Transcript Network Optimization - United International College

Network Optimization
Chapter 4
Minimum Cost Flow Problems
(MCF)
1
4.1 The upper-bounded transshipment problem

In Chapter 2, we learned uncapacitated
transshipment problem, i.e., the flows along
the arcs does not have capacity restriction.
We now consider capacitated transshipment
problem:
min cx
s.t. Ax=b
0xu
2
4.1 The upper-bounded transshipment problem


The cost vector c and supply/demand
vector b are defined as before, u is the
capacity vector, i.e., we request
xij  uij for each arc (i,j).
The capacitated transshipment problem
is usually called minimum cost flow
problem.
3
4.1 The upper-bounded transshipment problem




If in addition to the upper bound u, the flows
also have (nonzero) lower bound constraint x
 v, it is easy to change all lower bound to 0.
For example, if we ask
xij  vij
for each arc (i,j)
then we introduce new variables
x’ij = xij - vij
So, vij xij  uij  0  x’ij uij -vij.
Let vector u =u - v, then 0 x’  u .
4
4.1 The upper-bounded transshipment problem

A problem is defined as follows :
min cx
min cx+cv
x  x' v
s.t. Ax=b  s.t. Ax=b-Av
v xu
0x’  u
5
4.1 The upper-bounded transshipment problem


Let b =b-Av, then the above problem is
equivalent to
min cx’
s.t. Ax’= b
0 x’  u
So, we can always assume the lower
bound v is a 0 vector.
6
4.1 The upper-bounded transshipment problem


We now extend the network simplex
method to solve MCF problems.
To do so, we need to recall the simplex
method to the LP problems with upper
bound constraints in which non-basic
variables can have either 0 value (lower
bound value), or the upper bound value.
7
4.1 The upper-bounded transshipment problem





Let x be a feasible flow to the MCF
problem min{cxAx=b, 0 xu}.
For each arc (i,j),
if xij =0, we say (i,j) is free w.r.t. x
if xij =uij, we say (i,j) is saturated w.r.t. x
if 0< xij <uij, we say (i,j) is unsaturated
w.r.t. x
8
4.1 The upper-bounded transshipment problem
Feasible Tree Solution x (FTS):
(1) x is a feasible flow (meet nonnegativity, capacity and supply/demand
requests).
(2) there is a spanning tree T such that
any arc not in T is either free, or
saturated. In other words, only the arcs
in T can be unsaturated arcs.

9
4.1 The upper-bounded transshipment problem


Dual variables y w.r.t. a FTS x
Again, the variables y are determined
by the equations
yj - yi=cij,
(i,j) T
and set yn =0.
10
4.1 The upper-bounded transshipment problem


Profitable arc
An arc (i,j)T is called a profitable arc if
yj - yi< cij when (i,j) is a free arc, or yj yi > cij when (i,j) is a saturated arc.
11
4.1 The upper-bounded transshipment problem





Theorem 4.1 : An FTS x without profitable
arcs is an optimal solution.
Proof : Let x’ be any feasible solution.
We need to show that cxcx’.
Let d=c - yA (expressed as a row vector), i.e.,
dij =cij +yi - yj.
cx = (d+yA)x = dx+yb.
cx’= (d+yA)x’= dx’+yb.
We should compare dx and dx’.
12
4.1 The upper-bounded transshipment problem





(i) if (i,j) T, then dij = 0  dij xij = dij x’ij.
(ii) if (i,j) T and is a free arc.
dij 0 (non-profitable), xij = 0, x’ij  0
 dij xij  dij x’ij .
(iii) if (i,j) T and is a saturated arc,
dij  0 (non-profitable), xij =uij x’ij
 dij xij  dij x’ij .
Therefore, dx = dij xij  dij x’ij =dx’
 cx  cx’.
The proof is completed.
13
4.1 The upper-bounded transshipment problem




By the above theorem, if x is not optimal,
then there is at least one profitable arc, say
(p,q).
If (p,q) is free and profitable, when xpq
increases from 0 to t, the objective function
value shall decrease (the reason is similar to
the un-capacitated case in Ch.2).
If (p,q) is saturated and profitable, then xpq =
upq. If we obtain a new FTS x’ with x’pq = upq
- t (t>0) and x’ij =xij for other arcs (i,j)T, the
objective function value shall decrease.
The reason is below.
14
4.1 The upper-bounded transshipment problem




