST, AMBUSH
DATA
DATA
PATH
AMBUSH
/ 10.
.
(lA.EQ.
2.
25 ,9. 45
,
7.
45
,
13.
55/
/4. 0/
IF(WHOLD .GE.
WHOLD LT.
AMBUSH-RANGE/ 2.
AMBUSH
.OR.
DFANS =
!
.AND.
AND.
.
THEN
lA.EQ. 3))
1
108
BRDIST = AMBUSH - WHOLD
GO TO 1699
ELSE IF( WHOLD GE. PATH( lA) -RANGE .AND.
WHOLD .LT. PATH(IA)
THEN
1
)
DFANS = 2
BRDIST = PATH(IA) - WHOLD
GO TO 1699
END IF
1600 CONTINUE
DFANS =
1699
RETURN
END
.
* IV-5.
SUBROUTINE FOR COMPUTING CASUALTIES BY INDIRECT-FIRE
SUBROUTINE IDFIR2 ( lA, IT,RF,REDAVE ,POIF, IFCAS)
INTEGER NR0UND,IA,IT,REDAVE(3),P0IF
REAL
LETHAL, ROWSP,COLSP, MUX, MUY,RHOX,RHOY, RATIO, PKILL
REAL
RF( 4,0: 1001), IFCAS
NROUND LETHAL
MUX,MUY
RHOX,RHOY
DATA
DATA
DATA
/18,70. 8/
,
IF(REDAVE(POIF) EQ.
ROWSP =12.
COLSP = 12.
GO TO 1669
ELSE IF(REDAVE(POIF)
ROWSP =10.
COLSP = 10.
GO TO 1669
ELSE IF(REDAVE(POIF)
ROWSP = 8.
COLSP = 8.
GO TO 1669
END IF
.
/15. ,10.
/15. ,10.
THEN
1)
.
.
EQ.
2)
THEN
EQ.
3)
THEN
RATIO = (NROUND'^ LETHAL)
PHI = 3. 141592654
1669
A =
( (
LETHAL'>NROUND) / (
PKILL = RAT 1
EXP(
1
-
->
.
5--n
( A--'(
A/ (
/
(ROWSP'''COLSP*RF( lA, IT)
/
2''^PHI
(
MUX-->-'^2
)
)*-•>.
5
A^> A+RHOX-'"> 2
)
/
IFCAS = PKILL * RF(IA,IT)
(
)
•>
(
A''^A+RH0X*''^2
/
)
A -^ A+RHO Y''"'" 2 ) ) '^'^ 5 ) '^
+ ( MUY''^''^2 ) / ( A''^A+RH0Y*''^2 )
)
2.
RETURN
END
* IV-6.
SUBROUTINE FOR COMPUTING CASUALTIES BY MINEFIELDS
SUBROUTINE MFILD2
(
lA, IT, MEANS INCRE,NA,WID, TIME, RF,MFCAS)
,
INTEGER
REAL
MFANS,INCRE,NA,NOPM,NOSM,WID
TIME(NA),RF(4,0: 1001) ,PCOEF,SCOEF,MFCAS,TPM, IBM
DATA
DATA
NORM ,NOSM
TPM ,TSM
/300,200/
/20. 0,15.0/
109
)
DATA
PCOEF,SCOEF
/.
00020
00015/
,.
IF(MFANS .EQ. 1) THEN
MFCAS = (PCOEF''--RF(IA,IT)'^NOPM''^(TPM/60.
ELSE IF(MFANS EQ. 2) THEN
MFCAS = (SCOEF'>RF(IA,IT)*NOSM->(TSM/60.
END IF
RETURN
END
))
/
(TIME(WID)/INCRE)
))
/
(TIME(WID)/INCRE)
.
* IV- 7.
