7.3 Transportation Simplex Method
How to pivot
Step 1 Determine the variables that should enter the basis.
Step 2 Find the loop involving the entering variable and some of the
basic variables.
Step 3 Counting only cells in the loop. Label even cells and odd cells.
Step 4 Find the odd cell whose variable assume the smallest value. To
perform the pivot, decrease the value of each odd cell and increase the
value of each even cell.
0
1
2
3
4
5
(1,4) – (3,4) – (3,3) – (2,3) – (2,1) – (1,1)
35
10
45
*
20
20
20
10
30
35
50
30 40
30
35-20
10+20
0+20
20
20-20
10+20 30-20
45
20
30
30
35
50
40
How to determine whether a bfs is optimal
How to determine which nonbasic variable should enter the basis
1. First supply constraint dropped (u1 =0)
2. -ui : shadow price of the i th supply constraint
-vj : shadow price of the j th demand constraint
3. cij  ui  v j  0 or ui  v j  cij  0 for all nonbasic variables
Then current bfs is optimal
4. Otherwise nonbasic variables with most positive value of
u i  v j  cij should enter the basis
u1  0
c ij  ui  v j  cij
35 8
6 10 9 35
u1  v1  8
c12  u1  v2  c12
50
10 9 20122013 7
u2  v1  9
 0  11  6  5
14 9 101630 5 40
u2  v2  12
c13  2, c14  8
45 20 30 30
u2  v3  13
ui  v j  cij  0
u3  v3  16 c 24  5, c 31  2
c 32  6
for all BV
u3  v 4  5
How to determine whether a bfs is optimal
vj
1
35 8
6 10 9 35
10 9 20122013 7 50
14 9 101630 5 40
45 20 30 30
vj
3
ui
8
6 12 2
0 25 8 10 6 10 9 35
1 20 9 123013 7 50
3 1410 9 1630 5 40
45 20 30 30
8 11 12 7
2 0 35 8
6 10 9 35
ui 1 10 9 10123013 7 50
-2 1410 9 1630 5 40
45 20 30 30
6
4
ui
vj
6 10 2
0
8 10 6 2510 9 35
3 45 9 12 5 13 7 50
3 1410 9 1630 5 40
45 20 30 30
z  8 x11  6 x12  10 x13  9 x14  9 x21  12 x22  13 x23  7 x24
 14 x31  9 x32  16 x33  5 x34
 6(10)  10( 25)  9( 45)  13(5)  9(10)  5(30)
 1020
Solution of Exercise
7
3
1 0 12 7 6 3
0
9 14 3
8
7
2
12 20
6
4
2 10 18
8 6
8 22
8 8 18 6 26
16 18
2
7 -1
0 12 7
3
9 20 3
4
8
7
6
12 20
2
0
6 2 10 18
2 6
8 22
8 8 18 6 26
16 18
c13  4, c14  6, c21  2, c24  4, c31  1, c32  2 c12  4, c14  10, c21  2, c24  4, c31  5, c32  2
3
7 -1 2
5
0 4 7
3 14 2 10 18
4 9 20 3 2 6
8 22
7
8 18 6 26
1 8 8
12 20 16 18
4
7
1
2
5
0 2 7 3 16 2 10 18
2 2 9 20 3
6
8 22
1 8 8 7
8 18 6 26
12 20 16 18
c12  4, c14  5, c21  2, c24  1, c32  7, c33  5 c12  2, c14  5, c23  2, c24  1, c32  5, c33  5
z  7 x11  3 x12  2 x13  10 x14  9 x21  3 x22  6 x23  8 x24
 8 x31  7 x32  8 x33  6 x34
 7(2)  2(16)  9(2)  3(20)  8(8) 6(18) 296
All negative
7.5 Assignment Problems
Machine/Job
1
2
3
4
1
14
2
7
2
2
5
12
8
4
3
8
6
3
6
4
7
5
9
10
xij  1 : if machine i is assigned to meet the demand of job j
xij  0 : if machine i is not assigned to meet the demand of job j
min z  14 x11  5 x12  8 x13  7 x14  2 x21  12 x22  6 x23  5 x24
 7 x31  8 x32  3 x33  9 x34  2 x41  4 x42  6 x43  10 x44
s.t. x11  x12  x13  x14  1 Machine
constraints

x11  x21  x31  x41  1 Job
constraints

xij  0 or xij  1
Assignment Problem
All supplies and demands = 1
7.6 Transshipment Problems
150
25
8
13 28
MEM
16
NY
200
DEN
supply point
Total Supply excess
total demand
150  200  130  130
dummy demand point
130
17
6
15 26
12
25
LA
6
CHI
14
16
transshipment point
Optimal
MEM
Tableau
dummy 90
BOS
130
demand point
NY CHI
LA BOS Dum
130 8
13 25 28 20
DEN
15 12 2613025 70
NY 220 0
6 13016 17
CHI
6 350 0 14 16
350
350 130 130
0
0
0
0
90
150
200
350
350
Find bfs Minimum cost method
14 5
8
2 12 6
7
8
3
1 2
4
6
+
1
1
7
5
9
10
1
1
1
1
+
14 1 5
8 7 +
2 12 6 1 5 +
7
8 1 3 9 +
1 2
4
6 10 +
+
+
+
+
14 5
8
2 12 6
7
8 1 3
1 2
4
6
+
1
+
7
5
9
10
1
1 14 5
8
7 1
1
2 12 6 1 5 +
+
7
8 1 3
9 +
+ 1 2
4
6 10 +
+
1
+
+
Determine whether a bfs is optimal
3
5
7
7
0 14 1 5
8
7 Combinatorial
-2
2 12 6 1 5 Optimization
Problems
-4
7
8 1 3
9
-1 1 2
4
6 10 One of Integer
Programming
Z =5+5+3+2=15