ANSAL INSTITUTE OF TECHNOLOGY & MANAGEMENT Sushant Golf City, Lucknow

Course Code: NMBA-025 (Assignment) Course Title: Opration Research Instructor: Dr. V.K. Chaubey Semester and Session: MBA-IInd Semester Submitted By: Sarvesh Kumar Roll No.: 1374870038 Date of Submission: 06/04/2014

ASSIGNMENT-IV

A1.

Job Person 1 2 3 4

A 20 25 22 28 B 15 18 23 17 C 19 17 21 24 D 25 23 24 24

Find the minimum cost

According to the rules, selecting the smallest element in each row and subtracting it all the corresponding element of that row,

Job Person 1 2 3 4

A 0 5 2 8 B 0 3 8 2 C 2 0 4 7 D 2 0 1 1

Selecting the smallest element in each column and subtracting it all the corresponding element of that column

Job Person 1 2 3 4

A 0 5 1 7 B 0 3 7 1 C 2 0 3 6 D 2 0 0 0

The procedure of making assignment is a follows:-

Job Person 1 2 3 4

A 0 5 1 7 B 0 3 7 1 C 2 0 3 6 D 2 0 0 0

Selecting a smallest element which is not covered by horizontal and vertical line, and subtracting it the entire element, which are not covered by horizontal and vertical line the element which is covered by horizontal and vertical line are remained same and intersection of if smallest element will be added.

Job Person 1 2 3 4

A 0 5 0 6 B 0 3 6 0 C 2 0 2 5 D 3 1 0 0

Minimum cost is

Person Job Minimum

Cost A 1 2 B 4 17 C 2 17 D 3 24

78

A2. Solve following problem and find minimum cost

Job Person 1 2 3 4

A 20 25 22 28 B 15 18 23 17 C 19 17 21 24

Since above cost matrix is not square in this matrix 3 rows and 4 column so make a square matrix we add dummy row which all cost values are zero then we have,

Job Person 1 2 3 4

A 20 25 22 28 B 15 18 23 17 C 19 17 21 24 D 0 0 0 0

Selecting smallest element in each row and subtracting it all the corresponding element of that row,

Job Person 1 2 3 4

A 0 5 2 8 B 0 3 8 2 C 2 0 4 7 D 0 0 0 0

In Case of column the above matrix will be same the procedure of making assignment is as follows:-

Job Person 1 2 3 4

A 0 5 2 8 B 0 3 8 2 C 2 0 4 7 D 0 0 0 0

Selecting a smallest element which is not covered by horizontal and vertical line and subtracting it the entire element which are not covered by horizontal and vertical line. The element which is covered by horizontal and vertical line is remained same and in intersection of it smallest element will be added.

Job Person 1 2 3 4

A 0 5 0 6 B 0 3 6 0 C 2 0 2 5 D 2 2 0 0

Minimum Cost is:-

Person Job Minimum

Cost A 1 20 B 4 17 C 2 18 D 3 0

55

A3. Solve the minimum total cost

Person Counter A B C D E

1 30 37 40 28 40 2 40 24 27 21 36 3 40 32 33 30 35 4 25 38 40 36 36 5 29 62 41 34 39

Selecting the smallest element in each row and subtracting it all the corresponding element of that row,

Person Counter A B C D E

1 2 9 12 0 12 2 19 3 6 0 15 3 10 2 3 0 5 4 0 13 15 11 11 5 0 33 12 5 10

Selecting the smallest element in each column and subtracting it all the corresponding element of that column

Person Counter A B C D E

1 2 7 9 0 7 2 19 1 3 0 10 3 10 0 0 0 0 4 0 11 12 11 6 5 0 31 9 5 5

The procedure of making assignment is as follows:-

Person Counter A B C D E

1 2 7 9 0 7 2 19 1 3 0 10 3 10 0 0 0 0 4 0 11 12 11 6 5 0 31 9 5 5

Selecting a smallest element which not covered by horizontal and vertical line and subtracting it all the element which are not covered by horizontal and vertical line. The element which is not covered by horizontal and vertical line is remained same and in intersection of it smallest element will be added.

