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# Transportation problem

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### Transportation problem

1. 1. TRANSPORTATION PROBLEMS (TPs)WHAT IS TRANSPORTATION PROBLEM?A TRANSPORTATION PROBLEM (TP) CONSISTS OFDETERMINING HOW TO ROUTE PRODUCTS IN ASITUATION WHERE THERE ARE SEVERAL SUPPLYLOCATIONS AND ALSO SEVERAL DESTINATIONS INORDER THAT THE TOTAL COST OF TRANSPORTATION ISMINIMISED
2. 2. TRANSPORTATION PROBLEMS (TPs) SUPPLY DEMAND A D B E C F
3. 3. MATHEMATICAL STATEMENT OF A TRANSPORTAION PROBLEMLETai = QUANTITY OF PRODUCT AVAILABLE AT SOURCE i,bj = QUANTITY OF PRODUCT REQUIRED AT DESTINATION j,cij = COST OF TRANSPORTATION OF ONE UNIT OF THE PRODUCT FROM SOURCE i TO DESTINATION j,xij = QUANTITY OF PRODUCT TRANSPORTED FROM SOURCE i TO DESTINATION j.ASSUMING THAT, TOTAL DEMAND = TOTAL SUPPLY, ie, Sai = Sbj THENTHE PROBLEM CAN BE FORMULATED AS A LPP AS FOLLOWS. m nMINIMISE TOTAL COST Z = SS cij xij i=1 j=1 nSUBJECT TO Sxij = ai FOR i = 1,2,3…m j=1 m S xij = bj FOR j = 1,2,3…n i=1AND xij ≥ 0FULL STATEMENT OF TP AS LPP
4. 4. TRANSPORTATION MODEL –TABULAR FORM. SOURCES DESTINATIONS SUPPLY ai 1 2 3 … n 1 x11 x12 x13 x1n a1 c11 c12 c13 c1n 2 x21 x22 x23 x2n a2 c21 c22 c23 c2n 3 x31 x32 x33 x3n a3 c31 c32 c33 c3n . . . m xm1 xm2 xm3 xmn am cm1 cm2 cm3 cmn DEMAND bj b1 b2 b3 bn Sai= SbjMINIMISE TOTAL COST Z = SS cij xij FOR i=1 to m & j=1 to n
5. 5. TRANSPORTAION PROBLEMS (TPs)• TRANSPORTATION COST PER UNIT MATRIX• TRANSPORTATION DECISION VARIABLE MATRIX• SUPPLY COLUMN• DEMAND ROW• TOTAL TRANSPORTATION COST• SOLUTION OF THE TRANSPORTATION PROBLEM• OCCUPIED CELLS• EMPTY CELLS• CONSTRAINTS IN A TP• VARIABLES IN A TP:
6. 6. TRANSPORTATION PROBLEM–TABULAR FORM. SOURCES DESTINATIONS SUPPLY ai 1 2 3 1 x11 x12 x13 a1 c11 c12 c13 2 x21 x22 x23 a2 c21 c22 c23 3 x31 x32 x33 a3 c31 c32 c33 DEMAND bj b1 b2 b3 Sai= Sbj MINIMISE TOTAL COST Z = SS cij xij FOR i=1 to m & j=1 to n
7. 7. EXAMPLE 1 OF TPA STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL. CAPACITIES OF 45,15 AND 40 THOUSAND TONNES OF CR COILS. THE PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND W3 WITH ANNUAL OFFTAKE OF 25,55 AND 20 THOUSAND TONNES OF CR COILS. THE TRANSPORTATION COST (Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE. FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST. SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 10 7 8 45 P2 15 12 9 15 P3 7 8 12 40 DEMAND 25 55 20 100
8. 8. TP FOMULATED AS A LPPMin Z =10x11+ 7x12+ 8x13+ 15x21+ 12x22+ 9x23+ 7x31+ 8x32+ 12x33Subject tox11+ x12+ x13= 45x21+ x22+ x23= 15 SUPPLY CONSTRAINTSx31+ x32+ x33= 40x11+ x21+ x31= 25x12+ x22+ x32= 55 DEMAND CONSTRAINTSx13+ x23+ x33= 20 xij ≥ 0 FOR i = 1,2,3 AND j = 1,2,3
9. 9. TP REFOMULATED AS A LPP FOR SIMPLEX METHODMin Z =10x11+ 7x12+ 8x13+ 15x21+ 12x22+ 9x23+ 7x31+ 8x32+ 12x33 + MA1 + MA2 + MA3 + MA4 + MA5 + MA6Subject tox11+ x12+ x13+A1= 45x21+ x22+ x23+A2= 15 SUPPLY CONSTRAINTSx31+ x32+ x33+A3= 40x11+ x21+ x31+A4= 25x12+ x22+ x32+A5= 55 DEMAND CONSTRAINTSx13+ x23+ x33+A6= 20 xij ≥ 0 FOR i = 1,2,3 AND j = 1,2,3
10. 10. DUAL OF A TP FORMULATED AS A LPPMax G = u1+ u2+ u3+ v1+ v2+ v3+Subject tou 1+ v 1 +
11. 11. METHODS OF SOLVING A TP1. SIMPLEX METHOD TP CAN BE STATED AS LPP AND THEN SOLVED BY THE SIMPLEX METHOD2. TRANSPORTATION METHOD THIS INVOLVES THE FOLLOWING STEPS i) OBTAIN THE INITIAL FEASIBLE SOLUTION USING - NORTH WEST CORNER RULE - VOGEL’S APPROXIMATION METHOD ii) TEST FEASIBLE SOLUTION FOR OPTIMALITY USING - STEPPING STONE METHOD - MODIFIED DISTRIBUTION METHOD iii) IMPROVE THE SOLUTION BY REPEATED ITERATION
12. 12. NORTH WEST CORNER (NWC) RULE1. START WITH THE NW CORNER OF TP TABLE2. TAKE APPROPRIATE STEPS IF a1 > b1 a1 < b1 a1 = b13. COMPLETE INITIAL FEASIBLE SOLUTION TABLE
13. 13. VOGEL’S APPROXIMATION METHOD - STEPS1. FIND DIFFERENCE IN TRANSPORTATION COSTS BETWEEN TWO LEAST COST CELLS IN EACH ROW AND COLUMN.2. IDENTIFY THE ROW OR COLUMN THAT HAS THE LARGEST DIFFERENCE.3. DETERMINE THE CELL WITH THE MINIMUM TRANSPORTATION COST IN THE ROW/COL4. ASSIGN MAXIMUM POSSIBLE VALUE TO xij VARIABLE IN THE CELL IDENTIFIED ABOVE5. OMIT ROW IF SUPPLY EXHAUSTED AND OMIT COL IF DEMAND MET6. REPEAT STEPS 1 AND 5 ABOVE
14. 14. TESTING FEASIBLE SOLUTION FOR OPTIMALITY 1. STEPPING STONE METHOD 2. MODIFIED DISTRIBUTION METHOD
