SMU MBA II Sem Operation Research May 2014

Q.       A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper in a week. There are 160 production hours in a week. It requires 0.20 and 0.40 hours to produce a ton of grade X and Y papers. The mill earns a profit of Rs. 200 and Rs. 500 per ton of grade X and Y paper respectively. Formulate this as a Linear Programming

Let x1 for paper X and x2 for paper Y

Objective Function (Z): Profit of both X and Y Paper is given.

Max (Z) = 200×1+500×2

Constrains: two, one for production hours and second for raw material;

Subject to:

X1<=400

X2<=300

0.20×1+0.40×2<=160

x1, x2 >=0

Q.     A Company produces 150 cars. But the production rate varies with the distribution.

Production Rate 147 148 149 150 151 152 153
Probability 0.05 0.10 0.15 0.20 0.30 0.15 0.05

At present the track will hold 150 cars. Using the following random numbers determine the average number of cars waiting for shipment in the company and average number of empty space in the truck. Random Numbers 82, 54, 50, 96, 85, 34, 30, 02, 64, 47.

Solution:

Production Rate Probability Cumulative probability RN Range
147 0.05 0.05 00 –04
148 0.10 0.15 05 – 14
149 0.15 0.30 15-29
150 0.20 0.50 30 – 49
151 0.30 0.80 50 – 79
152 0.15 0.95 80 – 94
153 0.05 1.00 95 – 99

Simulation for 10 days using the given random numbers

Days RN Production Rate Car Waiting Space in the truck
1 82 152 2
2 54 150 –
3 50 150 –
4 96 153 3
5 85 152 2
6 34 150 –
7 30 150 –
8 02 147 – 3
9 64 151 1
10 47 150 2

Total                                                                                                            8                      3

Therefore, Avg number of cars waiting =8/10= 0.8 /day

Avg number of empty space =3/10= 0.3/day

Q.     Find an optimal solution to an assignment problem with the following cost matrix:

J1 J2 J3 J4

M1 10 9 7 8

M2 5 8 7 7

M3 5 4 6 5

M4 2 3 4 5

To illustrate this consider the following assignment problem
J1 J2 J3 J4
M1 10 9 7 8
M2 5 8 7 7
M3 5 4 6 5
M4 2 3 4 5
The solution is as follows.
First, the minimum element in each row is subtracted from all the elements in that row.
This gives the following reduced-cost matrix.
J1 J2 J3 J4
M1 3 2 0 1
M2 0 3 2 2
M3 1 0 2 1
M4 0 1 2 3

Since both the machines M2 and M4 have a zero cost corresponding to job J1 only, a
feasible assignment using only cells with zero costs is not possible. To get additional
zeros subtract the minimum element in the fourth column from all the elements in that
column. The result is shown in the table below.

J1 J2 J3 J4
M1 3 2 0 0
M2 0 3 2 1
M3 1 0 2 0
M4 0 1 2 2

Only three jobs can be assigned using the zero cells, so a feasible assignment is still not
possible. In such cases, the procedure draws a minimum number of lines through some
selected rows and columns in such a way that all the cells with zero costs are covered by
these lines. The minimum number of lines needed is equal to the maximum number of
jobs that can be assigned using the zero cells. 65
In the current example, this can be done with three lines as follows.
J1 J2 J3 J4
M1 3 2 0 0
M2 0 3 2 1
M3 1 0 2 0
M4 0 1 2 2

Now select the smallest element, which is not covered by the lines. In the current
example, it is 1. Subtract this number from all the elements, which are not covered. Then
add this number to all those covered elements that are at the intersection of two lines.
This gives the following reduced cost matrix.
J1 J2 J3 J4
M1 4 2 0 0
M2 0 2 1 0
M3 2 0 2 0
M4 0 0 1 1

A feasible assignment is now possible and an optimal solution is assigned
M1 J3
M2 J1
M3 J4
M4 J2
The total cost is given by: 7 +5+ 5+ 3 = 20
An alternate optimal solution is:
M1 J3
M2 J4
M3 J2
M4 J1
In case a feasible set could not be obtained at this step, one has to repeat the step of
drawing lines to cover the zeros and continue until a feasible assignment is obtained.

MBA IInd (2nd) Operation Research Solved Assignment 2014

For complete solution you can go to:
http://www.researchingtheworld.com/category/smu-assignment-2/smu-assignment/

Solution 2b)

 

Let us variables X1 & X2 respectively for black & white

The profit per unit on these products in Rs. 50 & Rs. 40 respectively

So this is the case for profit maximisation

Max(Z) = 50*X1 + 40*X2

Total capacity = 50,000 man hours

Taken time to produced one unit of black = 3 hours

Taken time to produced one unit of white = 2 hours

3*X1 + 2*X2 ≤ 50,000

Market max demand for black = 8,000 units

Market max demand for white = 10,000 units

X1 ≤ 8,000

X2 ≤ 10,000

Solution 3b)

1)    Identify the smallest and the next to smallest costs. Determine the difference between them for each row & column

2)    Identify the row or column with the largest difference among all the rows and columns.

