# Need an research paper on 4: case problem stateline shipping and transport company. Needs to be 2 pages. Please no plagiarism.

Need an research paper on 4: case problem stateline shipping and transport company. Needs to be 2 pages. Please no plagiarism. Case Study John Doe Case Study The scope is to study to create two models. one will show how to minimize the shipping cost of transportation of waste from six plants to three disposal sites, the other one will illustrate how to organize the same work using transshipment concept. The objects for transshipments are the six plants and three disposal sites.

Transportation Model

From the viewpoint of demand and supply, the six plants are considered as supply sources and three waste sites are demands sites. both of them have limitation, which is expressed as a quantity that can be supplied and stored. The objective is to achieve cost minimization under the given limitations (Reeb & Leavengood, 2002)

### Save your time - order a paper!

Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlines

Order Paper Now

The solution is approached by creating a 6 x3 matrix illustrated in Table 1. Each cell of transportation expressed through Xij depicts quantity from the supply source to waste disposal site. The matrix also shows that total supply quantity is 223 bbls per week and the total demand quantity is 250 bbls per week. Supply and demand are not balanced. the solution requires to implement a dummy supply source of 27 bbls. Decision variable, in this case, is the quantity for a site, and objective function is cost minimization. The model is represented through the following linear equations (“Linear programing”, n.d).

Minimization is solved using the following equation, subject to: Z=12X11+15X12+17X13+14X21+9X22+10X23+13X31+20X32+11X33+17X41+16X42+19X43+7X51+14X52+12X53+22X61+16X62+18X63.

X11 + X12 + X13 = 35X11 + X21 + X31 + X41 + X51 + X61 = 65

X21 + X22 + X 23 = 26X12 + X22 + X32 + X42 + X52 + X62 = 80

X31 + X32 + X33 = 42X13 + X23 + X33 + X43 + X53 + X63= 105

X41 + X42 + X43 = 53

X51 + X52 + X53 = 29

X61 + X62 + X63 = 38

The solution was obtained using the “Transportation” module of POM – OM software (“The Transportation model”, n.d.) The shipment from the supply source to waste the side is illustrated in Table 3. The POM – OM solution includes a dummy supply source for 27 bbls. The minimum cost is \$2,832. it does not include dummy supply source quantity.

Table 3. Shipment from supply sources to the disposal sites

Transportation Solution

Optimal solution value = \$2832

Whitewater

Los Canos

Duras

Kingsport

35

&nbsp.

&nbsp.

Danville

&nbsp.

&nbsp.

26

Macon

&nbsp.

&nbsp.

42

Selma

1

42

10

Columbus

29

&nbsp.

&nbsp.

Allentown

&nbsp.

38

&nbsp.

Dummy

&nbsp.

&nbsp.

27

Note: Quantity in barrels

Table 4. Cost of transportation from the plants to the disposal sites

Transportation Solution

&nbsp.

Whitewater

Los Canos

Duras

Kingsport

35/\$420

&nbsp.

&nbsp.

Danville

&nbsp.

&nbsp.

26/\$260

Macon

&nbsp.

&nbsp.

42/\$462

Selma

1/\$17

42/\$672

10/\$190

Columbus

29/\$203

&nbsp.

&nbsp.

Allentown

&nbsp.

38/\$608

&nbsp.

Dummy

&nbsp.

&nbsp.

27/\$0

Transshipment Model

The idea is based on the concept that shipping line will use an intermediary supply center, which could be either a plant or waste disposal site. This concept gives a 9 x 9 matrix where supply plus disposal sources together act as supply sources and disposal sources (Rajendran & Pandian, 2012). Table 5 displays the feed matrix to achieve a solution. The values for supply and demand quantities of the matrix are based on the following assumptions:

Table 5. Transshipment solution matrix

Demand Sites

&nbsp.

