MASTERS OF COMMERCE

APPLIED MATHEMATICS

ASSIGNMENT No. 2

Linear Programming

Prepared by: Rajab Ali Abdul Majeed

Section – B

Date: Dec 12th, 2014

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Linear Programming Problems; graph the region of feasible solution (if one exists) and solve by the corner point method.

Q 21)

Maximize Z = 4x1 + 8x2

subject to x1 + x2 ≤ 202x1 + x2 ≤ 32 x1, x2 ≥ 0

Find constraintsC1 = x1 + x2 = 20C1 = x1/20 + x2/20 = 1

C2 = 2x1 + x2 = 32C2 = 2x1/32 + x2 /32 = 1C2 = x1/16 + x2 /32 = 1

Put (0,0)C1 = x1 + x2 ≤ 20C1 = 0 ≤ 20 (True)

C2 =2x1 + x2 ≤ 32C2 = 0 ≤ 32 (True)

Point B intersecting C1 & C2

x1 + x2 = 202x1 + x2 = 32 subtract the equation

-x1 = -12

X1 = 12

Put the value of x1 in C1

x1 + x2 = 2012 + x2 = 20

X2 = 8

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Corners Z = 4x1 + 8x2

A (16, 0) 4 (16) + 8 (0) = 64B (12, 8) 4 (12) + 8 (8) = 112C (0, 20) 4 (0) + 8 (20) = 160D ( 0, 0) 4 (0) + 8 (0) = 0

Maximum Z = 160 at C (0, 20) where X1 = 0 and X2 = 20

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Q 23)

Maximize Z = 30x1 + 20x2

subject to 3x1 + x2 ≤ 18 x1 + x2 ≤ 12 x1 ≥ 2 x2 ≥ 5 x1, x2 ≥ 0

Find constraintsC1 = 3x1 + x2 = 18C1 = 3x1/18 + x2/18 = 1C1 = x1/6 + x2/18 = 1

C2 = x1 + x2 = 12C2 = x1/12 + x2 /12 = 1

C3 = x1 = 2C3 = x1/2 = 1

C4 = x2 = 5C4 = x2/5 = 1

Put (0,0)C1 = 3x1 + x2 ≤ 18C1 = 0 ≤ 18 (True)

C2 = x1 + x2 ≤ 12C2 = 0 ≤ 12 (True)

C3 = x1 ≥ 2C3 = 0 ≥ 2 (False)

C4 =x2 ≥ 5C4 =0 ≥ 5 (False)

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Point B intersecting C2 & C3

X1 + X2 = 12X1 = 2 subtract the equation

X2 = 10 X1 = 2

Point C intersecting C1 & C2

3x1 + x2 = 18 x1 + x2 = 12 subtract the equation 2x1 = 6

X1 = 3 X2 = 9

Point D intersecting C1 & C4

3x1 + x2 = 18Since X2 = 5 Putting the value of X2 in C1

3x1 + 5 = 18X1 = 13/3

Corners Z = 30x1 + 20x2

A (2, 5) 30 (2) + 20 (5) = 160B (2, 10) 30 (2) + 20 (10) = 260C (3, 9) 30 (3) + 20 (9) = 270D ( 13/3, 5) 30 (13/3) + 20 (5) = 230

Maximum Z = 270 at C (3, 9) where X1 = 3 and X2 = 9

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Q 33)Product A Product B Weekly Hour

Department 1 2 3 60Department 2 4 2 80Profit Margin $3 per unit $ 4 per unit

a) Formulate Linear Programing model for determining the Product Mix which maximize the Profit.

b) Solve using the corner point methodc) Fully interpret the results indicating the recommended Product Mix.

What percentage of daily capacity will be utilized in each department?

