20
Network Models
20.1. INTRODUCTIOON
There is a multitude of operation research sibvations that can be conveniently modeled and solved by
network models. In addition to the project planning, the product distribution problem, the transhipment
model, the
shortest route
models, equipment replacement models and scheduling models
can
also be
represented and solved by network models.
20.2. NETWORK DEFINITIONS
A network consists of a set of nodes linked by arcs (or
branches). The notation for describing a network is (N, A),
As an
where N is the set of nodes, and A is the set of ares.
illustration, the network in Fig. 20.l is described as
N={1,2, 3,4, 5}
Fig. 20.1.
(2,5), (3,4), (3,5), (4,5)}
(2,4),
(2,3),
(1,2),
{(1,3),
A
automobile
of flow (e.g., oil products flow in a pipeline and
Associated with each network is some type
which
its
flow in a network is limited by the capacity of arcs,
traffic flows in a road network). In general, the
may be finite
or
infinite.
direction and zero flow in the
allows positive flow in one
it
if
oriented
or
directed
An arc is said to be
has all directed branches.
opposite direction. A directed network
direction of flow in each
two nodes regardless of the
that
branches
join
distinct
A path is a sequence of
the branches
to itself. For example, in Fig. 20.1,
node
a
connects
ifit
a
are
branch. A path forms a loop or cylce
in which all the branches
a circuit) is a loop
directed
(or
A
loop
a
,3), (3, 4) and (4, 2) form loop.
oriented in the same direction.
least one path. The network
distinct nodes are linked by at
two
that
such
every
subset
A connected network is
that may involve only a
tree is a connected network
A
network with no loops
in Fig. 20.l demonstrates this type ofnetwork.
links all the nodes of the
a spanning tree
whereas
of all the nodes of the network,
tree
of a tree and a spanning
allowed. Fig. 20.2 provides exmaples
Q
O
Spanning Tree
Tree
Fig. 20.2.
995
Operations Research
996
20.3. MINIMAL SPANNING TREE PROBLEMS
Consider
a
notwork
problem. where our primary interest
is to select
a
set of arcs such
aas
path exists from any node to any other node
(i) there is no cycle in the network.
i)
a
Such a collection of nodes is called a spanning tree. For example, for the network shown in Fig. 20.3
the arcs (1, 2). (1, 3), (2, 6), (1, 4), (4, 5) fom a spanning tree.
Spanning tree has the property that the
addition of one more arc to the tree will result in a cycle. The length of the spanning tree is defined as the
sum of distances of all arcs in the tree. Since there is more than one spanning tree for a given network, the
one with the least distance is called minimal spanning tree. Procedural steps for the Minimal Spanning Tree
are given below.
Step 1 Select an arbitrary node initially. Identify a node that is 'closest' to the selected node and
include the arc connecting these two nodes in the spanning tree.
Step 2 Out of the remaining unconnected nodes, determine the one that is closest to a node already
selected in the spanning tree. Include the arc connecting these two nodes in the spanning tree. Whenever
there is a tie for the closest node, it is broken arbitrarily.
Step (2) is repeated until all the nodes in the network have been selected in the spanning t
Example 20.1 Determine the minimal spanning tree for
9
the folloing nerwork.
Solution: Let set G denote the arcs of minimal spanning
tree.
Step 1 Choose node (1) as the initial node. Since (2) is
the closest node, arc (1, 2) becomes part of the spanning tree.
Hence G {(1, 2)}
1
Step 2 Consider the arcs connected to nodes (1) and (2).
Choose the node that is closest to either node (1) or node
(2).
At this stage node (3) will be selected, arc (2, 3) will be added
to the spanning tree and G
{(1, 2), (2, 3)}
=
Step 3 Considering the ares connected to nodes (1), (2)
arc (3, 4) is
closest to node (3). Thus
and (3) we find that
G
(1, 2), 2,3),
6,4
Fig. 20.3.
Step 4 Both arcs (3, 5) and (4, 5) are closest to the arcs in
set G. Choosing arbitrarily arc
(3, 5) we get G {(1,2), (2,3),
(3,4), (3, 5)
=
Step 5 Arc (5, 6) is selected at this step and the minimal
spanning tree is given by G {(1, 2), (2, 3), (3, 4), (3, 5), (5,
=
6)
The
minimal spanning tree
is shown in
Fig.
20.4.
The minimal length
= 3+2+1 +3+3 =12.
Ans.
Fig. 20.4.