15. 15. TESTING FEASIBLE SOLUTION FOR OPTIMALITYSTEPPING STONE METHOD1. IDENTIFY THE EMPTY CELLS2. TRACE A CLOSED LOOP3. DETERMINE NET COST CHANGE4. DETERMINE THE NET OPPORTUNITY5. IDENTIFY UNOCCUPIED CELL WITH THE LARGEST POSITIVE NET OPPORTUNITY COST6. REPEAT STEPS 1 TO 5 TO GET THE NEW IMPROVED TABLES
16. 16. TESTING FEASIBLE SOLUTION FOR OPTIMALITYRULES FOR TRACING CLOSED LOOPS1. ONLY HORIZONTAL OR VERTICAL MOVEMENT ALLOWED2. MOVEMENT TO AN OCCUPIED CELL ONLY3. STEPPING OVER ALLOWED4. ASSIGN POSITIVE OR NEGATIVE SIGNS TO CELLS5. LOOP MUST BE RIGHT ANGLED6. A ROW OR COL MUST HAVE ONE CELL OF POSITIVE SIGN AND ONE CELL OF NEGATIVE SIGN ONLY7. A LOOP MUST HAVE EVEN NUMBER OF CELLS8. EACH UNOCCUPIED CELL CAN HAVE ONE AND ONLY ONE LOOP9. ONLY OCCUPIED CELLS ARE TO BE ASSIGNED POSITIVE OR NEGATIVE VALUES10. LOOP MAY NOT BE SQUARE OR RECTANGLE11. ALL LOOPS MUST BE CONSISTENTLY CLOCKWISE OR ANTICLOCKWISE
17. 17. TESTING FEASIBLE SOLUTION FOR OPTIMALITYMODIFIED DISTRIBUTION METHOD In case there are a large number of rows and columns, then Modified distribution (MODI) method would be more suitable than Stepping Stone methodStep 1Add ui col and vj row: Add a column on the right hand side of the TP table and title it ui. Also add a row at the bottom of the TP table and title it vj.Step 2 This step has four parts.i) Assign value to ui=0 To any of the variable ui or vj, assign any arbitrary value. Generally the variable in the first row i.e. u1 is assigned the value equal to zero.ii) Determine values of the vj in the first row using the value of u1 = 0 and the cij values of the occupied cells in the first row by applying the formula ui + vj = cijiii) Determine ui and vj values for other rows and columns with the help of the formula ui + vj = cij using the ui and vj values already obtained in steps a), b) above and cij values of each of the occupied cells one by one.iv) Check the solution for degeneracy. If the soln is degenerate [ie no. of occupied cells is less than (m+n-1)], then this method will not be applicable.
18. 18. TESTING FEASIBLE SOLUTION FOR OPTIMALITYMODIFIED DISTRIBUTION METHODStep 3Calculate the net opportunity cost for each of the unoccupied cells using the formula δij = (ui + vj) - cij. If all unoccupied cells have negative δij value, then, the solution is optimal.Multiple optimality: If, however, one or more unoccupied cells have δij value equal to zero, then the solution is optimal but not unique.Non optimal solutionIf one or more unoccupied cells have positive δij value, then the solution is not optimal.Largest positive dj value: The unoccupied cell with the largest positive δij value is identified.Step 4A closed loop is traced for the unoccupied cell with the largest δij value. Appropriate quantity is shifted to the unoccupied cell and also from and to the other cells in the loop so that the transportation cost comes down.Step 5The resulting solution is once again tested for optimality.If it is not optimal, then the steps from 1 to 4 are repeated, till an optimal solution is obtained
19. 19. EXAMPLE 1 OF TPA STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL. CAPACITIES OF 45,15 AND 40 THOUSAND TONNES OF CR COILS. THE PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND W3 WITH ANNUAL OFFTAKE OF 25,55 AND 20 THOUSAND TONNES OF CR COILS. THE TRANSPORTATION COST (Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE. FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST. SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 10 7 8 45 P2 15 12 9 15 P3 7 8 12 40 DEMAND 25 55 20 100
20. 20. EXAMPLE 1 OF TP. INITIAL BASIC FEASIBLE SOLUTION BY NORTH WEST CORNER METHOD SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 25 20 10 7 8 45 P2 15 15 12 9 15 P3 20 20 7 8 12 40 DEMAND 25 55 20 100
21. 21. EXAMPLE 1 OF TP. INTIAL BASIC FEASIBLE SOLUTION BY VOGEL APPROXIMATION METHOD SOURCES DESTINATIONS SUPPLY ITERATIONS W1 W2 W3 1 2 3 4 P1 40 5 1 1 3* X 10 7 8 45 P2 15 3* X X X 15 12 9 15 P3 25 15 1 1 1 1* 7 8 12 40 DEMAND 25 55 20 100 1st ITERATION 3 1 1 2nd ITERATION 3 1 4* 3rd ITERATION 3 1 X 1 4th ITERATION X X X
22. 22. EXAMPLE 1 OF TP. INITIAL BASIC FEASIBLE SOLUTION BY NORTH WEST CORNER METHOD SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 25 20 10 7 8 45 P2 15 15 12 9 15 P3 20 20 7 8 12 40 DEMAND 25 55 20 100
