# REVISED SIMPLEX EXAMPLE

A paint manufacturer produces two types of paint; water based and solvent based. The goal is to maximize the profit by producing right combination of paint types.

Profit margin for water based paint is 50 usd/ton while it is 30 usd/ton for solvent based type.

Material 1 and Material 2 are main raw materials; 6 gallons of Material 1 and 1 gallon of Material 2 are used for a ton of water based paint. 4 gallons of Material 1 and 2 gallons of Material 2 are used for a ton of solvent based paint.

The raw materials have limitation for daily supply; 24 gallons for Material 1 and 6 gallons for Material 2.

Another important criteria is market demand. If the products cannot be sold, the profit maximization will be just on a paper, although the production optimized. According to the sales reports, maximum of 2 tons of water based paint is sold per day. Solvent based paint is sold 1 ton less than water based paint.

Mathematical modeling and simplex solution of this case are as follows;

Maximize Z =  50.00x1 +40.00x2

Constraints:

6.00x1 + 4.00x2 <= 24.00

1.00x1 + 2.00x2 <=  6.00

-1.00x1 + 1.00x2 <=  1.00

Initial Table

Z     x1     x2     s1     s2     s3

Z    0.00  50.00  40.00   0.00   0.00   0.00

s1  24.00   6.00   4.00   1.00   0.00   0.00

s2   6.00   1.00   2.00   0.00   1.00   0.00

s3   1.00  -1.00   1.00   0.00   0.00   1.00

Iteration: 1

Incoming vector; x1 (Row 0 Value = 50.00), Outgoing Vector; s1 (Ratio = 4.00), Pivot Value = 6.00

New Table

Z      x1      x2      s1      s2      s3

Z  -200.00    0.00    6.67   -8.33    0.00    0.00

x1    4.00    1.00    0.67    0.17    0.00    0.00

s2    2.00    0.00    1.33   -0.17    1.00    0.00

s3    5.00    0.00    1.67    0.17    0.00    1.00

Iteration: 2

Incoming vector; x2 (Row 0 Value = 6.67), Outgoing Vector; s2 (Ratio = 1.50), Pivot Value = 1.33

New Table

Z      x1      x2      s1      s2      s3

Z  -210.00    0.00    0.00   -7.50   -5.00    0.00

x1    3.00    1.00    0.00    0.25   -0.50    0.00

x2    1.50    0.00    1.00   -0.12    0.75    0.00

s3    2.50    0.00    0.00    0.37   -1.25    1.00

The problem has optimum solution

***** RESULTS: *****

Zmax =   210.0000000000

x1   =     3.0000000000

x2   =     1.5000000000

Elapsed Time for Calculation: 0.03200000 seconds