,l .\f,,.itl.<
<J<J')
The L)p(im.um route is obt~ incd starting from node 7 nnd tracing backward through the nodes using the
ls' in ft1 m111t1 on. The fo llowm g sequence demon strates the procedu re :
li) _, [1 3, 5] ➔ (5) ➔ (7 , 2] ➔ (2) ➔ [2, 1] ➔ (I)
Tne Jlgorirhum, in fact, provides the shortest distance between node I un d node 7 and dista nce between
n,")tic from its nearest node.
20. 4.
[u1111plt!
. ,, 11 / 11
Fig. 20. 7 compute shortest distance and its designated route to each of the f ollowing nodes.
(l) Node 4
[An·s.
u◄
= 7,
l ➔ 3 ➔ 4.)
(: ) Node 6
[Ans. _u(, = 5, 1 -► 3 ➔ 6.)
i,i Rrcompute the shortest route in Fig. 20.7 when each .of the following changes is effected
1
. I
;1:kpen,h.' nt y.
·
(l) Node 4 is connected to 7 by an arc of length 5.
[Ans. u7 = 12, 1 ➔ 3 ➔ 4 ➔ 7.]
t1) Node 5 is reached from 6 by an arc of length 2.
[Ans. u7 = 13, 1 ➔ 2 ➔ 5 ➔ 7 or 1 ➔ 3 ➔ 6 ➔ 5 _;. 7.]
ycl ic (Dijkstra 's) Algorithm
The acyclic algorithm will not perform
'correct ly if the network happens to include
~y direered loops. To demonstrate this point,
tonsider the network in Fig. 20.8, where a
~ ctrd loop is formed by nodes 2, 3, and 4 .
_Vt~th the rules of the acyclic algorithm, it will
i>e impossible to evaluate any ofthe loop nodes
2. 3 and 4 ·because the algorithm requires
nodes leading into
computing u; for all
node) before u. can be evaluated.
100
20
the
.
J
15
30
10
50
The cyclic alg9rithm differs .from the
Z0 clic algorithm in that it allows_as niany
60
chan(:es as may be needed to reevaluate a node.
Wh:11 it becomes evident that the shortest
Fig. 20-8- ·
has been reached ' the node ·
di5rance to a node
.
.
.
~ exc luded from further consideration: The process ends when the destination or sink node has been evaluated.
The cycJic algorithm (also known as _Dijkstr:a's algorithm) uses two types of labels: temporary and
~~iTn anent. Both labels utilize the same fonnat used with the acychc algorithm : namely, [d, n], yvhere dis
·trie~hortest distance so far available for a current node, and n is the im~ediate predecessor node responsible
fo! realizing -che distanced. The algorithm starts with the source node carrying the permanent label (0, -1.
Xe\t we consider all the nodes that t~n be reached directly from the source node an<l then detem1ine their
?.S~oc iated labels ...The newly created labels are designated as temporary. The next permanet label is selected
~um among all current temporary labels as the one having the smallest din the label [d, 11] (ties are broken
?:~ itrar.ily) . The process ·is npw repeated for the last node that has been designated pelTilanent. In such a
:~s;:, a temp;rary l~bel of a node may be.changed only if then~~ ~abel yield:s a smaller dis_tance d.
Let us apply ttie procedure to the network in Fig. 20.8. A basic assumption of the.algorithm is that ail
L"'e distances in the network are non-negative.
.
1000
Operationv /(eTeorr..}; .
Iteration O : Node I carries the permannet label [O, - ].
Iteration I : Nodes 2 and 3, which can be reached directly from node I (the last permanently lc1b~let1
.
node), now carry the tempora,y labels [O + 100, I] and (0 + 30, I], or [ 100, 1] and [30, I], respective]y.
Among the current temporary labels, node 3 has the smalJest distanced = 30 (= min {J 00, 30} ). 'f11t:s·
·
.
node 3 is pennanently labeled.
Iteration 2 : Nodes 4 and 5 can be reached from the last permanently labeled node (node 3) and their .
temporary labels are [30 + 10, 3] and [30 + 60, 3] (or [40, 3] and [90, 3]), respectively. At this poi.nt, we have
the three temporary labels [100, I], [40, 3], and [90,_3] associated with nodes 2, 4 and 5, respectively.
Temporari!y labeled node 4 has the smallest d = 40 (= min {100, 40, 90}) and hence its label [40, 3] is
.
_ _. .
. . . ·
.·
.
. _
converted to the pennanent status.