Let F be the set of free arcs, S be the
set of saturated arcs w.r.t. x.
Since d = c - yA, cx’ - cx = dx’ - dx.
For the value of dx, as
dij = 0, (i,j)  T
xij = 0, (i,j)  F
xij = uij, (i,j)  S
we have dx =  dij uij .
S
(i, j)
15
4.1 The upper-bounded transshipment problem


For the value of dx’, as
x’ij = xij = 0, (i,j)  F
x’ij = xij = uij, (i,j) S & (i,j)≠(p,q)
x’pq = upq - t
we have dx’=  d ij x'ij = dij uij - dpqt
(i, j)S
(i,j)S
Therefore, dx’- dx = - dpqt. As (p,q) is
profitable, by definition, dpq = cpq + yp - yq >
0. Therefore, dx’ - dx < 0  cx’ < cx.
16
4.1 The upper-bounded transshipment problem
How to obtain an improved
solution:
 Adjoin the selected profitable arc e to
the spanning tree T to obtain a unique
cycle C(e).
 The arcs in C which have the same
direction as e are forward arcs and
otherwise backward arcs.

17
4.1 The upper-bounded transshipment problem
 If e is free, increase flows on e and each forward arc
by t, and reduce flows on each backward arc by t;
 If e is saturated, decrease flows in e and each forward
arc by t, and increase flows on each backward arc by
t.
if e is a free arc
if e is a saturated arc
18
4.1 The upper-bounded transshipment problem

The value of t is determined by the non-negativity
and capacity requests, i.e., until the flow along an arc
f in T is first reduced to 0, or increased to its upper
bound. This arc f is the leaving arc, and the new
spanning tree is T’= T + e – f.
if e is a free arc
if e is a saturated arc
19
4.1 The upper-bounded transshipment problem



Network simplex method for minimum
cost flow problems
Step 1. The current FTS is x with a tree T.
Any arc not in T is either free or saturated.
Step 2. Obtain a (unique) vector y with n
components, the last being zero, by solving
the n - 1 defining equations of the type yv - yu
= cuv, where (u,v) is an arc in T.
20
4.1 The upper-bounded transshipment problem


Step 3. Let dij = cij + yi - yj. A free arc (i,j) is
profitable if dij is negative. A saturated arc (i,j)
is profitable if dij is positive. If there are no
profitable arcs with respect to the current FTS
x, then x is optimal and we stop.
Step 4. Otherwise choose a profitable arc e
= (p,q) as the entering arc. Then a unique
cycle C is created. Arcs in C which have the
same direction as e are forward arcs, the
others being backward arcs.
21
4.1 The upper-bounded transshipment problem
Step 5.
(a) If the entering arc e is free, then add t units along e
and also to the flow along all forward arcs. At the
same time subtract t units from the flow in all
backward arcs.
(b) If the entering arc e is saturated, then subtract t
units from the flow in e and also from the flow in all
forward arcs. At the same time add t units to the flow
in all backward arcs.
(c) t is non-negative and the choice of t is subject to
capacity constraints and to the non-negativity
constraints of the flow vector.

22
4.1 The upper-bounded transshipment problem

Step 6. Choose t such that in the cycle C
there is at least one arc which becomes free
or saturated. Remove one such arc, say f,
from T, creating a new tree T’=T+e-f and an
updated flow x’. Go back to Step 2.
23
4.1 The upper-bounded transshipment problem


Example 4.1 Suppose in the network
below, the supply-demand vector bT =
[-9, 4, 17, 1, -5, -8], xT = [x12, x13, x15,
x23, x42, x43, x53, x54, x56, x62, x64],
cost vector c = [3, 5, 1, 1, 4, 1, 6, 1, 1,
1, 1], capacity vector uT = [2, 10, 10, 6,
8, 9, 9, 10, 6, 7, 8].
24
4.1 The upper-bounded transshipment problem


