1.
Types of Cotton
Raw Cotton Per yard
Denim
Processing Time
Profit
5
3
$2.25 per yard
7.5
3.2
$3.10 per yard
6500 lb/month
3000 hr/month
Corduroy
Maximum Demand for Corduroy= 510 yard/ month
A. Decision Variables
x1= Number of Denim
x2= Number of Corduroy
Objective Function
Maximize Z= 2.25x1+3.10x2
Subject to;
5x1 + 7.5x2 ≤ 6500
3x1 + 3.2x2 ≤ 3000
X2 ≤ 510
X1,x2 ≥ 0
B. Standard Form
Maximize Z=
2.25x1+3.10x2+0S1+0S2+0S3
Subject to;
5x1 + 7.5x2 + 0S1 = 6500
3x1 + 3.2x2 +0S2= 3000
X2 + 0S3 = 510
X1,x2 ≥ 0
C. Graphical Computation
5x1 + 7.5x2 = 6500
Let x1=0, 5(0) + 7.5x2 = 6500
x2 = 6500/ 7.5
x2= 2600/3
3x1 + 3.2x2 = 3000
Let x1=0, 3(0)+ 3.2x2 = 3000
x2 = 3000/3.2
x2= 1875/2
Let x2=0, 5x1+ 7.5(0)= 6500
x1 = 6500/5
x2= 1300
Let x2=0, 3x1 + 3.2(0)=3000
x1 = 3000/3
x2= 1000
X2 = 510, x1=0
Corner Points
x1
x2
Profit
Max. Z= 2.25x1+3.10x2
A
0
510
$1,581
B
0
2600/3
$2,686.6667
C
3400/13
9000/13
$ 2.734.6154
D
465
510
$ 2,627.25
Therefore, 3400/13 or 261. 5385 yards of Denim and 9000/13 or 692.3077 yards of Corduroy
must produce to maximize profit with $ 2,734. 6154.
I.
Extra Cotton Constraint:
5x1 + 7.5x2 = 6500
5(456)+7.5(510)= 6500
S1= 6500- 6105
S1= 395 pounds of cotton left over
Demand for Corduroy:
X2 ≤ 510
510=510
Therefore, the demand for corduroy is met
Processing Time Left:
3x1 + 3.2x2 ≤ 3000
3(456) + 3.2(510) =3000
S2= 3000-3000
S2= 0 - no left over processing
time
II.
Corner Points
x1
x2
Profit
Max. Z= 3x1+3.10x2
A
0
510
$1,581
B
0
2600/3
$2,686.6667
C
3400/13
9000/13
$ 2,930.7692
D
465
510
$ 2,976
Therefore, increasing profit per yard of denim from $2.25 to $3.00 will change the optimal
solution, due to the sensitivity analysis's allowed increase is lesser than the increase in the
objective function coefficient. The Irwin Textile Mills must produce 465 yards of Denim and 510
yards of Corduroy to maximize profit with $ 2,976.
Corner Points
x1
x2
Profit
Max. Z= 2.25x1+4x2
A
0
510
$2,040
B
0
2600/3
$3,466.6667
C
3400/13
9000/13
$ 3,357.6923
D
465
510
$ 3,086.25
Therefore, increasing profit per yard of Corduroy from $3.10to $4.00 will change the optimal
solution, due to the sensitivity analysis's allowed increase is lesser than the increase in the
objective function coefficient. The Irwin Textile Mills must not produce yards of Denim and
2600/3 or 866.6667 yards of Corduroy to maximize profit with $ 3,466.6667.
III.
