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Topic – 10 Transportation and Assignment Problems Transportation Problem Major Characteristics: Several sources (plant) with given supply capacity. Several destinations (warehouse) with given demand. Cost Matrix: unit transportation cost from each source to each destination. Assumptions: No transshipment between any two destinations/two sources. All demand/capacity constraints should be attempted to meet. Transportation Problem Decisions: How much should be shipped from each source to each destination so that total transportation cost is minimized? Why Develop a New Method? (LP can certainly be applied) LP is still NP-complete to large size problems. Computational efficient methods are desirable to large size problems. Transportation Problem Model: Tabular Presentation Transportation Problem Transportation Cost Transportation Table (Seek its initial solution on board) LP Formulation: Let: Xad – Amount to be shipped from D to A Xcf – Amount to be shipped from F to C (9 Variables) Objective Function: Min: Z = 5Xad + 4Xbd + 3Xcd + 8Xae + 4Xbe + 3Xce + 9Xaf + 7Xbf + 5Xcf Sub. To: Xad + Xbd + Xcd ≤ 100 Xae + Xbe + Xce ≤ 300 (Supply Constraints) Xaf + Xbf + Xcf ≤ 300 Xad + Xae + Xaf ≥ 300 Xbd + Xbe + Xbf ≥ 200 (Demand Constraints) Xcd + Xce + Xcf ≥ 200 (All Xij ≥ 0) Steps in Solving the Transportation Problem Solution Procedure 1. Balancing Table: Make ∑di = ∑Si – add a "Dummy" source when ∑ di > ∑ Si , or – add a "Dummy" destination when ∑ Si > ∑di. 2. Finding an initial solution: (several methods) – Upper-Left Corner (Northwest) Method – Least-Cost (Large-Profit) Method – VAM Method 3. Testing Optimality: (Can current solution be improved?) – Stepping-stone Procedure – Modified-Distribution Method 4. Improving current solution toward a better solution (Stepping-Stone method). Solution Procedure Special cases in Transportation Problem: – Prohibited transportation route (take large "M" as cost) – Multiple optimal solution is possible. – Transshipment allowed/Route shipment capacity limits (All special cases can be easily formulated and solved by LP.) Application Examples: – Production Planning/Scheduling/Routing Selection/.... Sup. P.10-18: Problem #6 Tabular form: (by northwest method-”left corner”) to W-1 W-2 W-3 W-4 from Mine-1 8 13 9 500 Mine-2 11 14 5 1300 Dummy 0 0 0 600 Demand 1200 500 700 (2400) Transportation problem Tabular form: (by northwest method-”left corner”) to W-1 W-2 W-3 W-4 from Mine-1 (1) 8 13 9 500 500 Mine-2 (2) 11 (3) 14 (4) 5 1300 700 500 100 Dummy 0 0 (5) 0 600 600 Demand 1200 500 700 (2400) TC=8*500+11*700+14*500+5*100+0*600=19,200 Transportation problem Tabular form: (by least-cost method) to W-1 W-2 W-3 W-4 from Mine-1 500 8 13 9 500 700 Mine-2 11 14 600 5 1300 500 100 Dummy 0 0 0 600 Demand 1200 500 700 (2400) TC=8*500+11*700+5*600+0*600=14,700 <19,200 Problem formulation of LP X11-amount to be shipped from M-1to W-1 X12-amount to be shipped from M-1to W-2 … …………………………………. ... … X23-amount to be shipped from M-2to W-3 Min z= Ci X ij =8X11+13X12+9X13+11X21+14X22+5X23 SUB TO: X11+X12+X13≤500 X21+14X22+5X23≤1300 X11+X21≥1200 X12+X22≥500 X13+X23≥700 (Xij≥0, i,j=1…3) Transportation Problem Maximization Objectives Problem Description Klein Chemicals, Inc. manufactures a product at two plants (Clifton Springs and Danville) and ships it to four different customers (denoted D1, D2, D3 and D4) Profit per unit for shipping from each plant to each customer. Profit Per Unit: Customers D1 D2 D3 D4 Plant Clifton $32 $34 $32 $40 Springs Danville $34 430 $28 $38 Plant Capacities and Customer Orders: Plant Capacities Customers Orders (Units) (Units) Clifton 5000 D1 2000 Danville 3000 D2 5000 D3 3000 D4 2000 Total 8000 Total 12000 Capacity Orders The Stepping-Stone Method 1. Select any unused square to evaluate. 2. Begin at this square. Trace a closed path back to the original; square via square that are currently being used (only horizontal or vertical moves allowed). 