Problems on Simplex Methods

1. Sample Problem of Simplex

2. Problem
• Objective function
• Maximization Z = 5x +4y
• Constraints
0x + 2y <_ 50
1x + 2y <_ 25
1x + 1y <_ 15
x>=0, y>=0

3. Solution
• Given data
Objective Function : Maximization
Z = 5x + 4y + 0S1 + 0S2 + 0S3
Subject to constraints
0x + 2y <_ 50
1x + 2y <_ 25
1x + 1y <_ 15
Let us convert the above constraints from inequalities to equality by adding
the slack variable
0x + 2y + S1 = 50
1x + 2y + S2 = 25
1x + 1y + S3 =15

4. Posting the values in the tabular form
Cj 5 4 0 0 0 ratio
Cb b Xb x y S1 S2 S3 Xb / x
0 S1 50 0 2 1 0 0 50/0 = 0
0 S2 25 1 2 0 1 0 25/1 =25
0 S3 15 1 1 0 0 1 15/1 =15
Zj 0 0 0 0 0
Cj – Zj 5 4 0 0 0
Zj = cb1*x1 + cb2*x2 +cb3*x3
0 * 0 + 0 * 1 + 0 * 1 = 0
cb1*y1 +cb2*y2 + cb3*y3
0 * 2 + 0 * 2 + 0* 1 = 0
Zj = cb1*S11 + cb2*S12 +cb3*S13
0 * 1 + 0 * 0 + 0 * 0 * 0 = 0
1

5. Iteration - 1
Cj 5 4 0 0 0
Cb b Xb x y S1 S2 S3
0 S1 50 0 2 1 0 0
0 S2 10 0 1 0 1 -1
5 x 15 1 1 0 0 1
Zj 5 5 0 0 5
Cj - Zj 0 -1 0 0 -5
R3 – R3/Key element
15 / 1 = 15
1 / 1 = 1
1/ 1 = 1
0 / 1 = 0
0 / 1 = 0
1 / 1 = 1
R2 = old value –(key corres column x key corres row)/Key
element
R1 = old value –(key corres column x key
corres row)/Key element

6. 25 = 25 – (1 x 15) = 25 – 15 = 10
1
1 = 1 – ( 1x1) = 1 – 1 = 0
1
2 = 2 – (1x 1) = 2 – 1 = 1
1
0 = 0 – (1x0) = 0 – 0 = 0
1
1 = 1 – (1 x 0) = 1 – 0 = 1
1
0 = 0 – (1 x 1) = - 1
1
R2 = old value –(corresponding key column x corresponding key row)
Key element

7. 50 = 50 – ( 0 x 15) = 50 – 0 = 50
1
0 = 0 – (0x1) = 0 – 0 = 0
1
2 = 2 - (0 x 1) = 2 – 0 = 2
1
1 = 1 – (0x0) = 1 – 0 = 1
1
0 = 0 – (0 x 0) = 0 – 0 = 0
1
0 = 0 – ( 0x 1) = 0 – 0 = 0
1
R1 = old value –(corresponding key column x corresponding key row)
Key element

8. x = 15, y = 0
Objective Function
Z = 5x + 4y
Substitute the x = 15, y =0 in the above function
Z = 5(15) + 4(0)
Z = 75 + 0
Z = 75
Interpretation : The maximum profit of the
organization can be 75