For the two numbers beside each arc, for example
(3,2) beside e1, the first is cost, & the second is
capacity.
Suppose we have an initial FTS xT = [2, 0, 7, 5, 0, 9,
3, 3, 6, 7, 7], z = cx =68, with the spanning tree T
shown below.
25
4.1 The upper-bounded transshipment problem
xT = [2, 0, 7, 5, 0, 9, 3, 3, 6, 7, 7]


Iteration 1 :
Step 1. using the formula yj - yi = cij,
(i,j)T we obtain y=[-1,5, 6, 1, 0, 0].
The arc (1,3) is free and
d13 = c13 + y1 - y3 = 5 + (-1) –6 = -2 < 0
26
4.1 The upper-bounded transshipment problem



Step 2. The arc (1,3) is free and
d13 = c13 + y1 - y3 = 5 + (-1) –6 = -2 < 0
(1,3) is a profitable arc ~ take it as the
entering arc e.
x13: 0→t, x53: 3→3-t, x15: 7→7-t
27
4.1 The upper-bounded transshipment problem




x13: 0→t, x53: 3→3-t, x15: 7→7-t
Choose the largest possible t to meet the
conditions:
t  u13 =10, 3 - t  0, 7 - t  0  t = 3
(5,3) ~leaving arc.
Iteration 2 :
Step 1. The new xT=[2, 3, 4, 5, 0, 9, 0, 3, 6,
7, 7], z=62 with the tree T shown below.
28
4.1 The upper-bounded transshipment problem



xT=[2, 3, 4, 5, 0, 9, 0, 3, 6, 7, 7]
y=[-1, 3, 4, 1, 0, 0]
The arc (5, 6) is saturated and d56 = c56 + y5
- y6 = 1 + 0 – 0 > 0.
29
4.1 The upper-bounded transshipment problem



Step 2. The arc (5, 6) is saturated and d56 =
c56 + y5 - y6 = 1 + 0 – 0 > 0.
So, (5,6) is profitable ~ take as the entering
arc e.
x56: 6→6-t, x64: 7→7-t, x54: 3→3+t
30
4.1 The upper-bounded transshipment problem


x56: 6→6-t, x64: 7→7-t, x54: 3→3+t.
Under the condition
x56 0, x64  0, x54  u54 = 10,
we take t=6 and (5,6) is the leaving arc f.
Note that e=f, and hence T does not change,
but the iterative solution x is changed.
31
4.1 The upper-bounded transshipment problem




Iteration 3 :
Step 1. xT=[2, 3, 4, 5, 0, 9, 0, 9, 0, 7, 1] with z=56.
y=[-1, 3, 4, 1, 0, 0].
(Note that y does not change because T is the same)
Step 2. There are no profitable arcs. So, the current
FTS is optimal.
32
4.2 Initialization




Just like the un-capacitated case, in order to
obtain an initial FTS, we need to locate a
vertex v satisfying three properties:
1. If i is a source, then there is an arc from i
to v whose capacity is not less than the
supply at i.
2. If j is a sink, then there is an arc from v to
j whose capacity is not less than the demand
at j.
3. If k is an intermediate vertex, then there is
an arc from v to k or from k to v.
33
4.2 Initialization


If we cannot find such a vertex v, we choose
an arbitrary vertex v and construct artificial
arcs if necessary to meet these requirements.
The only difference from the un-capacitated
case is that if there is an arc from a source to
v (or from v to a sink) in the network, but its
capacity is less than the supply of the source
(or the demand of the sink), then we still
need to construct an artificial arc with
unlimited capacity from the source to v (or
from v to the sink).
34
4.2 Initialization

For example, take vertex 2 as v. Even though
there is an arc (3,2) from source 3 to v, as its
capacity u32 = 1 < 9 (the supply), we need to
introduce another arc (artificial arc) (3,2) with
unlimited capacity.
35
4.2 Initialization

After introducing the artificial arcs, we solve an
MCF problem with
0 for originalarcs 
c’ij = 

1 for artificialarcs

If the optimal value is 0, the original MCF
problem is feasible and we obtain an FTS,
otherwise, the original problem is infeasible.
36
4.2 Initialization






Example 4.2
xT = [x12, x13, x24, x32, x34]
capacity vector uT =[8, 3, 16, 1, 8]
supply/demand vector bT = [-7, 10, -9, 6]
Take vertex 2 as v.
We use x’32 to denote the flow along the
artificial arc and consider the arc as the last
arc.
37
4.2 Initialization