23. 23. EXAMPLE 1 OF TP. INTIAL BASIC FEASIBLE SOLUTION BY VOGEL APPROXIMATION METHOD SOURCES DESTINATIONS SUPPLY ITERATIONS W1 W2 W3 1 2 3 4 P1 40 5 1 1 3* X 10 7 8 45 P2 15 3* X X X 15 12 9 15 P3 25 15 1 1 1 1* 7 8 12 40 DEMAND 25 55 20 100 1st ITERATION 3 1 1 2nd ITERATION 3 1 4* 3rd ITERATION 3 1 X 1 4th ITERATION X X X
24. 24. TESTING FEASIBLE SOLUTION FOR OPTIMALITY 1. STEPPING STONE METHOD 2. MODIFIED DISTRIBUTION METHOD
25. 25. TESTING FEASIBLE SOLUTION FOR OPTIMALITYSTEPPING STONE METHOD1. IDENTIFY THE EMPTY CELLS2. TRACE A CLOSED LOOP3. DETERMINE NET COST CHANGE4. DETERMINE THE NET OPPORTUNITY5. IDENTIFY UNOCCUPIED CELL WITH THE LARGEST POSITIVE NET OPPORTUNITY COST6. REPEAT STEPS 1 TO 5 TO GET THE NEW IMPROVED TABLES
26. 26. TESTING FEASIBLE SOLUTION FOR OPTIMALITYRULES FOR TRACING CLOSED LOOPS1. ONLY HORIZONTAL OR VERTICAL MOVEMENT ALLOWED2. MOVEMENT TO AN OCCUPIED CELL ONLY3. STEPPING OVER ALLOWED4. ASSIGN POSITIVE OR NEGATIVE SIGNS TO CELLS5. LOOP MUST BE RIGHT ANGLED6. A ROW OR COL MUST HAVE ONE CELL OF POSITIVE SIGN AND ONE CELL OF NEGATIVE SIGN ONLY7. A LOOP MUST HAVE EVEN NUMBER OF CELLS8. EACH UNOCCUPIED CELL CAN HAVE ONE AND ONLY ONE LOOP9. ONLY OCCUPIED CELLS ARE TO BE ASSIGNED POSITIVE OR NEGATIVE VALUES10. LOOP MAY NOT BE SQUARE OR RECTANGLE11. ALL LOOPS MUST BE CONSISTENTLY CLOCKWISE OR ANTICLOCKWISE
27. 27. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLNMODIFIED DISTRIBUTION METHOD
28. 28. EXAMPLE 1 OF TP. INITIAL BASIC FEASIBLE SOLUTION BY NORTH WEST CORNER METHOD SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 25 20 10 7 8 45 P2 15 15 12 9 15 P3 20 20 7 8 12 40 DEMAND 25 55 20 100
29. 29. UNBALANCEDTRANSPORTATION PROBLEMS • TOTAL SUPPLY EXCEEDS TOTAL DEMAND • TOTAL DEMAND EXCEEDS TOTAL SUPPLY
30. 30. EXAMPLE 2a OF TPA STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL. CAPACITIES OF 60, 20 AND 40 THOUSAND TONNES OF CR COILS. THE PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND W3 WITH ANNUAL OFFTAKE OF 25,55 AND 20 THOUSAND TONNES OF CR COILS. THE TRANSPORTATION COST (Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE. FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST. SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 10 7 8 60 P2 15 12 9 20 P3 7 8 12 40 120 DEMAND 25 55 20 100
31. 31. SOLUTION OF EXAMPLE 2a OF TPCREATE A DUMMY DESTINATION W4 WITH DEMAND = 20,000 TONNES SOURCES DESTINATIONS SUPPLY W1 W2 W3 W4 P1 10 7 8 0 60 P2 20 15 12 9 0 P3 40 7 8 12 0 120 DEMAND 25 55 20 20 120
32. 32. EXAMPLE 2b OF TPA STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL. CAPACITIES OF 55, 15 AND 40 THOUSAND TONNES OF CR COILS. THE PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND W3 WITH ANNUAL OFFTAKE OF 30, 70 AND 20 THOUSAND TONNES OF CR COILS. THE TRANSPORTATION COST (Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE. FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST. SOURCES DESTINATIONS SUPPLY W1 W2 W3 P1 10 7 8 55 P2 15 12 9 15 P3 7 8 12 40 100 DEMAND 30 70 20 120
33. 33. SOLUTION OF EXAMPLE 2b OF TPCREATE A DUMMY SOURCE P4 WITH SUPPLY = 20,000 TONNESSOURCES DESTINATIONS SUPPLY W1 W2 W3P1 10 7 8 55P2 15 15 12 9P3 40 7 8 12P4 20 0 0 0 120DEMAND 30 70 20 120
34. 34. . EXAMPLE 3 OF TP (DEGENERACY)AN ALUMINIUM MANUFACTURER HAS THREE PLANTS A, B, AND C WITHANNUAL CAPACITIES OF 60,100 AND 40 THOUSAND TONNES OFALUMINIUM INGOTS. THE PRODUCT IS DISTRIBUTED FROM FOURWAREHOUSES D, E, F, AND G WITH ANNUAL OFFTAKE OF 20, 50, 50,AND 80 THOUSAND TONNES OF AL INGOTS. TRANSPORTATION COST(Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE.FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST. SOURCES DESTINATIONS SUPPLY D E F G A 7 3 8 6 60 B 4 2 5 10 100 C 2 6 5 1 40 DEMAND 20 50 50 80 200
35. 35. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN MODIFIED DISTRIBUTION METHOD TP TABLE 1 (NON OPTIMAL) dj IS NET OPPOR. AVAIL D E F G ui (ui+vj)=cij EMPTY NET COSTA 20 40 u1 u1=0 ROW A CELL d =(u +v )-c dj CHANGE 60 AD: v1= 7 ij i j ij 7 3 8 6 0 AE::v2= 3 AF 0+6-8=-2 -2B 10 50 40 u2 ROW B AG 0+11-6=+5 +5 100 BE: u2= -1 BF: v3= 6 BD -1+7-4=+2 +2 4 2 5 10 -1 BG: v4= 11 CD -10+7-2=-5 -5C 40 u3 ROW C 40 CG: u3= -10 CE -10+3-6=-13 -13 2 6 5 1 -10 CF -10+6-5=-9 -9 25 55 20 200 - SELECT THE CELL WITH THE LARGEST POSITIVE dj VALUE (+5) ie CELL AG v1 v2 v3 V4 - TRACE LOOP AG-BG-BE-AE vj 7 3 6 11 - SHIFT 40 UNITS FROM HIGHER COST CELL BG TO LOWER COST CELL AG - SHIFT 40 UNITS FROM CELL AE TO BEZ=7x20+3x40+2x10+5x50 SO THAT DEMAND SUPPLY CONSTRAINTS ARE NOT AFFECTED+10x40+1x40 = 970 -THIS GIVES US THE NEXT TABLE 2
