transportation problem

Transportation Problem

Transportation ProblemIntroductionThe Transportation Problems are one of the type of Linear Programming ProblemThe objective is to minimize the cost of distribution a product from a number of sources or origins to a number of destinations in such a manner that the cost of transportation is minimum.

The Transportation Model CharacteristicsA product is transported from a number of sources to a number of destinations at the minimum possible cost.Each source is able to supply a fixed number of units of the product, and each destination has a fixed demand for the product.The linear programming model has constraints for supply at each source and demand at each destination.All constraints are equalities in a balanced transportation model where supply equals demand.Constraints contain inequalities in unbalanced models where supply does not equal demand.

33The Transportation ModelCharacteristicsTransportation modelling is an iterative procedure for solving problems that involve minimizing the cost of shipping products from a series of sources to a series of destinations.

Origin Points (or sources) can be factories, warehouses, car rental agencies like AVIS or any other points from which goods are shipped.

Destinations are any points that receive goods.

Transportation models are useful when considering alternative facility locations. The choice of a new location depends on which will yield the minimum cost for the entire system 44The Transportation Model CharacteristicsTo use the transportation model we need to know the following:

The origin points and the capacity or supply per period at eachThe destination points and the demand per period at eachThe cost of shipping one unit from each origin to each destination

The transportation problem can be regarded as a generalization of the assignment problem. In an assignment problem we have a number of origins each processing one item, and the same number of destinations each requiring one item and are asked to empty the origins and fill the destinations in such a way the Total Effectiveness is optimized. In a transportation problem we have m origins, with origin i processing ai items and n destinations (possibly different number from m) with destination j requiring bj items, and with ai = bj.55A Balanced Transportation ProblemThe total supply is equal to the total demand

A Balanced Transportation ProblemTotal Supply = 1000+1500+1200 = 3700Total Demand = 2300 + 1400 = 3700Total Supply = Total DemandThis is a balanced Transportation Problem.DenverMiamiSUPPLYLos Angeles802151000Detroit1001081500New Orleans102681200DEMAND23001400UnBalanced Transportation ProblemTotal Demand > Total SupplyAdd dummy column in matrix with zero costTotal Supply > Total DemandAdd dummy row in matrix with zero costDEMAND > SUPPLY

SUPPLY > DEMAND

Initial Basic Feasible SolutionCan determining using 3 type available methodNorth-West Corner MethodLeast Cost MethodVogels Approximation MethodNorth-West Corner MethodMost systematic and easiest method for obtaining initial feasible basic solution

North-West Corner Method AlgorithmStep1: Select the upper left (north-west) cell of the transportation matrix and allocate the maximum possible value to X11 which is equal to min(a1,b1).

Step2: If allocation made is equal to the supply available at the first source (a1 in first row), then move vertically down to the cell (2,1).If allocation made is equal to demand of the first destination (b1 in first column), then move horizontally to the cell (1,2).If a1=b1 , then allocate X11= a1 or b1 and move to cell (2,2).

Step3: Continue the process until an allocation is made in the south-east corner cell of the transportation table.North-West Corner ProblemCheck:Total Supply = 15+25+10 = 50Total Demand = 5+15+15+15= 50Total Supply = Total DemandThis is a balanced Transportation Problem.The feasible solution is exist.

abcdSUPPLYA102201115B12792025C414161810DEMAND5151515North-West Corner Problem (continue)Step1:Select cell (A1,a1)Allocate the maximum possible value to Supply1 which is equal to min Demand1

abcdSUPPLYA5102201115-5 =10B12792025C414161810DEMAND51515150North-West Corner Problem (continue)Step2:Select cell (A1,b1)Allocate the maximum possible value to Supply1 which is equal to min Demand2

abcdSUPPLYA510102201110B12792025C414161810DEMAND015-10 = 515150North-West Corner Problem (continue)Step3:Select cell (B2,b2)Allocate the maximum possible value to Supply2 which is equal to min Demand2