Person Counter A B C D E

1 2 6 8 0 6 2 19 0 2 0 9 3 11 0 0 1 0 4 0 10 11 10 5 5 0 30 8 5 4

Apply above procedure:-

Person Counter A B C D E

1 2 6 6 0 4 2 9 0 0 0 7 3 13 2 0 3 0 4 0 10 9 11 3 5 0 30 6 5 2

Apply same procedure:-

Person Counter A B C D E

1 2 4 4 0 2 2 21 0 0 2 7 3 15 2 0 5 0 4 0 8 7 11 1 5 0 28 4 5 0

Minimum Cost is:-

Counter Person Minimum

Cost 1 D 28 2 B 24 3 C 33 4 A 25 5 E 39

149

A4. Solve find minimum cost,

Job Person 1 2 3 4

A 1 8 15 22 B 13 18 23 28 C 13 18 23 28 D 19 23 27 31

Selecting the smallest element in each row and subtracting it all the corresponding element of that row,

Job Person 1 2 3 4

A 0 7 14 21 B 0 5 10 15 C 0 5 10 15 D 0 4 8 12

Selecting the smallest element in each column and subtracting it all the corresponding element of that column,

Job Person 1 2 3 4

A 0 3 6 9 B 0 1 2 3 C 0 1 2 3 D 0 0 0 0

The procedure making assignment is as follows:-

Job Person 1 2 3 4

A 0 3 6 9 B 0 1 2 3 C 0 1 2 3 D 0 0 0 0

Selecting a smallest element which is not covered by horizontal and vertical line and subtracting it the entire element which are not covered by horizontal and vertical line. The element which is covered by horizontal and vertical line is remained same and in intersection of it smallest element will be added.

Job Person 1 2 3 4

A 0 2 5 8 B 0 0 1 2 C 0 0 1 2 D 1 0 0 0

Apply above procedure:-

Job Person 1 2 3 4

A 0 2 4 7 B 0 0 0 1 C 0 0 0 1 D 2 1 0 0

Minimum Cost is:-

Person Job Minimum

Cost A 1 1 B 2 18 C 3 23 D 4 31

73

A5. Find the least cost allocation for the following data:-

Typist Rate Per

Hour No. of Pages Typed/Hour Job No. of Pages

A 5 12 P 199 B 6 14 Q 175 C 3 8 R 145 D 4 10 S 198 E 4 11 T 178

Develop the cost table from the given problem, in which entries represent the cost to be increased due to assignment of jobs to various typists on a one-to-one basis.

Job Typist P Q R S T

A 85 75 65 125 75 B 90 78 66 132 78 C 75 66 57 114 69 D 80 72 60 120 72 E 76 64 56 112 68

Selecting the smallest element each column and subtracting in all the corresponding element of that column

Job Typist P Q R S T

A 47 57 67 7 57 B 42 54 60 0 54 C 57 66 75 18 63 D 52 60 72 12 60 E 56 68 76 20 64

Select the smallest element in each row and subtract in all the corresponding element of that row.

Job Typist P Q R S T

A 40 20 60 0 50 B 42 54 60 0 54 C 39 48 57 0 45 D 40 48 60 0 48 E 36 48 56 0 44

Selecting doing same process with column:-

Job Typist P Q R S T

A 4 2 4 0 6 B 6 6 4 0 10 C 3 0 1 0 1 D 4 0 4 0 4 E 0 0 0 0 0

The procedure of making assignment is as follows:-

Job Typist P Q R S T

A 4 2 4 0 6 B 6 6 4 0 10 C 3 0 1 0 1 D 4 0 4 0 4 E 0 0 0 0 0

Selecting a smallest element which is not covered by horizontal and vertical line and subtracting it the entire element which are not covered by horizontal and vertical line. The element which is covered by horizontal and vertical line is remained same and in intersection of it smallest element will be added.

Minimum Cost is:-

Typist Job Minimum

Cost A S 125 B R 66 C T 69 D Q 72 E P 76

408

A6.

