Course Instructor: Dr. Swati Singh

Course: BBA III

Amity Business School

Linear Programming Problem

LPP is a mathematical modeling technique, used to

determine a level of operational activity in order to

achieve an objective, subject to restrictions.

It is a mathematical modeling technique, useful for

economic allocation of ‘scarce’ or ‘limited’ resources

like labor, material, machine, time, space, energy etc. to

several competing activities like product, services, jobs

etc. on the basis of a given criterion of optimality.

LPP Consists of: Decision Variables: Decision to produce no. of units of

different items.

Objective Function: Linear mathematical relationship used to describe objective of an operation in terms of decision variables.

Constraints: Restrictions placed on decision situation by operating environment.

Feasible Solution: Any solution of general LPP which also satisfies non negative restrictions.

Optimum Solution : The feasible solution which optimizes the objective function.

General Structure of LPP Maximize (or Minimize) Z = c1x1 + c2x2 + ---- + cnxn

Subject to,

a11x1 + a12x2 + -------- + a1nxn (≤, =, ≥ ) b1

a11x1 + a12x2 + -------- + a1nxn (≤, =, ≥ ) b2

an1x1 + an2x2 + -------- + annxn (≤, =, ≥ ) bn

where, x1 ≥ 0, x2 ≥ 0 ---- xn ≥ 0

Question 1. A dealer wishes to purchase a no. of fans and Air Conditioners. He has only Rs. 5760 to invest & space for at most 20 items.

A fan costs him Rs. 360 & AC Rs. 240. His expectation is that he can sell a fan at a profit of Rs. 22 & AC at profit of Rs. 18.

Assuming he can sell all items he can buy, how

should he invest money in order to maximize his profits?

Solution 1. Let us suppose, dealer purchases x1 Fans & x2 ACs.

Since no. of fans & ACs can’t be negative

So, x1 ≥ 0, x2 ≥ 0

Since cost of fan = Rs. 360 & AC = Rs. 240

& Total money to be invested = Rs. 5760

Thus, 360 x1 + 240 x2 ≤ 5760

Also, space is for at most 20 items

So, x1 + x2 ≤ 20

Again, if dealer can sell all his items

Profit is Z = 22 x1 + 18 x2, which is to be maximized

Thus, the required LPP is:

Maximize Z = 22 x1 + 18 x2

Subject to Constraints,

360 x1 + 240 x2 ≤ 5760

x1 + x2 ≤ 20

& x1 ≥ 0, x2 ≥ 0

Question 2. A company produces two articles R & S. Processing is done through assembly & finishing departments. The potential capacity of the assembly department is 60 hrs. a week & that of finishing department is 48 hrs. a week.

Production of one unit of R requires 4 hrs. in assembly & 2hrs. in finishing.

Each of the unit S requires 2 hrs. in assembly & 4hrs. in finishing.

If profit is Rs. 8 for each unit of R & Rs. 6 for each unit of S. Find out the no. of units of R & S to be produced each week to give maximum profit.

Solution 2.

Objective Function: Max. Z = 8x1 + 6x2

Subject to Constraints,

4 x1 + 2 x2 ≤ 60 (Time available in assembly dept.)

2 x1 + 4 x2 ≤ 48 (Time available in finishing dept.)

where, x1 ≥ 0, x2 ≥ 0

Products Time Required for Producing One Unit

Total hrs. available

x1 x2

Assembly Dept. 4 2 60 Finishing Dept. 2 4 48

Profit Rs. 8 Rs. 6