abcdSUPPLYA51010220110B125792025-5 = 20C414161810DEMAND0515150North-West Corner Problem (continue)Step4:Select cell (B2,b2)Allocate the maximum possible value to Supply2 which is equal to min Demand2

abcdSUPPLYA51010220110B12571592020-15 = 5C414161810DEMAND0015150North-West Corner Problem (continue)Step5:Select cell (B2,b2)Allocate the maximum possible value to Supply2 which is equal to min Demand3

abcdSUPPLYA51010220110B12571595205C414161810DEMAND00015-5 = 100North-West Corner Problem (continue)Step6:Select cell (B2,b2)Allocate the maximum possible value to Supply2 which is equal to min Demand3

abcdSUPPLYA51010220110B12571595200C41416101810-10 = 0DEMAND000100North-West Corner Problem (continue)Step7:Minimum Transportation Cost Z = (5*10)+(10*2)+(5*7)+(15*9)+(5*20)+(10*18)=$520Feasible Basic solution is X11 = 5 , X12 = 10 , X22 = 5 , X23 = 15 , X24 = 5 , X34 = 10abcdSUPPLYA51010220110B12571595200C4141610180DEMAND0000Least Cost MethodThis method takes into consideration the lowest cost and therefore takes less time to solve the problemLeast Cost Method AlgorithmStep1: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.

Step2: Then eliminate the row or column in which supply or demand is exhausted. If both the supply and demand values are same, either of the row or column can be eliminated.In case, the smallest unit cost is not unique, then select the cell where maximum allocation can be made.

Step3: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Least Cost MethodCheck:Total Supply = 15+25+10 = 50Total Demand = 5+15+15+15= 50Total Supply = Total DemandThis is a balanced Transportation Problem.The feasible solution is exist.

abcdSUPPLYA102201115B12792025C414161810DEMAND5151515Least Cost Method (continue)Step1: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.

abcdSUPPLYA102201115B12792025C414161810DEMAND5151515Arbitrary selection = random selection25Least Cost Method (continue)Step2: Then eliminate the row or column in which supply or demand is exhausted. If both the supply and demand values are same, either of the row or column can be eliminated.In case, the smallest unit cost is not unique, then select the cell where maximum allocation can be made.

abcdSUPPLYA10152201115-15 =0B12792025C414161810DEMAND51515150Arbitrary selection = random selection26Least Cost Method (continue)Step3: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Step4: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.

abcdSUPPLYA1015220110B12792025C414161810DEMAND501515Arbitrary selection = random selection27Least Cost Method (continue)Step5: Then eliminate the row or column in which supply or demand is exhausted. If both the supply and demand values are same, either of the row or column can be eliminated.In case, the smallest unit cost is not unique, then select the cell where maximum allocation can be made.abcdSUPPLYA1015220110B12792025C5414161810-5 = 5DEMAND5015150Arbitrary selection = random selection28Least Cost Method (continue)Step6: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Step7: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.


abcdSUPPLYA1015220110B1271592025-15 = 10C541416185DEMAND0015150Arbitrary selection = random selection30Least Cost Method (continue)Step9: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Step10: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.


abcdSUPPLYA10152200110B1271592025-15 = 10C541416185DEMAND00015-0 = 150Arbitrary selection = random selection32Least Cost Method (continue)Step12: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Step13: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.


abcdSUPPLYA10152200110B1271592010C5414165185DEMAND00015-5 = 100Arbitrary selection = random selection34Least Cost Method (continue)Step15: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Step16: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell.


abcdSUPPLYA10152200110B127159102010-10 = 0C5414165185DEMAND000100Arbitrary selection = random selection36Least Cost Method (continue)Step 6:Calculate total minimum cost

abcdSUPPLYA10152200110B12715910200C5414165180DEMAND0000Arbitrary selection = random selection37Least Cost Method (continue)Step 6:Total minimum cost = $475Feasible Basic solution is X12=2, X31=4, X23=9, X14=11, X34=18, X24=20Here, we can see that the Least Cost Method involves a lower cost than the North-West Corner Method.

abcdSUPPLYA10152200110B12715910200C5414165180DEMAND0000ShipmentUnitsShippedTrans. costPer unitTotal CostFromToAb15230Ca5420BC159135Ad0110Bd1020200Cd51890TOTAL475Arbitrary selection = random selection38Vogels Approximation MethodStep1: Calculate penalty for each row and column by taking the difference between the two smallest unit costs. This penalty or extra cost has to be paid if one fails to allocate the minimum unit transportation cost.