36. 36. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN MODIFIED DISTRIBUTION METHOD TP TABLE 2 (NON OPTIMAL) dj IS NET OPPOR. AVAIL D E F G ui (ui+vj)=cij EMPTY NET COSTA 20 e 40 u1 u1=0 ROW A CELL d =(u +v )-c dj CHANGE 60 AD: v1= 7 ij i j ij 7 3 8 6 0 AE::v2= 3 AF 0+6-8=-2 -2B 50 50 u2 AG:v4= 6 BD -1+7-4=+2 +2 100 ROW B BE: u2= -1 BG -1+6-10=-5 -5 4 2 5 10 -1 BF: v3= 6 CD -5+7-2=-0 0C 40 u3 ROW C 40 CG: u3= -5 CE -5+3-6=-8 -8 2 6 5 1 -5 CF -5+6-5=-4 -4 25 55 20 200 -SOLN IS DEGENERATE SINCE NO. OF xij VARIABLES (5) IS LESS THAN (m+n-1=6). TWO v1 v2 v3 v3 RECENTLY VACATED CELLS ARE AE & BG vj ASSIGN e VALUE TO AE SINCE IT HAS LOWER 7 3 6 6 cij VALUE. PROCEED LIKE EARLIER STEP 1 - CELL BD HAS LARGEST dj VALUE =+2 - TRACE LOOP BD-AD-AE-BEZ=7x20+3xe+6x40+2x50 - SHIFT 20 UNITS FROM AD T0 BD+5x50+1x40 = 770 (3xe=0) - SHIFT 20 UNITS FROM BE TO AE
37. 37. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN MODIFIED DISTRIBUTION METHOD TP TABLE 3 (OPTIMAL) dj IS NET OPPOR. AVAIL D E F G ui (ui+vj)=cij EMPTY NET COSTA 20 40 u1 u1=0 ROW A CELL d =(u +v )-c dj CHANGE 60 AE::v2= 3 ij i j ij 7 3 8 6 0 AG: v4= 6 AD 0+5-7=-2 -2B 20 30 50 u2 ROW B AF 0+6-8=-2 -2 100 BD: v1= 5 BE: u2= -1 BG -1+6-10=-5 -5 4 2 5 10 -1 BF: v3= 6 CD -5+5-2=-2 -5C 40 u3 ROW C 40 - CG: u3= -5 CE -5+3-6=-8 -8 -5 CF -5+6-5=-4 -4 2 6 5 1 25 55 20 200 - SINCE ALL dj VALUE ARE NEGATIVE v1 v2 v3 V4 THEREFORE THIS SOLUTION vj 5 3 6 6 IS AN OPTIMAL SOLUTIONZ=3x20+6x40+4x20+3x30+5x50+1x40 = 730
38. 38. . EXAMPLE 4 OF TP (MAXIMISATION)A FERTILIZER COMPANY HAS THREE FACTORIES A, B, AND C WITHANNUAL CAPACITIES OF 200, 500 AND 300 THOUSAND TONNES OFUREA. THE PRODUCT IS DISTRIBUTED FROM FOUR WAREHOUSESD, E, F, AND G WITH ANNUAL OFFTAKE OF 180, 320, 100,AND 400 THOUSAND TONNES OF UREA. PROFIT(Rs LAKH PER THOUSAND TONNES) IS AS PER FOMWING TABLE.FIND OPTIMUM TRANSPORTATION SCHEDULE TO MAXIIMISE PROFIT. SOURCES DESTINATIONS SUPPLY D E F G A 12 8 6 25 200 B 8 7 10 18 500 C 14 3 11 20 300 DEMAND 180 320 100 400 1000
39. 39. TRANS SHIPMENT IN TPSI
40. 40. PROHIBITED ROUTES INTPSI
41. 41. . EXAMPLE 5 OF TP PROHIBITED ROUTES IN THE TP SOURCES DESTINATIONS SUPPLY ITERATIONS W1 W2 W3 U 1 2 3 4 5 P1 40 40 5 5 0 1 1 1 X M 7 8 45 P2 15 15 1 3 3 3 X 15 12 9 15 P3 25 25 15 15 1 1 4* X X 7 8 12 40 DEMAND 25 55 20 100 1 25 V 6 7 8 2 15 1 M-7* 1 1 3 2 X 1 1 40 3 X 5* 1 4 4 X X 1* 5 5 X 5 15
42. 42. TP NUMERICALSS
43. 43. TP NUMERICALSQ. NO. 1. (NWC RULE,SSMI METHODS & VAM, MODI METHODS)A PVC MANUFACTURING COMPANY HAS THREE FACTORIESA, B, AND C AND THREE WAREHOUSES D, E, AND F. THEMONTHLY DEMAND FROM THE WAREHOUSES AND THEMONTHLY PRODUCTION OF THE FACTORIES, IN THOUSANDOF TONNES OF PVC AND THE TRANSPORTATION COSTS PERUNIT ARE GIVEN IN THE FOLLOWING TABLE. WAREHOUSES MONTHLYFACTORIES D E F PRODNA 16 19 22 14B 22 13 19 16C 14 28 8 12MONTHLY DEMAND 10 15 17DETERMINE THE OPTIMAL SHIPPING SCHEDULE SO THATTHE TRANSPORTATION COST IS MINIMIZED USINGi) NWCR AND SSMii) VAM AND MODIFIED DISTRIBUTION METHOD
44. 44. TP NUMERICALSQ. NO. 2SOLVE Q. NO. 1, BY USING VAM ANDMODI DISTRIBUTION METHOD IF IT IS GIVENTHAT, MONTHLY PRODUCTION OF FACTORIES A,B AND C IS 16, 20 AND 12 THOUSAND TONNESRESPECTIVELY MONTHLY DEMAND OFWAREHOUSES D, E AND F IS 15, 15 AND 20THOUSAND TONNES RESPECTIVELY.
45. 45. TP NUMERICALSQ. No. 3A LIGHTING PRODUCTS COMPANY HAS FOUR FACTORIES F1, F2, F3,AND F4, WHICH PRODUCE 125, 250, 175 AND 100 CASES OF 200-WATTLAMPS EVERY MONTH.THE COMPANY SUPPLIES THESE LAMPS TO FOUR WAREHOUSES W1,W2, W3 AND W4 WHICH HAVE DEMAND OF 100, 400, 90 AND 60 CASESPER MONTH RESPECTIVELY. THE PROFIT IN Rs PER CASE, AS CASESARE SUPPLIED FROM A PARTICULAR FACTORY TO A PARTICULARWAREHOUES, IS GIVEN IN THE FOLLOWING MATRIX. WAREHOUSES W1 W2 W3 W4FACTORIES F1 90 100 120 110 F2 100 105 130 117 F3 111 109 110 120 F4 130 125 108 113DETERMINE THE TRANSPORTATION SCHEDULE SO THAT PROFIT ISMAXIMIZED GIVEN THE CONDITION THAT WARE HOUSE W1 MUST BESUPPLIED ITS FULL REQUIREMENT FROM FACTORY F1. USE VAM ANDMODIFIED DISTRIBUTION METHOD.ALSO SOLVE THE TP WITHOUT THE CONDITION GIVEN ABOVE USINGNWCR AND STEPPING STONE METHOD.
