Transcript 2.1 The Network Simplex Method

Network Optimization
Chapter 2
Transshipment Problem
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2.1 The Network Simplex Method
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G = (V, E) is a digraph. |V|=n, |E|=m.
A type of commodity is shipped on the network.
For each arc i, j  , cij is the cost of sending one
unit of commodity from vertex i to vertex j
cij  0 .
Vector c =  cij  has m components, and is called
the cost vector.
There are three kinds of vertices:

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

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2.1 The Network Simplex Method
(1) Source
If the supply at source i is i ( >0 ), then
let bi =- i .
(2) Sink
If the demand at sink j is  j ( >0 ), then
let bj =  j .
(3) Intermediate vertex
Neither a source, nor a sink. If k is an
intermediate vertex, then let bk = 0.
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2.1 The Network Simplex Method
S denotes supply and D denotes demand
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2.1 The Network Simplex Method
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Balanced transshipment problem
Total supply = total demand:  i   j, i.e.
n
b
i 1
i
0
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2.1 The Network Simplex Method
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inflow at vertex i ~ the total commodity sent to i;
outflow at vertex i ~ the total commodity sent
out from i;
netflow at vertex i ~ inflow at i - outflow at i;
m
flow vector x ~ x  xij  R , where xij is the
amount of commodity from i to j xij  0 ;
feasible flow x ~ a flow x that makes the netflow
at each vertex i equal bi .
 
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2.1 The Network Simplex Method
A Feasible Flow
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2.1 The Network Simplex Method
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Single commodity uncapacitated
transshipment problem
Find a feasible flow x  0 such that its cost is
as low as possible.
Let A be the incidence matrix of the
network.
The supply-demand can be expressed by
Ax=b
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2.1 The Network Simplex Method
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For example, if at vertex i:
then
e1 e2 e3

 x1   b1 

 x    

 2   
i  1 1  1 0 ... 0 x3  bi
 



    

   

    bn 
A
x
b
 x1  x 2  x3  bi


netflow at i
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2.1 The Network Simplex Method
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The transshipment problem is now formulated as
min cx
s.t. Ax=b
x0
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c ~ row vector, x & b ~ column vector
c, x  R m , b  Rn
uncapacitated ~ all xij have no upper bound
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2.1 The Network Simplex Method
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Example 2.1
Consider the network shown below with 7
vertices and 10 arcs in which any flow vector x is
of the form
xT = [x13 x21 x32 x51 x52 x61 x64 x67 x74 x75].
The cost vector is given by
c = [1 1 1 1 5 11 20 5 5 6]
and the supply-demand vector by
bT = [0 2 3 5 -2 -3 -5].
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2.1 The Network Simplex Method
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2.1 The Network Simplex Method
e1 e2
e3 e4 e5 e6 e7
e8 e9 e10
1
1
 1 1




1
1
1


1

1


A
1
1


1 1
1



1

1

1



1  1  1

1
2
3
4
5
6
7
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2.1 The Network
Simplex Method
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1
1
 1 1



1 1
1


1

1


A
1
1


1 1
1


1 1 1



1  1  1

Note that the rows of A are linearly dependent
because the sum of all rows of A is a zero
vector.
So, rank (A) < n.
On the other hand, we will see later that A
contains an n  1  n  1 nonsingular submatrix.
Therefore, rank (A) = n-1.
We may delete one equation from Ax = b
without changing the solution set.
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Do you still remember?
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What is a basic?
What is a basic solution?
What is a basic feasible solution?
What is the main idea of Simplex
method? From a vertex (feasible basic
sol) to another vertex, obj. value keep
decreasing.
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2.1 The Network Simplex Method
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Feasible tree solution (FTS)
A feasible flow x is called an FTS if there
is a spanning tree T such that
xij  0 if i, j  T .
As each spanning tree has n-1 edges, an
FTS x has at most n-1 nonzero
components xij.
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2.1 The Network Simplex Method
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If an FTS x has n-1 nonzero components, then
the associated spanning tree is unique. It must
be the tree consisting of the n-1 edges eij that
correspond to the n-1 nonzero xij.
We call such an FTS a nondegenerate FTS.
An FTS x is called a degenerate FTS if the
number of nonzero components is less than n-1.
A degenerate FTS may associate with more than
one spanning tree.
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2.1 The Network Simplex Method
Example
1
 1
 1 1


A


1 1


1  1

b = (-3, 3, 0, 0)T
It is easy to see that x = ( 3, 0, 0, 0 )T is a
feasible solution of the equation Ax = b.
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2.1 The Network Simplex Method
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In fact x is a degenerate FTS, and the
associated spanning trees are not unique.
They can be one of the following trees:
Conversely, every spanning tree T is
associated with at most one feasible tree
solution.
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The above few pages told you
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FS can be classified into FTS/not FTS.
FTS can be further classified into DE/non-DE.
Given non-DE FTS, we can defined a unique ST.
Given DE FTS, there will be many ST.
The following few pages tell: Given a ST,
we can solve for a unique solution. The solution
may be a FS or not FS.
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2.1 The Network Simplex Method
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By relabeling vertices and arcs, we may let
the part of A corresponding to the arcs in T
contain a nonsingular n  1  n  1 matrix B.
For example, if we consider the following
spanning tree T (formed by solid lines) in
Example 2.1:
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2.1 The Network Simplex Method
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It is not clear if in A, the columns corresponding to
the arcs in T contain an n  1  n  1 nonsingular B.
1
1
 1 1