Delhi - Mumbai

Delhi - Mumbai Flight No. Departure Arrival

Flight No. Departure Arrival 1 7:00 AM 8:00 AM

101 8:00 AM 9:00 AM 2 8:00 AM 9:00 AM

102 9:00 AM 10:00 AM 3 1:00 PM 2:00 PM

103 12:00 NOON 1:00 PM 4 6:00 PM 7:00 PM

104 5:00 PM 6:00 PM

Job Typist P Q R S T

A 3 2 3 0 5 B 5 6 3 0 9 C 2 0 0 0 0 D 3 0 3 0 3 E 0 1 0 1 0

Job Typist P Q R S T

A 0 2 0 0 2 B 2 6 0 0 6 C 2 3 0 3 0 D 0 0 0 0 0 E 0 1 0 1 0

Crew Based Time at Delhi

* Crew Based Time at Mumbai Flight No. 101 102 103 104

Flight No. 101 102 103 104

1 24 25 28 9

1 22 21 18 13 2 23 24 27 8

2 23 22 19 14 3 18 19 22 27

3 28 27 24 19 4 13 14 17 22

4 9 8 5 24

Time Layer Minimum:-

Flight No. 101 102 103 104

Flight No. 101 102 103 104

1 22* 21* 18* 9

1 13* 12* 9* 0 2 23 22* 19* 8

2 15 14* 11* 0 3 18 19 22 19*

3 0 1 4 1* 4 9* 8* 5* 22*

4 4* 3* 0* 17

Flight No. 101 102 103 104

Flight No. 101 102 103 104

1 13* 11* 9* 0

1 11* 9* 9* 0 2 15 13* 4* 0

2 13 11* 11* 0 3 0 0 4 1*

3 0 0 6 3* 4 4* 2* 0* 17

4 2* 0* 0* 17 Flight No. 101 102 103 104 1 2* 0* 0* 0 2 4 2* 2* 0 3 0 0 6 12* 4 2* 0* 0* 26

1 - 102 = Mumbai 2 - 104 = Delhi 3 - 101 = Delhi 4 - 103 = Mumbai

A7. = 50 + 60

2 + 300

3 + 4 509

4 + 7 812

Simplify the equation,

2

300+

300300

300

150+

300 1

3

509+4

509509

509

170+

127 1

4

812+7

812812

812

203+

116 1

, 0

Plot the Graph

However we cannot find any particular point (, )in there shaded regions than cant satisfy all the constraint simultaneously, thus the Linear Programming Problem has an infeasible solution.

A8. = 400 + 500

6 + 3 300

2 + 6 250

6 + 4 120

8 + 5 100

Simplify the equation:-

6

300+3

300300

300

50+

100 1

2

250+6

250250

250

125+

42 1

6

120+4

120120

120

20+

30 1

8

100+5

100100

100

13+

20 1

Plot the Graph:-

The coordinate of extreme point of the feasible region are 0= (0, 0), A = (150, 0), B = (140, 30), C = (70, 75), D = (0, 116)

The value of objective function at each of the extreme point is as follows:-

Extreme Point Coordinate

Objective Function Value Z=50x1+60x2

O (0,0) 50*0+60*0=0 A (180,0) 50*150+60*0=7500 B (140,30) 50*140+60*30=8800 C (70,75) 50*70+75*60=8000 D (0,116) 50*0+116*60=6960

The maximum value of Z=8800, occurs at the extreme point (140, 30). Hence the optimal solution to be given L.P problem is = 140, = 30,. = 8800

A9. Solve the following LPP.

Maximize 1170 + 110

. . 9 + 5 500

7 + 9 300

5 + 3 1500

7 + 9 1900

2 + 4 1000

, 0

= 1170 + 110

Simplify the equation:-

56+

100 1

43+

33 1

300+

500 1

271+

211 1

500+

250 1

However we cannot find any particular point (, ) in these shaded regions than is cannot satisfy all the constraint simultaneously, thus the LP Problem has infeasible solution.