1
SUBROUTINE FOR COMPUTING CASUALTIES BY DIRECT FIRE
SUBROUTINE DRFIR2 (DFANS ,DFA,BRDIST, lA, IT,RF,BF,AR,RDFCAS,BDFCAS,
PLTN, RANGE, NNN,ACO,AMO)
INTEGER DFANS, DFA( 4, 1000), PLTN
RF(4,0: 100n,BF(4,0: 1001) ,PLBP ,FBF(0: 1001)
REAL
BRDIST,RANGE,RED,BLU,RDFCAS,BDFCAS,AR,BR,MU,PL
REAL
DATA
DATA
DATA
DATA
DATA
PLBP
AOMEGA.BOMEGA
AC, EC
AM,BM
MU
/
0.
0/
/O. 50,1.00/
/O. 0380 ,0. 0160/
/O. 23 , 0. 00010/
/l-O/
ACO = AC
AMO = AM
IF(NNN .GT. 0) GO TO 1810
PL = PLTN
1810 FBF(IT) = PL
C---FIND ENGAGED FORCES OF BOTH SIDES ON DIRECT FIRE BATTLE
RED = RF(IA,IT)
BLU = BF(IA,IT)
1
C---CASE-r'a (MIXED LAV)
IF( DFANS .EQ.
:
BLUE FORCE AMBUSHES THE RED FORCE
1) THEN
(1. 0-BRDIST/R.ANGEl'"'^MU
AR = AM
BR = BM '• ( 1. 0-BRDIST/RA.NGE)->'''MU
BLU = FBF(IT)
IF(BLU .LT. PLBP) THEN
RDFCAS = 0.
BLU = 0.
ELSE
RDFCAS = AR
BLU
END IF
BDFCAS = BR * BLU '^ RED
FBF(IT) = BLU - BDFCAS
PL = FBF(IT)
NNN = NNN + 1
GO TO 1900
-'•-
•>''
C---CASE-#2 (HEMBOLT EQLTION)
ELSE IF(DFANS EQ. 2) THEN
IF(RED LE. 0.
.OR.
BLU
BDFCAS = 0.
RDFCAS = 0.
GO TO 1900
!
.
.
.
LE.
110
0.)
TliEN
!
!
END IF
AR = AC * (1. 0-BRD I ST/RANGE )'"'--^MU
BR = BC * (1. 0-BRDIST/RANGE)'"^'VMU
RDFCAS = AR^>((RED/BLU)^'---"-(1.0-BOMEGA))'''BLU
BDFCAS = BR"((BLU/RED)-''"-(l. 0-AOMEGA))'-RED
END IF
RETURN
END
1900
* IV-8.
*
SUBROUTINE FOR COMPUTING THE ADJUSTED SPEED BASED ON
THE NUMBER OF RED UNITS (BATTALIONS) ON ONE AVENUE.
SUBROUTINE
(
I
,
IR,SPEED,ADSP)
INTEGER
REAL
IR,A0A1(4) ,AOA2(3) ,A0A3(6)
SPEED(44),ADSP(44)
DATA
DATA
DATA
AOAl
A0A2
A0A3
75,15,33,35/
/4,7,19/
/2 11 13,23,42,43/
,
,
GO TO (201,202,203,204)
IR
ADSP(I) = SPEED(I)
GO TO 2010
ADSP(I) = SPEED(I)
DO 2011 J = 1,3
ADSP(I) =
IF(I .EQ. A0A2(J))
GO TO 2010
ADSP(I) = SPEED(I)
DO 2012 J = 1,6
1F(I .EQ. A0A3(J))
ADSP(I) =
GO TO 2010
ADSP(I) = SPEED(I)
DO 2013 J = 1,3
IF(I .EQ. A0A2(J))
ADSP(I) =
201
202
2011
203
2012
204
2013
0. 9''-SPEED( I)
0. 9-'^SPEED( I)
0. 8^'-SPEED( I)
RETURN
END
2010
'••-
ASPED2
IV-9.
*
SUBROUTINE FOR COMPUTING THE REDUCED SPEED
BASED ON EACH WEAPON TYPE.
