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P. Maria Sheeba
PG Student,
Mepco Schlenk Engineering College
1. A company makes 2 kinds of leather belts. Belt A is a high quality belt, and belt
B is of lower quality. The respective profits are Rs.4.00 & Rs.3.00 per belt. Each
belt of type A requires twice as much time as a belt of type B the company
could make 1000 belts per day. The supply of leather is sufficient for only 800
belts per day (both A and B combined). Belt A requires a fancy buckle and only
400 buckles per day are available. There are only 700 buckles a day available
for belt B. Determine the optimal product mix.
Given data:
A B
Profit per belt (in
Rs.)
4 3
Production rate 2 1 1000(belts per day)
Leather Supply
constraint
1 1 800(per day)
Availability of
buckles(per day)
400 700
1. Key decision:
No of belts produced in each type per day.
2. Decision Variables:
x – no of belts of type A produced per day.
y – no of belts of type B produced per day.
3. Feasible alternatives:
x ≥ 0
y ≥ 0
4. Constraints:
2x + y ≤ 1000 (Production rate)
x + y ≤ 800 (Leather availability)
x ≤ 400 (Buckle availability)
y ≤ 700 (Buckle availability)
5. Objective: Maximize the profit.
z = 4x + 3y.
Mathematical Formulation
Graphical representation   copy

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Graphical representation copy

  • 1. P. Maria Sheeba PG Student, Mepco Schlenk Engineering College
  • 2. 1. A company makes 2 kinds of leather belts. Belt A is a high quality belt, and belt B is of lower quality. The respective profits are Rs.4.00 & Rs.3.00 per belt. Each belt of type A requires twice as much time as a belt of type B the company could make 1000 belts per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 buckles per day are available. There are only 700 buckles a day available for belt B. Determine the optimal product mix. Given data: A B Profit per belt (in Rs.) 4 3 Production rate 2 1 1000(belts per day) Leather Supply constraint 1 1 800(per day) Availability of buckles(per day) 400 700
  • 3. 1. Key decision: No of belts produced in each type per day. 2. Decision Variables: x – no of belts of type A produced per day. y – no of belts of type B produced per day. 3. Feasible alternatives: x ≥ 0 y ≥ 0 4. Constraints: 2x + y ≤ 1000 (Production rate) x + y ≤ 800 (Leather availability) x ≤ 400 (Buckle availability) y ≤ 700 (Buckle availability) 5. Objective: Maximize the profit. z = 4x + 3y. Mathematical Formulation