1
1
1


1

1


A
1
1


1 1
1



1

1

1



1  1  1

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2.1 The Network Simplex Method
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Relabel vertices and arcs as follows:
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2.1 The Network Simplex Method
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How to relabel?
First pick up an arbitrary vertex and label it as the
first one.
In general suppose i vertices and i-1 arcs in the
tree are already relabeled ( 1  i  n  1 ). Then
from all unlabeled vertices, choose one (say w)
which is adjacent to a relabeled vertex (say u),
and label w as the i+1th vertex, and also label
the arc between w and u in the tree (either
(u,w), or (w,u)) as the ith arc.
Repeat the procedure until we have labeled all n
vertices and n-1 arcs in the tree.
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2.1 The Network Simplex Method
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a7, a8, a9, a10 are the four arcs not in T, now
the incidence matrix
1
 1  1
1


1 1

A  
1
1

1

1 1


1

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
 v1
.  v 2
.  v3

.  v4
.  v5

.  v6
.  v 7
a1 a2 a3 a4 a5 a6 a7 a8 a9 a10
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2.1 The Network Simplex Method
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Let x = (x1, … , x10)T , where xi corresponds
to arc ai. Rows 2-7 and columns 1-6 form a
matrix B, which is upper triangular, and the
diagonal elements are 1 or -1. So, B is a
nonsingular n  1  n  1 matrix.
1
 1  1
1


1 1

A  
1
1

1

1 1


1

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
 v1
.  v 2
.  v3

.  v4
.  v5

.  v6
.  v 7
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2.1 The Network Simplex Method
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The feasible tree solution corresponding to T must
have
x7 = x8 = x9 = x10 = 0.
For xB= ( x1, x2, x3, x4, x5, x6 )T, it can be decided
by
 b2 
 
 b3 
BxB   

 
b 
 7

def
b
bi ~ the supply or demand, or zero corresponding
to vi.
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2.1 The Network Simplex Method
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Therefore, xB is uniquely determined:
(1) If x B  0 , then x  x ,0,0,0,0 is an FTS, i.e., T
is associated with a unique FTS.
(2) If some components of xB are negative,
then x  x ,0,0,0,0 is not a feasible solution, i.e.,
T is not associated with any FTS.
x B  B 1 b
T
T
T
B
T
B
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2.1 The Network Simplex Method
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Summary of the relationship between
FTS and ST
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2.1 The Network Simplex Method
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From the above analysis, we know that a Tree
Solution is actually a Basic Solution of the system
Ax = b, and the base B can be determined by an
associated spanning tree. An FTS is a feasible
basic solution (FBS).
In simplex method we move from one FBS to
another FBS. Now in network simplex method, we
move from one FTS to another FTS, and
geometrically, from one spanning tree to another
spanning tree.
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2.1 The Network Simplex Method
The dual solution defined by a spanning tree
We know that the dual problem of
min cx
s.t. Ax=b
x0
is
max yb
s.t. yA  c
where y is a row vector:  y1 , y 2 ,..., y n1 , y n  .
Also, the complementarity slackness condition
holds.
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2.1 The Network Simplex Method
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Now for a nondegenerate FTS x, the associated
spanning tree T is unique, and for any arc i, j 
in T, xij  0 . Let the column in A which
corresponds to arc i, j  be Aij , then by the
complementarity slackness condition,
xij  0  yAij  cij
i.e.
 
 
  1 i 
 y1 ,, yn    cij
1  j
 
 
 