Factories

C1

C2

C3

C4

Supply

F1

3

2(40)

7

6

50-40

1

F2

7

5

2

3

60

1

F3

2

5

4

5

25

2

Requirements

60

40

20

15

1

3

2

2

Factories

C1

C2

C3

C4

Supply

F1

3(10)

2(40)

7

6

10

1

3

F2

7

5

2

3

60

1

1

F3

2

5

4

5

25

2

2

Requirements

60-10

40

20

15

1

3

2

2

1

X

2

2

Factories

C1

C2

C3

C4

Supply

F1

3(10)

2(40)

7

6

10

1

3

x

F2

7

5

2

3

60

1

1

1

F3

2(25)

5

4

5

25

2

2

2

Requirements

50-25

40

20

15

1

3

2

2

1

X

2

2

5

X

2

2

Factories

C1

C2

C3

C4

Supply

F1

3(10)

2(40)

7

6

10

1

3

x

F2

7(25)

5

2

3

60-25

1

1

1

F3

2(25)

5

4

5

25

2

2

2

Requirements

25

40

20

15

1

3

2

2

1

x

2

2

5

x

2

2

Factories

C1

C2

C3

C4

Supply

F1

3(10)

2(40)

7

6

10

1

3

x

F2

7(25)

5

2(20)

3

35-20

1

1

1

F3

2(25)

5

4

5

25

2

2

2

Requirements

25

40

20

15

1

3

2

2

1

x

2

2

5

x

2

2

Factories

C1

C2

C3

C4

Supply

F1

3(10)

2(40)

7

6

10

1

3

x

F2

7(25)

5

2(20)

3(15)

15

1

1

1

F3

2(25)

5

4

5

25

2

2

2

Requirements

25

40

20

15

1

3

2

2

1

x

2

2

5

x

2

2

Total cost        = 3*10 + 2*40 + 7*25 + 2*20 + 3*15 + 2*25

= 30 + 80 + 175 + 40 + 45 + 50

= 420

Solution 4b)

1) 1)    Check the number rows equal to the number columns

Rows               = 4

Columns          = 4

Both rows & columns are equal, so move to next steps

2)  2) Row reduced matrix –

J1 J2 J3 J4
M1 10-7 9-7 8-7 7-7
M2 3-3 4-3 5-3 6-3
M3 2-1 1-1 1-1 1-1
M4 4-3 3-3 5-3 6-3

J1 J2 J3 J4
M1 5 2 1 0
M2 0 1 2 3
M3 1 0 0 0
M4 1 0 2 3

3)3)    Column reduced matrix –

J1 J2 J3 J4
M1 5 2 1 0
M2 0 1 2 3
M3 1 0 0 0
M4 1 0 2 3

444)    Assign zero row-wise if there is only one zero in the row and cross (X) or cancel other zeros in that column.

J1 J2 J3 J4
M1 5 2 1 [0]
M2 [0] 1 2 3
M3 1 (x) [0] (x)
M4 1 [0] 2 3

Since the number of assignments is 4 which is equal to the number of rows

M1 to J4          =>        7

M2 to J1          =>        3

M3 to J3          =>        1

M4 to J2          =>        3

Total cost       =>        14

Therefore, the optimum assignment schedule is M1 to J4, M2 to J1, M3 to J3, M4 to J2;

SMU Solved Assignment For MBA (2013-2014): Statistics (Numerical Part only)

For complete solved assignment in clear image please click:

http://www.researchingtheworld.com/smu-assignment-2/smu-assignment-statistics/smu-mba-assignment-statistics-numerical-part-only/

Q.2 (b) In a sample of 120 workers in a factory, the mean and standard deviation of wages
were Rs. 11.35 and Rs.3.03 respectively. Find the percentage of workers getting wages
between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally
distributed.

Q.4 (b) Calculate Karl Pearson’s coefficient of correlation between X series and Y series.
X 110 120 130 120 140 135 155 160 165 155
Y 12 18 20 15 25 30 35 20 25 10

Q.6 Construct Fisher’s Ideal Index for the given information and check whether Fisher’s
formula satisfies Time Reversal and Factor Reversal Tests.
Items  P0 Q0 P1 Q1
A            16 5 20 6
B             12 10 18 12
C             14 8 16 10
D            20 6 22 10
E            80 3 90 5
F            40 2 50 5
Formula of Fishers Ideal Index
Computation of Fisher’s Ideal Index
Fisher’s formula satisfies Time Reversal Test
Fisher’s formula satisfies Factor Reversal Test