Step2: Select the row or column with the highest penalty and select the minimum unit cost of that row or column. Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.

Step3: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

Step4: Repeat the process until all the supply sources and demand destinations are satisfied.

Vogels Approximation ProblemCheck:Total Supply = 15+25+10 = 50Total Demand = 5+15+15+15= 50Total Supply = Total DemandThis is a balanced Transportation Problem.The feasible solution is exist.

abcdSUPPLYA102201115B12792025C414161810DEMAND5151515Vogels Approximation Problem (continue)Step1: Calculate penalty for each row and column by taking the difference between the two smallest unit costs. This penalty or extra cost has to be paid if one fails to allocate the minimum unit transportation cost.

abcdSUPPLYRow diff.A10220111510-2 = 8B127920259-7 = 2C41416181014-4 = 10DEMAND5151515Col. Diff.10-4 = 67-2 = 516-9 = 718-11 = 7Vogels Approximation Problem (continue)Step2: Select the row or column with the highest penalty and select the minimum unit cost of that row or column.Between penalties 8, 2, 10, 6, 5, 7, 7 select 10Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.Between allocation costs 4, 14, 16, 18 select 4

abcdSUPPLYRow diff.A10220111510-2 = 8B127920259-7 = 2C41416181014-4 = 10DEMAND5151515Col. Diff.10-4 = 67-2 = 516-9 = 718-11 = 7Vogels Approximation Problem (continue)Step3: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

abcdSUPPLYRow diff.A102201115B12792025C5414161810-5= 5DEMAND5151515Col. Diff.-0Vogels Approximation Problem (continue)Step4: Repeat the process until all the supply sources and demand destinations are satisfied.

abcdSUPPLYRow diff.A10220111511-2 = 9B127920259-7 = 2C54141618516-14 = 2DEMAND0151515Col. Diff.-7-2 = 516-9 = 718-11 = 7Vogels Approximation Problem (continue)Step5:Between penalties 9, 2, 2, 5, 7, 7 select 9Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.Between allocation costs 2, 20, 11 select 2

abcdSUPPLYRow diff.A1015220111511-2 = 9B127920259-7 = 2C54141618516-14 = 2DEMAND015-15 = 01515Col. Diff.-7-2 = 516-9 = 718-11 = 7Vogels Approximation Problem (continue)Step6: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

abcdSUPPLYRow diff.A10152201115-B127920259-7 = 2C54141618516-14 = 2DEMAND015-15 = 01515Col. Diff.--16-9 = 718-11 = 70Vogels Approximation Problem (continue)Step7:Between penalties 11, 2, 7, 2 select 11Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.Between allocation costs 9, 20 select 9

abcdSUPPLYRow diff.A1015220110-B1279202520-9 = 11C54141618518-16 = 2DEMAND001515Col. Diff.--16-9 = 720-18 = 2Vogels Approximation Problem (continue)Step8: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

abcdSUPPLYRow diff.A1015220110-B1271592025-15=1020-9 = 11C54141618518-16 = 2DEMAND001515Col. Diff.--16-9 = 720-18 = 20Vogels Approximation Problem (continue)Step9:Between penalties 20, 18, 2 select 20Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.Between allocation costs 20 select 20

abcdSUPPLYRow diff.A1015220110-B127159201020 = 20C54141618518 = 18DEMAND00015Col. Diff.---20-18 = 2Vogels Approximation Problem (continue)Step10: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

abcdSUPPLYRow diff.A1015220110-B12715910201020-15 = 5C54141618518 = 18DEMAND00015-10=5Col. Diff.---20-18 = 20Vogels Approximation Problem (continue)Step11:Between penalties 18, 18 select 18Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.Between allocation costs 18 select 18

abcdSUPPLYRow diff.A1015220110-B12715910200-C54141618518 = 18DEMAND0005Col. Diff.---18 = 18Vogels Approximation Problem (continue)Step12: Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value. Any row or column having zero supply or demand, can not be used in calculating future penalties.

