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1. 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. 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. 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. 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. 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 For online classes in Mathematics at any level, please contact vattamattam@gmail.com References [1] J K Sharma,Operations Research Theory and Applications,4th Ed, Macmillan