THE SIMPLEX METHOD AND MINIMIZATION
 Summarize the relevant data in a table:
New York
San Francisco
Minimum needed
$425
$475
200
$525
$500
150
Most Shipped
120
250
 Determine the decision variables
x1 = Number of cars shipped from New York to Chicago
x1 = Number of cars shipped from New York to Saint Louis
x1 = Number of cars shipped from San Francisco to Chicago
x1 = Number of cars shipped from San Francisco to Saint Louis
 Form the dual problem
Minimize C = 425x1 + 525x2 + 475 x3 + 500x4
Subject to:
x1 + x2  120
x3 + x4  250
x1 + x3  200
x2 + x4  150
x1, x2, x3, x4  0
Solution of Minimization Problems
1. Write all problem constraints as  inequalities.
(This may introduce negative numbers on the right side of some problem constraints.)
-x1 - x2  -120
-x3 - x4  -250
x1 + x3  200
x2 + x4  150
2.
Form the dual problem.
Minimize
C = 425x1 + 525x2 + 475 x3 + 500x4
-x1
- x2
- x3
+ x3
x1
x2
- x4
+ x4
 -120
 -250
 200
 150
-1 -1 0
0 -120 O
L
M
0
0 -1 -1 - 250P
M
P
AM
1
0
1
0
200 P
M
P
0
1
0
1
150 P
M
M
425 525 475 500
1 P
N
Q
-1
L
M
-1
M
A  M0
M
0
M
M
-120
N
t
0
0
1
0
0
1
-1
1
0
-1
0
1
- 250 200 150
O
P
P
475P
P
500P
1P
Q
425
525
The dual problem is
Maximize P = -120y1 -250y2 + 200y3 + 150y4
Subject to
 y1  y3  425
 y1  y4  525
 y2  y3  475
 y2  y4  500
y1 , y2 , y3 , y4  0
3. Introduce slack variables x1, x2, x3, and x4, and form the initial system for the dual:
-y1
-y1
+
y3
+
y3
+ x1
+
120y1
y2
y2
+ 250y2
- 200y3
y4
+ x3
+ y4
- 150y4
*Note: Set the maximize function = 0
4.
+ x2
Form the simplex tableau and solve.
+ x4
+P
= 425
= 525
= 475
= 500
= 0*
y1
y2
y3
y4
x1
x2
x3
x4
P
x1
0
1
0
1
0
0
0
0 425
2
0
0
1
0
1
0
0
0
3
1
1
0
0
0
1
0
0
4
1
0
1
0
0
0
1
0
250 200 150 0
0
0
0
1
1
L
M
x
1
M
x M0
M
x M0
PM
N120
L
M
M1
x M
M
x M0
PM
80
N
y 3 1
x 2 1
3
4
L
M
M1
x M
M
y M0
PM
80
N
0
0
4
0
1
1
4
1
0
1 0
0 0
1
0
1 0 0  1
1 0 1 0
100 0 0 200
1
L
M
0
M
y M
1 1
M
y M
0 1
PM
N0 20
y3 0
x2 0
1
1
1 0
0
1
1 0
1
0
250 0 150 200
y 3 1
x 2 1
3
1
0
1 0
0 0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0
0
1
0 150
0
0
1
0
1
0
1
1
1
0
1
0
0
0 0 1
0 1 0
0 0 120
0
0
0
1
0
0
1
80 150
0
0
1
O
525P
P
475P -1R + R
P
500P
0 P
Q 200 R + R
1
1
4
3
1
3
2
3
R5
5
O
R +R
P
R +R
P
50 P
P
500 P
160000P
Q 80R + R
425
25
R2=
2
4
5
R5
5
O
P
-1R + R
P
50 P
P
500 P
85000P
Q150R + R
425
525
R3
3
R1
R2
R5
O
P
P
50 P
P
500 P
164000P
Q
475
75
You are now ready to read off the minimum cost and number of cars for x1, x2, x3, and x4 (column
headings have not changed).
Minimum cost is $164,000
cars from NY to Chicago
x1 = 120
cars from NY to St Louis
x2 = 0
cars from San Francisco to Chicago x3 = 80
cars from San Francisco to St Louis x4 = 150