Let c’ = [0 0 0 0 0 1]. We solve the MCF
problem.
The optimal solution is x12 = 7, x13 = x24 =
x32 = 0, x34 = 6, x’32 = 3.
Since the optimal value c’x = 3 > 0, the
original problem is infeasible. In fact, the flow
along the artificial arc is x’32 = 3, which is still
positive.
38
4.2 Initialization

Exercise Formulate the following problem as
a Minimum-Cost-Flow problem (but you do
not need to solve the problem). You may
formulate it by simply giving a suitable
network in which you specify (cost, capacity)
along each arc and the net supply (S) or
demand (D) beside some vertices.
39
4.2 Initialization


Oilco has oil fields in San Diego (SD) and Los Angeles
(LA). The SD field can produce up to 5 units of
unrefined oil (1 unit =100,000 barrels) per day, and
the LA field can produce up to 4 units per day. Oil is
sent from the fields to a refinery, either in Dallas (DA)
or in Houston (HO) (assume each refinery has
unlimited capacity).
To refine one unit costs＄800 at DA and＄1,000 at
HO. Refined oil is shipped to customers in Chicago
(CH) and New York (NY). CH customers require 4
units of oil per day, and NY customers require 3 units
per day. The costs of shipping one unit of oil (refined
or unrefined) between cities are shown in the table
below:
40
4.2 Initialization
TO
From
Dallas
Houston New York
Chicago
Los Angeles ＄300
＄110
-
-
＄420
＄100
-
-
San Diego
Dallas
-
-
＄450
＄550
Houston
-
-
＄470
＄530
41
4.2 Initialization



Also, assume that the volume of oil before
and after the refinement process has no
change.
Determine a plan to minimize the total cost.
If each refinery has a capacity limit of 5 units
per day, how would you revise the model?
42
4.2 Initialization

Answer 1 (incomplete):
43
4.2 Initialization

Answer 2 (incomplete):
44
4.3 The maximum flow problem





Let G (V,E) be an upper-bounded network, V
= {1, 2,…, n}, E= m,
Vertex 1 is the unique source with unknown
supply v.
Vertex n is the unique sink with unknown
demand v.
All other vertices are intermediate vertices.
So, the supply-demand vector b = [-v, 0,…, 0,
v]T (dimension n). Let the capacity vector be
u = [u1, u2,…, um]T.
45
4.3 The maximum flow problem


Maximum Flow Problem (MFP):
Find a feasible flow such that v is as
large as possible. Let the nm vertexarc incidence matrix be A. Then the
MFP can be expressed as
max v
s.t. Ax = b
0  x  u.
46
4.3 The maximum flow problem

We may assume that the indegree of
the source = 0, i.e., no arc enters
vertex 1.
47
4.3 The maximum flow problem



Similarly, we assume the outdegree of vertex
n is 0.
A flow x which satisfies the condition 0  x 
u, and Ax =b = [-v, 0,…, 0, v]T for a value v
 0 is called a feasible flow, and v is called
the value of the flow x.
What we want to find is the maximum flow
(MF) and its value v’.
48
4.3 The maximum flow problem



Let p = the sum of the capacities of all the arcs
directed from the source, and
q = the sum of the capacities of all the arcs directed
to the sink.
Then, obviously, the value of the maximum flow v’ 
min {p,q}.
49
4.3 The maximum flow problem




Transform an MFP to a minimum cost
flow (MCF) problem.
Introduce an artificial arc from n to 1: e =
(n,1) with unlimited capacity (or let min {p,q}
be its capacity). Arrange e as the last arc.
Expand G (V,E) to G’ (V,E’), where E’ = E + e.
Capacity vector of G’: u’ = (u, ∞)
50
4.3 The maximum flow problem





Let flow x’ in G’ be : x’ =(x, xe) where u, x
Rm, u’, x’ Rm+1.
Let the n(m+1) incidence matrix of G’ be A’.
In G’, each vertex becomes an intermediate
vertex:
flow in = flow out.
A feasible flow x’ of G’ satisfies the condition
A’x’ = 0, 0  x’  u’
Such a flow (inflow = outflow at any vertex)
is called a circulation.
51
4.3 The maximum flow problem