. Iteration 3 : From node 4, We now label node 2 with the new.temporary label [40 + 15, 4] = [55, 4]
which replaces the old temporary label [I00;-1]. Next node 5 is labeled temporarily with [40 + 50, 4] = [90,
4]. The temporary labels now include [55, 4] and [90, 4] associated with nodes 2 and 5, respectively. We ·
·
··
thus label node 2 permanently with [55, 4].
The only remaining node is the sink node 5, which converts its [90, 4] into a permanent label, thus
completing the procedure.
*
[55, 4](3)
[109, 1](1)
2
Legend : [ ] = label
( ) = iteration
60
.
*
[90, 4](3)
5
*.
[90, 3)(2)
Fig. 20.9.
The computational steps above are summarised graphically in Fig. 20.9. Observe that the calcula~ions
th
are based on the recursion concept employed_with the acyclic algorithm. The prime difference between ~
O
two algorithms occurs in that a node in the cyclic algorithm may be (temporarily) labeled regardless
whether or not all the nodes directly leading into if have been labeled. The solution in _Fig. 20.9 provides tbe
shortest distance to each .node in ~e network together wi~ its_r(?ute.
20.6: MAXIMAL.FLOW PROBLEM
Here we consider the problem of shipping a certain homogeneous commodity from a specifie~ po:t,
ca lled a source, to a particular point called the sink. There may not be a direct link or arc connecting e
source and the sink. The flow network will generally consist of some intermediate nodes, known as
tra,;hipment points, through which the fl~ws are routed.
_·
l.Ha,ft-/s
l 00 I
An e:-::unple of the maximal - flow problem is the situation where a number of refinerie s are
connected
distributio n tem1in~l through a network of pipelines. Booster and pumping stations are
mounted on the
e 10 move the oil products to the distribution terminals. The objective of analysing the situation
is to
~ isr the flow between the refineries and the terminals under the capacity restrictions of
the refinerie s
,.i'mc pipelines, Fig. 20.10 ?epicts the r~finery maximal-flow problem. Nodes 1, 2,
3 represent the refinerie s
.j noJe 5 7 and 8 represntmg the tenmnals. The remaining nodes represent the
pumping stations:
i
I
I
I
I
I
I
I
/
Q
)
!
I
Refineries
Pumping Stations
Terminals
Fig. 20.10.
Since the maximal-flow model requires only one source node and one destination node,
nodes O and 9
it added to represent such end points. The capacities of the arcs from the
source node Ocan
be taken equal
l!be outputs of the different refineries. On the other hand, the capacities from the distribution tenninals to
Wnode 9 may represent the demands for .the oil proudct. It is assumed in this situation
that the pipeline
;~ports only one type of oil product at a time.
· · The network in Fig. 20.10 has some arcs with positive capacities in one direction only.
These arcs are ·
~ 11 on the figure with arrows. As for the pumping stations 4,
5 and 6, flow may occur in either direction,
~ ibly with different capacities, depending on the design of the pipeline network .
.:_· · We use a special notati~n to represent the bidirectional flow on an arc. For an arc with
end nodes i an'd
~~ notation (a, b) signifies that the flow capacity from i ~o j i~ a, and that from} to i is b. For example
, in
~t 20.lO the capacities from node O(source) to the refineries is _represented _as (cl' 0), (c , 0) and (c ,
0)
2
3
2 and c3 are the capacities (per unit time) ofrefineries 1, 2 and 3. In tbe·case of the arcs
connecting
·Ptlmpmg stations, both a and b may be positive.
·
:·; The basic idea of the ·maximal-flow algorithm is to find a bre!)kthrough path that links
the source_node
\ lbt sink node such that the capacity of each arc on the path is positive. The maximum
flqw along this path ·
~ !hen equal the smallest capacity, c·, of all the arcs comprising the·path. We then modify the
arc capacities
~)-along the path to- (a - c', b + c') or (a+ c·, b - c') depending on whether the-flow on arc
(i, j) is i ➔ j or
1
., • respectively. The modification is int~nded to indicate
that the flow c* has been "committed". The
~ of finding breakthrough. paths from source to sink is repeated until it becomes evident that no further
~ :hrough are possible. The maximal flow then equals the sum of the c'-values determined in the
successive
~ 1ons
·
rec,, ~
+
~-t~e requirement that (a, b) be modifi:ed to. (a-c'; b c·) or (a+ c·, o~_c·) i~_crucial b~cau~e
it ~llo~s fu~re
_,a c1on of a previously committed flow c , when necessary. By adding c to the opposite d1rect1o
n of a
~ ltted flow we now have an instrument for "remembering" how much flow can be cancelle
d at a later
1
,
Operati
1002
Resean
later iteration. The networks in Fig. 20.11(a) illustrate this point. All arcs have the capacity (5, 0), indicatin.