abcdSUPPLYRow diff.A1015220110-B12715910200-C541416518518 = 18DEMAND0005-0 = 0Col. Diff.---18 = 180Vogels Approximation Problem (continue)Step13:Total minimum cost (5*4)+ (15*2)+ (15*9)+ (10*20)+ (5*18) = $475Feasible Basic solution is X31=4, X12=2, X23=9, X24=20, X34=18

abcdSUPPLYRow diff.A1015220110-B12715910200-C5414165180-DEMAND0000Col. Diff.----Optimal Solution Once an initial basic feasible solution (Nortwest Corner Rule, Least Cost or Vogels Approximation) is obtained, the next step is to check its optimality. An optimal solution is one where there is no other set of transportation routes (allocations) that will further reduce the total transportation cost. Thus, we have to evaluate each un-occupied cell in the transportation table in terms of an opportunity of reducing total transportation cost. If the following is not an optimal solution for the transportation problem then modify it to obtain the optimal solution.

Checking AcceptabilityTest the initial feasible solution for acceptability (checking whether this feasible solution can proceed for optimality test or not). It should be non-degenerate basic feasible solutionNon-degenerate has following two properties:Number of allocation should be equal with m+n-1Where m is number of rows and n is number of column All allocation should be in independent positions.(said to be in independent positions, if it is impossible to form a closed path)

Checking Acceptability (continue)Step1:Allocation = m+n-1 where m=3 , n=4 = 3 + 4 1 = 6All allocations are in independent positions. abcdSUPPLYA51010220110B12571595200C4141610180DEMAND0000Rows x columnm x n3 x 4Setting opportunity cost for un-occupied cellsStep2:Determine a set of m+n numbersui (i=1,2,....m) and vj (j=1,2,...n)such that for each occupied cells (r,s)crs = ur+vsStep3:Calculate cell evaluations (unit cost difference)dj for each empty cell (i,j) by using the formuladij = cij ( ui+vj )Setting opportunity cost for un-occupied cells (continue)Step2:There will be 7 variable and 6 basic equation.To solve, need to set ui = 0 and then solve other variable as followed. (For this consider the row or column in which the allocations are maximum)Basic VariableEquation (u,v)SolutionX22U2 + V2 = 7U2 = 0 V2 = 7

X23U2 + V3 = 9U2 = 0 V3 = 9X24U2 + V4 = 20U2= 0 V4= 20X12U1 + V2 = 2V2= 7 U1 = -5X11U1 + V1 = 10U1 = -5 V1 = 15X34U3 + V4 = 18V4 =20 U3 = -2U1=-5 , U2=0, U3=-2V1=15, V2=7, V3=9, V4=20 Ui where i is row 2V1=15V2=7V3=-9V4=20SUPPLYU1=-5510102201115U2=0125715952025U3=-241416101810DEMAND515151558Setting opportunity cost for un-occupied cells (continue)Step3:Calculate cell evaluations (unit cost difference)dj for each empty cell by using the formuladij = Cij ( Ui+Vj )

