BBUS 2302: Quantitative Analysis for Management Assignment Solution

This document provides a comprehensive solution to a Quantitative Analysis for Management assignment. The solution addresses two main problems. The first problem involves graphically solving a linear programming problem to maximize profit, including determining the optimal solution and analyzing the impact of changing a constraint. The second problem tackles an assignment problem, using the Hungarian algorithm to minimize the total time for job assignments across different machines. The solution includes detailed steps, calculations, and the final optimal assignments, as well as references to external resources like Excel Easy and Study.com for further understanding of the methods used.

3)
Minimize,
10 x1 A +4 x1 B +11 x1C +12 x2 A + 5 x2 B +8 x2 C +9 x3 A +7 x3 B +6 x3 C
Subject to:
x1 A +x1 B + x1C ≤70
x2 A +x2 B +x2C ≤ 50
x3 A +x3 B + x3 C ≤ 30
x1 A +x2 A +x3 A =40
x1 B + x2 B +x3 B=50
x1 C+x2 C + x3 C=60
xij ≥ 0 for all i∧ j
Report Created: 08-04-2019 15:46:46
Result: Solver has converged to the current solution. All Constraints are satisfied.
Solver Engine
Engine: GRG Nonlinear
Solution Time: 0.578 Seconds.
Iterations: 13 Subproblems: 0
Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001
Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Central
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume
NonNegative
Objective Cell (Min)
Cell Name Original Value Final Value
$K$16 unit used 0 1040.000011
Variable Cells
Cell Name Original Value Final Value Integer
$B$4:$J$4
$B$4 x1A 0 38.88888921 Contin
$C$4 x1B 0 31.11111141 Contin
$D$4 x1C 0 0 Contin
$E$4 x2A 0 0 Contin
$F$4 x2B 0 18.88888921 Contin
$G$4 x2C 0 31.11111141 Contin

$H$4 x3A 0 1.111111252 Contin
$I$4 x3B 0 0 Contin
$J$4 x3C 0 28.88888921 Contin
Constraints
Cell Name Cell Value Formula Status
Slac
k
$K$10 Constraint 4 used 40.00000046 $K$10=$M$10 Binding 0
$K$11 Constraint 5 used 50.00000061 $K$11=$M$11 Binding 0
$K$12 Constraint 6 used 60.00000061 $K$12=$M$12 Binding 0
$K$7 Constraint1 used 70.00000061 $K$7<=$M$7 Binding 0
$K$8 Constraint 2 used 50.00000061 $K$8<=$M$8 Binding 0
$K$9 Constraint 3 used 30.00000046 $K$9<=$M$9 Binding 0
x1A x1B x1C x2A x2B x2C x3A x3B x3C
38.88
889
31.11
111
0 0 18.88
889
31.11
111
1.111
111
0 28.88
889
used Avail
able
Constr
aint1
1 1 1 70 <
=
70
Constr
aint 2
1 1 1 50 <
=
50
Constr
aint 3
1 1 1 30 <
=
30
Constr
aint 4
1 1 1 40 = 40
Constr
aint 5
1 1 1 50 = 50
Constr
aint 6
1 1 1 60 = 60
unit 10 4 11 12 5 8 9 7 6 1040
Final Lagrange
Cell Name Value
Multiplie
r
$K$1 Constraint 4 40.0000004 0

0 used 6
$K$1
1
Constraint 5
used
50.0000006
1 -6
$K$1
2
Constraint 6
used
60.0000006
1 -3
$K$7
Constraint1
used
70.0000006
1 10
$K$8
Constraint 2
used
50.0000006
1 11
$K$9
Constraint 3
used
30.0000004
6 9
The yellow highlighted portion shows the optimal solution of the variables and the least cost
incurred is 1040 highlighted in Red. (Microsoft, n.d.)
1)
Objective Function: Maximize
z=8 x1 +5 x2
Constraint 1 drawn x1+ x2 ≤10
Consider it as x1+ x2=10
if x1=0
⇒ x2=10
if x2=0
⇒ x1=10
x1 0 10
x2
1
0 0
Constraint 2 drawn 2 x1 ≤ 6
Consider it as 2 x1 =6
then, x1=3

Here line is parallel to Y-axis
x1 3 3
x2 0 1
Extreme point coordinates (
x1 , x2 ¿
Lines through Extreme point Objective function value
z=8 x1 +5 x2
O (0, 0) 3 → x1 ≥0
4 → x2 ≥ 0
8(0) + 5 (0) = 0
A (3, 0) 2 →2 x1 ≤ 6
4 → x2 ≥ 0
8(3) + 5(0) = 24
B (3, 7) 1 → x1+ x2 ≤10
2 →2 x1 ≤ 6
8(3) + 5(7) = 59
C (0, 10) 1 → x1+ x2 ≤10
3 → x1 ≥0
8(0) + 5(10) = 50

The maximum value of the objective function z = 59 occurs at the extreme point (3, 7).
Hence, the optimal solution to the given LP problem is:
x1=3 , x2 =7∧z=59
2)
This is the original cost matrix:
10 14 16 13
12 13 15 12
9 12 12 11
14 16 18 16
Subtract row minima for each of the row and the matrix becomes
0 4 6 3 (-10)
0 1 3 0 (-12)
0 3 3 2 (-9)
0 2 4 2 (-14)
Subtract column minima from each of the column and the matrix becomes
0 3 3 3
0 0 0 0
0 2 0 2
0 1 1 2
(-1) (-3)
Cover all zeros with a minimum number of lines, 3 lines are required to cover all zeros

0 3 3 3
0 0 0 0 x
0 2 0 2 x
0 1 1 2
x
Create additional zeros
Since the number of lines is smaller than 4and the smallest number which is not covered is 1, so
we are going to subtract this smallest number from the uncovered portion and add the smallest
number to those elements in the array where the lines covers the elements twice so the matrix
becomes
0 2 2 2
1 0 0 0
1 2 0 2
0 0 0 1
Again, cover all zeros with a minimum number of lines, now 4 lines are required
0 2 2 2 x
1 0 0 0 x
1 2 0 2 x
0 0 0 1 x
The matrix now becomes,

0 2 2 2
1 0 0 0
1 2 0 2
0 0 0 1
Considering the corresponding cost matrix and taking the jobs for each machines corresponding to
zeros we get the following assignments
10 14 16 13
12 13 15 12
9 12 12 11
14 16 18 16
Machine W – A12
Machine X – B9
Machine Y – B2
Machine Z – A15
The optimal value equals 50.
(Gu, 2019)
References
Gu, A. (2019, 04). Using the Hungarian Algorithm to Solve Assignment Problems. Retrieved from
Study.com: https://study.com/academy/lesson/using-the-hungarian-algorithm-to-solve-
assignment-problems.html
Microsoft. (n.d.). Solver. Retrieved from Excel Easy:
https://www.excel-easy.com/data-analysis/solver.html