The MFP is equivalent to request the
flow along the arc e to be as large as
possible:
max xe.
But we need to change it to a MCF problem.
How? What is the cost vector?
52
4.3 The maximum flow problem



Define the cost vector c’ = (c, ce) as c = 0
(Rm), ce = -1.
Then we calculate the MCF problem
min c’x’
s. t. A’x’ = 0
(Q)
0  x’  u’
Once we solve the above MCF problem,
suppose the solution is
x ’= ( x , x e).
then x is the solution for MFP, and xe is the
value of the maximum flow.
53
4.3 The maximum flow problem

Question: Here ce is a negative number.
Will the minimum value of problem (Q)
become -∞?
54
4.3 The maximum flow problem



Answer: No! Because we know that the
MF from 1 to n is v’  min{p, q}.
The flow on arc e must be finite:
x e  min {p, q}.
Hence the optimal (minimum) value of
(Q)
c’ x ’= - x e  - min {p, q}.
55
4.3 The maximum flow problem


The max-flow & min-cut theorem:
If the network G contains only one
directed path P from 1 to n:
then MF = min {u1, u2,…, ul}. Each arc ei in P
is a cut. If we delete the arc, vertices 1 and n
will be disconnected, and ui is the capacity of
the cut.
56
4.3 The maximum flow problem



So, in this case,
value of the maximum flow = capacity
of the minimum cut.
We shall show that this result can be
extended to general case.
57
4.3 The maximum flow problem




Definition of a source-sink cut
Partition V = {1,…, n} into two subsets S and
T such that
1S, nT, S∩T =Ø , S∪T = V = {1,…, n}.
Let (S,T) be the set of arcs (i, j), either iS,
jT, or iT, jS.
Call (S,T) the source-sink cut, or simply a cut.
58
4.3 The maximum flow problem



An arc (i, j) such that iS, jT is called a
forward arc of the cut;
An arc (i, j) such that iT, jS is called a
backward arc of the cut.
The sum of the capacities of all the forward
arcs in the cut (S,T) is called the value of the
cut, or the capacity of the cut, denoted by
C(S,T).
59
4.3 The maximum flow problem
Example
 source ~ 1, sink ~ 4
(a) If S={1}, T={2, 3, 4}, (S,T)={(1, 2), (1, 3)},
C(S,T) = 10 + 7 = 17
(b) If S= {1, 2}, T={3, 4}, (S,T) = {(1, 3), (2, 3), (2, 4)},
C(S,T) = 7+7+8 = 22

60
4.3 The maximum flow problem
(c) If S= {1, 3}, T= {2, 4}, (S,T)={(1, 2), (3, 4), (2,3)},
C (S, T) = 10+6 =16
(d) If S={1, 2, 3}, T={4}, (S,T) = {(2, 4), (3, 4)},
C (S, T) = 8+6 =14
 So, the capacity of the minimum cut = min {17, 22, 16,
14} = 14.
61
4.3 The maximum flow problem

What is the maximum flow? See the solution
below.
62
4.3 The maximum flow problem


We will show that for any feasible flow and
any source-sink cut (S,T), the value of the
flow is no more than the capacity of the cut:
v  C(S,T).
Proof: Let x = (xij) be any feasible flow with
flow value v. As we know, for each vertex, net
flow = outflow - inflow and for MFP, the net
flow of vertex 1= v; the net flow of any
intermediate vertex = 0.
63
4.3 The maximum flow problem

So, for any cut (S,T),
v =  (  xij - xji )
iS
= 
ji
ji
 (xij - xji) +   (xij - xji)
iS jS
j i
iS jT
= 
(xij - xji)

= 
 xij -   xji
iS jT
iS jT
iS jT
total flow on forward arcs total flow on backward arcs
64
4.3 The maximum flow problem

 uij - 0
iS jT

= C(S,T)
Since for any feasible flow and any cut, the
above relationship holds,
max v  min C(S,T).
any feasible flow
any cut
65
4.3 The maximum flow problem


So, if there exists a cut (S’,T’) and a feasible flow
with value v’ such that v’= C(S’,T’), then the cut is a
minimum cut and the flow is a maximum flow, and
the value of maximum flow is equal to the
capacity of minimum cut. (*)
reason:
v’  max v  min C(S,T)  C(S’,T’)
66
4.3 The maximum flow problem


