Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
12th Edition
ISBN: 9781259580093
Author: William J Stevenson
Publisher: McGraw-Hill Education
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Chapter 19, Problem 2P

a)

Summary Introduction

To solve: The linear programming problem and answer the questions.

Introduction:

Linear programming:

Linear programming is a mathematical modelling method where a linear function is maximized or minimized taking into consideration the various constraints present in the problem. It is useful in making quantitative decisions in business planning.

a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Minimize Z=1.80S+2.20TSubject to:5S+8T200gr(Potassium)15S+6T240gr(Carbohydrate)4S+12T180gr(Protein)T10gr(T)S,T0(Nonnegativity)

Calculation of coordinates for each constraint and objective function:

Constraint 1:

5S+8T200gr(Potassium)

Substituting S=0 to find T,

5(0)+8T=2008T=200T=2008T=25

Substituting T=0 to find S,

5S+8(0)=2005S=200S=2005S=40

Constraint 2:

15S+6T240gr(Carbohydrate)

Substituting S=0 to find T,

15(0)+6T=2406T=240T=2406T=40

Substituting T=0 to find S,

15S+6(0)=24015S=240S=24015S=16

Constraint 3:

4S+12T180gr(Protein)

Substituting S=0 to find T,

4(0)+12T=18012T=180T=18012T=15

Substituting T=0 to find S,

4S+12(0)=1804S=180S=1804S=45

Constraint 4:

T10gr(T)

Therefore T=10.

Objective function:

The problem is solved with iso-cost line method.

Let 1.80S+2.20T=99

Substituting S=0 to find T,

1.80(0)+2.20T=992.20T=99T=992.20T=45

Substituting T=0 to find S,

1.80S+2.20(0)=991.80S=99S=991.80S=55

Graph:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 19, Problem 2P , additional homework tip  1

(1) Optimal value of the decision variables and Z:

The coordinates for the cost line is (45, 55). The cost line is moved towards the origin. The lowest at which the cost line intersects in the feasible region will be the optimum solution. The following equation are solved as simultaneous equation to find optimum solution.

5S+8T=200 (1)

15S+6T=240 (2)

Solving (1) and (2) we get,

S=8,T=20

The values are substituted in the objective function to find the objective function value.

Minimize Z=1.80(8)+2.20(20)=14.40+44=58.40

Optimal solution:

S=8T=20Z=58.4

(2)

None of the constraints are having slack. All ≤ constraints are binding.

(3)

Protein and T constraint have surplus.

Protein:

4(8)+12(20)180gr32+240180272180

The surplus is 92 (272 – 180).

T:

T102010

The surplus is 10 (20– 10).

(4)

The protein constraint is redundant because, it does not intersect at any point in the feasible region.

b)

Summary Introduction

To solve: The linear programming problem and answer the questions.

Introduction:

Linear programming:

Linear programming is a mathematical modelling method where a linear function is maximized or minimized taking into consideration the various constraints present in the problem. It is useful in making quantitative decisions in business planning.

b)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Minimize Z=2x1+3x2Subject to:4x1+2x220(D)2x1+6x218(E)x1+2x212(F)x1,x20(Nonnegativity)

Calculation of coordinates for each constraint and objective function:

Constraint 1:

4x1+2x220(D)

Substituting x1=0 to find x2,

4(0)+2x2=202x2=20x2=202x2=10

Substituting x2=0 to find x1,

4x1+2(0)=204x1=20x1=204x1=5

Constraint 2:

2x1+6x218(E)

Substituting x1=0 to find x2,

2(0)+6x2=186x2=18x2=186x2=3

Substituting x2=0 to find x1,

2x1+6(0)=182x1=18x1=182x1=9

Constraint 3:

x1+2x212(F)

Substituting x1=0 to find x2,

(0)+2x2=122x2=12x2=122x2=6

Substituting x2=0 to find x1,

x1+2(0)=12x1=12x1=121x1=12

Objective function:

The problem is solved with iso-cost line method.

Let 2x1+3x2=24

Substituting x1=0 to find x2,

2(0)+3x2=243x2=24x2=243x2=8

Substituting x2=0 to find x1,

2x1+3(0)=242x1=24x1=242x1=12

Graph:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 19, Problem 2P , additional homework tip  2

(1) Optimal value of the decision variables and Z:

The coordinates for the cost line is (12, 8). The cost line is moved towards the origin. The lowest at which the cost line intersects in the feasible region will be the optimum solution. The following equation are solved as simultaneous equation to find optimum solution.

4x1+2x2=20 (1)

2x1+6x2=18 (2)

Solving (1) and (2) we get,

x1=4.2,x2=1.6

The values are substituted in the objective function to find the objective function value.

Minimize Z=2(4.2)+3(1.6)=8.4+4.8=13.2

Optimal solution:

x1=4.2x2=1.6Z=13.2

(2)

Constraint F is having slack as shown below.

(4.2)+2(1.6)124.2+3.2127.412

The slack is 4.6 (12 – 7.4).

(3)

There are no surplus. D and E constraints with ≥ are binding.

(4)

There are no redundant constraints

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