 yi  y j  cij ,  (i, j) T
(*)
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2.1 The Network Simplex Method
 yi  y j  cij
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i, j  T
(*)
The above set (*) has n-1 equations for n
variables. So, the solutions are not unique. If
we fix yn = 0, then y1, … , yn-1 can be uniquely
determined. (If we fix yn to another value t,
then in the solution, each yi shall increase its
value by t.)
We call the row vector y   y1 , , y n1 ,,0
determined by the equations (*), the dual
solution defined by the spanning tree T.
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2.1 The Network Simplex Method
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Profitable arcs
Suppose x is an FTS that defines a spanning tree
T(x) and a dual solution y. An arc (i, j) which is
not in T(x) is said to be profitable with respect to
the FTS x if y j  yi  cij .
Note that if we change the value of yn in deciding
y, as all yi shall change by the same value, the
value of every y j  yi shall remain unchanged. So,
the determination of profitable arcs is independent
of the value of yn chosen.
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2.1 The Network Simplex Method
Theorem 2.2 An FTS u with no profitable
arcs is an optimal solution.
Proof Suppose u is associated with tree
T(u) and dual solution y. Define c  c  yA .
We know that c u  0 . In fact
(1) If arc i, j T u  , then by the definition of
y, cij  cij   yi  y j   cij  yi  y j   0
(2) If arc i, j   T u  , then uij  0.
So, for any arc i, j  , cij uij  0  cu  0.
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2.1 The Network Simplex Method
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Now for any feasible solution v: Av= b, v  0 ,
(1)
cv  c  yAv  cv  yb .
In particular, when v = u,
(2)
cu  c u  yb  yb .
(1) - (2): cv  cu  c v .
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2.1 The Network Simplex Method
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Since there is no profitable arc, for any
arc i, j  ,
y j  yi  cij  cij  cij  yi  y j   0 .
But v  0 , and hence
c v  0  cv  cu
 u is an optimal solution.
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2.1 The network simplex method
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The above theorem provides an optimality criterion.
Now we consider what happens if there is a
profitable arc e   p, q . Of course e is not in T,
and c pq  0 . If we can increase the flow on  p, q
from u pq  0 to v pq  t  0 , meanwhile keep vij  uij  0
for other arcs i, j  not in T, then as cpq v pq  0 , we
may have
cv  cu  cv  cpq t  0 (will explain in detail later),
i.e. we obtain a better solution v.
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2.1 The Network Simplex Method
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When arc e is added to the spanning tree
T, a unique cycle C(e) will be obtained.
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2.1 The Network Simplex Method
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In the cycle, we call each arc with the same
direction as e a forward arc (arcs e1 & e2 in
the above graph) and each arc with the
opposite direction a backward arc (arcs e3 &
e4 in the above graph).
If the flow along arc e is increased from 0
to t, then the flow along each forward arc ei
is from ui to ui+t, and the flow along each
backward arc ej is from uj to uj-t.
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2.1 The Network Simplex Method
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There must be some backward arcs in the
cycle. Why? Because if not, we may
increase t infinitely (Can this problem be
unbounded?).
But when t  , the new feasible flow v
shall let cv  cu  cpq t    cv   , which is
impossible as c  0 and v  0 .
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2.1 The Network Simplex Method
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Now since cycle C(e) has backward arcs,
when t is increased to certain amount, the
flow along a backward arc becomes zero,
and we cannot further increase t (for
otherwise some flow shall become negative,
violating the feasibility).
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2.1 The Network
Simplex Method
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In the above example, the largest value of t to
maintain a non-negative flow is t  minu3 , u4 .
Suppose t = u3. In this case, call e3 the leaving arc.
Generally, if an arc f in C(e) is the leaving arc, we
obtain a new tree T  by deleting f  T u  and
adjoining eT u  .
The updated flow vector v is associated with T , i.e.,
v is a new FTS, and the associated spanning tree
is T v   T .
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2.1 The Network Simplex Method
Theorem 2.3 If u is an FTS with a profitable arc
e and if v is the updated flow using this arc e, then
v is an FTS and cv  cu .
Proof We first show that v is still an FTS, because
(1) When we change from flow u to v, the net-flow
at each vertex remains unaffected, and hence all
supply / demand requirements are still satisfied,
i.e. Av = b.
(2) Obviously, v  0 .
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2.1 The Network Simplex Method
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We next show that cv  cu .
Let c  c  yA , where y is the dual solution of u.
By Theorem 2.2, cv  cu  c v .
Consider two types of arcs:
(a) If arc i, j  is in the tree T(u)
 cij  0  cij vij  0
(b) If arc i, j  is not in the T(u) and other than the
profitable arc  p, q .
 vij  0  cij vij  0
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2.1 The Network Simplex Method
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Therefore, cv   cij vij  cpq v pq  0  cv  cu .
In particular, if u is a nondegenerate FTS
 uij  0 ,  arc i, j T u 
 t = min { u | i, j  is a backward arc of
ij
T(u) } >0  v pq  t  0  cv  cu  v  u .
If all calculated FTS are non-degenerate, the
computation must be terminated after a
finite number of iterations (i.e. no cycling).
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2.1 The Network Simplex Method
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We now summarize the steps.
(1)
x
Initial Step: find an initial FTS
and an
associated spanning tree T (1). Let k = 1.
Iterative Steps: for k = 1, 2, …
Step 1 Calculate dual solution y (k ) on the
tree T (k.)
(k )
(k )
y

y
 cij , i, j  T ( k ) i.e., the
Step 2 If j
i
network has no profitable arc, terminate, x ( k ) is
the optimal FTS. Otherwise obtain a profitable
arc e   p, qT (k ) with yq(k )  y (pk )  c pq .
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2.1 The Network Simplex
Method
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Step 3 Form the cycle C(e)  e  T (k ) and determine
(k )
x
t = min { ij | i, j  is a backward arc in C(e) }
Suppose the minimum is attained at arc f.
Let
 xijk   t , If (i, j) is a forward arc of C(e)
 k 
 k 1  x ij  t , If (i, j) is a backward arc of C(e)
xij  
If (i, j) = (p, q)
 t,
 xijk  , For other arcs

( k 1)
and T
 T (k )  e  f .
k  k  1 and return to Step 1.
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2.1 The Network Simplex Method
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Let us now use Example 2.1 to explain the
algorithm.
S=5
S=2
D=2
D=5
S=3
D=3
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Tradition about arc labeling
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In the above example, there are totally
10 arcs: (1,3), (2,1), (3,2) (5,1), (5,2),
(6,1), (6,4) (6,7) (7,4) (7,5).
We normally, label them according to
the first number and then the second.
e1 = (1,3), e2 = (2,1), …
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2.1 The Network Simplex Method

Iteration 1
We have given an initial feasible tree solution with
which we start our first iteration.
Step 1. The initial FTS is
xT = [ 3 0 0 0 2 3 0 0 5 0 ]
and its cost is z = cx = 71. The arcs of a spanning
tree T (not unique since x is degenerate) with
which x can be associated are (1, 3), (2, 1), (5, 2),
(6, 1), (6, 7) and (7, 4).
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2.1 The Network Simplex Method
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For each arc i, j  in T
y j  yi  cij or yi  y j  cij
Start with y 7  0
.
So, y = [ 6, 5, 7, 5, 0, -5, 0 ] .
y 4  y7  c74  0  5  5
y6  y7  c67  0  5  5
y1  y 6  c61  5  11  6
y3  y1  c13  6  1  7
y2  y1  c21  6  1  5
y5  y2  c52  5  5  0
52
2.1 The Network Simplex Method
Step 2. We test whether there is a profitable
arc e with respect to x. In the present case
we see that e = (5, 1) is the only profitable
arc which comes. into T(x) as an entering arc.
For each arc i, j  not in T, calculate c  c
ij
 yi  y j
~ entering arc.
In the degenerate case, cycling might occur,
but we can use the methods learned in
53
Linear Programming to avoid cycling.
  c51  y5  y1  1  0  6  5  0
c51

