OPTIMISATION MODEL IN DECISION MAKING PROJECT (2021) TOPIC UNBALANCED TRANSPORTATION PROBLEM WITH MINIMIZATION SUBMITTED TO:MR.ALOK KUMAR \ SUBMITTED BY: GROUP 1: ALBHVYA ACKSHATT 301007 AMAN NAHATA 301009 ANVEET JOSHI 301013 GAURAV SINGH RAWAT 301022 SAHIL 301046 Introduction Transportation problem is a particular class of linear programming, which is associated with day to-day activities in our real life and mainly deals with logistics. It helps in solving problems on distribution and transportation of resources from one place to another. The goods are transported from a set of sources (e.g., factory) to a set of destinations (e.g., warehouse) to meet the specific requirement. There is a type of linear programming problem that may be solved using a simplified version of the simplex technique called transportation method. Because of its major application in solving problems involving several product sources and several destinations of products, this type of problem is frequently called the transportation problem. It gets its name from its application to problems involving transporting products from several sources to several destinations. Although the formation can be used to represent more general assignment and scheduling problems as well as transportation and distribution problems. The two common objectives of such problems are either (1) minimize the cost of shipping m units to n destinations or (2) maximize the profit of shipping m units to n destinations. A transportation problem is said to be unbalanced if the supply and demand are not equal Problem Statement Guinness Ghana LTD is one of the top five worldwide brewery companies. Transportation cost represents about 25% of the total production cost. The company has outsourced its transportation to external logistics services Providers. Guinness Ghana Ltd has registered about 20 transporters who operate with 97 trucks. Each of the plants at the various sites namely Achimota and Kaasi has its own constraint with respect to plant and warehouse capacity. Thus, there is a limited capacity at each plant. The total plant capacity for Kaasi (S1) and Achimota( S2) per day is 1736 and 2419 crates depending on the plant efficiency. This project is intended to minimize the total transportation cost from two production sites namely Kaasi (Kumasi), and Achimota (Accra) to its numerous key distributors geographically scattered all over Ghana. As each site has its limit that supply and each customer a certain demand at a time. The required data includes: A list of all products, sources, demand for each product by customer, the full truck transportation cost,. The study concerned the supply of Malta Guinness from two production sites Kaasi and Achimota to 8 key distributors geographically scattered in the regions of Ghana. The study covered data gathered on the periods July07-June08, and Sep 08-June09. The demand for each destination was also known in advance in table 1.1. The data included the transport cost per truck from each source to each destination. For example- the transportation cost from Kaasi to D1 is 90 and Kaasi to D2 is 88 and so on. All cost with total demand and supply is given in table 1.1 Unbalanced Transportation - Manual Method Table 1.1 D1 D2 D3 D4 D5 D6 D7 D8 Supply S1 90 88 82 69 30 424 30 13 1736 S2 228 37 176 73 114 173 73 38 2419 576 445 335 272 431 304 120 Demand 907 Total demand = 3390 Total supply= 4155 As total demand is not equal to total supply therefore it is a case of unbalanced transportation problem. Thus to make it balanced we are introducing dummy column D9. So the balanced table is- D1 D2 D3 D4 D5 D6 D7 D8 D9 Supply S1 90 88 82 69 30 424 30 13 0 1736 S2 228 37 176 73 114 173 73 38 0 2419 576 445 335 272 431 304 120 0 Demand 907 To get initial basic feasible solution we are using Vogel’s