SUBROUTINE
RSPED2
1
(
lA, IT, NA, SPEED, TIME, DIST,WID, RANGE, AR,RF,
IFANS, MEANS, DFANS,OSPEED,ACO,AMO)
INTEGER
REAL
REAL
NA,VID,IFANS,MFANS,DFANS
AR,DIST(NA),SPEED(NA),TIME(NA),MINSPD,RF(4,0: 1001)
OSPEED(NA)
DATA
MINSPD
/O.
500/
IFdFANS
.EQ. 1) THEN
SPEED(WID) = DIST(WID)*60./(TIME(WID)+10.
END IF
IF(MFANS .EQ. 1) THEN
SPEED(WID) = DIST(WID)''--60./(TIME(VID)+20.
111
)
END IF
IF(MFANS .EQ. 2) THEN
SPEED(WID) = DIST(WID)''^60. /(TIME(WID) + 15. )
END IF
IF(DFANS .EQ. 1) THEN
SPEED(WID) = SP£ED(WID) * (1. 0-AR/AMO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(WID) = OSPEED(WID)
END IF
IF(DFANS .EQ. 2) THEN
IF(RF(IA,IT) .EQ. 0. ) THEN
SPEED(WID) = 0.
ELSE
SPEED(WID) = SPEED(WID) * (1.0-AR/ACO) + MINSPD
IF(SPEED(WID) .GT. OSPEED(WID)) SPEED(VID) = OSPEED(WID)
END IF
END IF
RETURN
END
'>
IV- 10.
SUBROUTINE TO FIND THE TIME AND LOCATION
WHEN RED FORCE REACH BREAK-POINT
!
SUBROUTINE
BREAK2
(
lA, IT,NOC ,WHOLD,RF,BF,RRESER,BREAKP,BREAKT)
INTEGER RRESER
REAL
BREAKP(4),BREAKT(4),WH0LD,RF(4,0: 1001),BF(4,0: 1001)
RF(IA,IT) = 0.
BF(IA,IT) = BF(IA,IT-1)
IF(N0C .EQ. 0) THEN
RF(IA,1T) = RF(IA.IT-l) + RRESER
BREAKP(IA) = WHOLD
BREAKT(IA) = IT
END IF
NOC = NOC + 1
RETURN
END
C
nil
''
IV- 11.
•
SUBROUTINE TO PRINT THE CASUALTIES AND SURVIVORS
FOR BOTH SIDES IN BATTLE2 TYPE.
iV Vr iV sV iV -,V Vf Vr vV Vr v.- Vr Vr Vr Vr Vr vV Vr Vc Vr Vr Vr ^V V,- Vc v.- V,- V,- v.- v.- Vr v.-
SUBROUTINE
1
-.V
,
,
,
1
INTEGER
REAL
REAL
REAL
Vc Vr Vr V: Vr V,-
PRINT2 (RBN BCO BLUER 1 ,BLUEP2 RRESER, PLTN,NA,
IFSUM MFSUM RDFSUM BDFSUM
TRCAS,TBCAS,TRSUR,TESUR)
,
RBN,BC0,BLUEP1,BLUEP2,NA, RRESER, PLTN
IFSUM( 4 ) MFSUM( 4 ) RDFSUM( 4 ) BDFSUM( 4)
,
,
,
TIFSUM,TMFSUM,TRDF,TBDF
TRCAS,TRSUR,TBCAS.TBSUR
TIFSUM = 0.
TMFSUM = 0.
TRDF = 0.0
TBDF = 0.
TRCAS =
,
,
C.
112
TBCAS = 0.
TRSUR = 0.0
TBSUR = 0.0
DO 101 lA = 1,4
TIFSUM = IFSUM(IA) + TIFSUM
TMFSUM = MFSUM(IA) + TMFSUM
TRDF = RDFSUM(IA) + TRDF
TBDF = BDFSUM(IA) + TBDF
CONTINUE
+ RRESER
TRF = RBN*3.
+ PLTN
TBF = BC0'>3.