ij
2.1 The Network Simplex Method
Step 3. The cycle C(e) is

in which (5, 2) and (2, 1) are both backward arcs with
existing flow 2 and 0, respectively. We revise the flow in this
cycle as t units along (5, 1), 2 - t units along (5, 2) and 0 - t
units along (2, 1). Therefore t = 0. Since (5, 1) has just
entered it does not make sense to take it as a leaving arc.
Thus (2, 1) becomes the leaving arc. Since t = 0, the cost
remains unchanged.
54
2.1 The Network Simplex Method
Iteration 2
Step 1. The current FTS is
xT = [ 3 0 0 0 2 3 0 0 5 0 ]
with cost z = 71. The arcs of the spanning
tree are (1, 3), (5, 1), (5, 2), (6, 1), (6, 7)
and (7, 4). From the defining equations,
y = [ 6 10 7 5 5 -5 0 ]
Step 2. We see that e = (3, 2) is the only
profitable arc which enters T(x).
55
2.1 The Network Simplex Method
Step 3. The cycle C(e) is
We get t = 2 and (5, 2) leaves the tree.
56
2.1 The Network Simplex Method
Iteration 3
Step 1. Our FTS is now
xT = [ 5 0 2 2 0 3 0 0 5 0 ]
and z = 67. The spanning tree has arcs (1, 3), (3,
2), (5, 1), (6, 1), (6, 7) and (7, 4). Also
y = [6 8 7 5 5 -5 0 ]
Step 2. There are no profitable arcs for the current
FTS. At this stage the algorithm halts after
reporting that the current FTS is an optimal
solution.
57
Again to avoid having many
different answer


If there are two profitable arc, add the
arc having prior (smallest) label.
For example, if there are 3 profitable
arc (2,3), (3,4) and (4,1), Then please
add (2,3). Otherwise, I may mark your
answer wrong.
58
2.2 Initialization



As we have seen, the network simplex
method need to have a feasible tree solution
(FTS) to start with.
How to obtain the initial FTS and the
associated spanning tree?
 Use artificial arcs.
59
2.2 Initialization
The way to obtain an initial spanning
tree T
1. choose an arbitrary vertex v in the graph.
(Always pick vertex 1 in exam)
2. for each source i, if there is an arc from i
to v, then this arc (i, v) is an arc of T;
if there is no arc from i to v, construct an
artificial arc (i, v) and take it as an arc of T;
60
2.2 Initialization
3. for each sink j, if there is an arc from v to j,
then this arc (v, j) is an arc of T;
if there is no arc from v to j, construct an artificial
arc (v, j) and take it as an arc of T;
4. for each intermediate vertex k, if there is an arc
from v to k or from k to v, we take that arc as an
arc of T;
if there is no such an arc, construct an artificial
arc in either direction and take it as an arc of T.
This T must be a spanning tree.
61
2.2 Initialization


Suppose there are r artificial arcs, we denote
the enlarged network by G  which has m + r
arcs, but still n vertices.
Define a new cost vector c  R m r by taking
0
cij  
1
If (i, j) is an arc of the original network G
If (i, j) is an artificial arc
62
2.2 Initialization


If the incidence matrix of G  is A , we
solve the problem:
c x
Pmin
s.t. Ax  b
x0
Note that the vector x  R m r .
63
2.2 Initialization

Obviously,
xiv  bi , for all sources i;
xvj  b j , for all sinks j;
, for other arcs in G 
form a feasible tree solution (FTS) corresponding
to the tree T.
Starting with this (initial) FTS, we use the
network simplex method to solve problem P .
xij  0

64
2.2 Initialization



Suppose x* is an optimal FTS with an associated
spanning tree T*.
There are three possibilities:
Case 1 T* contains no artificial arc.
In this case T* is a spanning tree of the original
network G and the components of x* that
correspond to the artificial arcs must be zero. Thus
the original transshipment problem is feasible. We
delete all artificial arcs, replace cost vector c  by c
and use T* as the initial tree to solve the original
network problem.
65
2.2 Initialization
Case 2 T* has at least one artificial arc
along which the flow is positive.
 In this case, the optimal value of
problem P is positive which means
that the original problem has no
feasible solution.
66
2.2 Initialization

Case 3 T* has at least one artificial arc, but the
flow in every artificial arc is zero.
The original problem has a feasible solution x
obtained by the restriction of x* to the set of
original arcs. After deleting the artificial arcs in T*,
the remaining part of T* has less than n-1 arcs,
which cannot form a spanning tree. So, it must
consist of several disconnected components,
which implies that the original problem can be
decomposed into several smaller sub-problems
which may be solved separately.
67
2.2 Initialization

Example 2.3
x  x12 , x14 , x23 , x25 , x35 , x45 , x46 , x57 , x67 
T
b   6,4,8,0,0,11,7
T
68
2.2 Initialization
Iteration 1.
Introduce arcs a10 =(1,5) and a11 =(5,6).
Initial spanning tree T(1):
D = 11
S=6
S=8
D=7
S=4
69
2.2 Initialization







Let c  = [ 0 0 0 0 0 0 0 0 0 | 1 1 ] .
Initial FTS
x(1) = [ 0 0 0 4 8 0 0 7 0 | 6 11 ]T .
y7  0 ,
 1 ,
  0 , y6  y5  c56
y5  y7  c57
 0 ,
  0 , y3  y5  c35
y 4  y5  c45
  0 , y1  y5  c15  1 .
y 2  y5  c25
y (1)   1,0,0,0,0,1,0 .
70
2.2 Initialization

