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1.
Learn about applications of linear programming that are solved in practice.
2.
Develop an appreciation for the diversity of problems that can be modeled as linear programs.
3.
Obtain practice and experience in formulating realistic linear programming models.
4.
Understand linear programming applications such as:
data envelopment analysis
revenue management
portfolio selection
game theory
5.
Know what is meant by a two person, zero sum game.
6.
Be able to identify a pure strategy for a two person, zero sum game.
7.
Be able to use linear programming to identify a mixed strategy and compute optimal probabilities for
the mixed strategy games.
8.
Understand the following terms:
game theory
two person, zero sum game
saddle point
pure strategy
mixed strategy
The application problems of Chapter 5 are designed to give the student an understanding and appreciation of
the broad range of problems that can be approached by linear programming. While the problems are
indicative of the many linear programming applications, they have been kept relatively small in order to ease
the student's formulation and solution effort. Each problem will give the student an opportunity to practice
formulating a linear programming model. However, the solution and the interpretation of the solution will
require the use of a software package such as
,
's Solver or LINGO.
5 1
,
1.
5
a.
Min
s.t.
285.2
123.80
106.72
+
+ 34.62
+ 27.11
+
148
+
27
+ 162.3
+ 128.70
+ 64.21
48.14
43.10
253
41
+
285.2
+ 1123.80
+ 106.72
,
2.
+
+ 36.72
+ 45.98
+
175
+
23
+ 275.7
+ 348.50
+ 104.10
,
+
+
+
+
+
+
+
+
33.16
56.46
160
84
210.4
154.10
104.04
=
≥
≥
≥
≥
≤
≤
≤
1
48.14
43.10
253
41
0
0
0
≥0
,
b.
Since
= 1.0, the solution does not indicate General Hospital is relatively inefficient.
c.
The composite hospital is General Hospital. For any hospital that is not relatively inefficient, the
composite hospital will be that hospital because the model is unable to find a weighted average of the
other hospitals that is better.
a.
Min
s.t.
55.31
49.52
281
47
250 +310
316 +134.6
94.4 +116
+
+
+
+
+
+
+
+
37.64
55.63
156
3
278.5
114.3
106.8
+
+
+
+
+
+
+
+
,
b.
+
+
+
+
+
+
+
+
32.91
25.77
141
26
165.6
131.3
65.52
,
+
+
+
+
+
+
+
+
33.53
41.99
160
21
250
316
94.4
,
,
,
= 0.924
= 0.074
= 0.436
= 0.489
All other weights are zero.
c.
is relatively inefficient
Composite requires 92.4 of 's resources.
5 2
32.48
55.30
157
82
206.4
151.2
102.1
,
≥0
+
+
+
+
+
+
+
+
48.78
81.92
285
92
384
217
153.7
+
+
+
+
+
+
+
+
58.41
119.70
111
89
530.1
770.8
215
=
1
≥ 33.53
≥ 41.99
≥ 160
21
≥
0
≤
0
≤
0
≤
=
d.
>
=
34.37 patient days (65 or older)
41.99 patient days (under 65)
3.
e.
Hospitals A, C, and E.
a.
Make the following changes to the model in problem 27.
New Right Hand Side Values for
Constraint 2
32.48
Constraint 3
55.30
Constraint 4
157
Constraint 5
82
New Coefficients for
Constraint 6
Constraint 7
Constraint 8
b.
4.
= 1;
in
206.4
151.2
102.1
= 1; all other weights = 0
c.
No; = 1 indicates that all the resources used by Hospital
Hospital .
are required to produce the outputs of
d.
Hospital is the only hospital in the composite. If a hospital is not relatively inefficient, the hospital
will make up the composite hospital with weight equal to 1.
a.
Min
s.t.
110
22
1400
3800
25
8
+
96
+
16
+ 850
+
+ 4600
+
32
+
8.5
+ 110
+
22
+ 1400
,
,
+
+ 4400
+
35
+
8
+ 100
+
18
+ 1200
,
,
5 3
≥0
+
+
+
+
+
+
+
6500
30
10
125
25
1500
+
+
+
+
+
+
+
6000
28
9
120
24
1600
=
≥
≥
≥
≤
≤
≤
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4600
32
8.5
0
0
0
,
5
b.
$$$$$$$$$$$$$$
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c.
Yes; = 0.960 indicates a composite restaurant can produce Clarksville's output with 96% of
Clarksville's available resources.
d.
More Output (Constraint 2 Surplus) $220 more profit per week.
Less Input
Hours of Operation 110 = 105.6 hours
FTE Staff 22 1.71 (Constraint 6 Slack) = 19.41
Supply Expense 1400 129.614 (Constraint 7 Slack) = $1214.39
The composite restaurant uses 4.4 hours less operation time, 2.6 less employees and $185.61 less
supplies expense when compared to the Clarksville restaurant.
5.
e.
= 0.175,
restaurants.
= 0.575, and
= 0.250. Consider the Bardstown, Jeffersonville, and New Albany
a.
If the larger plane is based in Pittsburgh, the total revenue increases to $107,849. If the larger plane
is based in Newark, the total revenue increases to $108,542. Thus, it would be better to locate the
larger plane in Newark.
Note: The optimal solution to the original Leisure Air problem resulted in a total revenue of
$103,103. The difference between the total revenue for the original problem and the problem that has
a larger plane based in Newark is $108,542 $103,103 = $5,439. In order to make the decision to
change to a larger plane based in Newark, management must determine if the $5,439 increase in
revenue is sufficient to cover the cost associated with changing to the larger plane.
5 4
=
b.
>
=
Using a larger plane based in Newark, the optimal allocations are:
PCQ = 33
PCY = 16
NCQ = 26
NCY = 15
= 32
= 46
= 23
PMY= 6
= 56
NMY = 7
CMY = 8
COY = 10
= 43
POY = 11
NOQ = 39
NOY = 9
The differences between the new allocations above and the allocations for the original Leisure Air
problem involve the five ODIFs that are boldfaced in the solution shown above.
c.
Using a larger plane based in Pittsburgh and a larger plane based in Newark, the optimal allocations
are:
PCQ = 33
PCY = 16
NCQ = 26
NCY = 15
= 37
= 44
PMQ = 44
PMY= 6
= 56
NMY = 7
CMY = 8
COY = 10
= 45
POY = 11
NOQ = 39
NOY = 9
The differences between the new allocations above and the allocations for the original Leisure Air
problem involve the four ODIFs that are boldfaced in the solution shown above. The total revenue
associated with the new optimal solution is $115,073, which is a difference of $115,073 $103,103 =
$11,970.
6.
d.
In part (b), the ODIF that has the largest bid price is COY, with a bid price of $443. The bid price
tells us that if one more Y class seat were available from Charlotte to Myrtle Beach that revenue
would increase by $443. In other words, if all 10 seats allocated to this ODIF had been sold,
accepting another reservation will provide additional revenue of $443.
a.
The calculation of the number of seats still available on each flight leg is shown below:
!"
!"