Empty Cell Variable Equation (u,v) Ui + Vj

unit cost differencedij = Cij ( Ui+Vj )X13U1 + V3 = (-5)+9 = 4C13 - 4=20 - 4 = 16X14U1 + V4= (-5)+20 = 15C14 15=11 -15 = -4X21U2 + V1 = 0+15 = 15C21 - 15 =12 - 15 = -3X31U3 + V1 = (-2)+15 = 13C31 - 13=4 - 13 = -9X32U3 + V2 = (-2)+7 = 5C32 - 5=14 - 5 = 9X33U3 + V3 = (-2)+9 = 7C33 -7=16 - 7 = 9X13=16 , X14=-4, X21=-3, X31=-9, X32=9, X33=9 abcdSUPPLYA510102201115B125715952025C41416101810DEMAND5151515U1=-5 , U2=0, U3=-2V1=15, V2=7, V3=9, V4=20 Check Optimal or NotStep4:Examine the matrix of cell evaluation dij for negative entries and conclude that(i) If all dij > 0 => Solution is optimal and unique.(ii) If all dij 0 => At least one dij = 0 => Solution is optimal and alternate solution also exists.(iii) If at least one dij < 0 => Solution is not optimal. If it is so, further improvement is required by repeating the above process. See step 5 and onwards.BasicX11X12

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Z0016-4-3000-9990X13=16 , X14=-4, X21=-3, X31=-9, X32=9, X33=9 BasicX11X12

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Z0016-4-3000-9990Repeat Process to find optimal solutionStep5:See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.V1=10V2=2V3=4V5=15SUPPLYU1=05101021620-41115U2=5-3125715952025U3=3-94914916101810DEMAND5151515First LoopX13=16 , X14=-4, X21=-3, X31=-9, X32=9, X33=9 V1=10V2=2V3=4V5=15SUPPLYU1=0510102201115U2=5125715952025U3=3-941416101810DEMAND5151515Repeat Process to find optimal solutionStep5:See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.Min [5- , 10 - ] = 0 5 - = 0 = 5--+-++abcdSUPPLYA510102201115B125715952025C-941416101810DEMAND5151515Repeat Process to find optimal solution--+-++abcdSUPPLYA10152201115B1207159102025C54141651810DEMAND515151510 + =10 + 5=155 - =5 - 5=05 + =5 + 5=105 - =5 - 5=0Minimum Transportation Cost Z = (15*2)+(0*7)+(15*9)+(10*20)+(5*18)+(5*4)=$475V1=6V2=7V3=9V4=20SUPPLYU1=-510152201115U2=01207159102025U3=-254141651810DEMAND5151515Setting opportunity cost for un-occupied cells (continue)Step2:There will be 7 variable and 6 basic equation.To solve, need to set ui = 0 and then solve other variable as followed. (For this consider the row or column in which the allocations are maximum)Basic VariableEquation (u,v)SolutionX22U2 + V2 = 7U2 = 0 V2 = 7

X23U2 + V3 = 9U2 = 0 V3 = 9X24U2 + V4 = 20U2= 0 V4= 20X12U1 + V2 = 2V2= 7 U1 = -5X34U3 + V4 = 18V4 =20 U3 = -2X31U3 + V1 = 4U3 = -2 V1 = 6U1=-5 , U2=0, U3=-2V1=6, V2=7, V3=9, V4=20 Ui where i is row 2Setting opportunity cost for un-occupied cells (continue)Step3:Calculate cell evaluations (unit cost difference)dj for each empty cell by using the formuladij = Cij ( Ui+Vj )


unit cost differencedij = Cij ( Ui+Vj )X11U1 + V1 = (-5)+6 = 1C11 - 1=10 - 1 = 9X13U1 + V3= (-5)+9 = 4C14 4=11 -4 = 7X14U2 + V4 = 0+20 = 20C14 - 20 =11 - 20 = -4X21U2 + V1 = (-2)+6 = 4C21 - 4=12 - 4 = 8X32U3 + V2 = (-2)+7 = 5C32 - 5=14 - 5 = 9X33U3 + V3 = (-2)+9 = 7C33 -7=16 - 7 = 9X11=1 , X13=7 X14=-4, X21=8, X32=9, X33=9 abcdSUPPLYA10152201115B1207159102025C54141651810DEMAND5151515U1=-5 , U2=0, U3=-2V1=6, V2=7, V3=9, V4=20 V1=6V2=7V3=9V4=20SUPPLYU1=-5910152720-41115U2=081207159102025U3=-25491491651810DEMAND5151515BasicX11X12