a one-way flow in the direction i
with a maximum capacity of 5. The first iteration (arbitraril
identifies the breakthrough path 1 »2
3
4, which results in modifying the capacities of arcs (1.2
(2. 3) and (3, 4) from (5, 0) to (0. 5) because c' - 5, as shown in Fig. 20.11(b). Next, in Fig. 20.11(b) we
identity the breakthrough path 1 -> 3
2 > 4. Notice that arcs (1, 3) and (2, 4) have the capacity (5, a
and arc (3. 2) has the capacity (0, 5) which shows a positive flow in the direction 3 > 2. This path yields
S and results in the network of Fig. 20.11(c).
(5.0
S,,0)
Q
5.0)
(5.0)
(5,0)
Path 1 2 »3 4 , c* =5
(0, 5)
(5,0)
(0.
(0,5)
(O. 5)
(5, 0)
Path 1
(6, 0)
(0, 5)
(0, 5)
No breakthrough, optimum
3 2 4 , ch =5
c)
(b)
(a
(0, 5)
Fig. 20.11.
Observe carefully that what happened in the transition from (b) to (e) is nothing but a cancellation of
3. In essence, the algorithm recoginzes thatthe arc (2,3)
a previously committed flow in the direction 2
should not be used and that the maximum flow is achieved by usingthetwo paths l>2> 4 and 1 > 3
4. The algorithm is able to "remember" that a flow hasbeen committed previously only through the use ofthe
modifications (a -c*, b+c") and (a +c*, b-c) explained earlier.
Example 20.5. Determine the maximal flow in the network shown below.
Solution: It is obvious observing the network
that only arc (3, 4) has capacities in both directions. It
is 5
10
systmatize
the procedure for determining a breakthrough path as
from+%and from3- Wewill
4
(10,0)
20,0)
(10, 5)
follows.
Starting at node 1, we can choose either node 2,
(30, 0)
(30, 0)
3 or 4 as the next node to be linked to current node 1.
(20, 0)
(20, 0)
Thus from 1, we link to node 3 because it has the
highest arc capacity {= max (20, 30, 10)}. In case of a
tie we select the first ordered node in the tie set. Indicate
(40,0)
by labeling node 3 with [30, 1] which represents the
Fig. 20.12.
flow capacity (= 30) of the just added arc originating
at 1. At the source node 1 the label [o, -] indicate an infinite starting capacity with no predecessor node
Next at node 3 we can link to either 4 (tlow capacity= I0) or to 5 (flow capacity 20). Node 2 is excluded
becuase it shows a zero (0) flow capacity in the direction 3
2. Thus according to our heuristic, we link 5
to 5 and label 5 with [20, 3]. Since 5 is the sink node, the breakthrough path 1 -> 3
5 has been constructed.
The maximum flow along this path is determined directly from the labels as e* = min {o, 30, 20} = 20. See
Fig. 20.13.
l(H H
[30, 1]
c* = 20
(40, 0)
Fig. 20 .1 3.
[10, 3]
·
sts th e (a, b) the
\o w Fig. 20. 14 ad ju
.
h t fl
a committed
:t ➔ _,' ➔ )'t a re ec tsan ge (30, 0) of
ch
ufc• = 20; namely w e
)} , i.e., (IO ,
~ 3 to {(30 - 20), (0 + 20 -2 0) ,( 0
o {( 20
~ (20. 0) or ar c3 ➔ 5t
- c*, b + c*).
,) e. (0, 20)(Remember a
.,:. .,
for the
procedure is repeated
,;,~ext
.14. Start from L
•."'-~-- fl ows in Fig. 20
). Ch ao s;
~J_:rJ,~· froni l is 20, (max. 1030, 20
next 3
l~ 2, next 2 ➔ 3 (max4 --+ , 5:40),Keep in
_;Hl!lax. 0, 5) and then led, it cannot be
node is labe
-c:.- 1k1_r once a
during the same
:, ;, led (ia btled again)
node I is labeled
~, . Thus_ in Fig. 20.14
.
\.
the
,f~, -]
.
~
.,."'
.
.-
\-
O)'
(10, 0
(
[C X J, -] ~
.
.
~
,\ ~ '
\ .~
(2 0, 0) \
\
(1 0 20 )
.
.
.
(40, 0)
c*
Fi g. 20.14.
0, I]
[2
node 2 is labeled with
[40, 2)
node 3 is labeled \Vith
[10, 3]
node 4 is labeled with
.