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
PCQ
PMQ
POQ
PCY
PMY
POY
NCQ
NMQ
NOQ
NCY
NMY
NOY
CMQ
CMY
COQ
COY
33
44
22
16
6
11
26
36
39
15
7
9
31
8
41
10
25
44
18
12
5
9
20
33
37
11
5
8
27
6
35
7
5 5
8
0
4
4
1
2
6
3
2
4
2
1
4
2
6
3
,
5
Flight Leg 1: 8 + 0 + 4 + 4 + 1 + 2 = 19
Flight Leg 2: 6 + 3 + 2 + 4 + 2 + 1 = 18
Flight Leg 3: 0 + 1 + 3 + 2 + 4 + 2 = 12
Flight Leg 4: 4 + 2 + 2 + 1 + 6 + 3 = 18
Note: See the demand constraints for the ODIFs that make up each flight leg.
b.
The calculation of the remaining demand for each ODIF is shown below:
!"
!"
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
c.
PCQ
PMQ
POQ
PCY
PMY
POY
NCQ
NMQ
NOQ
NCY
NMY
NOY
CMQ
CMY
COQ
COY
33
44
45
16
6
11
26
56
39
15
7
9
64
8
46
10
25
44
18
12
5
9
20
33
37
11
5
8
27
6
35
7
8
0
27
4
1
2
6
23
2
4
2
1
37
2
11
3
The LP model and solution are shown below:
4 (). "56- . 56--. 563. "76+*
76*
76(
3.* "76+++ 76*.
76() " 563. " 76--+" 56*.-" 7
(8
-8
38
+8
*8
8
)8
.8
8
( 8
((8
(-8
(38
(+8
(*8
( 8
()8
(.8
( 8
- 8
( "56(
( "56(
( 56(
( 56(
( "59.
( 59(
( 59-)
( "79+
( 79(
( 79( "59
( 59-3
( 59( "79+
( 79( 79(
(" 593)
(" 79(" 59((
(" 793
56(
56(
76(
76(
56( "76(
56( "76(
56( 76("
56( 76("
76(
76(
56("
56("
5 6
79(
79(.
79(79(.
"56-+
563+
56
=
>
=
(*)3
$$$$$$$$$$$$$$
"5
5
5
"7
7
7
"5
5
5
"7
7
7
" 5
" 7
" 5
" 7
$$$$$$$$$$$$$$$
.
(
3
+
(
-
!
! " # #
$$$$$$$$$$$$$$$$$$
3
+
(
3
)
3
Note: The values shown above provide the allocations for the remaining seats available. The bid
prices for each ODIF are provided by the deal prices in the following output.
" #
$$$$$$$$$$$$$$
(
3
+
*
)
.
(
((
((3
(+
(*
(
()
(.
(
7.
a.
/0
1 #
$$$$$$$$$$$$$$$
2
#
$$$$$$$$$$$$$$$$$$
+
)
()
--+
()+
.*
-+
3)
-)3
33(**
3(*
( *
-.
3+
- (
+
3*.
Let CT = number of convention two night rooms
CF = number of convention Friday only rooms
CS = number of convention Saturday only rooms
RT = number of regular two night rooms
RF = number of regular Friday only rooms
RS = number of regular Saturday only room
5 7
,
5
b./c.The formulation and output obtained using
%
:
:
' %
4 --*" 6(-3" 6(3 " 6- *
(8 (" 9+
-8 (" 938 (" 9(*
+8 ( 9*8 ( 93
8 ( 9-*
)8 (" 6("
.8 (" 6("
8 (" 6("
( 8 (" 6("
is shown below.
6(+
;+.
;+.
6( 6(
6( 6(
6(*-
9
9
-*3(+
$$$$$$$$$$$$$$
"
"
"
$$$$$$$$$$$$$$$
3
((*
-.
-*
" #
$$$$$$$$$$$$$$
(
3
+
*
/0
1 #
$$$$$$$$$$$$$$$
+
.
!
! " # #
$$$$$$$$$$$$$$$$$$
2
#
$$$$$$$$$$$$$$$$$$
-.
+)
*
$-3
)
.
3
(+
( -
(
',%"
% " %
$$$$$$$$$$$$
"
"
"
" %
:%
<
=
$$$$$$$$$$$$$$$
(-3
*
( -+.
(-3
( -
"
$$$$$$$$$$$$$$$
--*
(-3
(3
- *
(+
(*-
5 8
11
=
$$$$$$$$$$$$$$$
-*3
(+
11
=
11
=
( 3
11
=
=
:>
>
2
" #
$$$$$$$$$$$$
(
3
+
*
)
.
(
d.
8.
2%
>
=
:%
<
=
$$$$$$$$$$$$$$$
3
(((
(.
-.
-(
+
<
=
.
3
"
$$$$$$$$$$$$$$$
+
(*
3
-*
+.
+.
11
=
$$$$$$$$$$$$$$$
11
=
11
=
-3
-3
11
=
-.
*
*(
.
(
The dual price for constraint 10 shows an added profit of $50 if this additional reservation is
accepted.
To determine the percentage of the portfolio that will be invested in each of the mutual funds we use
the following decision variables:
= proportion of portfolio invested in a foreign stock mutual fund
= proportion of portfolio invested in an intermediate term bond fund
= proportion of portfolio invested in a large cap growth fund
= proportion of portfolio invested in a large cap value fund
= proportion of portfolio invested in a small cap growth fund
= proportion of portfolio invested in a small cap value fund
a.
A portfolio model for investors willing to risk a return as low as 0% involves 6 variables and 6
constraints.
Max 12.03
+ 6.89
! 20.52
+ 13.52
+ 21.27
+ 13.18
s.t.
10.06
13.12
13.47
45.42
21.93
+
+
+
+
+
17.64
3.25
7.51
1.33
7.36
+
+
+
+
32.41
18.71
33.28
41.46
23.26
+
+
+
+
+
32.36
20.61
12.93
7.06
5.37
+
+
+
+
+
"
"
"
+
"
5 9
"
≥0
33.44
19.40
3.85
58.68
9.02
+
+
+
+
+
24.56
25.32
6.70
5.43
17.31
≥
≥
≥
≥
≥
=
0
0
0
0
0
1
,
5
b.
The solution obtained using
is shown.
(. +
$$$$$$$$$$$$$$
!
! " # #
$$$$$$$$$$$$$$$$$$
(3 - )
3+)
* (-*
-
$$$$$$$$$$$$$$$
'
:
:
" #
$$$$$$$$$$$$$$
(
3
+
*
*)
3+3
/0
1 #
$$$$$$$$$$$$$$$
3 3 .
-( +-.
-3
+ +3.
2
#
$$$$$$$$$$$$$$$$$$
$ 3 )
(. +
The recommended allocation is to invest 65.7% of the portfolio in a small cap growth fund and
34.3% of the portfolio in a small cap value fund. The expected return for this portfolio is 18.499%.
c.
One constraint must be added to the model in part a. It is
≥ .10
The solution found using
is given.
() ().
$$$$$$$$$$$$$$
$$$$$$$$$$$$$$$
(
'
:
:
" #
$$$$$$$$$$$$$$
(
3
+
*
!
! " # #
$$$$$$$$$$$$$$$$$$
3+)
* (-*
-
* .
3 /0
1 #
$$$$$$$$$$$$$$$
-) -*
-(
.(
3 * (
2
#
$$$$$$$$$$$$$$$$$$
$ 3 )
(. +
$(3 - )
)
The recommended allocation is to invest 10% of the portfolio in the foreign stock fund, 50.8% of the
portfolio in the small cap growth fund, and 39.2 percent of the portfolio in the small cap value fund.