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Z907-480000990Repeat Process to find optimal solutionStep5:See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.Second LoopX11=1 , X13=7 X14=-4, X21=8, X32=9, X33=9 Repeat Process to find optimal solutionStep5:See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.Min [15- , 10 - ] = 0 10 - = 0 = 10abcdSUPPLYA1015220-41115B1207159102025C54141651810DEMAND5151515--++abcdSUPPLYA105220101115B121071592025C54141651810DEMAND5151515abcdSUPPLYA1015220-41115B1207159102025C54141651810DEMAND5151515Repeat Process to find optimal solution--++15 - =15 + 10=50 + =0 +-10=10Minimum Transportation Cost Z = (5*2)+(10*11)+(10*7)+(15*9)+(5*4)+(5*18)=$435Setting opportunity cost for un-occupied cells (continue)Step2:There will be 7 variable and 6 basic equation.To solve, need to set ui = 0 and then solve other variable as followed. (For this consider the row or column in which the allocations are maximum)Basic VariableEquation (u,v)SolutionX12U1 + V2 = 2U1 = 0 V2 = 5

X14U1 + V4 = 11U1 = 0 V4 = 11X22U2 + V2 = 7V2 = 5 U2 = 2X23U2 + V3 = 9U2= 2 V3 = 7X34U3 + V4 = 18V4 =11 U3 = 7X31U3 + V1 = 4U3= 7 V1= -3U1=0 , U2=2, U3=7V1=-3, V2=5, V3=7, V4=11 Ui where i is row 2V1=-3V2=5V3=7V4=11SUPPLYU1=0105220101115U2=2121071592025U3=754141651810DEMAND515151569Setting opportunity cost for un-occupied cells (continue)Step3:Calculate cell evaluations (unit cost difference)dj for each empty cell by using the formuladij = Cij ( Ui+Vj )


unit cost differencedij = Cij ( Ui+Vj )X11U1 + V1 = 0+(-3) = -3C11 - 1=10 (-3) = 13X13U1 + V3= 0+7 = 7C13 7=20 -7 = 13X21U2 + V1 = 2+(-3) = -1C21 - 20 =12 (-3) = 15X24U2 + V4 = 2+11 = 13C24 - 4=20 - 13 = 7X32U3 + V2 = 7+5 =12C32 - 5=14 - 12 = 2X33U3 + V3 = 7+7 = 14C33 -7=16 - 14 = 2X11=13 , X13=13 X21=15, X24=7, X32=2, X33=2 abcdSUPPLYA105220101115B121071592025C54141651810DEMAND5151515U1=0 , U2=2, U3=7V1=-3, V2=5, V3=7, V4=11 V1=-3V2=5V3=7V4=11SUPPLYU1=01310521320101115U2=2151210715972025U3=75421421651810DEMAND5151515BasicX11X12

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X33

X34

Z130130150070220Repeat Process to find optimal solutionStep5:See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.FOUND OPTIMAL SOLUTION!!X11=13 , X13=13 X21=15, X24=7, X32=2, X33=2 optimal solution conclusionabcdSUPPLYA105220101115B121071592025C54141651810DEMAND5151515Minimum Transportation Cost Z = (5*2)+(10*11)+(10*7)+(15*9)+(5*4)+(5*18)=$435Algorithm for optimality testStep 1: Start with basic feasible solution consisting of m+n1 allocations in independent positions.Step 2: Determine a set of m+n numbersui (i=1,2,....m) and vj (j=1,2,...n)such that for each occupied cells (r,s)crs = ur+vsStep 3: Calculate cell evaluations (unit cost difference) dj for each empty cell (i,j) by using the formuladij = cij ( ui+vj )Step 4: Examine the matrix of cell evaluation dij for negative entries and conclude thatIf all dij > 0 => Solution is optimal and unique.If all dij 0 => At least one dij = 0 => Solution is optimal and alternate solution also exists.

