Question

In: Operations Management

Chemco produces two chemicals A and B. These chemicals are produced via two manufacturing processes. Process...

Chemco produces two chemicals A and B. These chemicals are produced via two manufacturing processes. Process 1 requires 2 hours of labor and 1 lb of raw materials to produce 2 oz of A and 1 oz of B. Process 2 requires 3 hours of labor and 2 lbs of raw material to produce 3 oz of A and 2 oz of B. Sixty hours of labor and 40 lbs of raw material are available. Demand for Chemical A is unlimited, but only 20 oz of B can be sold. Chemical A sells for $16/oz, and B sells for $14/oz. Any chemical B that is unsold must be disposed of at a cost of $2/oz. Formulate an LP model to maximize Chemicos revenue and solve.

1. Maximum Revenue

2. Amount of chemical A produced (oz)

3. Amount of chemical B produced and sold (oz)

4. Amunt of chemical B destroyed (oz)

Solutions

Expert Solution

Solution:-

Consider Variable as:-

X1 = Process 1

X2  = Process 2

A = Chemical A produced

B = Chemical B produced

B2  = Chemical unsold

Now,

To maximize chemicos revenue,

Z= ( Cost of A) (Units of process 1) + (cost of B) ( units of process 2)- (cost of B2)(units of B unsold)

Z = 16 (2x1 + 3x2) + 14 (x1 + 2x2 - B2)  - 2B2

Maximize are at the Z value

Z = 16 (2x1 + 3x2) + 14 (x1 + 2x2 - B2)  - 2B2

Let the equation are:-

Constraint 1:- At most, 60 hours of labor are available

[( hour of labor) (Units of chemical produced by process 1) + (hour of labor) (Units of a chemical produced by process 2)]

= 2x1 + 3x2

Constraint 2:- At most, 40 lbs of raw materials are available

[(pounds of raw materials ) (Units of a chemical produced by process 1) + (pounds of materials) (Units of a chemical produced by process 2)]
= x1 + 2x2

Constraint 3:- At most, 20 ounces of chemical B can be sold.

[(ounces of a chemical produced by process 1) + (ounces of a chemical produced by process 2) - (ounces of a chemical unsold)]

x1 + 2x2 - B2

Thus, the formulation of linear programme model is shown as

Maximize Z = 16 (2x1 + 3x2) + 14 (x1 + 2x2 - B2)  - 2B2

Z= 32x1 + 48x2 + 14x1 + 28x2 - 14 B2 - 2B2

Mamimize Z= 46x1 + 76x2 - 16B2

Depending on the constraint

2x1 + 3x2

x1 + 2x2

x1 + 2x2 - B2

Thus all variables x1, x2, B2 are  


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