Question-3

A company wants to purchase at most 1800

units of a product. There are two types of

the product M1 and M2 available. M1

occupies 2ft3, costs Rs. 4 and the company

makes the profit of Rs. 3. M2 occupies 3ft3,

costs Rs. 5, and the company makes the

profit of Rs. 4. If the budget is Rs. 5500/=

and warehouse has 3000 ft3for the products.

Requirement

Formulate the problem as linear programming

problem.

Solution

M1 (x1) M2 (x2) MaximumAvailability

Area 2 3 3000Cost 4 5 5500Profit 3 4AdditionalCondition

Company wants to purchase at most

1800 units of a product.

Let

X1 = Product M1

X2 = Product M2

Formulation of Linear Programming (Cont.)

Step 1

- Key decision to be made

Maximization of Profit

- Identify the decision variables of the problem

Let

X1 = Product M1

X2 = Product M2

Formulation of Linear Programming (Cont.)

Step 2

- Formulate the objective function to be optimized

- For maximization the objective function is based on profit

- Profit from M1 = 3X1

- Profit from M2 = 4X2

so our objective function will be like this

Maximize Z = 3X1+4X2

Formulation of Linear Programming (Cont.)

Step 3

- Formulate the constraints of the problem

- For area the maximum availability is 3000 ft3, and area required for M1 (X1) is 2 ft3 where for product M2 (X2) is 3 ft3. so the constraint become as under

- 2X1 + 3X2 < 3000- For cost the maximum availability is Rs. 5500,

Product M (X1) required Rs. 4 per unit and product M2 (X2) required Rs. 5 per unit so the constraint become as under

- 4X1 +5 X2 < 5500

Formulation of Linear Programming (Cont.)

Third condition:

Company wants to purchase at most 1800

units of a product

This is the production constraints that the

company must produce at most 1800 of the

product and the product is composed of X1

and X2, so the mixture of these two

X1 + X2 1800

Formulation of Linear Programming (Cont.)

Step 4

- Add non-negativity restrictions or constraints

The decision variables should be non negative, which can be expressed in mathematical form as under;

X1, x2 > 0

Formulation of Linear Programming (Cont.)

The whole Linear Programming model is

as under;

Maximize Z = 3X1+4X2 (Objective Function)

Subject to

2X1 + 3X2 < 3000 (AreaConstraint)

4X1 + 5 X2 < 5500 (CostConstraint)

X1 + X2 1800 (ProductConstraint)

X1,X2 > 0 (Non-Negative Constraint)

Question 4 ((Note: pl also see in Hillier Sol.Manual, page 33-

34(Prob3.1-11))

3.1-11 The Omega Manufacturing Manufacturing Company has

discontinued the production of a certain unprofitable product line. This

act created considerable excess production capacity. Management is

considering devoting this excess capacity to one ore more of three

products; call them products 1, 2, and 3. The available capacity on the

machines that might limit output is summarized in the following table:

Machine Type

Milling machine

Lathe

Grinder

Available Time

(Machine Hours per Week)

500

350

150

The number of machine hours required for each unit of the

respective products is

Productivity coefficient (in machine hours per unit)

Machine Type

Milling machine

Lathe

Grinder

Product 1

9

5

3

Product 3

5

0

2

Product 2

3

4

0

The sales department indicates that the sales potential for products 1

and 2 exceeds the maximum production rate and that the sales

potential for product 3 is 20 units per week. The unit profit would

be 30, 12, and 19, respectively, on products 1, 2, and 3. The

objective is to determine how much of each product Omega should

produce to maximize profit.

(a) Formulate a linear programming model for this problem.

(b) Use a computer to solve this model by the simplex method.

3-1-11(a) Let

.0 ,0 ,0

20

150 2 3

350 45

500539subject to

,191230Maximize

3product of units ofnumber

2product of units ofnumber

1product of units ofnumber

321

3

31

21

321

321

3

2

1

xxx

x

xx

xx

xxx

xxxZ

x

x

x

(b)

7619.2904*

20*

7619.54*

1965.26*

3

2

1

Z

x

x

x