For the arc (1, 2) not in T(1),

  y1  y 2  0   1  0  1  0 .
c12  c12

arc a1  1,2 is profitable.
.
T
(1)
 a1 contains cycle
C (1)
71
2.2 Initialization

The only backward arc is a10  1,5 which
is the leaving arc. t = 6.
x(2) = [6 0 0 10 8 0 0 7 0 0 11 ]T
72
2.2 Initialization
Iteration 2.
y2  y7 don’t change, y1  y 2  c12  0  0  0.
So, y(2) = [ 0 0 0 0 0 1 0 ].
For arc a7  4,6 ,
  y 4  y6  0  0  1  1  0.
c46  c46
 a7  4,6 is a profitable arc.
Backward arcs in C ( 2) : (4, 5) and (5, 6)
Since min {x45, x56} = min {0, 11} = 0,
a6  4,5 is the leaving arc.
73
2.2 Initialization
x
( 3)
x
( 2)
, but T(3) is different from T(2).
74
2.2 Initialization
Iteration 3.
y7 , y6 , y5 , y3 , y 2 , y1 don’t
change,
  1  0  1.
y 4  y6  c46
So, y(3) = [ 0 0 0 1 0 1 0 ].
For arc a2  1,4 , c14  c14  y1  y 4  0  0  1  1  0
 a2  1,4 is a profitable arc.
75
2.2 Initialization

Since min {x12, x25, x56} = min {6, 10,
11} = 6,
a1  1,2 is the leaving arc.
x(4) = [ 0 6 0 4 8 0 6 7 0 0 5 ]T.
76
2.2 Initialization
Iteration 4.
y2  y7 don’t change, y1  y 4  c14  1  0  1
.
So, y(4) = [ 1 0 0 1 0 1 0 ].
For the 5 arcs not in T(4),
So, there are no profitable
arcs, and x(4) is the optimal
solution of problem P .
But the flow on the artificial
arc (5, 6) is positive, which
means the original problem is
infeasible.
c67  0  y6  y7  1
c15  0  y1  y5  2
c45  0  y 4  y5  1
c23  0  y 2  y3  0
c12  0  y1  y 2  1
77
2.2 Initialization


Integer Solutions
If all supplies / demands are integer values,
i.e. b is an integer vector, then the optimal
solution x* of the transshipment problem,
obtained by the network simplex method,
must be an integer vector.
Why? because the incidence matrix A is
unimodular.
78
2.2 Initialization



Let A be the n 1  m matrix obtained by deleting the first
row of A; and b be the n  1  1 vector obtained by deleting
the first component of b.
Then the equation Ax  b is equivalent toA x  b (since rank(A)
= n-1). Any feasible tree solution x corresponds to a nonsingular n  1  (n  1) submatrix B of A such that x T  xBT , x TN  ,
where x N  0 and BxB  b .
p
x
x
x

So, each component i of B has the form i
,
det B
where p is an integer. Since A is unimodular, as a
nonsingular square submatrix of A, B must satisfy det B  1 .
So, x i must be an integer.
79
2.2 Initialization




Unbalanced Transshipment Problem
So far we consider balanced transshipment problem,
i.e. total supply = total demand.
Now if
total supply > total demand
we let r = total supply - total demand
and introduce one dummy sink v with the demand r
and an arc from each source vertex s i to v with
zero cost.
Hence the unbalanced problem becomes a
balanced problem in the enlarged network.
80
2.2 Initialization

Note that if in the optimal solution, the
supply of the source i is  bi and the flow from
i to the dummy vertex v is xiv  0 , then the
source i sends out only  bi   xiv units of the
commodity and remains xiv units at its place.
81
2.4 Transportation Problems
Motivating Example:
 A number of warehouses must ship a single
commodity to a number of manufacturing
plants.
 Shipping costs are proportional to distance
and the amount shipped.
 What shipping pattern will minimize total
shipment costs?
82
2.4 Transportation Problems
83
2.4 Transportation Problems


A transportation problem is a special
transshipment problem in which
(1) there is no intermediate vertices (i.e.
every vertex is either a source, or a sink).
(2) every arc is from a source to a sink.
The underlying graph of a transportation
problem is a bipartite graph.
84
2.4 Transportation Problems

G = (S, D, E).
S ~ the set of sources, denoted by S i (i =
1, … , m);
D ~ the set of sinks, denoted by D j (j =
1, …, n);
E ~ the set of arcs i, j  from a source S i to
a sink D j .
Flow matrix: m n matrix  x  ~ xij is the flow
on the arc i, j  ;
ij
85
2.4 Transportation Problems
Cost matrix: m n matrix c  ~ cij is the
cost to send one unit of the commodity from
S i to D j .
 Note that if actually there is no arc from S i
to D j , in order to express the problem easily,
we assume that there is an arc i, j  , but the
cost is very expensive, i.e. let cij  M , where
M is a very large number.
ij
86
2.4 Transportation Problems

Now the transportation problem can be
expressed mathematically as
m n
min  cij xij
s.t.
i 1 j 1
n
x
j 1
ij
m
x
i 1
xij  0
ij
 bi ,
for i = 1, … , m (1)
 b j , for j = 1, … , n
(2)
for i = 1, … , m; j = 1, … , n (3)
87
2.4 Transportation Problems
The constraint equations (1) and (2) can be jointly
expressed as:
A x = b,
where A is an (m+n) x mn matrix, which is in fact the
incidence matrix of the associated bipartite graph
(hence is a unimodular one according to Corollary
1.3b);
b is an m+n vector by combining the right-handside of the equations (1) and (2); and
x is an m∙n vector consists of all xij .

