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4: Case Problem Stateline Shipping and Transport Company - Assignment Example

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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…
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Assignment 4: Case Problem Stateline Shipping and Transport Company
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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) 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 = 35 X11 + X21 + X31 + X41 + X51 + X61 = 65X21 + X22 + X 23 = 26 X12 + X22 + X32 + X42 + X52 + X62 = 80X31 + X32 + X33 = 42 X13 + X23 + X33 + X43 + X53 + X63= 105X41 + X42 + X43 = 53X51 + X52 + X53 = 29X61 + X62 + X63 = 38The 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 sitesTransportation SolutionOptimal solution value = $2832WhitewaterLos CanosDurasKingsport35  Danville  26Macon  42Selma14210Columbus29  Allentown 38 Dummy  27Note: Quantity in barrelsTable 4. Cost of transportation from the plants to the disposal sitesTransportation Solution WhitewaterLos CanosDurasKingsport35/$420  Danville  26/$260Macon  42/$462Selma1/$1742/$67210/$190Columbus29/$203  Allentown 38/$608 Dummy  27/$0Transshipment ModelThe 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 Supply Q-tySupply Sites123456ABCKingsportDanvilleMaconSelmaColumbusAllentownWhitewaterLos CanosDuras1285KingsportX11X12X13X14X15X16X1AX1BX1C2276DanvilleX21X22X23X24X25X26X2AX2BX2C3292MaconX31X32X33X34X35X36X3AX3BX3C4403SelmaX41X42X43X44X45X46X4AX4BX4C5279ColumbusX51X52X53X54X55X56X5AX5BX5C6288AllentownX61X62X63X64X65X66X6AX6BX6CA250WhitewaterXA1XA2XA3XA4XA5XA6XAAXABXACB250Los CanosXB1XB2XB3XB4XB5XB6XBAXBBXBCC250DurasXC1XC2XC3XC4XC5XC6XCAXCBXCC2502502502502502503153303551.

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 matrixDemand Sites123456ABCSupply sitesFrom SitesTo SitesKingsportDanvilleMaconSelmaColumbusAllentownWhitewaterLos CanosDuras1Kingsport0649781215172Danville601110127149103Macon511037151320114Selma910303161716195Columbus71273014714126Allentown871516140221618AWhitewater1214131772201210BLos Canos1592016141612015CDuras17101119121810150The solution is achieved by solving the 9x9 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  Kingsport DanvilleMaconSelmaColumbusAllentownWhitewaterLos CanosDurasKingsport16/$9619/$76      Danville      80/$720 Macon       78/$858Selma  17/$5136/$108    Columbus     65/$455  Allentown 38/$266      Whitewater        Table 8. Transshipment solution Transshipment SolutionOptimal solution value = $2630Kingsport DanvilleMaconSelmaColumbusAllentownWhitewaterLos CanosDurasKingsport2501619      Danville 196     80 Macon  214     78Selma  1725036    Columbus    214 65  Allentown 38   250   Whitewater      250  Los Canos       250 Duras        250Dummy        27POM-QM for WindowsConclusionIn 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. ReferencesLinear Programming: Introduction. (n.d.). Retrieved from http://www.

purplemath.com/modules/linprog.htmRajendran, 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.pdfReeb, 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.pdfhttp://www.prenhall.com/weiss_dswin/html/trans.htmThe Transportation Model. (n.d.). Retrieved from http://www.prenhall.com/weiss_dswin/html/trans.htm

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