Flow augmenting path
In the directed network G, take an undirected
path between the unique source and the
unique sink, and then reintroduce the
directions in G to the edges in the path. Call
such a path P.
67
4.3 The maximum flow problem


An arc in P directed towards the sink is called
a forward arc, otherwise it is a backward arc.
In the above graph, (1,3), (5,6) & (6,n) are
forward arcs, (5,3) is a backward arc. A path
P is a flow augmenting path (FAP) with
respect to a feasible flow x, if
xij < uij for each forward arc (i,j) in P;
xij > 0 for each backward arc (i,j) in P.
68
4.3 The maximum flow problem


In other words, along an FAP, all forward arcs
are not saturated, and all backward arcs are
not free.
For an FAP, each forward arc has a positive
increment capacity tij = uij - xij, and the flow
along each backward arc can have a positive
decrement tij = xij.
69
4.3 The maximum flow problem

Let t = min {tij(i, j)  P}> 0 and
x  t (i, j) is a forwardarc of P
x’ij = x -t (i, j) is a backward arc of P
x
(i, j) is not in P
 ij


 ij

ij


70
4.3 The maximum flow problem
x’ is still a feasible flow, because:
(a) all vertices not in P are unaffected,
and hence the flow balance condition
inflow = outflow
still holds.

71
4.3 The maximum flow problem
(b) for the intermediate vertices in P (such as
vertices 3, 5, 6 in above graph).
the change of inflow = the change of outflow
and hence, just like x, the flow x’ still satisfies
inflow = outflow
(c) the non-negative (lower bound) and
capacity (upper bound) requests are satisfied.
72
4.3 The maximum flow problem


But under flow x’, the source 1 sends out t
more units, and the sink n receives t more
units, because the first and the last arc of P
must be forward arcs. So, the value of flow x’
is increased by t units.
We have seen that for a feasible flow x, if
there is an FAP from 1 to n, then x is not a
maximum flow, and we are able to find a
better flow x’.
73
4.3 The maximum flow problem




Now, if there is no FAP from 1 to n, must x be
a maximum flow?
YES. Why? The reason is as follows.
Let S be the set of vertices i such that there
is an FAP from 1 to i.
Let T = V \ S. Then 1S, nT, and we have a
cut (S,T).
74
4.3 The maximum flow problem
(a) For each arc (i, j) with iS, jT,

If xij < uij, then an FAP from 1 to i can be
extended to an FAP from 1 to j by adjoining
the arc (i, j) which is against the fact that
jS.

So, xij = uij for each forward arc of (S,T).
S: the set of vertices i such that
there is an FAP from 1 to i
T=V\S
75
4.3 The maximum flow problem
(b) Similarly, for each arc (i, j) with iT, jS, if
xij > 0, then there would be an FAP from 1
to i, which is wrong.

So, xij = 0 for each backward arc of (S,T).
76
4.3 The maximum flow problem
If there exists a cut (S,T) and a flow value v such that v= C(S,T),
then the cut is a minimum cut and the flow is a maximum flow, and
the maximum flow value is equal to the minimum cut value. (*)

Now for the feasible flow x, its value
v =   xij -   xji
iS
= 
iS

jT
iS
jT
 uij – 0
jT
= C(S,T)
Therefore, x must be a maximum flow. (by (*)
on page 66).
77
4.3 The maximum flow problem


It is found that we have a way to guarantee
that after getting flow augmenting paths for a
finite number of times, there will be no FAP
(we omit this proof). Therefore, by the above
reasoning, we obtain
Theorem 4.4 In a capacitated network, the
value of maximum flow is equal to the
capacity of minimum cut.
78
4.3 The maximum flow problem



The labeling algorithm for MFP
The main idea of the method is to find FAP by
propagating labels from the source until we
reach the sink or get stuck.
A vertex i is a labeled vertex if there is an FAP
from the source to i, and its label L(i) is a pair
[t,k] of two numbers:
79
4.3 The maximum flow problem



label L(i) is a pair [t,k] of two numbers:
t ~ the extra flow that the vertex i can
obtain from the source using the FAP under
consideration;
k ~ the last arc in the FAP is (k, i) or (i, k)
80
4.3 The maximum flow problem