The expected return for this portfolio is 17.178%. The expected return for this portfolio is 1.321%
less than for the portfolio that does not require any allocation to the foreign stock fund.
5 10
=
9.
>
=
To determine the percentage of the portfolio that will be invested in each of the mutual funds we use
the following decision variables:
= proportion of portfolio invested in a large cap stock mutual fund
= proportion of portfolio invested in a mid cap stock fund
= proportion of portfolio invested in a small cap growth fund
= proportion of portfolio invested in an energy sector fund
# = proportion of portfolio invested in a health sector fund
= proportion of portfolio invested in a technology sector fund
$ = proportion of portfolio invested in a real estate sector fund
a.
A portfolio model for investors willing to risk a return as low as 0% involves 7 variables and 6
constraints.
Max 9.%&
+ 5.91
! 15.20
32.3
23.2
0.9
49.3
22.8
+ 20.8
+ 22.5
+ 6.0
+ 33.3
+ 6.1
+
+ 11.74
+ 7.34# + 16.97
+ 15.44$
s.t.
35.3
20.0
28.3
10.4
9.3
+
+
+
+
+
+
+
+
+ 49.1#
+ 5.5#
+ 29.7#
+ 77.7#
24.9#
+
#
+
"
b.
25.3
33.9
20.5
20.9
2.5
"
"
"# "
The solution obtained using
"$
+
+
+
+
+
46.2
21.7
45.7
93.1
20.1
+ 20.5$
+ 44.0$
21.1$
+ 2.6$
+ 5.1$
+
$
≥0
is shown.
(* *3
$$$$$$$$$$$$$$
$$$$$$$$$$$$$$$
!
! " # #
$$$$$$$$$$$$$$$$$$
* )
(( *+.
*
%
>
+ -)
(
*
(+3
3*)
" #
$$$$$$$$$$$$$$
(
3
+
*
/0
1 #
$$$$$$$$$$$$$$$
-- 3--. *
2
#
$$$$$$$$$$$$$$$$$$
$
-. ..)
$
))
(* ) +
5 11
≥
≥
≥
≥
≥
=
2
2
2
2
2
1
,
5
The recommended allocation is to invest 50% of the portfolio in the small cap stock fund, 14.3% of
the portfolio in the technology sector fund, and 35.7% of the portfolio in the real estate sector fund.
The expected portfolio return is 15.539%.
c.
The portfolio model is modified by changing the right hand side of the first 5 constraints from 2 to 0.
d.
The solution obtained using
is shown.
(* ) +
$$$$$$$$$$$$$$
!
! " # #
$$$$$$$$$$$$$$$$$$
* )
(( *+.
$$$$$$$$$$$$$$$
-**
%
>
+ -)
(
*
-(*33
" #
$$$$$$$$$$$$$$
(
3
+
*
/0
1 #
$$$$$$$$$$$$$$$
3.
33 ).3
2
#
$$$$$$$$$$$$$$$$$$
$
-
*3
$
))
(* ) +
The recommended allocation is to invest 25.5% of the portfolio is the small cap stock fund, 21.2% of
the portfolio in the technology sector fund, and 53.3% of the portfolio in the real estate sector fund.
The expected portfolio return is 15.704%. This is an increase of .165% over the portfolio that limits
risk to a return of at least 2%. Most investors would conclude that the small increase in the portfolio
return is not enough to justify the increased risk.
10.
#
1
#
$
2
3
Minimum
1
8
5
7
5
2
2
4
10
2
8
5
7
Maximum
The maximum of the row minimums is 5 and the minimum of the column maximums is 5. The game
has a pure strategy. Player A should take strategy 1 and Player B should take strategy 2 . The value
of the game is 5.
11.
By definition, a pure strategy solution means each player selects a single strategy with probability 1.
Thus, if a linear programming solution has a probability decision variable equal to 1, the game has a
pure strategy solution.
5 12
=
12.
=
The expected payoffs for Company A are as follows:
Strategy
1
: Expected Payoff = 0.5(3) +0.5(2) = 2.5
Strategy
2
: Expected Payoff = 0.5(4) +0.5(1) = 2.5
Strategy
3
: Expected Payoff = 0.5( 2)+0.5(5) = 1.5
2 with
If Company B were to implement strategy
0.50, Company A can select strategy 1 or
in market share.
13.
>
2
probability 0.50 and strategy 3 with probability
to obtain an expected payoff providing a 2.5% increase
The row minimums are 5, 6, and 4. Station A prefers the maximin strategy 2 to obtain a gain of at
least 6. The column maximums are 10, 8 and 7. Station B prefers the minimax strategy 3 to limit its
maximum loss to no more than 7. However, because the maximum of the row minimums is not equal
to the minimum of the row maximums, the game does not have a pure strategy. A mixed strategy
solution with a value of the game between 6 and 7 exists.
The linear programming formulation and solution for Station A follows.
Max
' ('
s.t.
10)'1
+
8)'2
+
4)'3
' ('
> 0
$
(Strategy
1)
#
5)'1
+
7)'2
+
8)'3
' ('
> 0
(Strategy
2
)
3)'1
+
6)'2
+
7)'3
' ('
> 0
(Strategy
3
)
)'1
+
)'2
+
)'3
= 1
)'1, )'2, )'3 > 0
+
$$$$$$$$$$$$$$
(
3
:
$$$$$$$$$$$$$$$
" #
$$$$$$$$$$$$$$
(
3
+
/0
1 #
$$$$$$$$$$$$$$$
!
! " # #
$$$$$$$$$$$$$$$$$$
-
+
+
2
#
$$$$$$$$$$$$$$$$$$
$ -
(
$
The optimal strategy is for Station A to implement strategy
2
.
+
with probability 0.6 and strategy
3 with probability 0.4. Using the absolute value of the dual prices, we see that it is optimal for
Station B to implement strategy 1 with probability 0.2 and strategy 3 with probability 0.8. The
expected value of the game is 6.4. This is an expected increase of 6400 viewers for Station A.
5 13
,
5
14.
The row minimums are 15, 10, 25, and 10. The Republican candidate prefers the maximin
strategy 4 to obtain a gain of at least 10. The column maximums are 30, 20, 10, and 20. Station B
prefers the minimax strategy 3 to limit its maximum loss to no more than 10. The maximum of the
row minimums is equal to the minimum of the row maximums. The game has a pure strategy. The
Republican candidate goes to South Bend and the Democratic candidate goes to Fort Wayne. The
value of the game shows a 10,000 voter increase for the Republican candidate.
15. a.
The row minimums are 1, 3, and 4. Player A prefers the maximin strategy Red to obtain a payoff
of at least 1. The column maximums are 5, 4 and 2. Player B prefers the minimax strategy Blue to
limit its maximum loss to no more than 2. However, because the maximum of the row minimums is
not equal to the minimum of the column maximums, the game does not have a pure strategy. A
mixed strategy solution with a value of the game between 1 and 2 exists.
The linear programming formulation and solution for Player A is as follows.
Max
' ('
s.t.
)'1
2)'1
)'1
+
+
5)'2
4)'2
3)'2
)'2
+
+
+
2)'3
3)'3
4)'3
)'3
' ('
' ('
' ('
>
>
>
=
0
0
0
1
# $
#
(Red Chip)
(White Chip)
(Blue Chip)
)'1, )'2, )'3 > 0
*
$$$$$$$$$$$$$$
(
3
:
$$$$$$$$$$$$$$$
)
3
" #
$$$$$$$$$$$$$$
(
3
+
/0
1 #
$$$$$$$$$$$$$$$
(
!