Solution:a)Maximize Z = 3x1 + 4x2

subject to 2x1 + 3x2 ≤ 60 4x1 + 2x2 ≤ 80 x1, x2 ≥ 0

b)Find constraintsC1 = 2x1 + 3x2 = 60C1 = 2x1/60 + 3x2/60 = 1C1 = x1/30 + x2/20 = 1

C2 = 4x1 + 2x2 = 80C2 = 4x1/80 + 2x2/80 = 1C2 = x1/20 + x2 /40 = 1

Put (0,0)C1 = 2x1 + 3x2 ≤ 60C1 = 0 ≤ 60 (True)

C2 = 4x1 + 2x2 ≤ 80C2 = 0 ≤ 80 (True)

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Point B intersecting C1 & C2

2x1 + 3x2 = 604x1 + 2x2 = 80 (Multiply Eq.1 by 4 and Eq.2 by 2 and then subtract the Equation)

8x1 + 12x2 = 2408x1 + 4x2 = 160 (subtract the equation)

8x2 = 80X2 = 10

Putting the value of X2 in C1

2x1 + 3x2 = 602x1 + 3(10) = 60

X1 = 15

Corners Z = 3x1 + 4x2

A (20, 0) 3 (20) + 4 (0) = 60B (15, 10) 3 (15) + 4 (10) = 85C (0, 20) 3 (0) + 4 (20) = 80D ( 0, 0) 3 (0) + 4 (0) = 0

b) Maximum Z = 85 at B (15, 10) where X1 = 15 and X2 = 10

c) Since profit will be maximized at $85/week; if 15 units of Product A and 10 units of Product B will be produced and sold. In addition to this weekly labor capacity of both the departments will be consumed in totality.

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Q) 35Food 1 Food 2 Min. daily Req.

Vitamin 1 2 mg/oz 3 mg/oz 18 mgVitamin 2 4 mg/oz 2 mg/oz 22 mgCost per oz $0.12 $0.15

a) Formulate Linear Programing model for determining the quantities of two foods which will minimize the cost of the meal while ensuring that at least minimum levels of both vitamins will be satisfied.

b) Solve ensuring corner-point method, indicating what the minimum-cost meal will consist of and its cost. What percentage of the minimum daily requirements for each vitamin will be realized?

Solution:a)Minimize Z = 0.12x1 + 0.15x2

subject to 2x1 + 3x2 ≥ 18 4x1 + 2x2 ≥ 22 x1, x2 ≥ 0

b) Find constraintsC1 = 2x1 + 3x2 = 18C1 = 2x1/18 + 3x2/18 = 1C1 = x1/9 + x2/6 = 1

C2 = 4x1 + 2x2 = 22C2 = 4x1/22 + 2x2/22 = 1

C2 = x1/112 + x2 /11 = 1

Put (0,0)C1 = 2x1 + 3x2 ≥ 18C1 = 0 ≥ 18 (False)

C2 = 4x1 + 2x2 ≥ 22

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C2 = 0 ≥ 22 (False)

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Point B intersecting C1 & C2

2x1 + 3x2 = 184x1 + 2x2 = 22 (Multiply Eq.1 by 4 and Eq.2 by 2 and then subtract the Equation)

8x1 + 12x2 = 728x1 + 4x2 = 44 (subtract the equation)

8x2 = 28X2 = 7/2

Putting the value of X2 in C1

2x1 + 3x2 = 182x1 + 3(7/2) = 18

X1 = 15/4

Corners Z = 0.12x1 + 0.15x2

A (9, 0) 0.12 (9) + 0.15 (0) = 1.08

B (154 ,72) 0.12 (15/4) + 0.15 (7/2) = 0.975

C (0, 11) 0.12 (0) + 0.15 (11) = 1.65

b) Minimum Z = 0.975 at B (154 ,72) where X1 = 15/4 or 3.75 and X2 = 7/2 or

3.5Since cost will be minimized at $0.975 per meal; if 3.75 ounces of food 1 and 3.5 ounces of food 2 are served. 100 % of Minimum daily requirement will be realized for both vitamins.

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