88
2.4 Transportation Problems

Transportation problem can be solved on a
particularly designed tableau, called the
transportation tableau.
89
2.4 Transportation Problems

Finding an initial FTS ~ the northwest
corner method.
90
2.4 Transportation Problems


Note that the m+n-1 = 6 filled xij entries,
together with other xij  0 , form an
initial FTS.
The arcs corresponding to the filled xij
entries form a spanning tree T.
91
2.4 Transportation Problems




Calculating dual solution y associated
with the tree T
We use the equation yi  cij  y j for any arc i, j   T
to determine y.
In the above example, let y7  0 .
y3  c37  y7  y3  0  8  8
 y5  ...
y3  c36  y6  y6  8  9  1
 y1  ...
y2  c26  y6  y2  1  7  6
 y4  ...
We fill in y1 ~ y3 to the right-most column
and y4 ~ y7 to the bottom row.
92
2.4 Transportation Problems
Checking if there is a profitable arc
 For each arc i, j   T, calculate dij  cij  yi  y j and
check if all such dij  0 . If yes, an optimal solution
is obtained. But d16  c16  y1  y6  3   8  1  6  0  1,6
is profitable.
93
2.4 Transportation Problems


Deciding the leaving arc
Starting from the cell (1, 6) in the tableau,
move horizontally or vertically until return
to the cell (1, 6) to form a cycle.
1,6  2,6  2,5  1,5  1,6
It is seen that in the cycle
1 6  2  5 1
the arcs are forward and
backward alternatively.
94
2.4 Transportation Problems

Put a + in the tableau for every forward arc;

Put a - in the tableau for every backward arc;

Decide t = min { xij | i, j  is a backward arc}.
95
2.4 Transportation Problems
Moving from a non-optimal tableau to
the next tableau
Let
xijnew
 xij  t

  xij  t

 xij
If (i, j) is a forward arc;
If (i, j) is a backward arc;
Other arcs
96
2.4 Transportation Problems



Introduce the entering (profitable) arc into the
new tree; remove the leaving arc from the tree.
Note that in the example the two backward arcs
(1, 5) and (2, 6) both have x15  x26  t  10.
In this case, we can choose either one as a
leaving arc (in the example (2,6) is chosen to
leave the tree), but not both, because to maintain
a spanning tree need to have m+n-1 = 6 arcs.
So far we completed one iteration and switch to a
new tableau.
97
2.4 Transportation Problems
Example 2.7 If an FTS is given by the circled
numbers in the tableau below, and suppose the
entering cell is (3, 7), how to form a cycle in
the tableau?
98
2.4 Transportation Problems

Note that when we are at (1, 8) and move horizontally, we
have to skip the cell (1, 6) and reach (1, 4). Then we move
vertically. If we change the moving direction at (1, 6), no
cycle can be formed.
99
2.5 Some Applications to Combinatorics
Definition A matching in a graph is a set
M of edges of the graph such that no two
edges in M have a vertex in common.
graph G
a matching of G
100
2.5 Some Applications to Combinatorics

In a bipartite graph G = (X, Y, E) where
X and Y both have n vertices (it is
possible that between some pair of
vertices x  X and y  Y , there exists no
edges, i.e., G may not be a complete
bipartite graph), a matching M is called
a perfect matching if there are n edges
in M.
101
2.5 Some Applications to Combinatorics
This bipartite graph has
no perfect matching
This graph has a
perfect matching


M  u1 , v3 , u 2 , v2 , u3 , v1  


102
2.5 Some Applications to Combinatorics
A graph is called
a regular graph of
degree k if the
degree of each
vertex is k.

2-regular graph
Theorem 2.5 (Konig’s marriage theorem)
A regular bipartite graph G(U, V, E) of
degree k (where k is a positive integer)
with U  V  n has a perfect matching.
103
2.5 Some Applications to Combinatorics



Proof We formulate a transportation
problem as follows:
Let the vertices in U be sources;
the vertices in V be sinks;
For each edge {u, v}  E , give an arc from
u  U to v  V .
Let the supply at each source be 1;
the demand at each sink be 1;
the unit cost cuv on each arc (u, v) be 1.
104
2.5 Some Applications to Combinatorics





This problem is feasible because if we let the flow
along each arc be 1 k , it is a feasible flow.
Also, the supply-demand vector is an integer
vector as the incident matrix of a bipartite graph is
unimodular. Therefore, the transportation problem
must have an integer optimal solution x*.
In fact as all supplies / demands are 1, x* must be
a 0-1 vector.
*
x
Let M = { (u, v) | uv  1 , u, v = 1, … , n }.
It is easy to see that |M|=n, i.e., M must be a
perfect matching.
105
2.5 Some Applications to Combinatorics
Example Does the following bipartite
graph have a perfect matching? If yes,
give a perfect matching.
As this is a regular graph
of degree 2, it has a
perfect matching.
Find a perfect matching
by yourself.
106
2.5 Some Applications to Combinatorics
Definition A covering in a graph is a
set W of vertices such that every edge
of the graph is incident to at least one
of the vertices in W.
v6, v1, v3, v4 (or v5) is a
covering of the left
graph
107
2.5 Some Applications to Combinatorics

For any matching M and any covering C of a
graph, suppose M  k
 the graph has k edges e  u ,v  , e  u ,v  ,… ,
e  u , v , such that all u i and vi are different
 to cover these k edges, we need to use at
least k vertices from u1 ,...,uk , v1 ,...,vk 
 a covering C must have at least k vertices
1
k
k
1
1
2
2
2
k
 C k
108
2.5 Some Applications to Combinatorics

Therefore, M  C for any matching M
and any covering C
 max{|M | : for all M }
 min{|C | : for all C}