! " # #
$$$$$$$$$$$$$$$$$$
(
*
2
#
$$$$$$$$$$$$$$$$$$
$
$
*
*
*
The optimal mixed strategy for Player A is to select a red chip with a 0.7 probability and a white chip
with a 0.3 probability. Using the absolute value of the dual prices, the optimal mixed strategy for
Player B is to select a white chip with a 0.5 probability and a blue chip with a 0.5 probability.
b.
The value of the game is 0.5. This is an expected gain of 50 cents for Player A.
c.
Player A is the preferred player. Over the long run, Player A average winning 50 cents per game. To
make the value of the game 0 and thus, fair for both players, Player A should pay Player B to play the
game. The game would be considered fair if Player A pays Player B 50 cents per game.
5 14
=
16.
>
=
The row minimums are 0, 2, 2 and 2. Company A prefers the maximin strategy 3 to obtain a
payoff of at least 2. The column maximums are 4, 6, 5 and 6. Player B prefers the minimax strategy
1 to limit its maximum loss to no more than 4. However, because the maximum of the row
minimums is not equal to the minimum of the row maximums, the game does not have a pure
strategy. A mixed strategy solution with a value of the game between 2 and 4 exists.
The linear programming formulation and solution for Player A is as follows.
Max
*
s.t.
3) 1
2) 1
+
2) 3
2) 2
+
1) 3
+
5) 3
4) 1
+
2) 2
2) 1
+
6) 2
) 1
+
) 2
+
+
4) 4
6) 4
1) 3
+
) 3
+
*
≤ 0
#
(Strategy
1)
*
≤ 0
(Strategy
2
)
*
≤ 0
(Strategy
3
)
*
≤ 0
(Strategy
4
)
) 4
#
= 1
) 1, ) 2, ) 3, ) 4 > 0
- .
$$$$$$$$$$$$$$
'(
''3
'+
'
$$$$$$$$$$$$$$$
+
" #
$$$$$$$$$$$$$$
(
3
+
*
/0
1 #
$$$$$$$$$$$$$$$
(
3 -
!
! " # #
$$$$$$$$$$$$$$$$$$
(
-
- .
2
#
$$$$$$$$$$$$$$$$$$
.
$- .
The optimal mixed strategy solution for Company B is to select strategy
strategy
2
with probability 0.4 and
with probability 0.6. Using the dual prices, the optimal mixed strategy for Company A is
to select strategy 3 with a probability 0.8 and a strategy
gain for Company A is 2.8%.
17.
1
4
with a probability 0.2. The expected
The payoff table is as follows:
%$
$
&
Run Defense
Pass Defense
Minimum
Run
2
6
2
Pass
11
1
1
Maximum
11
6
5 15
,
5
The Bears prefer the maximin strategy run to obtain a payoff of at least 2 yards. The Packers prefer
the minimax strategy pass defense to limit its maximum loss to no more than 6 yards. However,
because the maximum of the row minimums is not equal to the minimum of the row maximums, the
game does not have a pure strategy. A mixed strategy solution with a value of the game between 2
and 6 exists.
The linear programming formulation and solution for the Bears is as follows.
Max
'(
'$
s.t.
%
2)'1
6)'1
)'1
+ 11)'2
1)'2
+
)'2
' ('
' ('
≥ 0
≥ 0
= 1
$ #
#
(Run Defense)
(Pass Defense)
+ -*
$$$$$$$$$$$$$$
(
:
'%
$$$$$$$$$$$$$$$
)*
-*
+ -*
" #
$$$$$$$$$$$$$$
(
3
/0
1 #
$$$$$$$$$$$$$$$
!
! " # #
$$$$$$$$$$$$$$$$$$
2
#
$$$$$$$$$$$$$$$$$$
$ +3)
$ * 3
+ -*
The optimal mixed strategy is for the Bears to run with a 0.75 probability and pass with a 0.25
probability. Using the absolute value of the dual prices, the optimal mixed strategy for the Packers is
to use a run defense with a 0.437 probability and a pass defense with a 0.563 probability. The
expected value of the game shows that with the mixed strategy solution, the Bears average 4.25 yards
per play.
5 16
1.
Be able to identify the special features of the transportation problem.
2.
Become familiar with the types of problems that can be solved by applying a transportation model.
3.
Be able to develop network and linear programming models of the transportation problem.
4.
Know how to handle the cases of (1) unequal supply and demand, (2) unacceptable routes, and (3)
maximization objective for a transportation problem.
5.
Be able to identify the special features of the assignment problem.
6.
Become familiar with the types of problems that can be solved by applying an assignment model.
7.
Be able to develop network and linear programming models of the assignment problem.
8.
Be familiar with the special features of the transshipment problem.
9.
Become familiar with the types of problems that can be solved by applying a transshipment model.
10
Be able to develop network and linear programming models of the transshipment problem.
11.
Know the basic characteristics of the shortest route problem.
12.
Be able to develop a linear programming model and solve the shortest route problem.
13.
Know the basic characteristics of the maximal flow problem.
14.
Be able to develop a linear programming model and solve the maximal flow problem.
15.
Know how to structure and solve a production and inventory problem as a transshipment problem.
16.
Understand the following terms:
network flow problem
transportation problem
origin
destination
capacitated transportation problem
assignment problem
transshipment problem
capacitated transshipment problem
shortest route
maximal flow
source node
sink node
arc flow capacities
6 1
Chapter 6
1.
The network model is shown.
Atlanta
1400
Dallas
3200
Columbus
2000
Boston
1400
2
6
Phila.
5000
6
2
1
2
New
Orleans
3000
5
7
2.
a.
Let
11
12
:
:
•
•
•
Min 14 11
s.t.
11
Amount shipped from Jefferson City to Des Moines
Amount shipped from Jefferson City to Kansas City
+
9 12
+
7 13
+
12
+
13
+
+
11
8 21
+
21
21
12
13
10 22
+
+
22
+
23
+
22
+
23
11, 12, 13, 21, 22, 23, ≥ 0
b.
Optimal Solution:
Amount
5
15
10
20
Jefferson City Des Moines
Jefferson City Kansas City
Jefferson City St. Louis
Omaha Des Moines
Total
6 2
Cost
70
135
70
160
435
5 23
≤
≤
=
=
=
30
20
25
15
10
Distribution and Network Models
3.
a.
Hamilton
1
400
Butler
2
200
Clermont
3
300
10
Southern
1
500
20
15
12
400
b.
Let
Min
s.t.
15
Northwest
2
18
= amount shipped from supply node to demand node .
10 11
+
20 12
+
15 13
11
+
12
+
13
+
+
11
12
12 21
21
21
+ 15 22
+
22
+
22
13
+ 18 23
+
23
+
23
≤
≤
=
=
=
500
400
400
200
300
≥ 0 for all ,
c.
Optimal Solution
Amount
200
300
200
200
Southern Hamilton
Southern Clermont
Northwest Hamilton
Northwest Butler
Total Cost
d.
