Simplex Method

Example 2

Luminous Lamps produces three types of lamps - A, B, and C. These lamps are processed on three machines - X, Y, and Z. The full technology and input restrictions are given in the following table.

Product Machine Profit per unit
X Y Z
A 10 7 2 12
B 2 3 4 3
C 1 2 1 1
Available Time 100 77 80  

Find out a suitable product mix so as to maximize the profit.

Solution.

The decision problem can be formulated as

Maximize z = 12x1 + 3x2 + x3

subject to

10x1 + 2x2 + x £ 100
7x1 + 3x2 + 2x3  £ 77
2x1+ 4x2 + x3  £ 80

x1, x2, x3 ³ 0

Converting inequalities to equalities

10x1 + 2x2 + x3 + x4 = 100
7x1 + 3x2 + 2x3 + x5 = 77
2x1+ 4x2 + x3 + x6 = 80
x1, x2, x3, x4, x5, x6 ³ 0

Where x4, x5 and x6 are slack variables.

Including these slack variables in the objective function, we get

Maximize z = 12x1 + 3x2 + x3 + 0x4 + 0x5 + 0x6

Initial basic feasible solution

x1 = 0, x2 = 0, x3 = 0, z = 0
x4 = 100, x5 = 77, x6 = 80

Table 1

    
cj 12 3 1 0 0 0
    
cB Basic variables
B
x1 x2 x3 x4 x5 x6 Solution values
b (=XB)
0 x4 10 2 1 1 0 0 100
0 x5 7 3 2 0 1 0 77
0 x6 2 4 1 0 0 1 80
zj-cj
    
-12 -3 -1 0 0 0
    

Key column = x1 column.
Minimum (100/10, 77/7, 80/2) = 10
Key row = x4 row
Pivot element = 10
x4 departs and x1 enters

Now we are assuming that you can easily calculate the values yourself.

"One that would have the fruit must climb the tree." -Thomas Fuller

Table 2

  
cj 12 3 1 0 0 0
  
cB Basic variables
B
x1 x2 x3 x4 x5 x6 Solution values
b (= XB)
12 x1 1 1/5 1/10 1/10 0 0 10
0 x5 0 8/5 13/10 -7/10 1 0 7
0 x6 0 18/5 4/5 -1/5 0 1 60
zj-cj
  
0 -3/5 1/5 6/5 0 0
  

Final Table

  
cj 12 3 1 0 0 0  
cB Basic variables
B
x1 x2 x3 x4 x5 x6 Solution values
b (= XB)
12 x1 1 0 -1/16 3/16 -1/8 0 73/8
3 x2 0 1 13/16 -7/16 5/8 0 35/8
0 x6 0 0 -17/8 11/8 -9/4 1 177/4
zj-cj
  
0 0 11/16 15/16 3/8 0
  

An optimal policy is x1 =73/8, x2 = 35/8, x3 = 0.
The associated optimal value of the objective function is z = 12 X (73/8) + 3 X (35/8) + 1 X 0 = 981/8.

 


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