46. 46. TABLE 1 TP NUMERICALS ANS TO Q NO 3 W1 W2 W3 W4 Dj F1W3 -5F1 90 100 120 110 125 F1W4 -2F2 100 105 130 117 250 F2W1 IGNORE F3W1 IGNOREF3 111 109 110 120 175 F4W1 IGNOREF4 130 125 108 113 100 F3W3 -24 F3W4 -1 100 400 90 60 F4W3 -42 TABLE 2 OPTIMAL SOLN F4W4 -24 W1 W2 W3 W4 1 2 3 4 ui 1ST W1 GETS FULL QTY FROM F1F1 40 100 30 25 10 20 125 X 10 10 10 0F2 30 25 100 0 90 13 60 250 X 13 13* 12* -5 2ND SUPPLY FROM F4 EXHAUSTEDF3 19 21 175 20 10 175 X 10 10 11 -9F4 0 5 100 22 17 100 X 12 X X -25 3RD DEMAND FROM W3 MET 100 400 90 601 X X X X 4TH DEMAND FROM W4 MET2 X 16* 10 3 NOTE: In the first iteration for VOGEL we put an X for all rows and columns3 X 4 10 3 because the constraint is that warehouse 5th SUPPLY FROM W1 is to be supplied entire quantity from F1 EXHAUSTED4 X 4 X 3 factory F1 6th SUPPLY FROMvj 40 30 5 18 F2 EXHAUSTED 7TH DEMAND FROM W2 MET
47. 47. TP NUMERICALSQ. NO 4 ( DEGENERACY)SOLVE THE FOLLOWING TRANSPORTATION PROBLEM. D E F G SUPPLYA 7 3 8 6 60B 4 2 5 10 100C 2 6 5 1 40DEMAND 20 50 50 80
48. 48. ANS FOR Q NO. 4 TABLE 1 TABLE 2 D E F G Sup Ui D E F G Sup UiA 20 40 -2 +5 60 0 A 20 e -2 40 60 0B +2 10 50 40 100 -1 B +2 50 50 -5 100 -1C -5 -13 -9 40 40 - C 0 -8 -4 40 40 -5 10 Dmd 20 50 50 80Dmd 20 50 50 80 Vj 7 3 6 6Vj 7 3 6 11 TABLE 3 OPTIMAL D E F G Sup UiA -2 20 -2 40 60 0B 20 30 50 -5 100 -1C -2 -8 -4 40 40 -5Dmd 20 50 50 80Vj 5 3 6 6
49. 49. TP NUMERICALSQ. NO. 5SOLVE THE FOLLOWING TRANSPORTATION PROBLEMUSING VOGEL’S APPROXIMATION METHOD. TEST THISSOLUTION FOR OPTIMALITY USING THE MODI METHOD. DESTINATIONS SUPPLYSOURCES D E F GA 6 4 1 5 14B 8 9 2 7 16C 4 3 6 2 5DEMAND 6 10 15 4
50. 50. ANS FOR Q NO. 5 by VAM TABLE 1 Optimal D E F G Sup 1 2 3 UiA 4 10 -1 -1 14 3 1 2* 0B 1 -3 15 -1 16 5* 1 1 2C 1 -1 -8 4 5 1 1 1 -2Dmd 6 10 15 41 2 1 1 3 1 152 2 1 X 3* 23 2 1 X X 4Vj 6 4 0 4 3 10 4 1,4,1
51. 51. TP NUMERICALSQ. NO. 6A COMPANY MANUFACTURING PUMPS FOR DESERT COOLERS SELLS THEM TO ITS FIVE WHOLE-SELLERS A, B, C, D & E AT RS 250 EACH AND THEIR DEMAND FOR THE NEXT MONTH IS 300,300, 1000, 500 AND 400 UNITS RESPECTIVELY. THE COMPANY MAKES THESE PUMPS AT THREE FACTORIES F1, F2 & F3 WITH CAPACITIES OF 500, 1000 AND 1250 UNITS RESPECTIVELY. THE DIRECT COSTS OF PRODUCTION OF A PUMP AT THE THREE FACTORIES F1, F2 & F3 ARE RS 100, 90 AND 80 RESPECTIVELY. THE COSTS OF TRANSPORTATION FROM EACH FACTORY TO EACH WHOLE- SELLER ARE AS GIVEN IN THE FOLLOWING TABLE. WHOLESELLERSFACTORIES A B C D EF1 5 7 10 25 15F2 8 6 9 12 14F3 10 9 8 10 15DETERMINE THE MAXIMUM PROFIT THAT THE COMPANY CAN MAKE USING VOGEL APPROXIMATION METHOD AND MODI METHOD FOR CHECKING OPTIMALITY.
52. 52. ANS FOR Q NO. 6 PROFIT MATRIX A B C D EF1 250-100-5 250-100-7 250-100-10 250-100-25 250-100-15 145 143 140 125 135F2 250-90-8 250-90-6 250-90-9 250-90-12 250-90-14 152 154 151 148 146F3 250-80-10 250-80-9 250-80-8 250-80-10 250-80-15 160 161 162 160 155 A B C D E FDUMMY F1 17 19 22 37 27 0 500 F2 10 8 11 14 16 0 1000 F3 2 1 0 2 7 0 1250 300 300 1000 500 400 250
53. 53. ANS FOR Q NO. 6 by VAM TABLE 1 Optimal1 A B C D E F Sup 1 2 3 4 5 6 Ui2 F 250 -4 -4 -17 -4 250 500 -250 17* 2 2 2 5 X 03 1 17 19 22 37 27 04 F 50 300 250 400 -7 1000 -300 8 2 2 2 1 X -7 2 10 8 11 14 16 0 -250-400 F -3 -4 750 500 -2 -18 1250 -500 0 1 1 X X X -18 3 2 1 0 2 7 0 -750 D 300 300 1000 500 400 2505 1 8 7 11 12 9 0 CIRCLED NUMERALS SHOW dj VALUES 2 8 7 11 12* 9 X6 3 8 7 11* X 9 X We choose B and not C or E because B has 4 7 11* 11 X 11 X lower cost cell (1) compared to C or E 5 7 X 11* X 11 X We choose C and not E because B has lower cost cell (11) compared to E (16,27) 6 7 X X X 11* X Vj 17 15 18 20 23 0