Cost
$ 2000
4500
2400
3000
$11,900
To answer this question the simplest approach is to increase the Butler County demand to 300 and to
increase the supply by 100 at both Southern Gas and Northwest Gas. The new optimal solution is:
Amount
300
300
100
300
Southern Hamilton
Southern Clermont
Northwest Hamilton
Northwest Butler
Total Cost
Cost
$ 3000
4500
1200
4500
$13,200
From the new solution we see that Tri County should contract with Southern Gas for the additional
100 units.
6 3
Chapter 6
4.
a.
1
Pittsburg
3000
10
20
9000
2
1
5
Seattle
Mobile
5000
9
10
2
10
2
3
8
4000
Columbus
Denver
4000
30
6
1
3
8000
New York
20
7
4
10
Los Angeles
6000
4
5
Washington
b.
3000
The linear programming formulation and optimal solution as printed by The Management Scientist
are shown. The first two letters of the variable name identify the “from” node and the second two
letters identify the “to” node. Also, The Management Scientist prints “<” for “≤.”
6 4
Distribution and Network Models
!
!
!
!
!
"
#$%&'()% *+,&'(-, ./0+% !
./1(/#0%
44444444444444
c.
./0+%
444444444444444
%2+&%2 -3'3
444444444444444444
The new optimal solution actually shows a decrease of $9000 in shipping cost. It is summarized.
Optimal Solution
Units
Seattle Denver
Seattle Los Angeles
Columbus Mobile
New York Pittsburgh
New York Los Angeles
New York Washington
4000
5000
5000
4000
1000
3000
6 5
Cost
$ 20,000
45,000
50,000
4,000
10,000
12,000
Total: $141,000
Chapter 6
5.
a.
100
180
B
2
75
C
3
100
D
4
85
125
Avery
1
160
A
1
115
100
120
135
Baker
2
160
115
120
155
150
140
Campbell
3
140
130
b.
Let
= number of hours from consultant assigned to client .
Max 100 11 + 125 12 + 115 13 + 100
+ 120 24 + 155 31 + 150
s.t.
11 +
13 +
12 +
21 +
+
11
21
+
12
13
14 + 120 21 + 135
32 + 140 33 + 130
14
22 +
23 +
31 +
+
31
+
22
+
23
+
14
≥ 0 for all ,
6 6
22 + 115 23
34
24
32 +
33 +
32
+
24
33
+
≤
≤
34 ≤
=
=
=
34 =
160
160
140
180
75
100
85
Distribution and Network Models
Optimal Solution
Hours
Assigned
40
100
40
35
85
140
Avery Client B
Avery Client C
Baker Client A
Baker Client B
Baker Client D
Campbell Client A
Total Billing
c.
New Optimal Solution
Hours
Assigned
40
100
75
85
140
Avery Client A
Avery Client C
Baker Client B
Baker Client D
Campbell Client A
Total Billing
6.
Billing
$ 5,000
11,500
4,800
4,725
10,200
21,700
$57,925
Billing
$ 4,000
11,500
10,125
10,200
21,700
$57,525
The network model, the linear programming formulation, and the optimal solution are shown. Note
that the third constraint corresponds to the dummy origin. The variables 31, 32, 33, and 34 are
the amounts shipped out of the dummy origin; they do not appear in the objective function since they
are given a coefficient of zero.
6 7
Chapter 6
Demand
Supply
D1
2000
D2
5000
D3
3000
D4
2000
32
5000
34
C.S.
32
40
34
30
3000
D.
28
38
0
0
4000
0
Dum
0
Note: Dummy origin has supply of 4000.
Max
32 11 + 34 12 + 32 13 + 40 14 + 34 21 + 30 22 + 28 23 + 38 24
s.t.
11 +
12 +
13 +
31 +
+
11
14
32 +
+
21
+
12
21 +
22 +
33 +
34
+
+
24
31
22
13
23 +
32
23
+
+
14
≥ 0 for all
6 8
24
33
+
34
≤
5000
≤
3000
≤
4000
=
2000
=
5000
=
3000
= 2000
Dummy
Distribution and Network Models
Optimal Solution
Units
Cost
Clifton Springs D2
Clifton Springs D4
Danville D1
Danville D4
4000
$136,000
1000
40,000
2000
68,000
1000
38,000
Total Cost: $282,000
Customer 2 demand has a shortfall of 1000
Customer 3 demand of 3000 is not satisfied.
7.
a.
1
Boston
50
7
1
100
11
Denver
8
13
2
20
Dallas
70
3
Los
Angeles
60
17
100
2
Atlanta
12
10
8
18
3
150
13
Chicago
16
4
St. Paul
b.
There are alternative optimal solutions.
Solution #1
Solution # 2
Denver to St. Paul: 10
Atlanta to Boston: 50
Atlanta to Dallas: 50
Chicago to Dallas: 20
Chicago to Los Angeles: 60
Chicago to St. Paul: 70
Denver to St. Paul: 10
Atlanta to Boston: 50
Atlanta to Los Angeles: 50
Chicago to Dallas: 70
Chicago to Los Angeles: 10
Chicago to St. Paul: 70
6 9
80
Chapter 6
Total Profit: $4240
If solution #1 is used, Forbelt should produce 10 motors at Denver, 100 motors at Atlanta, and 150
motors at Chicago. There will be idle capacity for 90 motors at Denver.
If solution #2 is used, Forbelt should adopt the same production schedule but a modified shipping
schedule.
8.
The linear programming formulation and optimal solution are shown.
Let
Min
1A
1B
•
•
•
=
=
Units of product A on machine 1
Units of product B on machine 1
3C
=
Units of product C on machine 3
1A + 1.2 1B + 0.9 1C + 1.3 2A + 1.4 2B + 1.2 2C + 1.1 3A +
3B + 1.2 3C
s.t.
1A +
1B +
1C
2A +
2B +
2C
3A +
+
1A
+
2A
+
1B
+
1C
3A
+
2B
1
1
2
3
3
A
C
A
A
B
Units
Cost
300
1200
1200
500
500
$ 300
1080
1560
550
500
Total: $3990
Note: There is an unused capacity of 300 units on machine 2.
6 10
3B
+
2C
≥ 0 for all
Optimal Solution
3B +
≤
1500
≤
1500
3C ≤
1000
=
2000
=
500
3C =
1200
Distribution and Network Models
9.
a.
10
1
1
1
16
Jackson
Client 1
1
32
14
2
1
2
22
Client 2
Ellis
1
40
22
24
3
1
Smith
3
34
Client 3
1
b.
Min 10 11 + 16 12 + 32 13 + 14 21 + 22 22 + 40 23 + 22 31 + 24 32 + 34 33
s.t.
≤ 1
11 + 12 + 13
≤ 1
21 + 22 + 23
31 + 32 + 33 ≤ 1
+
+
= 1
11
21
31
+ 22
+ 32
= 1
12
+ 23
+ 33 = 1
13
ij ≥ 0 for all ,
Solution 12 = 1, 21 = 1, 33 = 1
Total completion time = 64
6 11
Chapter 6
10. a.
30
44
1
Red
1
31
25
38
47
1
1
1
2
W hite
2
1
1
3
Blue
3
1
1
4
Green
4
1
5
1
34
44
26
43
5
Brown
1
28
b.
Min
s.t.
