1. PROBLEMS IN LINEAR PROGRAMMING 1
SEBASTIAN VATTAMATTAM
Problem 0.1. 2.1, p.33 in [1]
A company produces 3 types of products, A, B, C. It pro-
duces each day 50 units of A, 25 units of B and 30 units of
C. Only 100 man-hours are available daily for assembling
the products. Also the following additional information is
available.
Product Profit per unit Assembly Time
A 12 0.8
B 20 1.7
C 45 2.5
The company has a daily order commitment of 20 units
of A, and a total of 15 units of B and C. Formulate this as
an LP model so as to maximize the total profit.
Solution
The given data can be tabulated as follows:
Products A B C Total
Production Capacity(Units) 50 25 30
Man-hours per unit 0.8 1.7 2.5
Order Commitments(Units) 20 15(for B & C)
Profit per unit(Rs) 12 20 45
Decision variables: x1 , x2 , x3 - number of units of A, B, C
Objective Function(total profit)
 
x1
z : R3 → R,  x2  → 12x1 + 20x2 + 45x3
x3
1

2. 2 SEBASTIAN VATTAMATTAM
The problem is to maximize the objective function z(x1 , x2 , z3 )
subject to the constraints,
(1) 0.8x1 + 1.7x2 + 2.5x3 ≤ 100
(2) x1 ≤ 50
(3) x2 ≤ 25
(4) x3 ≤ 30
(5) x1 ≥ 20
(6) x2 + x3 ≥ 15
(7) x1 ≥ 0
(8) x2 ≥ 0
(9) x3 ≥ 0
Problem 0.2. P rob : 1, p.60 in [1]
A company sells 2 different products, A, B, making a
profit of Rs 40 and Rs 30 per unit respectively. The pro-
duction process has a total capacity of 30,000 man-hours.
It takes 3 hours to produce a unit of A and 1 hour for B.
The maximum number of A that can be sold is 8000 and
for B it is 12,000. Formulate this as an LP model so as to
maximize the total profit.
Solution
Table 1
Product Profit per unit Production time
A 40 3
B 30 1
Table 2
A B Total
Production capacity 8,000 12,000
Man-hours 3 1 30,000
Profit per unit 40 30
x1 , x2 - number of units produced.
x1
z : R2 → R, → 40x1 + 30x2
x2

3. LINEAR PROGRAMMING 3
The problem is to maximize the objective function z(x1 , x2 )
subject to the constraints,
(1) 3x1 + x2 ≤ 30, 000
(2) x1 ≤ 8, 000
(3) x2 ≤ 12, 000
(4) x1 ≥ 0
(5) x2 ≥ 0
Problem 0.3. Ex : 2.2, p.33 in [1] A company has two
plants, producing 2 products, A, B. Each plant can work
for 16 hours a day. In plant 1, it takes 3 hours to produce
1000 gallons of A and 1 hour to produce 1 quintal of B.
In plant 2, it takes 2 hours to produce 1000 gallons of A
and 1.5 hours to produce 1 quintal of B. In plant 1, it costs
Rs 15,000 to produce 1000 gallons of A and Rs 28,000 to
produce 1 quintal of B. In plant 2, it costs Rs 18,000 to pro-
duce 1000 gallons of A and Rs 26,000 to produce 1 quintal
of B. The company is obliged to to produce daily at least
10,000 gallons of A and 8 quintals of B. Formulate this as
an LP model so as to minimize the cost.
Solution
Let 1000 gallons be one unit of A, and 1 quintal be one unit
of B.
Table 1
Time(hr/unit) A B
Plant 1 3 1
Plant 2 2 1.5
Table 2
Cost(Rs) A B
Plant 1 15,000 20,000
Plant 2 18,000 28,000
Table 3
A B
Least Number of units 10 8

4. 4 SEBASTIAN VATTAMATTAM
Maximum Time: 16 hours/day
x1 , x2 - units of A in Plant 1 & Plant 2
x3 , x4 - units of B in Plant 1 & Plant 2
 
x1
x 
z : R4 → R,  2  → 15, 000x1 +18, 000x2 +28, 000x3 +26, 000x4
 x3 
x4
The problem is to minimize the total cost z(x1 , x2 , x3 , x4 )
subject to the constraints,
(1) 3x1 + x3 ≤ 16
(2) 2x2 + 1.5x4 ≤ 16
(3) x1 + x2 ≥ 10
(4) x3 + x4 ≥ 8
(5) x1 ≥ 0
(6) x2 ≥ 0
(7) x3 ≥ 0
(8) x4 ≥ 0
Problem 0.4. Omega leather goods company manufactures
two types of leather soccer balls X and Y. Each type of ball
requires work by both types of employees: semi skilled and
skilled. The semi skilled employees employ machines in the
manufacture of balls, while the skilled employees hand sew
the balls. The available time (per week) for each type of
employee and the time requirement for each type of ball are
given below:
Type of Employee Ball X Ball Y Time Available
Semiskilled 2 3 80
Skilled 4 6 150
The cost of an hour of semi skilled labor is Rs.5.50 and for
skilled labor is Rs.8.50 per hour. To meet weekly require-
ment, at least 15 balls of type X and at least 10 balls of
type Y must be manufactured. Formulate and solve this LP
problem.

5. LINEAR PROGRAMMING 5
Solution
x1 , x2 - number of units of X and Y produced per week
Production cost of one unit of X in a week
= 2 × 5.5 + 4 × 8.5 = 45.00
Production cost of one unit of Y in a week
= 3 × 5.5 + 6 × 8.5 = 67.50
x1
z : R2 → R, → 45x1 + 67.5x2
x2
The problem is to minimize z(x1 , x2 ) subject to the con-
straints,
(1) 2x1 + 3x2 ≤ 80
(2) 4x1 + 6x2 ≤ 150
(3) x1 ≥ 0
(4) x2 ≥ 0
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References
[1] J K Sharma,Operations Research Theory and Applications,4th Ed, Macmillan