Development of LP Problems
A firm produces two kinds of products A and B and sells them at a gain of Rs. 2 on class A and Rs. 3 on class B. All products are processed on two machines G and H. Class A needs 1 minute of processing time on G and 2 minutes on H; Class B needs 1 minute on G and 1 minute on H. The machine G is obtainable for not more than 6 hours 40 minutes while machine H is obtainable for 10 hours for any working day. Prepare the problem as a linear programming problem.
x1 be the number of products of class A
x2 be the number of products of class B
After comprehending the problem, the known information can be methodically arranged in the form of the table given below.
Class of products (minutes)
Class A (x1 units)
Class B (x2 units)
Available time (mins)
Profit per unit
As the gain on class A is Rs. 2 per product, 2 x1 will be the gain on selling x1 units of class A. Likewise, 3x2 will be the gain on selling x2 units of class B. Thus, total gain (profit) on selling x1 units of A & x2 units of class B is given through
Maximize Z = 2 x1+3 x2 (objective function)
As machine G takes 1 minute time on class A and 1 minute time on class B, the totality of minutes needed on machine G is given by x1+ x2.
Likewise, the total number of minutes needed on machine H is given by 2x1 + 3x2.
But, machine G is obtainable for not more than 6 hours 40 minutes (400 minutes). Consequently,
x1+ x2 ≤ 400 first constraint
In addition, the machine H is obtainable only for 10 hours that is 600 minutes, thus,
2 x1 + 3x2 ≤ 600 second constraint
As it is not feasible to manufacture negative quantities
x1 ≥ 0 and x2 ≥ 0 non-negative restrictions
Maximize Z = 2 x1 + 3 x2
Subject to restrictions
x1 + x2 ≤ 400
2x1 + 3x2 ≤ 600
& non-negativity constraints
x1 ≥ 0 , x2 ≥ 0
A firm manufactures two products A and B which have raw materials 400 quintals and 450 labour hours. It is identified that 1 unit of product A needs 5 quintals of raw materials and 10 man hours and gives a gain of Rs 45. Product B needs 20 quintals of raw materials, 15 man hours and gives a gain of Rs 80. Generate the LPP.
x1 - number of units of product A
x2 - number of units of product B
Maximize Z = 45x1 + 80x2
5x1+ 20 x2 ≤ 400
10x1 + 15x2 ≤ 450
x1 ≥ 0 , x2 ≥ 0
A firm produces 3 products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 correspondingly. The firm has 2 machines and below is given the necessary processing time in minutes for each machine on each product.
Machine X and Y have 2000 and 2500 machine minutes. The firm must produce 100 A's, 200 B's and 50 C's type, but not more than 150 A's.
x3 - number of units of product C
Max Z = 3x1 + 2x2 + 4x3
4x1 + 3x2 + 5x3 ≤ 2000
2x1 + 2x2 + 4x3 ≤ 2500
100 ≤ x1 ≤ 150
x2 ≥ 200
x3 ≥ 50
A company possesses 2 oil mills A and B which have varied production capacities for low, high and medium grade oil. The company signs a contract to deliver oil to a firm every week with 12, 8, 24 barrels of each grade correspondingly. It costs the company Rs 1000 and Rs 800 per day to run the mills A and B. On a day A generates 6, 2, 4 barrels of each grade and B generates 2, 2, 12 barrels of each grade. Formulate an LPP to find out number of days per week each mill will be operated so as to meet the contract cost-effectively.
x1 - no. of days a week the mill A has to work
x2 - no. of days per week the mill B has to work
Cost per day
Minimize Z = 1000x1 + 800 x2
6x1 + 2x2 ≥ 12
2x1 + 2x2 ≥ 8
4x1 +12x2 ≥ 24
A company has 3 operational divisions processing, weaving and packing with the capacity to manufacture 3 different kinds of clothes that are suiting, shirting and woolen yielding with the gain of Rs. 2, Rs. 4 and Rs. 3 per meters correspondingly. 1m suiting needs 2 mins in processing 3mins in weaving and 1 min in packing. In the same way 1m of shirting needs 1 min in processing 4 mins in weaving and 3 mins in packing whereas 1m of woolen requires 3 mins in each division. In a week total run time of each department is 60, 40 and 80 hours for weaving, processing and packing department correspondingly. Develop a LPP to find the product to maximize the profit.
x1 - number of units of suiting
x2 - number of units of shirting
x3 - number of units of woolen
Maximize Z = 2x1 + 4x2 + 3x3
3x1 + 4x2 + 3x3 ≤ 60
2x1 + 1x2 + 3x3 ≤ 40
x1 + 3x2 + 3x3 ≤ 80
x1≥0, x2 ≥0, x3≥0
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