MAT 540 Homework Week 7: Linear Programming and Sensitivity Analysis

This document presents a comprehensive solution to a MAT 540 homework assignment focusing on linear programming and sensitivity analysis. The assignment addresses several scenarios, including resource allocation for product manufacturing (basketballs and footballs), production optimization across different assembly lines, and cotton and denim production. The solution provides optimal production quantities, maximized profit calculations, and analyses of sensitivity ranges for profit coefficients and resource constraints. Furthermore, the document includes a computer solution and a sensitivity report, which helps determine the impact of changes in resource availability and profit margins on the optimal solution. The analysis also considers government restrictions and Lagrange multipliers to understand the effects of changes in land and budget allocation. The provided solution uses the simplex method to solve linear programming problems and offers insights into the impact of changing variables on the outcomes.

Running head: HOMEWORK 1
MAT 540 Homework Week 7
Student Name
Institution

HOMEWORK 2
Question 1
a)
Resource Requirements per Unit
Product Rubber (lb.) Leather (ft2)
Basketball 2.8 3.7
Football 1.5 5.2
600 900 input
Constraints
Total resources
available 600 900
Basketball Football
Profits ($) 11 15
Basketball 196.45 Decision
variablesFootball 33.30
Maximized profits 2660.38
Objective
function
The optimal solution:
Number of basket balls = 196.45
Number of football = 33.30
Maximized profit = $2660.38
b)
Resource Requirements per Unit
Product Rubber (lb.) Leather (ft2)
Basketball 2.8 3.7
Football 1.5 5.2
600 900 input
Constraints
Total resources
available 600 900

HOMEWORK 3
Basketball Football
Profits ($) 12 15
Basketball 196.45 Decision
variablesFootball 33.30
Maximized profits 2856.83
Objective
function
There is no change in the optimal solution, but the objective function value or profit has
increased due to an increase in unit profit of basketball
c) (i)
Resource Requirements per Unit
Product Rubber (lb.) Leather (ft2)
Basketball 2.8 3.7
Football 1.5 5.2
681.0810811 900
Total resources
available 1000 900
Basketball Football
Profits ($) 11 15
Basketball 243.24
Football 0.00
Maximized
profits 2675.68
After adding 400 pounds of rubber, the optimal solution and function values changed.
Extra 81 units of rubber were produced and the profits increased by $415.3
C(ii)
Resource Requirements per Unit
Product Rubber (lb.) Leather (ft2)
Basketball 2.8 3.7

HOMEWORK 4
Football 1.5 5.2
600 1500
Total resources
available 600 1500
Basketball Football
Profits ($) 11 15
Basketball 96.56
Football 219.76
Maximized
profits 4358.49
The optimal solution and objective function values are changed. The profit is maximized
by two times.
Question 2
a)
Hours/ Unit
Product Line 1 Line2
A 11 5
B 6 9
65 40 Constraints
Total Hours 65 40
A B
Profits ($) 11 15
Product A 5 decision variables
Product B 1.666666667
Maximized profits 80.00 Objective function
b)

HOMEWORK 5
Sensitivity range
Product A: 11
Product B: 15
c)
Shadow price for additional hours
Product A: 0.347826087
Product B: 1.434782609
The company would prefer line 1 for 2 hours since it is more economical than line 2
Question 3
a)
The standard form of an LP model:
Maximize: Z = 3.25x1+2.5x2
Subject to:
8 x1+6 x2 ≤ 6500 (Available cotton)
4x1+3x2 ≤ 3000 (processing time)
Corduroy denim
Profits for each
product 3.25 2.5
Resources
Available
Resources
Cotton 8 6 6000 <= 6500
Labor 4 3 3000 <= 3000
Demands
Corduroy 0 <= 510
denim 1000 <=
100000
0
Maximized profits 2500

HOMEWORK 6
The demand for corduroy is not met. There were zero demands for corduroy.
Extra cotton left = 500
Processing time left = 0
b)
Irwin mills should obtain additional cotton 500 pounds of cotton
Question 4
a)
Let x be the acres of corn
Let y be acres of tobacco
Maximize
Z= 300x+250y
St.
x+y ≤ 410
105x+210y≤ 52500
y<100
x,y≥ 0
Corn Tobacco Resources
Available
300 520
Land 1 1 410 <= 410
Budget 105 210 52500 <= 52500
Govt. Restriction 0 1 90 <= 100
X1 X2
Decision
Variables 320 90
Objective
function 142800

HOMEWORK 7
The optimal solution:
Number of acres of corn = 320
Number of acres of tobacco = 90
Maximized profit = $142800
b)
Acres of farmland will not be cultivated at the optimal solution = 0
Bradleys does not use the entire 100-acre tobacco allotment, i.e. (100-90) = 10 acres
c)
No, Bradley should not lease at 110$. Based on the sensitivity report, Lagrange multiplier
is 80 for land. It means, with each unit increase of land, Profit will increase by 80$. Thus,
the maximum price should be 80$ per acre. In a sensitivity report, the lagrange multiplier
show the amount of profit increase for each unit increment (Taha,2017; Yadav & Malik,
2014).
Microsoft Excel 15.0 Sensitivity
Report
Worksheet: [MAT540HomeWorkweek7answersheet.xlsx]P4
Report Created: 5/19/2019 9:47:41
AM
Variable Cells
Final Reduced Objective
Allowabl
e
Allowabl
e
Cell Name
Valu
e Cost
Coefficie
nt Increase
Decreas
e
$C$1
2
Decision Variables
X1 320 0 300 220 40
$D$1
2
Decision Variables
X2 90 0 520 80 220
Constraints
Final Shadow
Constrain
t
Allowabl
e
Allowabl
e

HOMEWORK 8
Cell Name
Valu
e Price R.H. Side Increase
Decreas
e
$E$7 Land 410 80 410 90 10
$E$8 Budget
5250
0
2.09523809
5 52500 1050 9450
$E$9 Govt. Restriction 90 0 100 1E+30 10
d)
Based on the LaGrange multiplier for Budget, with each unit increase in Budget, Profit
will increase by 1.1$.

HOMEWORK 9
References
Taha, H. (2017). Operations Research (8th ed., pp. 82-146). Harlow, United Kingdom:
Pearson Education Limited.
Yadav, S., & Malik, A. (2014). Operations research (1st ed., pp. 80-145). New Delhi:
Oxford university press.