BFCASl = BDFSUM(l) + BDFSUM(2)
BFCAS2 = BDFSUMO) + BDFSUM(4)
BFSURl = BLUEFl^'^BCO - BFCASl
BFSUR2 = BLUEP2'>BC0 - BFCAS2
101
TRCAS
TRSUR
TBCAS
TBSUR
=
=
=
=
TIFSUM + TMFSUM + TRDF
TRF - TRCAS
TBDF
TBF - TBCAS
RETURN
END
*
V.
SUBROUTINE TO COMPUTE THE WEIGHTED UTILITY VALUES OF BLUE
SURVIVOR FOR EACH UTILITY FUNCTIONS
!
SUBROUTINE
UTIL (MR,MB IB ,VALUE ,LIN,SQU,ROO)
INTEGER
REAL
REAL
MR, MB
DATA
DATA
WGT
ALIN
,
VALUE(4),UL(4),US(4),UR(4)
LIN,SQU,R00,ALIN,WGT(4)
/O. 05,0. 70,0. 05,0. 20/
/340. 0/
TEMPI = 0.
TEMP2 = 0.
TEMPS = 0.
C---UTILITY VALUES FOR THE BLUE SURVIVORS
DO 2301 K = 1,4
UL(K)=(9. /55. )'>VALUE(K) + 10.
IF(VALUE(K) .LE. ALIN) UL(K)=(9. /55. )*VALUE(K)
US(K)=(1. /3025. )'WALUE(K)**2.
UR(K)=(4. 264014327)'WALUE(K)**0. 5
!
2301
TEMPI = WGT(K)''fUL(K) + TEMPI
TEMP2 = WGT(K)''^US(K) + TEMP2
TEMPS = WGT(K)'>UR(K) + TEMPS
CONTINUE
LIN = TEMPI
SQU = TEMP2
ROO = TEMPS
RETURN
END
113
APPENDIX
This
is
E.
EXEC FOR ADVANCED MODEL
a execution code of Fortran-77 for the advanced
FILEDEF 12 DISK THESIS DATA
EXEC WF77 ADVAN (NOEXT NOWAR
114
model
in
Appendix D.
K
LIST OF
1.
Craig.
Dean
REFERENCES
"A Model for ihe Planning of Maneuver Unit and Engineer Asset
E.,
Placement", M.S. Thesis, Naval Postgraduate School, Monterey, California, Sep-
tember 1985.
2.
R..
"The Extension of Unit Allocation and Countermobility
in the
Airland Rearch Model", M.S. Thesis, Naval Postgraduate
McLaughlin, Joseph
Planning Algorithm
School, Monterey, California,
3.
March
1986.
Choi, Seok Chul. "Determination of iWetwork Attributes from A High Resolution
Terrain
Data
Base",
.VI.S.
Naval
Thesis,
Postgraduate
School,
Monterey,
Cahfornia. September 19S7.
4.
Engineering Design Handbook. "Army Weapon Systems Analysis. Part Two".
Army
5.
Material Development and Readiness
Taylor, James G.,
"
Command, October
US
1979.
Lanchcster-Type Models of Warfare, Volume IF, Naval Post-
graduate School. Monterey. California, 1980.
6.
Page, Alfred N.. "Utility Theory a Book of Reading" John Wiley
7.
Text book for the course {OA-3601
,
:
Airland
&
Sons, Inc.. 1968.
Combat Model). Sotes on Measures
of Effectiveness, Naval Postgraduate School, Monterey, California, 1987.
8.
Text book for the course (OA-3601
:
Airland
Combat Model), Notes on Game
Theory, Naval Postgraduate School, .Monterey, California, 1987.
9.
110.
McKinsey,
J.
C.
C,
"Introduction to the Theory of
Games" The R-^ND Corpo,
ration, 1952.
Schrage, Linus. "Linear, Integer and Ouadradic Programming with
Edition/',
The
Scientific Press,
LIN DO
(Third
540 University Avenue, Palo Alto. California. 1986.
115
11.
DeGroot, Morris H..
" Probahiliiy
Wesley Publishing Company.
12.