54. 54. TP NUMERICALSQ. No. 7A COMPANY HAS FOUR FACTORIES F1, F2, F3, F4, MANUFACTURING THE SAME PRODUCT. PRODUCTION COSTS AND RAW MATERIALS COST DIFFER FORM FACTORY TO FACTORY AND ARE GIVEN IN THE FOLLOWING TABLE (FIRST TWO ROWS).THE TRANSPORTATION COSTS FROM THE FACTORIES TO SALES DEPOTS S1, S2, S3 ARE ALSO GIVEN.THE SALES PRICE PER UNIT AND REQUIREMENT AT EACH DEPOT ARE GIVEN IN THE LAST TWO COLUMNS. THE LAST ROW IN THE TABLE GIVES THE PRODUCTION CAPACITY AT EACH FACTORY.DETERMINE THE MOST PROFITABLE PRODUCTION AND DISTRIBUTION SCHEDULE AND THE CORRESPONDING PROFIT. THE SURPLUS PRODUCTION SHOULD BE TAKEN TO YIELD ZERO PROFIT. F1 F2 F3 F4 SALES REQUIRE PRICE MENTPRODN COST/UNIT 15 12 14 13 AT DIFF AT DIFFRAW MATL COST 10 9 12 9 DEPOTS DEPOTSTRANSPORT(TO S1) 3 9 5 4 34 80-ATION (TO S2) 1 7 4 5 32 120COSTS (TO S3) 5 8 3 6 31 150PRODN. CAPACITY 100 150 50 100
55. 55. ANS FOR Q NO. 7 PROFIT MATRIX S1 S2 S3F1 34-(15+10+3) 32-(15+10+1) 31-(15+10+5) =6 =6 =1F2 34-(12+9+9) 32-(12+9+7) 31-(12+9+8) =4 =4 =2F3 34-(14+12+5) 32-(14+12+4) 31-(14+12+3) =3 =2 =2F4 34-(13+9+4) 32-(13+9+5) 31-(13+9+6) =8 =5 =3 NEGATIVE PROFIT MATRIX S1 S2 S3 S4DUMMY) SUPPLY NOTE: SINCE SURPLUSF1 2 2 7 0 100 PRODUCTION 4 4 6 0 150 YIELDS ZEROF2 PROFIT, THEREF3 5 6 6 0 50 FORE, IN THE PROFIT MATRIXF4 0 3 5 0 100 S4 IS ASSIGNEDDEMAND 80 120 150 50 400 ZERO VALUE S IN THE CELLS
56. 56. TP NUMERICALSQ. NO. 8.AN OIL COMPANY HAS THREE REFINERIES R1, R2, R3 AND FOUR REGIONAL OIL DEPOTS D1, D2, D3 D4. THE ANNUAL SUPPLY AND DEMAND IN MILLION LITRES IS GIVEN BELOW ALONG WITH THE TRANSPORTATION COSTS IN TERMS OF RS THOUSANDS PER TANKER OF 10 KILOLITRES. SOURCES DESTINATIONS SUPPLY D1 D2 D3 D4 R1 5 10 5 20 5 7 2 4 R2 5 20 25 7 2 8 6 R3 10 10 4 5 10 5 DEMAND 15 5 10 25 55
57. 57. TP NUMERICALSQ. NO. 8 contdANSWER THE FOLLOWING QUESTIONS.i. IS THE SOLUTION FEASIBLE?ii. IS THE SOLUTION DEGENERATE?iii. IS THE SOLUTION OPTIMAL?iv. DOES THIS PROBLEM HAVE MULTIPLE OPTIMAL SOLUTIONS? IF SO DETERMINE THEM.v. IF THE TRANSPORTATION COST OF ROUTE R2 D1 IS REDUCED FROM RS 7 TO RS 6, WILL THERE BE ANY CHANGE IN THE SOLUTION?
58. 58. ANS FOR Q NO. 8ANSWER THE FOLLOWING QUESTIONS.i) IS THE SOLUTION FEASIBLE?Yes because it satisfies all supply and demand constraints.x11+x13+x14 = 20; x11+x31=15 and so on.ii) IS THE SOLUTION DEGENERATE?No because No. of occupied cells = (m+n-1)iii) IS THE SOLUTION OPTIMAL?Yes soln is optimal since one dij value is zero and other all dij values are negative. Z= 235 (SEE NEXT SLIDE)iv) DOES THIS PROBLEM HAVE MULTIPLE OPTIMAL SOLUTIONS? IF SO DETERMINE THEM.Yes it has multipal optimal soludtions since one dij value is zero. Trace the loop: R2D1-R1D1-R1D4-R2D4. Shift 5 units from R1D1to R1D4. Shift 5 units from R2D4 to R2D1. The new solution has the same Z value ie 235. (SEE SLIDE AFTER THE NEXT)v) IF THE TRANSPORTATION COST OF ROUTE R2 D1 IS REDUCED FROM RS 7 TO RS 6, WILL THERE BE ANY CHANGE IN THE SOLUTION?Yes. The cost will come down by Rs 5 to Rs 230. (SEE THIRD SLIDE FROM THIS)
59. 59. ANS FOR Q NO. 8 OPTIMALITY CHECK BY MODI Optimal Table D1 D2 D3 D4 Supply Ui CIRCLED NUMERALS SHOW dj VALUESR 5 -7 10 5 0 201 5 7 2 4R 0 5 -4 20 2 252 7 2 8 6 FOR FINDING THE SECOND OPTIMAL SOLN, TRACER 10 -6 -9 -2 -1 LOOP FROM R2D1 AS 103 4 5 10 5 SHOWN AND SHIFT CELLS AS SHOWN IN THE NEXTDemand 15 5 10 25 55 SLIDEVj 5 0 2 4 Z = 235
60. 60. ANS FOR Q NO. 8 MULTIPLE OPTIMALITY CHECK BY MODI 1st Optimal Table 2nd D1 D2 D3 Optimal Soln D4 Supply Ui CIRCLED NUMERALS SHOW dj VALUES R 0 -7 10 10 0 20 1 5 7 2 4 R 5 5 -4 15 2 25 2 7 2 8 6 R 10 -6 -9 -2 -1 10 3 4 5 10 5 Demand 15 5 10 25 55 Vj 5 0 2 4 Z = 235
61. 61. ANS FOR Q NO. 8 - TPT COST OF R2D1 CHANGED FROM 7 YO 6 Optimal Table D1 D2 D3 D4 Supply Ui CIRCLED NUMERALS SHOW dj VALUES R 0 -7 10 10 0 20 1 5 7 2 4 R 5 5 -4 15 2 25 2 6 2 8 6 R 10 -6 -9 -2 -1 10 3 4 5 10 5 Demand 15 5 10 25 55 Vj 5 0 2 4 Z = 230
