# Mrs Rb

Topics: Optimization, Operations research, Constraint Pages: 7 (1313 words) Published: January 4, 2013
1 DEFINITION OF LINEAR PROGRAMMING (LP)
Linear programming (LP) is a mathematical technique used in finding the best possible allocation of resources to achieve the best outcome, which is maximising profit or minimising cost. However, it is only applicable where there is a linear relationship between the variables. For example, the linear relationships between hours of labour and output in a textiles factory means an increase or decrease in labour force has a direct impact on production, which is the output. Due to the constraints, it is essential to solve the linear programming problem and find the feasible region of the objective function.

2 ANSWER (A) OPTIMAL PROFIT
Calculations
Brass Ltd produces two products, the Masso product and the Russo product.
Masso (\$)Russo (\$)
Selling Prices150100
Materials8030
Salesmen's Commission3020
Profit4050

Let ‘M’ represent the Masso products
‘R’ represent the Russo products
‘Z’ represent the objective function

Profit, Z = 40M + 50R
Constraint: MachiningM + 2R 700
Assembly 2.5M + 2R 1000
Non- negativity M0
R0
Machine constraints
Use equation 1 if Brass Ltd will produce M = 0 product
M + 2R = 700
0 + 2R = 700
R = 700/2
R= 350
(Brass Ltd can produce 350 units of Russo products if not producing Masso products)

Use equation 1 if Brass Ltd will produce R = 0 products
M + 2R = 700
M + 2(0) = 700
M = 700
(700 units of Masso products can be produced if Brass Ltd is not producing Russo products) Therefore, (M, R) = (700, 350)

Assembly Constraint
Use equation 2 if Brass Ltd will produce M = 0 product
2.5M + 2R=1000
2.5(0) + 2R=1000
2R=1000
R =500
(Brass Ltd can assemble 500 units of Russo products in 2 hours if not producing Masso products) Use equation 2 if Brass Ltd will produce R = 0 products
2.5M + 2R=1000
2.5M + 2(0)=1000
2.5M=1000
M=400
(Brass Ltd can assemble 400 units of Masso products in 21/2 hours if not producing Russo products) Therefore, (M, R) = (400, 500)

Graph 1
R
M ≤ 400
800

600

P (400,400)
400 R≤ 400
(0,350)
(200,250) 200
Feasible
region (400,0)M
0
200 400 600 800 Simultaneous equation
- (M + 2R=700)
2.5M + 2R=1000

-M - 2R=-700
2.5M + 2R=1000
1.5M=300
M=300/1.5
M=200
Substituting M = 200 in equation 1
M + 2R=700
200 + 2R=700
2R=700 – 200
R=500/2
R=250
Therefore, (M, R) = (200, 250)

These are the four co-ordinates
Profit, Z=40M + 50R
(0,0), P=0
(400, 0), P=40(400) + 50(0)=16000
(200, 250), P=40(200) + 50(250)=20500 (Optimal profit)
(0, 350), P=40(0) + 50(350)=17500

From the calculations above, the optimal profit from the objective function is \$20500 which occurs at the co-ordinate (200, 250). The corner points (200, 250) indicate that the maximum profit can be obtained when manufacturing 200 units of Masso products and 250 units of Russo products. With reference to the government special requirement, Brass Ltd is permitted to have a maximum of 800 units, 400 units of each, of the product in August, which signifies that output is still considered according to the rules. The feasible region is the shaded part on graph 1 (see p.5) with corners (0, 0), (0, 350), (200, 250) and (0, 400).

The scenario demonstrates linearity as there is a relationship between the constraints and the objective...

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