approximation method (VAM) S1 D1 90 D2 88 D3 82 445 D4 69 D5 30 272 D6 424 D7 30 D8 13 D9 0 Supply 1736 S2 907 228 37 176 73 114 173 112 73 38 0 2419 907 576 576 445 335 335 272 431 431 192 304 120 120 765 765 Demand 138 138 - 51 51 51 51 51 - 94 94 94 - 4 4 4 4 4 4 84 84 84 84 - 258 - 43 43 43 43 43 43 25 25 25 25 25 25 13 13 13 13 37 37 37 37 0 0 0 0 0 0 13 13 37 38 Initial basic feasible solution = 90*907+ 37*576+82*445+73*335+30*272+173*431+30*112+73*192+38*120+0*765 = 2,68,546 Total no of allocation = 10 No of rows(m)= 2 No of column(n)=9 So total allocation= m+n-1= 10 To further improve the solution, we used MODI Metho Step1The most number of allocation are in row 2 therefore we are taking value of U2=0 Step2Now calculating all Vj and Ui by using allocated cells i.e Cij values Cij= Ui+Vj C22= U2+V2 37=0+V2 Therefore= V2=37 Similarly, calculating all the values of Vj and Ui U1=-43 V1= 115 V6= 173 V2= 37 V7= 73 V3= 107 V8= 38 V4= 73 V9= 0 V5=55 D2 88 S2 D1 90 907 228 Demand 907 576 S1 Vij 37 576 D3 82 445 176 D4 69 445 335 73 335 D5 30 272 114 D6 424 D8 13 D9 0 Supply Uij 1736 U1=-43 173 431 D7 30 112 73 192 38 120 0 765 2419 272 431 304 120 765 V1=133 V2=37 V3=125 V4=73 V5=73 V6=173 V7=73 V8=38 V9=0 U2=0 Calculating Dij values from unoccupied cellsDij=Cij-(Ui+Vj) D12= +81 D14= +26 D16= +281 D18= +5 D21=+90 D23=+23 D25=+41 Since all Dij values are positive thus the optimality condition is satisfied and the solution is optimal solution. Final optimal tableD2 88 S2 D1 90 907 228 Demand 907 576 S1 37 576 D3 82 445 176 D4 69 445 335 73 335 D5 30 272 114 D6 424 D8 13 D9 0 Supply 1736 173 431 D7 30 112 73 192 38 120 0 765 2419 272 431 304 120 765 The minimized cost is- 90*907+ 37*576+82*445+73*335+30*272+173*431+30*112+73*192+38*120+0*765 = 2,68,546 Solution in Excel with Sensitivity Report Let 1 Y =plant site at ACH 2 Y =plant site at KAS IJ X = the units shipped in crates from plant i to distribution centre j i =1, 2, 3... 9. and j =1, 2, 3..., 9. Objective function to Minimize cost of Transportation from Achitoma and Kassi plant 90x11 + 88x12 + 82x13 + 69x14 + 30x15 + 424x16 + 30x17 + 13x18 + 0x19 + 228x21 + 37x22+ 176x23+ 73x24 + 114x25 + 173x26+ 73x27 + 38x28 +0x29 Consider capacity constraint x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 ≤ 1736 x21 + x22 + x23 + x24 + x25 + x26 + x27 + x28 + x29 ≤ 2419 Demand constraint x11 + x21 =907 x12 + x22 =576 x13 + x23 =445 x14 + x24 =335 x15 + x25 =272 x16 + x26 =431 x17 + x27 =304 x18+ x28 =120 x19 + x29 =765 xij ≥ o for all i and j Z(Objective function) 268546 Achimota Kaasi Demand D1 90 228 907 D2 88 37 576 D3 82 176 445 D4 69 73 335 D5 30 114 272 D6 424 173 431 D7 30 73 304 D8 13 38 120 D9 0 0 765 xij Achitoma Kaasi 907 0 0 576 445 0 0 335 272 0 0 431 112 192 0 120 0 765 907 = 907 576 = 576 445 = 445 335 = 335 272 = 272 431 = 431 304 = 304 120 = 120 765 = 765 Supply 1736 2419 1736 2419 = = 1736 2419 D1 Achimota 90 Kaasi 228 Demand 907 D2 88 37 576 D3 82 176 445 D4 69 73 335 D5 30 114 272 D6 424 173 431 D7 30 73 304 D8 13 38 120 D9 0 0 765 Supply 1736 2419 x i j 907 0 445 0 272 0 112 0 0 0 576 0 335 0 431 192 120 765 =SU M(C1 5:C1 6) = 907 =SU M(D1 5:D1 6) = 576 =SU M(E1 5:E1 6) = 445 =SU M(F1 5:F16 ) = 335 =SU M(G1 5:G1 6) = 272 =SU M(H1 5:H1 6) = 431 =SU M(I1 5:I16 ) = 304 =SU M(J1 5:J16 ) = 120 =SU M(K1 5:K1 6) = 765 Cell $C$15 $D$15 $E$15 $F$15 $G$15 $H$15 $I$15 $J$15 $K$15 $C$16 $D$16 $E$16 $F$16 Final Name Value 907 D2 0 D3 445 D4 0 D5 272 D6 0 D7 112 D8 0 D9 0 0 D2 576 D3 0 D4 335 Reduced Cost 0 94 0 39 0 294 0 18 43 95 0 51 0 Objective Coefficient 90 88 82 69 30 424 30 13 0 228 37 176 73 Allowable Increase 95 1E+30 51 1E+30 41 1E+30 18 1E+30 1E+30 1E+30 94 1E+30 39 =SU = 1 M(C1 7 5:K1 3 5) 6 =SU = 2 M(C1 4 6:K1 1 6) 9 Allowable Decrease 1E+30 94 1E+30 39 1E+30 294 41 18 43 95 1E+30 51 1E+30 $G$16 $H$16 $I$16 $J$16 $K$16 D5 D6 D7 D8 D9 0 431 192 120 765 41 0 0 0 0 114 173 73 38 0 TRANSPORTATION OUTPUT TABLE PLANT DISTRIBUTO FULLTRUCK SITE(SOURC R(DESTINAT PER E) ION) CASE(0000) ACH ACH ACH D1= FTA 907 D2= RICKY 0 D3=OBIBA 445 JK ACH D4=KADOM 0 ACH D5=NAATO 272 ACH D6=LESK 0 ACH D7=DCEE 112 ACH D8=JOEMA 0 ACH D9=KBOA 0 KAS FTA 0 KAS RICKY 576 KAS OBIBA JK 0 KAS KADOM 335 KAS NAATO 0 KAS LESK 431 KAS DCEE 192 KAS JOEMA 120 KAS KBOA 765 The total transportation cost is ,: 268546 1E+30 294 41 18 43 41 1E+30 18 1E+30 1E+30 COST PER TOTAL FULL COST(GH TRUCK 000) LOAD 90 81630 88 0 82 36490 69 30 424 30 13 0 228 37 176 73 114 173 73 38 0 0 8160 0 3360 0 0 0 21312 0 24455 0 74563 14016 4560 0 Interpretation The GGL problem was solved with the linear programming module and the transportation module of The Management Scientist. The results from both the linear programming module and that of the transportation module of The Management Scientist yielded the same values, in terms of the optimal solution obtained. The excel solution shows the minimum total transportation cost is 268546. The value for the decision variables shows the optimal amount of drinks to be ship over each route. To minimize the transportation cost the management of Guinness Ghana Ltd should make the following shipments: Ship 907,000 cases of Malta Guinness from Plant ACH to Distributor FTA Ship 445,000 cases of Malta Guinness from Plant ACH to distributor OBIBA JK Ship 272,000 cases of Malta Guinness from Plant ACH to distributor NAATO Ship 112,000 cases of Malta Guinness from Plant ACH to distributor DCEE Ship 576,000 cases of Malta Guinness from Plant KAS to distributor RICKY Ship 335,000 cases of Malta Guinness from Plant KAS to distributor KADOM Ship 431,000 cases of Malta Guinness from Plant KAS to distributor LESK Ship 192,000 cases of Malta Guinness from Plant KAS to distributor DCEE Ship 120,000 cases of Malta Guinness from Plant KAS to distributor JOEMAN Ship 765,000 cases of Malta Guinness from Plant KAS to distributor KBOA Managerial Implication Cost Minimization: The transportation problem involves the distribution of goods from the manufacturing units to the warehouse, retailers, or the customers. The minimization of cost of supplying through optimum solution will help the manager to reduce the expenditure and thus increase the cash balance to the company which can be further used for the operations of the company distribution. This will also help to reduce the wastage and extra cost expenditure which is a liability to the company Flexible Practical Solution to the problem: The range found through sensitivity analysis will help the manager to take flexible plausible solutions which are to be taken in a practical situation thus avoiding any rigid solution which cannot be implemented on scale. Conclusion The minimization of cost to transport units from Achitoma plant and Kassi plant is done by finding the optimal solution through both excel method and manual method of Transportation. The transportation problem was formulated as a Linear Programming and solved with the standard LP solvers such as the Management scientist module to obtain the optimal solution. The sensitivity analysis was done to find the range where there will be no effect on the optimum solution of the problem. The transportation cost is an important element of the total cost structure for any business Through the use of this Transportation Model the business (GGBL) can identify easily and efficiently plan out its transportation, so that it not only minimizes the cost of transporting goods and services but also create time utility by reaching the goods and services at the right place at right time. This intend will enable them to meet the corporative objective such as education fund, entertainment and other support they offered to people of Ghana.