1. Operations Management MBA Sem II Module IV Transportation

2. …the physical distribution of goods and services from several supply centers to several demand centres.

3. Transportation The structure of transportation problem involves a large no. of shipping routes from several supply origins to several demand centres.

4. Objective : To determine the number of units of an item that should be shipped from an origin to a destination in order to satisfy the required quantity of goods or services at each destination centre.

5. Method 1.Formulate the problem and arrange the data in matrix form 2.Obtain an initial basic feasible solution by - North West Corner Method - Least Cost method - Vogel’s Approximation Method

6. The solution must satisfy all the supply and demand constraints The number of positive allocations must be m+n-1, where m is the no. of rows and n is the no. of columns 3. Test the initial solution for optimality (MODI) 4. Update the solution

7. Initial basic solution by North west corner method

8. Q A company has three production facilities S1,S2, S3 with production capacity of 7,9 and 18 units (in 100’s) per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5,6,7 and 14 units (in 100’s) per week, respectively.

9. The transportation costs(in rupees) per unit between factories to warehouses are given in the table in the next slide. Minimize the total transportation cost.

10. D1D2D3D4CAPACI TY S1193050107 S2703040609 S3408702018 Demand5871434

11. The no. of positive allocations (occupied cells) = m + n -1 = 6 D1D2D3D4CAPACI TY S119 530 250107 S27030 640 3609 S340870 420 1418 Demand5871434

12. Total cost = 5 *19 + 2 * 30 + 6*30 + 3 * 40 + 4 * 70 + 14* 20 = Rs. 1,015

13. Initial basic solution by VAM

14. Choose largest value  22 in columnD 2 Choose cell with lowest cost Satisfy it D1D1 D2D2 D3D3 D4D4 capacityRow differences S1S1 1930501079 S2S2 70304060910 S3S3 40870201812 Demand 5871434 Column differences 212210 ghj

15. D1D1 D2D2 D3D3 D4D4 capRow diff S1S1 1930501079 S2S2 70304060920 S3S3 40870201820 Demand 5871434 Column difference s 212210 8 5

16. D1D1 D2D2 D3D3 D4D4 capac ity Row diff S1S1 19305010740 S2S2 70304060920 S3S3 40870201850 Demand 5871434 Column differences 10 8 5

17. D1D1 D2D2 D3D3 D4D4 capa city Row diff S1S1 19305010740 S2S2 70304060920 S3S3 408702018 Demand 5871434 Column difference s 1050 5 8 10 2

18. D1D1 D2D2 D3D3 D4D4 capa city Row diff S1S1 19305010740 S2S2 70304060920 S3S3 408702018 Demand 5871434 Column differences 1050 5 2 2 7 8 10

19. Total cost : 5*19 + 2*10 + 7 *40 + 2*60 + 8 *8 + 10 *20 = Rs 779

20. Q consider the following transportation problem involving 3 sources and four destinations. The cell entries represent the cost of transportation per unit. Obtain the initial basic feasible soln using 1.NWCM 2.VAM Find the optimal soln (after NWCM) using UV method

21. Dest source 1234supply 13174300 22659400 38332500 Demand2503504002001200

22. By VAM

23. Dest source 1234supply 13174300 22659400 38332500 Demand2503504002001200 300 250 150 50 250 200

24. m+n-1 = 6 Hence this is a feasible soln

25. Total Cost = Rs. 2850

26. By NWCM Dest source 1234supply 13174300 22659400 38332500 Demand2503504002001200 250 50 300100 300200

27. Total cost = Rs. 4400

28. Optimal soln Row 1,2,3 are assigned values U1,U2,U3 AND Col 1, col2, col3 and col 4 are assigned variables V 1,V 2,V 3,V 4 FOR BASIC cells Ui + Vj = c ij Take U1 = 0

29. V1=3V2=1V3=0V4=-1 1234supply U1=013174300 U2=522659400 U3=338332500 2503504002001200 250 50 300 100 200 300

30. If all pij<=0; optimality is reached Compute Pij (penalties) for the non basic cells by using the formula : Pij = Ui + Vj - cij

31. V1=3V2=1V3=0V4=-1 1234supply U1=013174300 U2=522659400 U3=338332500 2503504002001200 250 50 300 100 300 200 6 -ve 1

32. If all Pij are <= 0; then optimality is reached; Cell (2,1) with penalty 6, has the most positive penalty. So, there is scope for improving soln. Construct a loop using one basic cell and other non basic cells. Assign + and – signs alternatively to the basic cells.

33. Choose the minimum among the –vely assigned basic cells = 250 Add to all +vely cells. Subtract from all –vely assigned cells.

34. V1=3V2=1V3=0V4=-1 1234supply U1=01-3-3 + 174300 U2=522 +  - 6 59400 U3=338332500 2503504002001200 250 6 -ve 50 300 1 -ve 100 300 -ve 200

35. V1=-3V2=1V3=0V4=-1 1234supply U1=013 174300 U2=522-6-6 5+5+ 9400 U3=338+ 3 X3 -2500 2503504002001200 -ve 300 -ve 250 50 100 -ve 1 300200

36. V1=-2V2=1V3=1V4=0 1234supply U1=01 3 1 7 4300 U2=422659400 U3=238332500 2503504002001200 -ve 300 -ve 250 150 -ve 50 -ve 250 200

37. All penalties <=0; Hence optimum solution is reached Total = 2850

38. Q Consider the transportation problem as shown in the table. Find the initial basic solution using 1.NWCM 2.VAM APPLY UV method to find the optimal solution.

39. 12345supply 1102161410300 2618121316500 384141210825 4142220818375 Deman d 350400250150400

40. The given problem is unbalanced because the total demand(1550) < total supply (2000) Convert to a balanced transportation problem.

41. 123456Supply 11021614100300 26181213160500 3841412100825 41422208180375 De man d 3504002501504004502000

42. Initial basic soln using NWCM

43. 123456supply 11021614100300 26181213160500 3841412100825 41422208180375 Dem and 350400250150400450 300 50 400 50 200150 400 75 375

44. Total cost = Rs. 19,700

45. Initial basic solution using VAM

46. 123456Supply 11021614100300 26181213160500 3841412100825 41422208180375 Demand 3504002501504004502000 300 350 75 100 175150400 375

47. Total cost = Rs. 12250 Asssign Ui & Vj

48. V1=6V2=2V3=12V4= 10 V5= 8 V6= 0 123456Supply U1=011021614100300 U2=0261812+13160 -500 U3=238414 -12100 +825 U4=041422208180375 D 3504002501504004502000 300 35075 100 175 150400 2 375 - ve 0 2

49. V1=6V2=2V3=12V4=10 V5=8V6=- 2 123456 S U1=011021614100300 U2=026181213160500 U3=238414- 12 10+ 0825 U4=241422208+8+ 180-0- 375 D 3504002501504004502000 -ve 300 350 150 0 100 150 400 75 4375

50. V1=6V2=2V3=12V4= 6 V5=8V6=-2 123456 S U1=011021614100300 U2=026181213160500 U3=238414- 12 10 0825 U4=241422208180-0- 375 D 3504002501504004502000 -ve 300 350 150 0100 400 225 150 225

51. Total cost = Rs. 11,500