Minitab Release
5.1.
and
Siatiiistics
''Second Ediiionj", Addison-
Inc., 196S.
"Minitab Reference Manual", Naval Postgraduate School,
Monterey, California, 19S5.
116
BIBLIOGRAPHY
Jensen, Paul A., Barnes,
J.
Wesley, "Network Flow Programming" John Willey
,
&
Sons, Inc., 1980.
Engineering Design Handbook, "Army Weapon Systems Analysis, Part One",
Army
Material Development and Readiness
117
Command, October
1979.
US
INITIAL DISTRIBUTION LIST
No. Copies
1.
Defense Technical Information Center
2
Cameron
Station
22304-6145
Alexandria.
VA
2.
3.
Librar}-. Code 0142
Naval Postcraduate School
Monterey. CA 93943-5002
Department Chairman. Code 55
Department of Operations Research
Naval Postgraduate School
Monterey.
4.
2
CA
1
93943-5000
Professor Samuel H. Parry. Code 55Py
Department of Operations Research
3
Naval Postcraduate School
Monterey. CA 93943-5U()()
5.
LTC Bard Mansager. Code 55Ma
Department of Operations Research
Naval Postcraduate School
Monterey.
6.
CA
1
93943-5000
Deputy Undersecretary of the Army
2
Operations Research
Room 2E261. Pentason
Washmgton. D.C. 20310
for
7.
Director
Attn: Mr. E. B. Vandiver 111
U.S. Armv Concepts Analvsis
1
Aeencv
Bethesda.'MD 20814
8.
9.
Bell Hall Library
U.S. Army Combined Arms Center
Fort Leavenworth, KS 66027
1
_-
-
~'^'--
Director
Requirements
1
&
Programs Director, Headquarters
TRAC
Attn: ATRC-RP (COL. Brinkley)
Fort Monroe, VA 23651-5143
10.
Director
1
TRAC-FLVN
Attn: ATRC-F
(Dr. LaRocque)
Fort Leavenworth. KS 66027-5200
118
11.
Director
TR.AC-WSMR
Attn:
ATRC-W
(Dr. Collier)
White Sands Missile Range,
12.
NM
88002-5502
Director
Research Directorate
ATRC-RD (Hue McCoy)
White Sands Missile Range,
Attn:
NM
13.
88002-5502
Commander
U.S. Armv TR.A.DOC
Analysis Command
Attn: ATRC-ZD (Mr. Bauman)
Fort Leavenworth, KS 66027-5200
14.
Systems Analysis Department
ROK Army Headquarters
Vonssan-Gu Yonssan-Dong
CPO
Box #2
Seoul. 140-799
Repubhc of Korea
15.
Hveon Soo
Lee,
SMC
^'2337
Naval Postgraduate School
Monterey, CA 93943
16.
Choi.
Mun
SMC
^1840
Soo
Naval Postgraduate School
Monterey, CA 93943
17.
Park.
Hun Keun
SMC
t.2997
Naval Postsraduate School
Monterey, CA 93943
18.
^'u.
Moo Bone
SMC?? 1845
Naval Postgraduate School
Monterey, CA 93943
19.
Yeone
SimGok-1-Dong 585, 10
BuCheon Citv, Nam-Gu
Lee, Jae
6
KyungGi-Do', 422-011
Republic of Korea (Seoul)
20.
Lee, Chong Ho
NamMun-1-Ri, 3-Ban
YangYang-Eup. YangYang-Gun
KangWon-Do, 210-21
Republic of Korea (Seoul)
119
21.
Song.
Won
Chong Chul
Sang Dae Burak. SangDac-Ri
PoDu-Moeyn. KoMcong-Kun.
Cheon-Nam. 543-21
Republic of Korea (Seoul)
120
DTTDLE";
•
Thesis
Th^ ^38635
L3J c.l
/
Lee
Airland
etric
con^K
^"^iysis of
/
Thesis
L38635
c.l
Lee
Parametric analysis of
airland combat model in
high resolution.
ui'.AinVATE SCFOOT,