62. 62. TP NUMERICALSQ. NO.9A LARGE BREAD-MANUFACTURING UNIT CAN PRODUCE SPECIALBREAD IN ITS TWO PLANTS P AND Q WITH MANUFACTURING CAPACITYOF 5000 AND 4200 LOAVES OF BREAD PER DAY RESPECTIVELY ANDCOST OF PRODUCTION OF Rs10 AND Rs 12 PER LOAF OF BREADRESPECTIVELY.FOUR RETALING CHAINS A,B,C,AND D PURCHASE BREAD FROM THISCOMPANY. THEIR DEMAND PER DAY IS RESPECTIVELY3600,4600,1100,AND 3500 LOAVES OF BREAD AND THE PRICES THATTHEY PAY PER LOAF OF BREAD ARE RESPECTIVELY Rs 19,17,20 AND18.THE COST OF TRANSPORTATION AND HANDLING IN Rs PER LOAF FORDELIVERY TO VARIOUS STORES OF THE RETAILING CHAINS IS ASFOLLOWS.PLANT RETAILING CHAINS A B C DP 1 2 3 2Q 4 1 2 1DETERMINE THE DELIVERY SCHEDULE FOR THE BREADMANUFACTURING COMPANY THAT WILL MAXIMIZE ITS PROFITS. WRITEA DUAL OF THE TP
63. 63. TP Q. NO. 9 – FOR INFO SUMMARYQ. NO.9A LARGE BREAD-MANUFACTURING UNIT CAN PRODUCE SPECIAL BREAD IN ITS TWOPLANTS AS PER DETAILS GIVE BELOW.PLANT Mfg CAP COST OF PRODN. LOAVES/DAY Rs PER LOAF OF BREADP 5000 10Q 4200 12FOUR LARGE RETALING CHAINS PURCHASE BREAD FROM THIS COMPANY. THEIRDEMAND AND THE PRICES THAT THEY PAY ARE GIVEN BELOW.RETAILING MAX DEMAND PRICE LOAVES/DAY RS PER LOAFA 3600 19B 4600 17C 1100 20D 3500 18THE COST OF TRANSPORTATION AND HANDLING IN Rs PER LOAF FOR DELIVERY TOVARIOUS STORES OF THE RETAILING CHAINS IS AS FOLLOWS.PLANT RETAILING CHAINS A B C DP 1 2 3 2Q 4 1 2 1DETERMINE THE DELIVERY SCHEDULE FOR THE BREAD MANUFACTURING COMPANYTHAT WILL MAXIMIZE ITS PROFITS. WRITE A DUAL OF THE TP
64. 64. ANS FOR Q NO. 9 PROFIT MATRIX A B C D SUPPLYP 19-10-1= 17-10-2= 20-10-3= 18-10-2= 5000 8 5 7 6Q 19-12-4= 17-12-4= 20-12-2= 18-12-1= 4200 3 4 6 5R DUMMY 3600SOURCE 0 0 0 0DEMAND 3600 4600 1100 3500 12800 NEGATIVE PROFIT MATRIX A B C D Supply P 0 3 1 2 5000 Q 5 4 2 3 4200 R 8 8 8 8 3600 Demand 3600 4600 1100 3500 12800k
65. 65. ANS FOR Q NO. 9 by VAM TABLE 1 Optimal A B C D Sup 1 2 3 4 5 6 Ui 1 3600P 3600 3600 0 1100 1100 300 5000- 300 3600 1 1* 1* X X X 0. 0 3 1 2 1100-300 2 1100Q -4 1000 1000 0 3200 4200- 3200 3200- 1 1 1 1 X X 1. 5 4 2 3 1000 3R -3 3600 3600 -2 -1 3600 0 0 0 0 X X 5 300. 8 8 8 8D 3600 4600 1100 3500 In 2nd iteration, we choose P row and 4 not other row or cols because P has 32001 5* 1 1 1 lowest cost cell (1) compared to all others2 X 1 1 1 5 In 3rd iteration, we choose P row and 10003 x 1 x 1 not other row or cols because P has4 x 4 x 5* lowest cost cell (2) compared to all others 65 X X X X 36006 X X X XVj 0 3 1 2 CIRCLED NUMERALS SHOW dj VALUES
66. 66. TP NUMERICALSQ. NO. 10 (TRANSSHIPMENT PROBLEM)A TRANSPORTER HAS DETERMINED THE COST OF TRANSPORTATIONPER PACKAGE FOR A CUSTOMER’S PRODUCT IS AS PER TABLE GIVENBELOW. EVERY WEEK HE HAS TO PICK UP300 PACKAGES FROMSOURCE S1 AND 200 PACKAGES FROM SOURCE S2 AND DELIVER 100PACKAGES TO DESTINATION D1 AND 400 PACKAGES TO DESTINATIOND2. THE TRANSPORTER HAS THE OPTION OF EITHER SHIPPINGDIRECTLY FROM THE SOURCES TO THE DESTINATIOS OR TOTRANSSHIP IF ECONOMICAL. DETERMINE THE OPTIMUM SHIPPINGSCHEDULE, WITH WOULD MINIMISE COST OF TRANSPORTATION. SOURCES DESTINATIONS S1 S2 D1 D2 S1 0 18 5 10 S2 18 0 8 16 D1 5 8 0 3 D2 10 16 3 0
67. 67. ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI TABLE 1 Optimal S1 S2 D1 D2 Sup 1 2 3 4 5 6 Ui 1 500S1 500 500 -20 300 300 -9 300+500 5 5 5* 5* X X 0 0 18 5 10 2S2 500 500 200 200 -5 200+500 8* 8* X X X X 2 200 -16 18 0 8 16D1 -16 100 100 400 500 400 3 3 3 3 3* X -6 3 -11 500 5 8 0 3D2 500 500 500 3 3 3 3 3 X -9 -19 -27 -6 CIRCLED NUMERALS SHOW dij VALUES 4 10 16 3 0 The total number of units trans 300D 500 500 100+500 400+500 2500 ported from all the sources to1 5 8 3 3 all the destinations is 500. This 5 qty is added to each supply and 1002 5 X 3 3 each demand and TP is solved3 5 X 3 3 The interpretation of this is that S1 will4 X X 3 3 transport300 units to D1 and S2 will 6 transport 200 units to D1. D1 will trans 5005 X X 3 3 thip 400 units to D2. This means that6 X X X 3* D2 will not get its packages from S1 or S2 but will get 400 units trans shippedVj 0 -2 6 9 trom D1.