[20, 4]
node 5 is labeled with
hich results
0, 40, 1 20} = l Ow
{2
in
m
=
c*
is
·
5
➔
2➔ 3➔ 4
o;
,.
= 10
path I ➔
Th e max. flow along the
s in Fig. 20.15.
t mod1fied flow capacitie
10)
10, 0 + i 0~, i.e., (10,
'• .
0
-(2
fo
2
➔
l
c
ar
We chan ge (2D, 0) of
0 + 10), r"e., (30, l 0)
,
10
40
(
to
3
-+
2
(40, 0) of arc
., (0, 15)
(1 0- 10 , 5 + 10), i.~
to
4
➔
3
c
ar
of
5)
(10,
.: (10: 10)
(2 0- 10 , 0 + 10), i.e
to
-5)
same
(3
c
ar
of
(20, 0)
3, 1 ➔ 2 each has the
.
➔
1
4,
➔
1
c
ar
e
as
~u
tie be
namely
t nod~ I there exists a
red node in the tie set,
de
or
st
fir
e
th
ct
le
,.Sr:e Fig. 20.15 now. A
se
e
b~w
me flow
ding ~o our rule of thum
➔ 5 each has the sa
2
d
an
3
➔
2
c
ar
•:apa_c ity ( = 10). Accor
label
n a tie as
"m node 3 we cannot
t node 2, there is agai
fro
er
ev
ow
H
·
3.
➔
2
~:- ~· ii ,. arc 1 ~ 2. A
nnot
s to node 3, ·arc
y labeled and hence ca
1
ad
re
same thumb rule lead
al
is
1
e
od
(n
0
~· (=== 30 ). The
=
g
4, 3 ➔ 5 each
be labeled again durin
t
no
4cir 1: od~ 5 because flow capacity of 3 ➔
ill
w
it
at
th
so
3
out
~
from 3 to 2 and cross
path 1 ➔ 2 ➔ 5
1-1 h
gh
ou
hr
3 ied)_ Vie thus BACKTRACK
k~
ea
br
e
th
us
·•i t
hich gives
n_ode 5 with . (30, 2] w
l
be
.la
e
w
2,
t
A
n.
· c-urrtn t ittratio
t
l
Operarir" IS
1()0,1
with c • =' min ( IO, 30) = I 0. The modified
flow is gives in Fig. 20.16 . Jt is 11npo rt1:1nl to
notice that, in genera l. the back track ing
process should be app lied repeated ly until we
either secure a breaktl1ro11 gh pnlh Clr until Hie
back1111cki11g step takes us back to the source
fn the latte r cnse, no furth er
node.
breakthrough paths are possible, signifying
th e end of the pr~cedure. Herc c .. = .10. '
We change (10, IO) of arc l ➔ 2 to
( 10- 10. 10+10) i.e., (0. 20)
(30, 0) of:Jrc 2 ➔ 5 to (J0 - 10, 0 + IO)
i. e., (20, 10).
Reu~ .r-
, l 0)
, :j
[JO,
.l
:1.
.-j,.
2 )-__ ::==::::t.- ::i
l
f
.
[10, 1]
c* = 10
Fig. 20.15.
·
[20, 2]
c* = 10
Fig. 20.16.
[ 10, 1]
See Fig. 20.16. Start at node J. Go
1 ➔ 3 as flow max is 10 (choosing 1 ➔ 3
· not 1 ➔ 4 same thumq rule). At node -3 we
have 3 ➔ 5 = 0, 3 ➔ 2 = 10 and 3 ➔ 4 = 0
flow, hence turn to node 2, then go to 2 ➔ 5.
Here c * =min (10, l 0, 20) = IO. We change
(10, 20) of 1 ➔ 3 to (10 - 10, 20 + 10), i.e.:
(0, 30) and (20, l 0) of 2 ➔ 5 to (20 ~ 10),
(10 + 10), i.e., (10,.20) and (10, 30) of3 ➔ 2
to (10 - 10, 30 + 10), i.e., (0, 40) (Note .this
step carefully)
[ 10, 4]
c* = 10
(40,
oy
t
(note carefully)
Fig. ·20.17~ .
4
4 has the maximum flow = (10). At ,
➔
I
it
as
4
node
o
·Got
1.
node
at
Start
.