30
11
+ 44
12
+ 38
13
+ 47
14
+ 31
15
11
+
12
+
13
+
14
+
15
21
+
22
+
23
+
24
+
25
31
+
32
+
33
+
34
+
35
41
+
42
+
43
+
44
+
45
53
+
54
+
21
+
31
+
+
+
22
+
32
13
+
23
+
14
+
24
15
+
51
11
+
12
≥ 0,
+ 25
21
+ ⋯
41
+
+
+
42
+
52
33
+
43
+
53
+
34
+
44
+
54
25
+
35
+
45
+
52
= 1, 2,.., 5;
55
51
55
+ 28
55
≤
≤
≤
≤
≤
=
=
=
=
=
1
1
1
1
1
1
1
1
1
1
= 1, 2,.., 5
Optimal Solution:
Green to Job 1
Brown to Job 2
Red to Job 3
Blue to Job 4
White to Job 5
$26
34
38
39
25
$162
Since the data is in hundreds of dollars, the total installation cost for the 5 contracts is $16,200.
6 12
Distribution and Network Models
11.
This can be formulated as a linear program with a maximization objective function. There are 24
variables, one for each program/time slot combination. There are 10 constraints, 6 for the potential
programs and 4 for the time slots.
Optimal Solution:
NASCAR Live
Hollywood Briefings
World News
Ramundo & Son
5:00 – 5:30 p.m.
5:30 – 6:00 p.m.
7:00 – 7:30 p.m.
8:00 – 8:30 p.m.
Total expected advertising revenue = $30,500
12. a.
This is the variation of the assignment problem in which multiple assignments are possible. Each
distribution center may be assigned up to 3 customer zones.
The linear programming model of this problem has 40 variables (one for each combination of
distribution center and customer zone). It has 13 constraints. There are 5 supply (≤ 3) constraints
and 8 demand (= 1) constraints.
The problem can also be solved using the Transportation module of
optimal solution is given below.
Plano:
Flagstaff:
Springfield:
Boulder:
13.
Assignments
Kansas City, Dallas
Los Angeles
Chicago, Columbus, Atlanta
Newark, Denver
Total Cost
. The
Cost ($1000s)
34
15
70
97
$216
b.
The Nashville distribution center is not used.
c.
All the distribution centers are used. Columbus is switched from Springfield to Nashville. Total cost
increases by $11,000 to $227,000.
A linear programming formulation and the optimal solution are given. For the decision variables, ,
we let the first subscript correspond to the supplier and the second subscript correspond to the
distribution hub. Thus, = 1 if supplier is awarded the bid to supply hub and = 0 if supplier is
not awarded the bid to supply hub .
6 13
Chapter 6
Min 190
11
+175 12 + 125 13 + 230 14 + 150 21 + 235 22 + 155
+ 135 33 +260 34 + 170 41 + 185 42 + 190 43 + 280
+ 240 54 + 270 61 + 200 62 + 130 63 + 260 64
23
44
+ 220
+ 220
24
51
+ 210
+ 190
31
52
+ 225
+ 140
32
53
s.t.
+
11
+
+
12
+
21
+
31
+
13
22
+
23
+
42
+
32
41
+
51
14
+
21
+
31
12
+
+
+
+
41
22
51
32
13
+
+
+
24
+
43
33
52
+
61
11
+
+
34
+
53
62
+
44
+
63
+
54
+
64
+
46
61
42
23
+
+
52
33
14
+
+
+
62
43
24
+
+
53
34
+
+
63
44
+
45
≤
≤
≤
≤
≤
≤
=
=
=
=
1
1
1
1
1
1
1
1
1
1
≥ 0 for all ,
14.
Optimal Solution
Bid
Martin – Hub 2
Schmidt Materials – Hub 4
D&J Burns – Hub 1
Lawler Depot – Hub 3
175
220
170
130
695
A linear programming formulation of this problem can be developed as follows. Let the first letter of
each variable name represent the professor and the second two the course. Note that a DPH variable
is not created because the assignment is unacceptable.
Max 2.8AUG
+ 2.2AMB
+ 3.3AMS
+ 3.0APH
+
+
+
AMS
BMB
CUG
+
+
+
CUG
BMB
AMS
+
+
+
+ 3.2BUG
+
OOO
+ 2.5DMS
s.t.
AUG
AUG
+
AMB
BUG
BUG
AMB
+
+
APH
BMS
CMB
DUG
DUG
CMB
BMS
APH
+
+
+
BPH
CMS
DMB
+
+
CPH
DMS
+
+
+
DMB
CMS
BPH
+
+
DMS
CPH
All Variables ≥ 0
Optimal Solution:
A to MS course
B to Ph.D. course
C to MBA course
D to Undergraduate course
Max Total Rating
6 14
Rating
3.3
3.6
3.2
3.2
13.3
≤
≤
≤
≤
=
=
=
=
1
1
1
1
1
1
1
1
Distribution and Network Models
15. a.
Min 150 11 +
+
210 12
170 21
+
+
+
270 13
230 22
180 31
+
+
+
220 23
230 32 +
160 41 +
225 33
240 42
+ 230 43
s.t.
11 +
12
21
+
+
+ 21
11
13
22
31
+
+
+ 31
+ 22
12
13
+ 32
+ 23
23
32 +
41 +
+ 41
+ 42
+ 33
33
42
+
+ 43
43
≤
≤
≤
≤
=
=
=
1
1
1
1
1
1
1
≥ for all ,
Optimal Solution: 12 = 1, 23 = 1, 41 = 1
Total hours required: 590
Note: statistician 3 is not assigned.
b.
The solution will not change, but the total hours required will increase by 5. This is the extra time
required for statistician 4 to complete the job for client A.
c.
The solution will not change, but the total time required will decrease by 20 hours.
d.
The solution will not change; statistician 3 will not be assigned. Note that this occurs because
increasing the time for statistician 3 makes statistician 3 an even less attractive candidate for
assignment.
16. a.
The total cost is the sum of the purchase cost and the transportation cost. We show the calculation
for Division 1 Supplier 1 and present the result for the other Division Supplier combinations.
Division 1 Supplier 1
Purchase cost (40,000 x $12.60)
Transportation Cost (40,000 x $2.75)
Total Cost:
6 15
$504,000
110,000
$614,000
Chapter 6
Cost Matrix ($1,000 )
Supplier
1
2
3
4
5
6
1
614
660
534
680
590
630
2
603
639
702
693
693
630
3
865
830
775
850
900
930
4
532
553
511
581
595
553
5
720
648
684
693
657
747
Division
b.
Optimal Solution:
Supplier 1
Supplier 2
Supplier 3
Supplier 5
Supplier 6
17. a.
Division 2
Division 5
Division 3
Division 1
Division 4
$ 603
648
775
590
553
Total $3,169
Network Model
Demand
Supply
450
1
P1
600
4
4
W1
8
300
8
C3
300
9
C4
400
4
8
5
3
5
W2
5
380
7
C2
6
6
3
P3
300
4
7
2
P2
6
C1
7
7
6
6 16
Distribution and Network Models
b. & c.
The linear programming formulation and solution as printed by The Management Scientist is shown.
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
4 5
4 5
4 5
4 5
4 5
4 5
!
!
!
!
!
!
"
#$%&'()% *+,&'(-, ./0+% !
./1(/#0%
44444444444444
5
5
5
5
5
5
5
5
5
5
5
5
5
5
./0+%
444444444444444
There is an excess capacity of 130 units at plant 3.