This result can be strengthened to
become an equality if bipartite graphs
are concerned. (Just like Primal and
Dual)
109
2.5 Some Applications to Combinatorics
Theorem 2.8 (Konig’s Theorem)
In any bipartite graph, the size of a
maximum matching equals to the size of a
minimum covering.
110
2.5 Some Applications to Combinatorics




Konig’s Theorem can be translated equivalently to
the context of matrices. We use the term a line to
denote either a row, or a column.
Suppose P is a property that an element of a
matrix may or may not have. We call an element
of a matrix a P-element if it has the property.
A set of P-elements is said to be P-independent if
no two of them lie on the same line.
The size of the largest P-independent set is called
the P-rank of the matrix.
111
2.5 Some Applications to Combinatorics



Example 2.10
Let matrix
1
0

A  0

0
0
1 1 0
0 1 0
0 1 0

0 1 0
0 1 1
An element of A is said to have property P if it is
positive, i.e., if its value is 1.
Obviously, the P-rank of A is 3 as a largest Pindependent set contains 3 elements, for example,
the (1, 1)th, (2, 3)th and (5, 4)th element of A.
112
2.5 Some Applications to Combinatorics
Theorem 2.9 (Konig-Egervary Theorem)
The P-rank of a matrix A is equal to the
minimum number of lines that contain all
P-elements of the matrix.
113
2.5 Some Applications to Combinatorics
Proof We construct a bipartite graph. Let each
row i of A correspond to a left vertex i; each
column j of A correspond to a right vertex j.
th


i
,
j
Introduce edge i, j  if and only if the
element
of A is a P-element.
1
0

A  0

0
0
1 1 0
0 1 0
0 1 0

0 1 0
0 1 1
114
2.5 Some Applications to Combinatorics



By Theorem 2.8,
Size of a max. matching = Size of a min. covering.
Now, a matching is a set of edges, all with different
(left and right) endpoints.
(correspond to)  a set of P-elements in
different rows/columns, i.e. a P-independent set.
So,
Size of a max. matching = Size of the largest Pindependent set = P-rank of A.
115
2.5 Some Applications to Combinatorics



A covering is a set of vertices such that every edge
of the graph is incident to at least one of these
vertices.
(correspond to)  a set of rows and columns
such that each P-element is contained in one of
these lines.
So,
Size of a min. covering = minimum number lines
of A that contain all P-elements of the matrix.
Thus, the theorem is proved.
116
2.6 Assignment Problem

The assignment problem is a special
kind of transportation problem in which
(1) the supply at each source and the
demand at each sink are equal to 1.
(2) the number of sources = the
number of sinks.
117


2.6 Assignment Problem
Let cij be the cost to send one unit of
product (or sevice) from source i to sink j.
The mathematical formulation of the
assignment problem
is
n
min  cij xij
i , j 1
n
x ij  1,
s.t. 
j 1
n
x ij  1 ,

i 1
x 0 ,
ij
i  1,...,n
j  1,...,n
i, j  1,..., n
118
2.6 Assignment Problem

Note that actually we are interested in
integer solutions, i.e. xij should be 0 or
1, but this request does not need to be
imposed explicitly, because the
unimodular property for bipartite graph
guarantees an integer optimal solution.
119
2.6 Assignment Problem

Example A mill has 4
machines and 4 jobs to
Time (hours)
be completed. Each
machine must be
Job 1
Job 2
Job 3
assigned to complete
Machine 1
14
5
8
one job. The times
Machine 2
2
12
6
required to complete
Machine 3
7
8
3
each job on different
Machine 4
2
4
6
machines are shown in
the table:
Decide an assignment scheme to minimize the total
time needed to complete the 4 jobs.
Job 4
7
5
9
10
120
2.6 Assignment Problem


Method 1 use the method for transportation
problems.
Method 2 the Hungarian method.
Theorem 2.17
Suppose C is the cost matrix and C' is the matrix
obtained by adding a number t to each element in
a row or to each element in a column. Then a
solution is optimal with respect to C' if and only if
it is optimal with respect to C.
121
2.6 Assignment Problem
Proof Suppose the k th row is added by t,
i.e. ckj  ckj  t , j  1,...,n and cij  cij , for i  k and
j  1,...,n

Then
for any feasible solution [ xij ] ,
n
n
n
n
n
n
 cij xij =  cij xij   ckj xkj =  cij xij   ckj  t xkj
i 1 j 1
i  k j 1
n
n
n
j 1
c x  t x

=
=  c x  t
Therefore, under C and C’ , the optimal
solutions are the same.
i 1 j 1

i  k j 1
j 1
ij
ij
j 1
n
n
kj
i 1 j 1
ij
ij
122
2.6 Assignment Problem

Based on this theorem, we can change C
to C’ such that each row and each
column of C’ has at least one zero entry
by the following operations:
Step 1
Subtract each row by its smallest entry;
Subtract each column by its smallest
entry.
123
2.6 Assignment Problem
2 examples
4
6

A  1

5
7
4 9
9 8

B7 5

9 5
10 11
1 3 2 4 (1)
3
4
2 2 4 5 (2)

3 4 1 1  (1)  A  0


2 3 4 1  (1)
4
4
6 5 3 3 (3)
3 11 4  (3) 1
3 10 8  (3) 6
3 8 6  (3)  4


3 4 6  (3) 6
7 10 11 (7) 3
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0
6 0 8 1
0
5
5 0 7 5

2 0 5 3  B    3


2 0 1 3
5
2
4 0 3 4
(-1) (-2)
(-1) (-1)
4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3
124
2.6 Assignment Problem

