sensitivity analysis-case ii
TRANSCRIPT
-
8/11/2019 Sensitivity Analysis-Case II
1/15
Case 2
Knitting Department
Submitted to- Dr. Hitesh Arora
By-
Abhijeet Agarwal 231005
Abhinav Joshi 231007
Atin Garg 231036
-
8/11/2019 Sensitivity Analysis-Case II
2/15
X1= Units of Product 1
X2 = Units of Product 2
X3 = Units of Product 3
Maximize Z= 7x1 + 3x2 + 9x3 (Profit)
S.t.
Labour 4x1 + 5x2 + 6x3
-
8/11/2019 Sensitivity Analysis-Case II
3/15
7 3 9 0 0 0
B.V Cb X1 X2 X3 S1 S2 S3 Sol. Ratio
S1 0 4 5 6 1 0 0 360 60
S2 0 2 4 6 0 1 0 300 50
S3 0 9 5 6 0 0 1 600 100
Cj-Zj 7 3 9 0 0 0
B.V Cb X1 X2 X3 S1 S2 S3 Sol. Ratio
S1 0 2 1 0 1 -1 0 60 30
X3 9 1/3 2/3 1 0 1/6 0 50 150
S3 0 7 1 0 0 -1 1 300 300/7
Cj-Zj 4 -3 0 0 -3/2 0
-
8/11/2019 Sensitivity Analysis-Case II
4/15
7 3 9 0 0 0
B.V Cb X1 X2 X3 S1 S2 S3 Sol. Ratio
X1 7 1 0 -1/2 0 30 -
X3 9 0 1 -1/6 1/3 0 40 120
S3 0 0 -5/2 0 -7/2 5/2 1 90 36
Cj-Zj 0 -5 0 -2 0
B.V Cb X1 X2 X3 S1 S2 S3 Sol.
X1 7 1 0 0 -1/5 0 1/5 48
X3 9 0 5/6 1 3/10 0 -2/15 28
S2 0 0 -1 0 -7/5 -1 2/5 36
Cj-Zj 0 -9/2 0 -13/10 0 -1/5
-
8/11/2019 Sensitivity Analysis-Case II
5/15
Maximize Z 588
Units of Product I 48
Units of Product II 0
Units of Product III 28
Labor Usage 360
-
8/11/2019 Sensitivity Analysis-Case II
6/15
-
8/11/2019 Sensitivity Analysis-Case II
7/15
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$E$4 x1 48 0 7 6.5 1
$E$5 x2 0 -4.5 3 4.5 1E+30
$E$6 x3 28 0 9 1.5 4.333333333
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$D$8 Labor Usage x3 360 1.3 360 25.71428571 93.33333333
$D$9
Machine Usage
x3 264 0 300 1E+30 36
$D$10
Material Usage
x3 600 0.2 600 210 90
-
8/11/2019 Sensitivity Analysis-Case II
8/15
-
8/11/2019 Sensitivity Analysis-Case II
9/15
-
8/11/2019 Sensitivity Analysis-Case II
10/15
As P1 is a basic variable,so for basic variable
Maxykj>0{Cj-Zj /ykj}
-
8/11/2019 Sensitivity Analysis-Case II
11/15
As allowable increase is 6.5 so increasing by 3would not impact the optimal solution
But as coefficient of x1 will change
Hence new Z=732
Which is shown by excel solver.
Q4-What would the values in the optimal solution be if
the objective coefficient of x1 were to increased by 3?
-
8/11/2019 Sensitivity Analysis-Case II
12/15
For labour constraint
Maxbik>0{-Xbi / bik}
-
8/11/2019 Sensitivity Analysis-Case II
13/15
Q6- What would the values in the optimal solution be if
the amount of labor available decreased by 10 hours?
From the excel solver solution
Z= 575
x1=50x2=0
X3=25
-
8/11/2019 Sensitivity Analysis-Case II
14/15
For material constraint
Max bik>0{-Xbi / bik}
-
8/11/2019 Sensitivity Analysis-Case II
15/15