A10. Solve the following L.P.P

Maximize = 18 + 16. . 15 + 25 375

24 + 11 264

, 0

= 18 + 16

Simplify the equation:-

15

375+25

375375

375

25+

15 1

24

264+11

264264

264

11+

24 1

, 0

Plot the graph,

The coordinate of extreme point of the feasible region are 0= (0, 0), A = (11, 0), B = (5, 12), C = (0, 15).

The value of objective function at each of the extreme point is as follows:-

Extreme Point Coordinate(, ) Objective Function Value = 18 + 16

O (0,0) 18*0+16*0=0 A (11,0) 18*11+16*0=198 B (5,12) 18*5+16*12=282 C (0,15) 18*0+15*15=225

The maximum value of Z=282, occurs at the extreme point (5, 12). Hence the optimal solution to be given L.P problem is = 5, = 12, . = 282

A11. Solve the following LPP.

Maximize = 6 2

. . 2 2

3

, 0

= 6 2

Simplify the equation:-

2

2+

22

2

1+

2 1

33

3

3 1

, 0

Plot the Graph:-

The coordinate of extreme point of the feasible region are 0= (0, 0), A= (1, 0), B= (3, 0), C= (3, 4).

The value of objective function at each of the extreme point is as follows:-

Extreme Point Coordinate(, ) Objective Function Value = 18 + 16

O (0,0) 6*0-2*0=0 A (1,0) 6*1-2*0=6 B (3,0) 6*3-2*0=18 C (3,4) 6*3-4*2=10

The maximum value of Z=18, occurs at the extreme point (3, 0). Hence the optimal solution to be given L.P problem is = 3, = 0, . = 18

A12. Solve the following LPP.

Maximize = 2

. . 80

60

5 + 6 600

+ 2 160

, 0

= 2

Simplify the equation:-

80

60

5

600+6

600600

600

120+

100 1

160+2

160160

160

160+

80 1

Plot the Graph:-

The coordinate of extreme point of the feasible region are 0= (0, 0), A = (80, 0), B = (80, 30), C = (60, 50), D = (35, 60), E= (0, 60)

The value of objective function at each of the extreme point is as follows:-

Extreme Point

Coordinate (, ) Objective Function Value = + 2

O (0,0) 0+0*0=0 A (80,0) 80+2*0=80 B (80,30) 80+2*30=140 C (60,50) 60+50*2=160 D (35,60) 35+60*2=155 E (0,60) 0+60*2=120

The maximum value of Z=160, occurs at the extreme point (60, 50). Hence the optimal solution to be given L.P problem is = 60, = 50,. = 160

A13. Solve the following LPP.

Maximize = 200 + 300. . 2 + 3 1200

+ 400

2 + 1.5 900

+ 0

= 200 + 300

Simplify the equation:-

2

1200+

3

12001200

1200

600+

400 1

400+

400400

400

2

900+1.5

900900

900

450+

600 1

+ 0

Plot the Graph:-

However we cannot find any particular point (, )in there shaded regions than cant satisfy all the constraint simultaneously, thus the Linear Programming Problem has an infeasible solution.

A14. Solve the following LPP.

Maximize = 40 + 60. . 2 + 70

+ 40

+ 3 90

, 0

= 40 + 60

Simplify the equation:-

35+

70 1

40+

40 1

90+

30 1

Plot the Graph:-

The coordinate of extreme point of the feasible region are 0= (0, 0), A = (30, 10), B = (40, 0), C = (90, 0), D = (25, 25)

The value of objective function at each of the extreme point is as follows:-

Extreme Point

Coordinate (, ) Objective Function Value = 40 + 60

O (0,0) 40*0+60*0=0 A (30,10) 40*30+600*10=1800 B (40,0) 40*40+60*0=1600 C (90,0) 40*90+60*0=3600 D (25,25) 40*25+60*25=2500

The maximum value of Z=1600, occurs at the extreme point (30, 10). Hence the optimal solution to be given L.P problem is = 30, = 10,. = 1600