68. 68. TP NUMERICALSQ. NO. 11 (TRANSSHIPMENT PROBLEM)A COMPANY HAS TWO FACTORIES F1 AND F2 HAVING PRODUCTIONCAPACITY OF 200 AND 300 UNITS RESPECTIVELY. IT HAS THREEWAREHOUSES W1,W2 AND W3, HAVING DEMAND EQUAL TO 100, 150AND 250 RESPECTIVELY. THE COMPANY HAS THE OPTION OF EITHERSHIPPING DIRECTLY FROM THE FACTORIES TO THE WAREHOUSES ORTO TRANSSHIP IF ECONOMICAL. DETERMINE THE OPTIMUM SHIPPINGSCHEDULE, WITH MINIMUM COST OF TRANSPORTATION. FACTORIES WAREHOUSES F1 F2 W1 W2 W3 F1 0 8 7 8 9 F2 6 0 5 4 3 W1 7 2 0 5 1 W2 1 5 1 0 4 W3 8 9 7 8 0
69. 69. . ANS TO Q. NO. 11TRANSSHIPMENT PROBLEM) FACTORIES WAREHOUSES F1 F2 W1 W2 W3 SUPPLY F1 0 8 7 8 9 200+500 =700 F2 6 0 5 4 3 300+500 =800 W1 7 2 0 5 1 500 W2 1 5 1 0 4 500 W3 8 9 7 8 0 500 DEMAND 500 500 100+ 150+ 250+ 2500 500 500 500
70. 70. TP NUMERICALSQ. NO. 12 (PROHIBITED ROUTES)A TOY MANUFACTURER HAS DETERMINED THAT DEMAND FOR A PARTICULAR DESIGN OF TOY CAR FROM VARIOUS DISRIBUTORS IS 500, 1000, 1400 AND 1200 FOR THE 1ST , 2ND , 3RD , AND 4TH WEEK OF THE NEXT MONTH WHICH MUST BE SATISFIED.THE PRODUCTION COST PER UNIT IS RS 50 FOR THE FIRST TWO WEEKS AND RS 60 PER UNIT FOR THE NEXT TWO WEEKS DUE TO EXPECTED INCREASE IN COST OF PLASITIC. THE PLANT CAN PRODUCE MAXIMUM OF 1000 UNITS PER WEEK. THE MANUFACTURER CAN ASK EMPLOYEES TO WORK OVER TIME DURING THE 2ND AND THE 3RD WEEK WHICH INCREASES THE PRODUCTION BY ADDITIONAL 300 UNITS BUT ALSO IT INCREASES THE COST BY RS 5 PER UNIT. EXCESS PRODUCTION CAN BE STORED AT A COST OF RS 3 PER UNIT PER WEEK.DETERMINE THE PRODUCTION SCEHEDULE SO THAT TOTAL COST IS MINIMISED.
71. 71. . ANS TO Q. NO. 12 PROHIBITED ROUTE TPPRODUCTION WEEK COST OF PRODUCTION PER UNIT WK1 WK2 WK3 WK4 DUMMY SUPPLY DEMANDWEEK1 (NORMAL) 50 53 56 59 0 1000WEEK2 (NORMAL) M 50 53 56 0 1000WEEK2 (OVERTIME) M 55 58 61 0 300WEEK3 (NORMAL) M M 60 63 0 1000WEEK3 (OVERTIME) M M 65 68 0 300WEEK4 (NORMAL) M M M 60 0 1000DEMAND 500 1000 1400 1200 500 4600
72. 72. ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI 10 TABLE 1 OPTIMAL SOLN. 200A W1 W2 W3 W4 DUM Sup Ui 1 2 3 4 5 6 7 8 9 1W 500 500 500 500 1000 0 50 50 50 3 3 X X X 500 0 0 -4 . *1 50 53 56 59 0W -M 500 500 500 500 1000 -3 50 50 50 3 3 3 3 X 2 0 -72 M +47 50 53 56 0 . 300W -M 100 200 100 200 300 2 55 55 55 3 3 3 3 32OT M +52 55 58 61 0 -2 . 3 200W -M -M 800 800 200 200 1000 4 60 60 60 M- 60 X X X X X3 M +54 M +57 60 63 0 0 . * * 4W -M -M -5 -5 300 300 300 4 65 65 X X X X X X X 8003OT M +54 M +57 65 68 0 . *W -M -M -M 1000 1000 1000 1 60 60 60 M- M- X X X X 5 -3 60 604 M +51 M +54 M +57 60 0 * 1000 4600 6D 500 1000 1400 1200 500 CIRCLED NUMERALS SHOW dij VALUES 500Vj 50 53 56 59 -4 71 M – 50* 3 3 3 0 Interpretation: -Co. will make 1000 units in 1 st week though dmd2 X 3 3 3 0 5003 X 3 3 3 0 is only 500 units. It will sell 500 of these in the4 X 3 3 3 X 1st week and 500 in the 2nd week. 8 -I will produce 300 by running OT in the 2 nd week5 X 3 3 3 X 5006 X 3 3 3 X It will sell 100 of these in the 3rd week and 200 of7 X 5* 5 5 X these in the 4th week. 9 -It will not run OT in the 3 rd week.8 X X 5* 5 X 100
73. 73. TP NUMERICALSQ. NO. 13 (PROHIBITED ROUTES)A COMPANY IS PLANNING ITS NEXT FOUR WEEKS’ PRODUCTION. THE PER UNIT PRODUCTION COST IS RS 10 FOR THE FIRST TWO WEEKS AND RS 15 FOR THE NEXT TWO WEEKS. DEMAND IS 300, 700, 900 AND 800 FOR THE 1ST, 2ND, 3RD, AND 4TH WEEK, WHICH MUST BE MET.THE PLANT CAN PRODUCE MAXIMUM OF 700 UNITS PER WEEK. THE COMPANY CAN ASK EMPLOYEES TO WORK OVER TIME DURING THE 2ND AND THE 3RD WEEK WHICH INCREASES THE PRODUCTION BY ADDITIONAL 200 UNITS BUT ALSO IT INCREASES THE COST BY RS 5 PER UNIT. EXCESS PRODUCTION CAN BE STORED AT A COST OF RS 3 PER UNIT PER WEEK
74. 74. . TP NUMERICALS Q. NO. 14 A FMCG COMPANY HAS THREE WARE HOUSES W1,W2 AND W3 AND SUPPLIES PRODUCTS FROM THESE WAREHOUSES TO THREE DISTRIBUTORS D1,D2 AND D3. FMCG COMPANY HAS DETERMINED THAT DURING THE NEXT MONTH, THERE WILL BE A SHORT FALL IN SUPPLY AGAINST THE PROJECTED DEMAND. IT HAS AGREED TO PAY A PENALTY PER UNIT AS PER THE TABLE GIVEN BELOW TO DISTRIBUTORS FOR DEMAND THAT IS NOT MET. FIND THE DELIVERY SCHEDULE THAT THE COMPANY SHOULD FOLLOW TO MINIMISE TRANSPORTATION COSTS AND PENALTY COST AND DETERMINE VALUES OF BOTH COSTS. SOURCES DESTINATIONS SUPPLY D1 D2 D3 W1 5 1 7 100 W2 6 4 6 800 W3 3 2 5 150 DEMAND 1050 750 200 500 1450 PENALTY 5 3 2
75. 75. . ANS TO Q. NO. 14SOURCES DESTINATIONS SUP D1 D2 D3 Sup Ui 1 2 3 4 5 6 7 8 9 W1 100 100 5 1 7 W2 600 100 100 800 6 4 6 W3 150 150 3 2 5 W4 400 400 DUMMY 5 3 2 DEMAND 750 200 500 1450 The transportation is Rs 5150. The penalty cost is Rs 800 V 1 There is a shortfall of 400 units. We 2 create a dummy warehouse (source) 3 with a supply capability of 400 units. The penalty cost per unit payable to 4 the distributors is put in the cells in row.
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