Answer to Question #190690 in Operations Research for arya soman

Question

A factory has decided to diversify its activities. The data collected for the sales and production departments are summarized below: Potential demand exists for two products A and B. Market can absorb any quantity of A, whereas the share of B for this organization is expected to be not more than 400 units per month. For every three units of B produced there is one unit of a by - product which sells at K3 per unit and only 100 units of this by - product can be sold per month. Contribution per unit of products A and B is expected to be K6 and K8 respectively. These products require three different processes and the time required per unit of product is given in the table below: Process Product Product B Available hours 900 600 1200 III Find the product mix to optimize the contribution by using Simplex Method.

Answer 1

x₁ is units of product A

x₂ is units of product B

maximize contribution:

$Z=6x_1+8x_2$

subject to:

$2x_1+3x_2\le 900$ - process 1

$x_1+2x_2\le 600$ - process 2

$2x_1+2x_2\le 1200$ - process 3

$x_2\le 400$

After introducing slack variables:

Max Z=6x₁+8x₂+0S₁+0S₂+0S₃+0S₄

subject to

2x₁+3x₂+S₁=900

x₁+2x₂+S₂=600

2x₁+2x₂+S₃=1200

x₂+S₄=400

and 

x₁,x_2,S₁,S₂,S₃,S₄≥0

Negative minimum Z_j-C_j is -8 and its column index is 2. So, the entering variable is x₂.

Minimum ratio is 300 and its row index is 2. So, the leaving basis variable is S₂.

∴ The pivot element is 2.

Entering =x₂, Departing =S₂, Key Element =2

Negative minimum Z_j-C_j is -2 and its column index is 1. So, the entering variable is x₁.

Minimum ratio is 0 and its row index is 1. So, the leaving basis variable is S₁.

∴ The pivot element is 0.5.

Entering =x₁, Departing =S₁, Key Element =0.5

Negative minimum Z_j-C_j is -2 and its column index is 4. So, the entering variable is S₂.

Minimum ratio is 150 and its row index is 2. So, the leaving basis variable is x₂.

∴ The pivot element is 2.

Entering =S₂, Departing =x₂, Key Element =2

Since all Z_j-C_j≥0

Hence, optimal solution is arrived with value of variables as :

x₁=450,x₂=0

Max Z=2700