6 17
%2+&%2 -3'3
444444444444444444
5
5
Chapter 6
18. a.
Three arcs must be added to the network model in problem 23a. The new network is shown.
Demand
Supply
1
P1
450
300
7
C2
300
8
C3
300
9
C4
400
4
6
7
4
4
W1
8
4
8
2
P2
600
6
C1
2
5
2
3
6
5
W2
5
7
6
3
P3
380
7
7
b.&c.
The linear programming formulation and optimal solution as printed by The management Scientist
follow:
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
!
!
!
5
!
6 18
5
5
5
5
5
4 5
4 5
4 5
4 5
4 5
4 5
4 5
4 5
!
!
Distribution and Network Models
"
#$%&'()% *+,&'(-, ./0+% !
./1(/#0%
44444444444444
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
./0+%
444444444444444
%2+&%2 -3'3
444444444444444444
The value of the solution here is $630 less than the value of the solution for problem 23. The new
shipping route from plant 3 to customer 4 has helped ( 39 = 380). There is now excess capacity of
130 units at plant 1.
19. a.
300
1
Augusta
7
100
150
6
NewYork
100
7
Philadelphia
150
8
3
Albany
5
5
2
Tupper Lake
5
Boston
7
5
3
4
4
Portsmouth
6
10
6 19
Chapter 6
b.
Min
s.t.
7 13 + 5 14 + 3 23 + 4 24 + 8 35 + 5 36 + 7 37 + 5 45 + 6 46 + 10 47
13
+
14
23
23
13
+
14
24
+
35
+
36
+
37
+
+
24
35
+
36
37
≥ 0 for all and
c.
Optimal Solution:
Variable
13
14
23
24
35
36
37
45
46
47
Value
50
250
100
0
0
0
150
150
100
0
Objective Function: 4300
6 20
45
45
+
46
+
46
+
47
+
47
≤
≤
=
=
=
=
=
300
100
0
0
150
100
150
Distribution and Network Models
20.
3
1
Muncie
34
4
Louisville
34
4
8
Concord
3
9
Chatham
3
32
3
8
57
35
5
Cincinnati
28
9
5
7
Greenwood
44
2
Brazil
3
Xenia
2
8
6
6
6
Macon
24
3
A linear programming model is
Min
8 14 + 6 15 + 3 24 + 8 25 + 9 34 + 3 35 + 44 46 + 34 47 + 34 48 + 32 49 + 57 56 + 35 57 + 28 58 + 24 59
s.t.
14 +
≤ 3
15
24 +
≤ 6
25
34 +
24
14
15
+
34
25
≤ 5
35
46 +
47 +
48 +
35
46
56 +
+
56
57 +
49
58 +
59 = 0
= 2
= 4
57
+
48
6 21
+
+
47
≥ 0 for all
= 0
49
= 3
58
+
59 = 3
Chapter 6
Optimal Solution
Units Shipped
Muncie to Cincinnati
1
Cincinnati to Concord
3
Brazil to Louisville
6
Louisville to Macon
2
Louisville to Greenwood
4
Xenia to Cincinnati
5
Cincinnati to Chatham
3
Cost
6
84
18
88
136
15
72
419
Two rail cars must be held at Muncie until a buyer is found.
21.
+3
2
10
12
6
+5
11
1
3
11
5
4
8
9
12
4
+2
The positive numbers by nodes indicate the amount of supply at that node. The negative numbers by
nodes indicate the amount of demand at the node.
22. a.
Min 20 12 + 25 15 + 30 25 + 45 27 + 20 31
+ 30 42 + 25 53 + 15 54 + 28 56
s.t.
31
12
31 +
= 8
15
25 +
27
36
53
54
53
+
54 +
56
15
12
27
67
≥ 0 for all ,
6 22
= 5
42
= 3
+
74
= 3
42
= 2
25
36
74
+ 35 36
+ 12 67 + 27 74
+
56
67
= 5
= 6
Distribution and Network Models
b.
12
15
25
27
31
36
42
=
=
=
=
=
=
=
0
0
8
0
8
0
3
=
=
=
=
=
=
53
54
56
67
74
56
5
0
5
0
6
5
Total cost of redistributing cars = $917
23.
Origin – Node 1
Transshipment Nodes 2 to 5
Destination – Node 7
The linear program will have 14 variables for the arcs and 7 constraints for the nodes.
1 if the arc from node to node is on the shortest route

=
0 otherwise

Let
Min 7
+6
12
+9
57
+2
+ 18
14
65 + 3
67
13
+3
23
+5
25
+3
32
+4
35
+3
46
+5
52
+4
53
+2
56
s.t.
Flow Out
Node 1
12
Node 2
23
Node 3
32
+
+
+
Node 4
46
Node 5
52 +
Node 6
65
+
13
+
Flow In
=1
14
−
25
12
−
35
53 +
56 +
57
13
−
−
23
−
−
52
=0
53
=0
=0
25 −
46
+
Node 7
−
32
14
−
67
−
57
−
+
35 −
65
=0
56
=0
67
=1
> 0 for all and
Optimal Solution:
12
=1,
25
=1,
56
= 1 , and
67
=1
Shortest Route 1 2 5 6 7
Length = 17
24.
The linear program has 13 variables for the arcs and 6 constraints for the nodes. Use same six
constraints for the Gorman shortest route problem as shown in the text. The objective function
changes to travel time as follows.
6 23
Chapter 6
Min 40
+ 36
12
+8
45
+6
13
+8
23
+ 11
54
Optimal Solution:
12
+6
46
=1,
32
+ 23
24
+ 12
24
+ 12
46
=1
42
+ 25
26
+ 15
35
+ 15
53
56
= 1 , and
Shortest Route 1 2 4 6
Total Time = 63 minutes
25. a.
Origin – Node 1
Transshipment Nodes 2 to 5
Destination – Node 6
The linear program will have 13 variables for the arcs and 6 constraints for the nodes.
1 if the arc from node to node is on the shortest route

=
0 otherwise

Let
Min 35
+16
12
+ 30
46
13
+15
+ 12
53
+ 18
23
+12
54
+30
24
+ 39
26
+ 12
32
+ 15
35
+ 18
42
+ 12
45
56
s.t.
Flow Out
+
Node 1
12
Node 2
23
Node 3
32
Node 4
Node 5
=1
13
+
+
42
+
Flow In
+
53 +
24
+
−
26
−
35
45
+
54
−
−
46
+
56
+
Node 6
12
13
−
−
−
−
35
32
23
−
42
=0
53
=0
54
=0
45
=0
24
26
−
+
46
+
56
=1
> 0 for all and
b.
Optimal Solution:
12
=1,
24
= 1 , and
46
=1
Shortest Route 1 2 4 6
Total time = 69 minutes
c.
Allowing 8 minutes to get to node 1 and 69 minutes to go from node 1 to node 6, we expect to make
the delivery in 77 minutes. With a 20% safety margin, we can guarantee a delivery in 1.2(77) = 92
minutes. It is 1:00 p.m. now. Guarantee delivery by 2:32 p.m.
6 24
Distribution and Network Models
26.
Origin – Node 1
Transshipment Nodes 2 to 5 and node 7
Destination – Node 6
The linear program will have 18 variables for the arcs and 7 constraints for the nodes.
