Thanks a lot. I had a lot of notes and books explaining this, but I couldnt understand it. I found this slide and you've explained it really well. I passed the exam and I know I wont forget this method again. Thank you so much :)
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Transportation Problem By : Alvin G. Niere Misamis University
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Aim of Transportation ModelTo find out optimum transportation schedule keeping in mind cost of transportation to be minimized.
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What is a Transportation Problem?• The transportation problem is a special type of LPP where the objective is to minimize the cost of distributing a product from a number of sources or origins to a number of destinations.• Because of its special structure the usual simplex method is not suitable for solving transportation problems. These problems require special method of solution.
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The Transportation Problem• The problem of finding the minimum-cost distribution of a given commodity from a group of supply centers (sources) i=1,…,m to a group of receiving centers (destinations) j=1,…,n• Each source has a certain supply (si)• Each destination has a certain demand (dj)• The cost of shipping from a source to a destination is directly proportional to the number of units shipped
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Simple Network Representation Sources Destinations Supply s1 1 Demand d1 1 Supply s2 2 2 Demand d2 … … xij n Demand dn Supply sm m Costs cij Transportation-5
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Application of Transportation Problem Minimize shipping costs Determine low cost location Find minimum cost production schedule Military distribution system
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Two Types of Transportation Problem• Balanced Transportation Problem where the total supply equals total demand• Unbalanced Transportation Problem where the total supply is not equal to the total demand
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Phases of Solution of Transportation Problem• Phase I- obtains the initial basic feasible solution• Phase II-obtains the optimal basic solution
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Initial Basic Feasible SolutionNorth West Corner Rule (NWCR)Row Minima MethodColumn Minima MethodLeast Cost MethodVogle Approximation Method (VAM)
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Optimum Basic SolutionStepping Stone MethodModified Distribution Method a.k.a. MODI Method
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Optimum Basic Solution: Stepping-Stone Method1. Select any unused square to evaluate2. Beginning at this square, trace a closed path back to the original square via squares that are currently being used3. Beginning with a plus (+) sign at the unused corner, place alternate minus and plus signs at each corner of the path just traced
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Stepping-Stone Method4. Calculate an improvement index by first adding the unit-cost figures found in each square containing a plus sign and subtracting the unit costs in each square containing a minus sign5. Repeat steps 1 though 4 until you have calculated an improvement index for all unused squares. If all indices are ≥ 0, you have reached an optimal solution.
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Problem Illustration TO A. B. C. FACTORYFROM ALBUQUERQUE BOSTON CLEVELAND CAPACITYD. DES MOINES 5 4 3 100E. EVANSVILLE 8 4 3 300F. FORT 9 7 5LAUDERDALE 300WAREHOUSEDEMAND 300 200 200 700
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Initial Feasible Solution using Northwest Corner Rule TO A. B. C. FACTORYFROM ALBUQUERQUE BOSTON CLEVELAND CAPACITYD. DES MOINES 5 4 3 100 100E. EVANSVILLE 8 4 3 200 100 300F. FORT 9 7 5LAUDERDALE 100 200 300WAREHOUSEDEMAND 300 200 200 700IFS= DA + EA +EB + FB + FC = 100(5) + 200(8) + 100(4) + 100(7) + 200(5)= 500 + 1600 + 400 + 700 + 1000 = 4200
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Optimizing Solution using Stepping-Stone Method To (A) (B) (C) Factory From Albuquerque Boston Cleveland capacity $5 $4 $3 (D) Des Moines 100 - 100 Des Moines- + $8 $4 $3 Boston index (E) Evansville 200 100 300 + - = $4 - $5 + $8 - $4 $9 $7 $5 (F) Fort Lauderdale 100 200 300 = +$3 Warehouse requirement 300 200 200 700 99 $5 1 $4 100 - + + - 201 $8 99 $4Figure C.5 200 100
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Stepping-Stone Method1. If an improvement is possible, choose the route (unused square) with the largest negative improvement index2. On the closed path for that route, select the smallest number found in the squares containing minus signs3. Add this number to all squares on the closed path with plus signs and subtract it from all squares with a minus sign
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Stepping-Stone Method To (A) (B) (C) Factory From Albuquerque Boston Cleveland capacity $5 $4 $3 (D) Des Moines 100 100 $8 $4 $3 (E) Evansville 200 100 300 - + $9 $7 $5 (F) Fort Lauderdale 100 200 300 + - Warehouse requirement 300 200 200 700 1. Add 100 units on route FA 2. Subtract 100 from routes FB 3. Add 100 to route EB 4. Subtract 100 from route EAFigure C.7
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Special Issues in Modeling Demand not equal to supply Called an unbalanced problem Common situation in the real world Resolved by introducing dummy sources or dummy destinations as necessary with cost coefficients of zero
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Total Cost Special50($8) + 200($4)in50($3) + 150($5) + 150(0) = 250($5) + Issues + Modeling = $3,350 To (A) (B) (C) Factory Dummy capacity From Albuquerque Boston Cleveland $5 $4 $3 0 (D) Des Moines 250 250 $8 $4 $3 0 (E) Evansville 50 200 50 300 $9 $7 $5 0 (F) Fort Lauderdale 150 150 300 Warehouse requirement 300 200 200 150 850 New Figure C.9 Des Moines capacity
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Special Issues in Modeling Degeneracy To use the stepping-stone methodology, the number of occupied squares in any solution must be equal to the number of rows in the table plus the number of columns minus 1 If a solution does not satisfy this rule it is called degenerate
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Special Issues in Modeling To Customer Customer Customer Warehouse From 1 2 3 supply $8 $2 $6 Warehouse 1 100 100 $10 $9 $9 Warehouse 2 0 100 20 120 $7 $10 $7 Warehouse 3 80 80 Customer demand 100 100 100 300 Initial solution is degenerate Place a zero quantity in an unused square andFigure C.10 proceed computing improvement indices
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