20.17
Fig.
make
Now
10. (Breakthrough path = I ➔ 4 ➔ 5). We
next tum is to node 5 obviously. Here c* = min(I 0, 10) =
(10, 10) of 4 ➔ 5 to (IO - 10, 10 + 10) i. e., (O,
change (10, 0) of 1 ➔ 4 to (l O- 10, 0 + 10) i.e., (0, 10) and
20). Now construct Fig. 20.18 as shown below.
1
1005
(40, 0) .
Fig. 20.18.
+
Max. Flow :;::: 60 = 20
..i /
(5-4 )
20
_j,
(5-2 )
+
20
_j,
(5-3 )
30), 1 ~ 2 (0, 20).
The sour ce is I ➔ 4 (0, I 0), I ➔ 3 (0,
5 ➔ 3 (20, 0).
The sink is 5 ➔ 4 (20, 0), 5 ➔ 2 (20, I 0),
ible,
Since no more further breakthrough is poss
:. procedure ends here.
in the
We can now obta in the opti mal flow
(a*, b*) in
network by subtracting the modified flow
20. 12
Fig. 20. J8 from the original flow (a, b) in Fig.
-lo get Fig. 20.19.
- d) in
lf (a - a*) > 0, a flow of amo unt (a
of (bdirection i ➔ j otherwise if (b- b.) > 0, a flow
ble to
_h') occurs in the direction} ➔ i. It is impossi
.o (zero)
~ve both (a...:. a·) and (b - b.) greater than
flow
simultaneously. The use of this rule yields the
ork
given .in Fig. 20.19. The maximum flow in the netw
the
~ the sum of flow out of the source node or into
Fig. 20. 19.
sink node (= 60).
for all the arcs in Fig. 20. I 9.
Example 20.6. Determine the swplus capacity
[Ans. arc 2 ➔
3
=
40
0
=40 ,
-1,
-l,
in original Fig. in Fig. 2_0.19]
through nodes 2, 3 and ./ in Fig. 20. 19.
Exnmple 20. 7. Determine the amount of flow
[Ans. node 2 == 20, node 3 = 30, node 4= 201
us (from the stand point of increas ing
geo
anta
adv
ba
if
ld
wou
2,
J
20.
Fig:
inal
orig
the
Example 20. 8. In
.
.
ies in rhc direction 3 ➔. 5 and 4 ➔ - 5 ?.
lhe max imal flow) to increase the/ low capacit
[A ns. No. node l still represents n bort\eneck.)
\V PROBLE l\'t
l0.7_ MIN IMUM- COST CA PAC lTA TED FLO
rnli scs the muximn \ tlow moctc\ in four i,spec\s
The minimu m .. cos t capo citaLed flow prob lem scnc
1. All .arcs are unidirec tional
()p er a t ion s f? i>st! nrch
1006
c:iC' li ~ire.
2. A (no11- neg;1 tivc) un it flow cos t is assoc int cd with
its.
3 . Arcs may have posit iv e lowe r c;ipaci ty lim
;i sink .
4. Any node in th e network may act as rl so urce or
t· ·
while.: '"at
ize th e total cost
I '; y I 1\ g
"
'
l determi nes the nows. in th e different arcs th at minim new. mode
The
be
c;-in
•
em
probl
·
~cmand amounts a l th e nod es. The
the flow _r'cslrict1?ns on the arcs _::me! th e supp ly and
l11t1 on.
fo nnul ated as a l11~ear prog ra m111 111 g prob lem for so
Capa citat ed Network Simp lex Algo rithm
ar simp lex method. H.ow ever it is designed 10
The algor ithm is bas ed on lh e exac t steps of th e regul
- cost tfow mode l.
explo it the special netw ork struc ture of the n}in imum
ith m stipulate:;
defin ed in the LPP, th e capac itated simplex algor
Gi ven/ is the net flow at node
•
ias
II
that th e netw ork must satisfy
_L f
=0.
ork equa ls the
The condition says that the total supply in th e netw
i =I
tion,
nt by addin g a balan cing dum my sourc e or de stina
total dema nd. We can alwa ys satis fy this requi reme
ever, the
by zero unit cost and infinite capa city arcs. How
which we conn ect to all other node s in the netw ork
icting
restr
the
ble solut ion as this may be precl uded by
balancing of the netw ork does not guar antee a feasi
capacties of the arcs.
and
d algor ithm . Fami liarit y with simp lex m ethod
We will now prese nt the steps of the capa citate
meth od
Also, know ledge of the uppe r-bou nded simr lex
duali ty theor y (Cha pters 3 and 6) are essential.
·
(Cha pter 7) shou ld be ~elpf ul.
ion (set of arcs) for the netw ork. G~ to step l .