The ways to choose a FAP may not be
unique, and they will affect the
efficiency of the method.
81
4.3 The maximum flow problem

Example 4.5 Find the maximum flow from
vertex 1 to vertex 4 in the network below.
82
4.3 The maximum flow problem



The value of flow in this network is increased
from 0 to M.
Then find another FAP 1
3
4.
The value of flow in the network is increased
from M to 2M ~ maximum flow. It needs only
2 iterations to obtain the maximum flow.
83
4.3 The maximum
flow problem




Way 2: Start with flow x = 0.
Find an FAP: 1
2
3
4
labels:
[M,1] [1,2] [1,3]
v: 0
1
Find another FAP: 1
3
2
4
v: 1
2
Find again the FAP: 1
2
3
4
v: 2
3
84
4.3 The maximum flow problem


Repeat the process until the remaining
capacities at arcs (1, 2), (2, 4), (1, 3),
and (3, 4) become 0.
We again obtain the maximum flow with
value v = 2M, but need 2M iterations.
85
4.3 The maximum flow problem




Note that in Way 1, each FAP consists of 2
arcs, whereas in Way 2, each FAP has 3 arcs.
It is found that generally, the efficiency of the
procedure depends upon the number of arcs
in the FAP (the fewer the better).
So, we try to choose an FAP with as fewer
arcs as possible.
If we assign each arc a unit weight, then the
number of arcs in an FAP = the weight of the
FAP. So, we may change the problem to find
a shortest FAP.
86
4.3 The maximum flow problem

We know that in an FAP, there are forward
and backward arcs, such as
but in the methods for finding shortest path, it is assumed
that all arcs in the path have same direction, i.e., all are
forward arcs.
87
4.3 The maximum flow problem

We will explain a method to solve this
problem, but first without loss of generality,
we assume that, between every pair of
vertices p, q in the network, there is only one
arc: either (p, q), or (q, p), but not both.
88
4.3 The maximum flow problem

In fact if the network has both (p, q) and (q,
p) with capacity upq and uqp, respectively, we
can make changes as shown in the graph
below. Now there is only one arc between
every pair of vertices.
89
4.3 The maximum flow problem
For a given feasible flow x, we construct a
directed network G(x) (called the residual
network of flow x) from the original network
G(V,E) as follows: for each arc (i,j) in G,
(a) If (i,j) is a free arc, then let this (i,j)
be an arc of G(x) with capacity uij.
(b) If (i,j) is a saturated arc, i.e., xij = uij, then
let (j,i) be an arc of G(x) with capacity uij.
(c) If (i,j) is an arc which is neither free, nor
saturated, i.e., 0< xij < uij, then we construct
two arcs in G(x), one is (i,j) with capacity uij –
xij, and the other is (j,i) with capacity xij.

90
4.3 The maximum flow problem

Example In the network G(V,E) below, (xij ,
uij) is shown beside each arc:
91
4.3 The maximum flow problem


Then, any directed path in G(x) from vertex 1
to 4 (all arcs are forward arcs) is an FAP, and
if there are several paths, choose the one
with the minimum number of arcs.
We obtain an FAP:
92
4.3 The maximum
flow problem

This augmented flow should have a value min
{6, 5, 3} = 3. Adding this additional flow to x,
we obtain x’:
93
4.3 The maximum flow problem


There is no path from vertex 1 to 4 in G(x’),
i.e., there is no FAP from the source to the
sink, and the labeling process gets stuck after
reaching vertex 2: 1 3 2.
So, x’ is a maximum flow and we stop.
94
4.3 The maximum flow problem


We see that the advantage of using residual
networks is that every FAP in the original
network (some are forward arcs and some
backward arcs) will be a directed path in the
residual network (all are forward arcs).
In order to obtain an FAP with minimum
number of arcs, in addition to using shortest
path methods, we may also use a simpler
way: by a breadth-first search starting from
the source until reaching the sink or getting
stuck.
95
4.3 The maximum flow problem