Let
1 if the arc from node to node is on the shortest route

=
0 otherwise

Min
35
+9
12
+ 30
13
+ 20
43
+ 15
47
+ 12
14
52
+8
+ 10
23
53
+ 12
+5
25
+8
+ 20
56
32
+9
57
34
+ 15
+ 10
74
35
+ 20
+ 20
75
+5
36
76
s.t.
Flow Out
Node 1
12
+
Node 3
+
32 +
Node 4
43 +
Node 5
52
Node 2
13
23
+
+
Flow In
=1
14
25
34 +
35 +
36
47
53
+
56
+
57
74 +
75 +
12
−
14 −
−
+
−
Node 6
Node 7
−
−
76
−
−
13
−
−
23
32
34 −
−
25
+
47 −
36
35
56
−
+
52
=0
43 −
=0
53
74
=0
75
=0
76
=1
57
=0
> 0 for all and
Optimal Solution:
14
=1,
47
= 1 , and
76
=1
Shortest Route 1 4 7 6
Total Distance = 40 miles
27.
Origin – Node 1
Transshipment Nodes 2 to 9
Destination – Node 10 (Identified by the subscript 0)
The linear program will have 29 variables for the arcs and 10 constraints for the nodes.
Let
1 if the arc from node to node is on the shortest route

=
0 otherwise

6 25
Chapter 6
Min
8 12 + 13 13 + 15 14 + 10 15 + 5 23 + 15 27 + 5 32 + 5 36 + 2 43 + 4
+3 46 + 4 54 + 12 59 + 5 63 + 3 64 + 4 67 + 2 68 + 5 69 + 15 72 + 4
+2 78 + 4 70 + 2 86 + 5 89 + 7 80 + 12 95 + 5 96 + 5 98 + 5 90
45
76
s.t.
Flow Out
+
Node 1
12
Node 2
23 +
Node 3
32
Node 4
+
43 +
Node 5
54
+
Node 7
63 +
72 +
Node 8
86
Node 9
95
Node 6
13
+
+
+
Flow In
14
+
=1
15
27
36
45
+
46
64 +
76 +
96
12 −
−
−
13
−
59
89
−
+
+
67 +
78 +
68 +
−
−
69
70
−
80
98
+
−
90
+
Node 10
−
14 −
15
−
36 −
27 −
68
59
−
−
70 +
32 −
−
54 −
23
45
46
−
−
=0
72
43
−
=0
63
=0
64
=0
95
76 −
86
−
96
=0
67
78
69
=0
−
−
80 +
=0
98
89
=0
90
=1
> 0 for all and
Optimal Solution:
15
=1,
54
= 1,
46
= 1,
67
= 1 , and
70
=1
Shortest Route 1 5 4 6 7 10
Total Time = 25 minutes
28.
Origin – Node 0
Transshipment Nodes 1 to 3
Destination – Node 4
The linear program will have 10 variables for the arcs and 5 constraints for the nodes.
Let
1 if the arc from node to node is on the minimum cost route

=
0 otherwise

6 26
Distribution and Network Models
Min
600
+ 1000
01
+800
02
+ 2000
+ 1600
23
24
03
+ 2800
+ 700
04
+ 500
12
+ 1400
34
s.t.
Flow Out
Node 0
01
+
+
Node 1
12
Node 2
23 +
Node 3
34
+
02
13
+
Flow In
03
+
=1
04
−
14
24
Node 4
=0
01
−
02 −
−
−
03
−
04 −
=0
12
−
14 −
13
=0
23
24
−
34
=1
> 0 for all and
Optimal Solution:
02
=1,
23
= 1 , and
34
=1
Shortest Route 0 2 3 4
Total Cost = $2500
29.
The capacitated transshipment problem to solve is given:
Max
s.t.
61
+
+
34 +
42 +
54 +
12
13
24
25
61
≤2
≤1
34 ≤ 3
42 ≤ 1
54 ≤ 1
+
36
43
61
12
42
13
+
56
36
=0
=0
43 = 0
14
+
45
+
46
25
45
46
56
≤6
≤4
36 ≤ 2
43 ≤ 3
56 ≤ 6
12
13
24
25
14
24
14
≤3
45
≤1
34
46
54
≤3
≥ 0 for all ,
6 27
=0
=0
=0
13
+ 2100
14
Chapter 6
3
2
2
5
1
3
3
1
4
1
4
6
2
4
Maximum Flow
9,000 Vehicles
Per Hour
2
3
The system cannot accommodate a flow of 10,000 vehicles per hour.
30.
4
2
3
5
1
3
3
1
4
6
3
5
6
2
11,000
2
3
31.
The maximum number of messages that may be sent is 10,000.
32. a.
10,000 gallons per hour or 10 hours
b.
33.
Flow reduced to 9,000 gallons per hour; 11.1 hours.
Current Max Flow = 6,000 vehicles/hour.
With arc 3 4 at a 3,000 unit/hour flow capacity, total system flow is increased to 8,000 vehicles/hour.
Increasing arc 3 4 to 2,000 units/hour will also increase system to 8,000 vehicles/hour. Thus a 2,000
unit/hour capacity is recommended for this arc.
34.
Maximal Flow = 23 gallons / minute. Five gallons will flow from node 3 to node 5.
35. a.
Modify the problem by adding two nodes and two arcs. Let node 0 be a beginning inventory node
with a supply of 50 and an arc connecting it to node 5 (period 1 demand). Let node 9 be an ending
inventory node with a demand of 100 and an arc connecting node 8 (period 4 demand to it).
6 28
Distribution and Network Models
b.
+ 2 15
Min
s.t.
+ 5 26
+ 3 37
+ 3 48
+ 0.25 56
+ 0.25 67
+ 0.25 78
+ 0.25 89
=
05
15
26
37
48
05
+
15
+
26
56
56
+
37
+
48
≥ 0 for all and
Optimal Solution:
05 = 50
15 = 600
26 = 250
37 = 500
48 = 400
67
67
56 = 250
67 = 0
78 = 100
89 = 100
Total Cost = $5262.50
6 29
78
78
89
89
≤
≤
≤
≤
=
=
=
=
=
50
600
300
500
400
400
500
400
400
100
Chapter 6
36. a.
Let R1, R2, R3
O1, O2, O3
D1, D2, D3
represent regular time production in months 1, 2, 3
represent overtime production in months 1, 2, 3
represent demand in months 1, 2, 3
Using these 9 nodes, a network model is shown.
275
100
O1
200
R2
50
O2
100
R3
50
b.
R1
D1
150
D2
250
D3
300
O3
Use the following notation to define the variables: first two letters designates the "from node" and the
second two letters designates the "to node" of the arc. For instance, R1D1 is amount of regular time
production available to satisfy demand in month 1, O1D1 is amount of overtime production in month
1 available to satisfy demand in month 1, D1D2 is the amount of inventory carried over from month
1 to month 2, and so on.
6 30
Distribution and Network Models
≤
≤
≤
≤
≤
≤
4
!
4
!
!
c.
Optimal Solution:
./1(/#0%
44444444444444
./0+%
444444444444444
Value = $46,750
Note: Slack variable for constraint 2 = 75.
d.
The values of the slack variables for constraints 1 through 6 represent unused capacity. The only
nonzero slack variable is for constraint 2; its value is 75. Thus, there are 75 units of unused overtime
capacity in month 1.
6 31
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