If at least one dij < 0 Solution is not optimal.If it is so, further improvement is required by repeating the above process. See step 5 and onwards.Step 5: See the most negative cell in the matrix [ dij ].Allocate q to this empty cell in the final allocation table. Subtract and add the amount of this allocation to other corners of the loop in order to restore feasibility.The value of q, in general is obtained by equating to zero the minimum of the allocations containing q(not + q) only at the corners of the closed loop.Substitute the value of q and find a fresh allocation table.

Step 6: Again, apply the above test for optimality till you find all dij 0Assignment Problem : Minimization ProblemA works manager has to allocate four different jobs to four workmen. Depending on the efficiency and the capacity of the individual the times taken by each differ as shown in the table 2. How should the tasks be assigned one jot to a worker so as to minimize the total man-hours?

Job\workerABCD11020181421525925330191712419242010Solution: The following steps are followed to find an optimal solution.

Step 1: Consider each row. Select the minimum element in each row. Subtract this smallest element form all the elements in that row. ABCD11020181421525925330191712419242010ABCD10108426160163187504914100-10-9-12-10Step 1: Consider each column. Select the minimum element in each column. Subtract this smallest element form all the elements in that column. ABCD10108426160163187504914100-0-7-0-0ABCD10384269016318050497100Step 3:In this way we make sure that in the matrix each row and each column has atleast one zero element.Having obtained at least one zero in each row and each column, we assign starting from first row. In the first row, we have a zero in (1, A). Hence we assign job 1 to the worker A. This assignment is indicated by a square . All other zeros in the column are crossed (X) to show that the other jobs cannot be assigned to worker A as he has already been assigned. In the above problem we do not have other zeros in the first column A.Proceed to the second row. We have a zero in (2, C). Hence we assign the job 2 to worker C, indicating by asquare . Any other zero in this column is crossed (X).Proceed to the third row. Here we have two zeros corresponding to (3, B) and (3, D). Since there is a tie for the job 3, go to the next row deferring the decision for the present. Proceeding to the fourth row, we have only one zero in (4, D). Hence we assign job 4 to worker D. Now the column D has a zero in the third row. Hence cross (3, D). All the assignments made in this way are as shown

ABCD10384269016318050 X497100ABCD11020181421525925330191712419242010STEP 4: Now having assigned certain jobs to certain workers we proceed to the column 1. Since there is an assignment in this column, we proceed to the second column. There is only one zero in the cell (3, B); we assign the jobs 3 to worker B. Thus all the four jobs have been assigned to four workers. Thus we obtain the solution to the problem as shown below.

ABCD10384269016318050 X497100ABCD10384269016318050 X497100Step5: The assignments areABCD11020181421525925330191712419242010Job toWorkerHours worked1A102B193C94D10Total working time = 10+19+9+10=48summaries of the procedure as a set of following rules:Subtract the minimum element in each row from all the elements in its row to make sure that at least we get one zero in that row.Subtract the minimum element in each column from all the elements in its column in the above reduced matrix, to make sure that we get at least one zero in each column.Having obtained at least one zero in each row and at least one zero in each column, examine rows successively until a row with exactly one unmarked zero is found and mark ( ) this zero, indicating that assignment in made there. Mark (X) all other zeros in the same column, to show that they cannot be used to make other assignments. Proceed in this way until all rows have been examined. If there is a tie among zeros defer the decision.Next consider columns, for single unmarked zero, mark them ( ) and mark (X) any other unmarked zero in their rows.Repeat (c) and (d) successively until one of the two occurs

Before proceed to the normal procedure, must first convert the maximization problem to minimization problem.Find the largest element in the matrixSubtract all the element in that matrix with this largest element.Then continue with the normal procedure.Assignment Problem : Maximization ProblemP1P2P3P4M158122M21728161M3202235M441079Example: 28-5=23ABCD123201626211012273862523424182119

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