For example, if G(x) is as follows, then, we
have the paths from vertex 1 to 5:
1 3 5
1 2 3 5
1 2 4 5
1 2 4 3 5
and the “shortest” one is 1
3
5 which consists of only two arcs.
96
4.3 The maximum flow problem

We now use breadth-first search (BFS)
starting from vertex 1:
it is clear that the “shortest” path from 1
to 5 is 1
3
5.
Now we are ready to give the flow chart for
the labeling algorithm:

97
98
4.3 The maximum flow problem






According to the flow chart, each iteration
consists of 5 steps:
Step1. Use the current flow x to display G,
with each arc showing both the flow and the
capacity.
Step 2. Construct the residual network G(x).
Step 3. Construct the BFS tree rooted at the
source in G(x) with labeled and unlabeled
vertices. Stop if the sink cannot be labeled, in
which case x is optimal. Otherwise go to step
4.
Step 4. Construct an FAP.
Step 5. Update the flow and go to step 1.
99
4.3 The maximum flow problem

Example 4.6: Find the maximum flow from
vertex 1 to 6 in the given network by the
labeling algorithm.
100
4.3 The maximum flow problem

Iteration 1: (start with x = 0)
Step 1:
Show x
&u
(v=0)
Step 2:
Construct
G(x)
101
4.3 The maximum flow problem
Step 2:
Step 3:
Construct
BFS tree
102
4.3 The maximum flow problem
Step 3:
Step 4:
Find FAP
The sink is labeled.
103
4.3 The maximum flow problem
Step 5:
Update
the flow
Flow is updated and
104
4.3 The maximum flow problem

Iteration 2:
Step 1:
Show x & u
(v = 4)
Step 2:
Construxt
G(x)
105
4.3 The maximum flow problem
Step 2:
Step 3:
Find BFS
tree
106
4.3 The maximum flow problem
Step 3:
Step 4:
Find FAP
The sink is labeled.
107
4.3 The maximum flow problem
Step 4:
The sink is labeled.
Step 5:
Update
the flow
The flow is updated and
108
4.3 The maximum flow problem

Iteration 3:
Step 1:
Show x & u
(v = 5)
Step 2:
Give G(x)
109
4.3 The maximum flow problem
Step 2:
Step 3:
Find BFS
tree
110
4.3 The maximum flow problem
Step 3:
Step 4:
Find FAP
The sink is labeled.
111
4.3 The maximum flow problem
Step 4:
The sink is labeled.
Step 5:
Update
the
flow
Flow is updated and
112
4.3 The maximum flow problem

Iteration 4:
Step 1:
Show x & u
(v = 7)
Step 2:
Give G(x)
113
4.3 The maximum flow problem
Step 2:
Step 3:
BFS tree
114
4.3 The maximum flow problem
Step 4:
The sink has no label. No FAP.
Current flow in Step 1 is optimal.
S = {1, 3} and T = {2, 4, 5, 6}.
Min-cut is (S, T) ={(1, 2), (3, 4)}.
Min-cut value = max-flow value = 7.
115
Use NETSOLVE to calculate
Min-Cost Flow Problem




Specify that it is a directed network.
Need to enter node data: node name and
supply. For the supply entry, type positive
value if it is a source; negative value if it is a
sink; and 0 if it is an intermediate.
Need to enter arc data. For each arc, enter
the starting node name, the ending node
name, the cost, the lower bound of flow
(usually 0), and the upper bound (i.e. the
capacity) of the arc by this order.
The command to solve such problems is
MINFLOW.
116
Use NETSOLVE to calculate
Min-Cost Flow Problem
For Example 4.1, for node data, type (6 lines):
19
2 -4
……
68
and for edge data, type (11 lines):
12302
1 3 5 0 10
……
64108

117
Use NETSOLVE to calculate
Maximum Flow Problem




Specify that it is a directed network.
Do not need to enter node data.
Need to enter arc data: for each arc, type the
starting node name, the ending node name, 1
(as the cost even though this parameter is
not used), 0 (the lower bound of flow), and
the capacity by this order.
If you want to find the maximum flow, say,
from node 1 to node 9, type command:
maxflow 1 9
118
Use NETSOLVE to calculate
Maximum Flow Problem

For Example 4.6, the arc data should be
typed as (7 lines):
12104
13108
25104
……
56106
119