Step 0. Dete nnin e a start ing basic feasible solut
·
the simp lex method optimality cond-ition . lf the
Step 1. Dete rmin e an entering arc (variable) using
solut ion is optim al, stop; other wise , go to step 2.
simplex meth od feasibili_ty cond ition . Change
Step 2. Determine the leaving arc (variable) using the
the basis , then go to step 1.
fi = 0) consjsts of n - 1 independen t constraint
Ann-node netv;ork with zero net flow (i.e .• ~+ Ji+·.. : +
de n - 1 arcs. It can be proved that a basic solution
equations. Thus, an asso.ciated basic solution inust inclu
always co'rresponds to a spanning tree of the network.
g the qbje_ctive coefficients c ;i - z ij fo r-all t~e
·The enter ing a:-c (step 1) is deterrnin~d by computin
n-v isc, y\ c scic-~~ l:,. _ ;1:::- :'_c..,.,"r
Orhe
basis is opl~m um.
'
•
current nonbasic arcs (i,j). If all clj.. - zlj.. s; 0, the current
arc with the most positive clj.. - zlj.. to enter the basis.
as we did with the transportation model.. Using.
The computation· of;lj.. - zlj.. is based ori .duality, exactly
1
ble associated with the constraint of node ,
varia
dual
the
be
w;
let
3
Ch.
in
ed
defin
ing
ramm
prog
the linear
given as:
then the dual problem (excluding the upper bounds) is
11
Maximise · z ~
If, w, . ·
i=I
subject to
_c I).., c·t,J-)
w.- w. <
I
A
}
w; unrestricted in sign, i = I, 2, ... n
From the theory of linear programming, we have
\. ri. Jrk Mc-dels
1007
,t
w; - ,-11= c/J, for bas ic arc (i,J)
For i:.r:invenience, we will set w1 = O. We then solve the (basic) equations w, - w1 ,,_ c 9. to determine the
ifmJining d:.i al values. Next, we compute cu - ziJ fo r the nonbas ic variab les as :
c 1, , - z I/.. == c 1/. - (wI - w)
j
\,\t)i F•i1
,
a ll clj.. - zI/.. ~ 0, the optim a l solution is reached.
Exoi,,ple 20.9. A network 0/ pipelines connects two water desalination plants to two cities. The daily
J1J{)l'~l' ali~ ,, 1m ts at the two plants are 40 and 5 0 m;//ion gallons and the daily demand amounts at cities J and
, ~~i' JO and 60 million_ gallons. Both plants 1 and 2 have direct links to each of cities J and 2. Desalinated
~h' r ji-o:n p lants 1 and 2 may als_o be ~ra~sproted t~ city 2 through a_special pumping station. Additionally,
ri:mi I is Jinked to plant 2, and crty I 1s lmked to ctty 2. ~he model is already balanced because the supply
al nocles J ond 2 equals the demand at nodes 4 and 5. Frg. 20.20 gives the associated network.
Plant 1 [40]
Arc capacity
Unit cost
--....$5
City 1
1 i---------,,,c.--+{
(35)
$4 (oo)
$'9
(~
Plant 2 [50] -2 ,..___ _ _$-'--I_ _ _._
City 2
(30)
Fig. 20.20.
$ 490 ]
x = 5-' x IJ = 0' x 14 = 35 ' x 23 = 25 ' x 25 = 30,· x 35 = 25, x45 = 5, Total cost=
[Ans
• 12
·
EXERCISES
4'.
ie s2me
Solve example 20.2starting at node 5 (instead of node I) and sho'w that thO algorithm prOduces
result.
2. For each network in Fig.. 20.21 determin~ the following (a) a path. (b) a loop (c) a dire~te~ loop or
·
·
·
:eituit (cl) a tree (e) a spanning tree.
../ (
1
1.
Fig. 20.21.
[Ans. A (a) path I - 3 - 4 - 2 (b) loop 1 ~ 5 - 4 - 3 "'"" 1 (c) circuit 1 - 3 - 4 - 5 - l
1008
OperationsRese
.
S)} (i~
,, •.,
4)
_
(J
(d) a tree {(I , 3), (3 , 4), (3 , 5) ( 4 '
J)
_
e a spannin g tree {(I,
'
4 - 2), (2 - 5) :~
,
Calculate for (B) sin, .1 }
. .
.
v;
·
.
tree of th
, arly]
the foll . .
Q.3. Fm d the mmm1al spa nni ng
e network ,n Fig. 20.5 under each of
.c
conditions.
,
(b)
ile cable.