Supply Q-ty

Supply Sites

1

2

3

4

5

6

A

B

C

Kingsport

Danville

Macon

Selma

Columbus

Allentown

Whitewater

Los Canos

Duras

1

285

Kingsport

X11

X12

X13

X14

X15

X16

X1A

X1B

X1C

2

276

Danville

X21

X22

X23

X24

X25

X26

X2A

X2B

X2C

3

292

Macon

X31

X32

X33

X34

X35

X36

X3A

X3B

X3C

4

403

Selma

X41

X42

X43

X44

X45

X46

X4A

X4B

X4C

5

279

Columbus

X51

X52

X53

X54

X55

X56

X5A

X5B

X5C

6

288

Allentown

X61

X62

X63

X64

X65

X66

X6A

X6B

X6C

A

250

Whitewater

XA1

XA2

XA3

XA4

XA5

XA6

XAA

XAB

XAC

B

250

Los Canos

XB1

XB2

XB3

XB4

XB5

XB6

XBA

XBB

XBC

C

250

Duras

XC1

XC2

XC3

XC4

XC5

XC6

XCA

XCB

XCC

250

250

250

250

250

250

315

330

355

1. Each plant may absorb total demand quantity 250 bbls. in addition to its own supply quantity,

2. Each waste site may absorb total demand quantity 250 bbls in addition to its own demand quantity.

Table 6. Transshipment cost matrix

Demand Sites

1

2

3

4

5

6

A

B

C

Supply sites

From Sites

To Sites

Kingsport

Danville

Macon

Selma

Columbus

Allentown

Whitewater

Los Canos

Duras

1

Kingsport

0

6

4

9

7

8

12

15

17

2

Danville

6

0

11

10

12

7

14

9

10

3

Macon

5

11

0

3

7

15

13

20

11

4

Selma

9

10

3

0

3

16

17

16

19

5

Columbus

7

12

7

3

0

14

7

14

12

6

Allentown

8

7

15

16

14

0

22

16

18

A

Whitewater

12

14

13

17

7

22

0

12

10

B

Los Canos

15

9

20

16

14

16

12

0

15

C

Duras

17

10

11

19

12

18

10

15

0

The solution is achieved by solving the 9×9 matrix for cost minimization. Each Xij of the matrix depicts the quantity it may contain in determining the minimum transportation cost. The supply and demand constraints are obtained, in the same way as shown in the previous example. The summation of each row of the matrix presents a supply constraint equation. The summation of each column of the matrix presents a demand constraint. The minimization solution is achieved using the “Transportation” module of POM – OM software. The results are presented in Tables 7 and 8. The results illustrate the required shipment directions and associated cost.

Table 7. Transshipment cost from one place to another

&nbsp.

Kingsport

Danville

Macon

Selma

Columbus

Allentown

Whitewater

Los Canos

Duras

Kingsport

16/\$96

19/\$76

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Danville

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

80/\$720

&nbsp.

Macon

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

78/\$858

Selma

&nbsp.

&nbsp.

17/\$51

36/\$108

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Columbus

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

65/\$455

&nbsp.

&nbsp.

Allentown

&nbsp.

38/\$266

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Whitewater

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Table 8. Transshipment solution

Transshipment Solution

Optimal solution value = \$2630

Kingsport

Danville

Macon

Selma

Columbus

Allentown

Whitewater

Los Canos

Duras

Kingsport

250

16

19

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Danville

&nbsp.

196

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

80

&nbsp.

Macon

&nbsp.

&nbsp.

214

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

78

Selma

&nbsp.

&nbsp.

17

250

36

&nbsp.

&nbsp.

&nbsp.

&nbsp.

Columbus

&nbsp.

&nbsp.

&nbsp.

&nbsp.

214

&nbsp.

65

&nbsp.

&nbsp.

Allentown

&nbsp.

38

&nbsp.

&nbsp.

&nbsp.

250

&nbsp.

&nbsp.

&nbsp.

Whitewater

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

250

&nbsp.

&nbsp.

Los Canos

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

250

&nbsp.

Duras

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

250

Dummy

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

&nbsp.

27

POM-QM for Windows

Conclusion

In this assignment, the demand quantity is 250 bbls whereas the supply quantity is 223 bbls. In approaching cost minimization, software POM –QM used a dummy supply source in the quantity of 27 bbls. The unbalanced demand and supply quantity is considered to be a limitation of the study. The study shows that using transshipment the company management can reduce the shipment cost. The shipment cost of shipment of 223 bbls without the transshipment option is \$2832 whereas with the option is \$2630. Hence, transshipment, in this case is a better solution.

References

Rajendran, P., & Pandian, P. (2012). Solving Fully Interval Transshipment Problems. Retrieved from http://www.m-hikari.com/imf/imf-2012/41-44-2012/pandianIMF41-44-2012.pdf

Reeb, J., & Leavengood, S. (2002). Transportation Problem: A Special Case for Linear Programming Problems. Retrieved from http://ir.library.oregonstate.edu/xmlui/bitstream/handle/1957/20201/em8779-e.pdf

http://www.prenhall.com/weiss_dswin/html/trans.htm