Types of Cotton
Denim
Corduroy
Raw Cotton Per yard
Processing Time
Profit
5
3
$2.25 per yard
7.5
3.2
$3.10 per yard
6000 lb/month
3000 hr/month
Maximum Demand for Corduroy= 510 yard/ month
A. Decision Variables
x1= Number of Denim
x2= Number of Corduroy
Objective Function
Maximize Z= 2.25x1+3.10x2
Subject to;
5x1 + 7.5x2 ≤ 6000
3x1 + 3.2x2 ≤ 3000
X2 ≤ 510
X1,x2 ≥ 0
B. Standard Form
Maximize Z=
2.25x1+3.10x2+0S1+0S2+0S3
Subject to;
5x1 + 7.5x2 + 0S1 ≤ 6000
3x1 + 3.2x2 +0S2≤ 3000
X2 + 0S3 ≤ 510
X1,x2 ≥ 0
C. Graphical Computation
5x1 + 7.5x2 = 6500
Let x1=0, 5(0) + 7.5x2 = 6000
x2 = 6000/ 7.5
x2= 800
3x1 + 3.2x2 = 3000
Let x1=0, 3(0)+ 3.2x2 = 3000
x2 = 3000/3.2
x2= 1875/2
Let x2=0, 5x1+ 7.5(0)= 6000
x1 = 6000/5
x2= 1200
Let x2=0, 3x1 + 3.2(0)=3000
x1 = 3000/3
x2= 1000
X2 = 510, x1=0
Corner Points
x1
x2
Profit
Max. Z= 2.25x1+3.10x2
A
0
510
$1,581
B
0
800
$2,480
C
435
510
$ 2,559.75
Therefore, decreasing pounds of cotton per month from 6500 to 6000 will change the optimal
solution, due to the allowed decrease in the sensitivity analysis being lower that the decrease of
the constraint on the right hand side. The Irwin Textile Mills must produce 435 yards of Denim
and 510 yards of Corduroy to maximize profit with $ 2,559.75.
D.
If Irwin Mills has the opportunity to acquire extra resources, additional processing time would be
the optimal choice. The sensitivity analysis indicates that the constraints can be increased the
processing time to 237, the result would be an additional profit of 177.75.
II. Sensitivity Ranges
objective function coefficients:
0≤c1≤2.91
2.6≤c2≤ ꝏ
Constraint; 6105≤b1≤ ꝏ
1632≤ b2≤ 3237
0≤ b3≤ 692.32
Range for corduroy = (Lower Bound, Upper Bound)= (0; 692.31)
The range for corduroy implies that within the range of 0 and 692.31 it will increase the profit by $0.7
due to the additional increase in demand
2.
A.
Objective Function
Minimize cost ($): 40x1 +35X2
Subject to constraints :
2x1+ 2X2 ≥ 12
5x1 + 3X2 ≥ 15
x,y ≥ 0
Decision Variables
x= equals the quantity of vitamin in mix 1
y= equals the quantity of vitamin in mix 2
B.
I.
Determine an optimal solution using Excel Solver or another software.
II.
Propose a graphical solution.
2x1+ 2X2 =12
Let x1=0, 2(0) + 2x2 = 12
x2 = 12/2
x2= 6
5x1 + 3x2 =15
Let x1=0, 5(0) + 3x2 = 15
x2 = 15/3
x2= 5
Let x2=0, 2x1+ 2(0) = 12
x1 = 12/2
x1= 6
Let x2=0, 5x1 + 3(0)=15
x1 = 15/5
x1= 3
III.
Find out the corner point solution.
Corner Points
x1
x2
A
3
0
B
0
5
C
0
6
D
6
0
C.
II & III.
Objective Function
Minimize cost ($):Z= 40x1 +35x2
Subject to constraints :
2x1+ 2X2 ≥ 12
5x1 + 3X2 ≥ 9
x,y ≥ 0
Graphical Computation
2x1+ 2X2 =12
Let x1=0, 2(0) + 2x2 = 12
x2 = 12/2
x2= 6
Let x2=0, 2x1+ 2(0) = 12
x1 = 12/2
x1= 6
x2 = 9/3
x2= 3
5x1 + 3x2 =9
Let x1=0, 5(0) + 3x2 = 9
Let x2=0, 5x1 + 3(0)=9
x1 = 9/5
Corner Points
x1
x2
Cost
Min. Z=40x1 +35X2
A (infeasible)
3
0
$120
B (infeasible)
0
5
$175
C
0
6
$210
D
6
0
$240
Therefore, reducing Vitamin C from 15 to 9 units will not change the optimal solution. The
manager must not produce Vitamin mix 1 and 6 Vitamin mix 2 to minimize cost with $ 210.