No des 5 and 6 arc link ed by a 2-m
Nod es 2 and 5 can not be link ed.
(d)
link ed by a If -,n,·I e ca bl e.
Nod es 2 and
. 6 are ·
2 ·s - .
The ca ble bet wee n nod es .J and 1 8 m1 1es 1ong .
(e)
ile cab le.
No des 3 and 5 are link ed by 'a 2-m
(I)
ctly to nod es 3 and 5 _
No de 2 can not be con nec ted dire
(a)
owing independenf
-.~
'
,
[(1- 3) or (3-4)], Len gth =- 14
[Ans. (a) (1-2 ), (2-5), (5-6), (4-6),
= 13]
), Lenoth
0
· (e) (1- 2 ), (2 - 5 ), ( 5 - 3), (2-4), (4-6
.·J
stu den t to solv e.
Res t of the sol utio ns is left to the
I
·
..
.
.
th
k
1
,
tw
ig
f
ati
in
nic
wn
mu
sho
es
com
c1t1
le
or
cab
at 1rnk the maJ
·
~
·
.
· on ne or
.
"'-Q.4. It is des ired. toh esta blis h. a.
•
·
·
ted suc h th at th e t ota I used cab le mll eag e 1s mm1m1zed. _..._.
20 .22 be Iow. D ete nm ne ow the c1t1es are con nec
Fzg. 20.22.
NY -
DC Length= 5080 miles}
[~4.ns. LA - SE - DE - DA- CH ads .
necting nine offshore natural gas wel lhe
con
s
link
le
sib
fea
the
of
e
eag
mil
the
e~ ·
Q.5. Fig. 20.23 shows
1 is the closest to the shore; it is equipp
ad
llhe
we
of
n
atio
l<?c
the
e
aus
Bec
very
:W~th an inshore delivery point.
of the remaining eight wells to the deli
put
out
the
p
·
pum
to
ty
aci
cap
e
rag
sto
with sufficient pumping and
livery point..
work that links the wellheads to the .de
net
e
elin
pip
um
.un
min
the
ine
term
point. De
, \iodels
"'" '
Fig. 20.23.
Q.6. In Fig. 20.23 of question (5) suppose that the wellheads can be divided into two groups depending
gas pressure ·: a high-pressure group that includes 2, 3, 4 and 6, and a low-pressure group that includes
r,!!s 5, 7, 8 and 9. Because of the pressure difference it is not possible to link the wellheads of the two
~ ps. At the same time, both groups must be connected to the delivery point through wellhead 1. Determine
k minimum pipeline network for this situation.
·:- fAns. High-pressure: (6 - 4), (4 - 3), (3 -2), (2- l); Low-pressure: (1- 5), (5 ~ 7), (5 - 9), (9 - 8))
,t
Q.7. A mobile-ph~ne company services six geographical areas. The satellite distances (in miles) among
~ six areas are given in Fig. 20.24 below. Detennine the _most e·fficient message routes that should be
ltab!ished between each t:wo areas in the network.. .
.
.
I-
t
100')
Fig. ·20.24:
[Ans. 1 - 2, route, 1 - 3 ~ 2, distance= 500 miles, 1 - 4 rout,e, l - 3 - 2 - 4~ distance== 700 ~iles
1 :--- 5 r6ute, 1 - 3 - 5, distance= 800 miles)
1014
Q.16. Complete the maximal flow f from s to n in the following network where the
numbers of the arcs represent their capacities.
,.
1-i'µ. 20.32,
Q• 17• De
tenl1ine the ma ximurn fl ow between nodes I and 5 for the network . .,.
IA11,.
in fig. 20.J1 .
15
t1n11,1
2
(4, 0)
Fig. 20.33.
[Ans. Max. flow== IOj
Q.18. Three refineries send their gasolene product to two terminals. The capacities of refineries a.-:
estimated at 2,00,000, 2,50,000 and 3,00,000 bbl per day. The demands at the terminals are known to be
4,00,000 and 4,50,000 bbl per day. Any demand that cannot be satisfied from the refineries is acquired from
other sources. The gasoJene product is transported to the terminals via a network of pipelines that are
boosted by 3 pumping stations Fig. 20.34 summarises the links of the network together with the capacity of
each pipeline. How much flow should be passing through each pumping station ?
.
Terminals ,
•:•
Pumping Station
l
l
, Terminals
►
►'•
20 II'
Fig: 20.34.
6
[Ans. 6 iterations, Max. flow= 110, Pump 4 = 30, Pump 5 = 50, Pump
701
===