If in such modified matrix C’ , there are n zero
entries and not two of them in the same row or
same column, for example
y
0
c  
y

y
y
y
0
y
0
y
y
y
y
y 
y

0
(all y are non-negative
numbers)
cx *  0

then obviously
other x  0
i, j  1,2,3,4 is an optimal solution, because for any
125
feasible solution x,
*
*
*
*
x13
 x21
 x32
 x44
1 ,
*
ij
2.6 Assignment Problem
c'ij , xij  0  c'ij xij  0,

*
c
'
x
but  ij ij  0.
Question: How to check if C’ has zero entries
at all different rows and columns?
This is in the method of step 2.
Step 2 Check whether C’ has n zero entries at
different rows and columns. If the answer is yes,
then an optimal assignment is obtained by taking
the assignment corresponding to the n zero
entries and we stop.
Otherwise we go to Step 3.
126
2.6 Assignment Problem


This step can be fulfilled by the following
sub-steps.
Step 2a Locate a row or a column in C’ with
exactly one 0, and the 0 is selected.
If every line (row or column) has more than
one zero, a line with the minimum number of
0s is located, and we select anyone of the 0s
in this line. This line should not be located
again (we can put a check mark beside it)
127
2.6 Assignment Problem



Put a circle
around the selected 0,
and draw a line through (covering) the
0. This line should be
a vertical line if the line located is a row,
and
a horizontal line if the line located is a
column.
128
2.6 Assignment Problem
Step 2b Repeat the previous step (but do
not consider any 0 which has been covered
by a line) till every 0 has at least one line
through it.
Step 2c If the number of lines drawn is n,
the n selected (circled) 0s will point out an
optimal solution. If the number of lines is
less than n, we go to Step 3.
129
n = 5 (lines), we obtain an optimal solution which
*
*
*
*
*
xis12
 x23
 x45
 x54
 x31
 1 and other xij*  0 i, j  1,...5
2.6 Assignment Problem

Examples
A 

3
4

0

3
4
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0
3
4

0

3
4
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0


3
4

0

3
4
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0
3
4

0

3
4
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0


3
4

0

3
4
3
4

0

3
4
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0
0 2 1 3
0 0 2 3
2 3 0 0

1 2 3 0
3 2 0 0
130
2.6 Assignment Problem
B 

0
5

3

5
2
0
5

3

5
2
4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3
4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3

0
5

3

5
2

4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3
0
5

3

5
2

0
5

3

5
2
4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3
4 0 7 0
3 0 6 4
0 0 4 2

0 0 0 2
2 0 2 3
n = 4 < 5, so, we are unable to obtain an optimal
assignment by using the modified matrix B  .
131
2.6 Assignment Problem




Generally, if the number of lines needed to cover
the zeros is k and k<n, we call the 0 element in
the matrix as P-element and by Th. 2.9, we know
that the P-rank of the matrix is k, i.e., the maximal
number of 0 elements in different rows and
columns is k.
In this case, we need to modify the matrix c 
further to create more 0s in the right place.
Let m be the smallest value of the uncovered
elements.
We subtract m from all uncovered rows ~ so that
more 0s appear in the previously uncovered part.
132
2.6 Assignment Problem

Example
0
5

B   3

5
2
m=2
4 0 7 0
0 7 0
0 4
3 0 6 4 (2) 3 1  2 4 2
0 0 4 2 (2)  1  2  2 2 0



0 0 0 2
5
0
0
0
2


2 0 2 3 (2) 0 0  2 0 1 
However, now in the
covered column, some
0 become –m which are
negative.
We now add m to the
covered columns.
(+2) (+2)
0
3

 1

5
0
6 2 7 0
3 0 4 2
0 0 2 0  B 

2 2 0 2
2 0 0 1 
133
2.6 Assignment Problem

The above operations can be further
simplified. We divide all elements in the
matrix obtained at Step 2 into 3 groups.
(1) Uncovered elements.
(2) Singly covered elements ~ covered
by only one line (a row or a column)
(3) Doubly covered elements ~ covered
by two lines: a row and a column.
134
2.6 Assignment Problem
Step 3a Find the value of the smallest
uncovered element. Suppose it is m. We
(a) subtract m from all uncovered elements;
(b) add m to all doubly covered elements.
(and keep the singly covered elements
unchanged)
135
2.6 Assignment Problem

For example, if we apply the operations (a)
and (b) to B  :
0
5

B   3

5
2
4 0 7 0
0
3
3 0 6 4

0 0 4 2   1


0 0 0 2
5
0
2 0 2 3
6 2 7 0
3 0 4 2
0 0 2 0

2 2 0 2
2 0 0 1 
.
which is just B 
136
2.6 Assignment Problem


Note that the 0s to be selected to draw
the lines (i.e. the 0s with
) are
singly covered, hence they are still 0.
But some elements in the uncovered
part become 0 ~ new 0, and thus need
to use more lines to cover all 0.
137
2.6 Assignment Problem
Step 3b Delete all covering lines and
go back to Step 2.
0
3

B   1

5
0
6 2 7 0  0
3 0 4 2 3
0 0 2 0   1
 
2 2 0 2  5
2 0 0 1  0
6 2 7 0
3 0 4 2
0 0 2 0

2 2 0 2
2 0 0 1 
138
2.6 Assignment Problem


Now the number of lines to cover all 0s
= 5 = n.
So, we obtain an optimal assignment:
*
*
*
*
*
*
x
x15  x23  x32  x44  x51  1 and other ij  0 ,
i.e. source 1 sink 5, source 2  sink 3,
source 3  sink 2, source 4  sink 4,
source 5  sink 1.
139