3. Place + in unused square; alternate – and + on each corner square of the closed path. 4. Calculate improvement index: add together the unit figures found in each square containing a +; subtract the unit cost figure in each square containing a -. 5. Repeat steps 1-4 for each unused square. Stepping-Stone Method: Tracing a Closed Path for the Des Moines to Cleveland route An Initial Solution by Northwest-Corner (VP- Corner Method) Warehouse Plant Capacity Miami Denver Lincoln Jackson 7 2 4 5 Chicago 70 30 100 3 1 5 2 Houston 60 15 75 6 9 7 4 Buffalo 30 50 80 Requirem 255 70 90 45 50 ent 255 Total cost = 70($7) + 30($2) + 60($1) + 15($5) + 30($7) + 50($4) = $1,095 Initial Solution by the “Least Cost Method” Warehouse VAM Plant Capacity Miami Denver Lincoln Jackson Costs 7 2 4 5 Chicago 40 15 45 100 $2 3 1 5 2 Houston 75 75 $1 6 9 7 4 Buffalo 30 50 80 $2 Require- 255 70 90 45 50 255 ments VAM $3 $1 $1 $2 Costs Total Cost = 40*7+15*2+45*4+75*1+30*6+50*4=945<(1095) Optimal Solution Warehouse Plant Capacity Miami Denver Lincoln Jackson 7 2 4 5 Chicago +$3 55 45 $3 100 3 1 5 2 Houston 40 35 +$2 +$1 75 6 9 7 4 Buffalo 30 +$5 +$1 50 80 Requirem 255 70 90 45 50 ent 255 Total Cost = 40($3) + 30($6) + 55($2) + 35($1) + 45($4) + 50($4) = $825 Assignment Problem: A Special Case 100=1 track load (can’t be split) warehouse W-1 W-2 W-3 W-4 supply plant P-1 3 1 2 100 4 P-2 5 9 2 1 100 P-3 6 6 3 4 100 P-4 7 100 9 4 2 demand 100 100 100 100 (400) Assignment Problem Major Characteristics: – Finite number of Team-Job (or the like) to be assigned. – Assignment must be made based on one-to-one basis. – Payoff matrix: cost (profit) for each possible assignment. Available Solution Techniques to Assignment Problem: – Linear Programming – Dynamic Programming – Integer IP (Branch-and-Bound) – Complete Enumeration – Transportation Model (special case with all Si = di = 1) – Hungrain Method (A "smart" method for large size problems) Assignment Problem Special Cases in Assignment Problem: – Prohibited Assignment (assign a large "M" as cost) – It is possible to have multiple optimal solutions. – Special arrangement for a fixed assignment or One-to-More assignment. Special Solution Procedure Hungrain Method: (Tabular Form) – Principle: Add (or subtract) a constant to all cells of a row (or column) in the table will not change final optimal solution. – Four-Step Procedure – For Max Problem: (Transforming into Min Problem) Change the sign ("+" or "-") of each cell in the table. Add the largest cell value to all cells. (Examples) Special Solution Procedure Summary: – Both Transportation Problem and Assignment Problem are special cases of LP problem, and can be solved by Simplex method. – Tabular solution technique mainly provides a more efficient way for some large size practical problems. – Many practical managerial problems can be formulated and solved by Transportation or Assignment problem techniques. The Assignment Method Algorithm 1. Subtract the smallest number in each row from every number in that row. Subtract the smallest number in each column from every umber in that column. 2. Draw the minimum number of vertical and horizontal straight lines necessary to cover zeros in the table. If the number of lines equals the number of rows or columns, then one can make an optimal assignment (Step-4). 3. If the number of lines does not equal the number of rows or columns: Subtract the smallest number not covered by a line from every other uncovered number. Add the same number to any number lying at the intersection of any two lines. Return to step 2 4. Make optimal assignments at locations of zeros within the table. Service costs of different team assignments ($ in thousand) Service zone Z1 Z2 Z3 team S1 20 15 31 S2 17 16 33 S3 18 19 27 Service costs of different team assignments ($ in thousand) Service zone Z1 Z2 Z3 team S1 20 5 15 0 31 16 S2 17 1 16 0 33 17 S3 18 0 19 1 27 9 Service costs of different team assignments ($ in thousand) 5 0 16 5 0 7 1 0 17 1 0 8 0 1 9 0 1 0 Problem 7- Supplement, p.10-18 mach (M) (N) (O) (P) job 1 2 3 4 A 6 7 5 9 B 8 5 6 7 C 10 8 6 6 dummy 0 0 0 0 Problem 7- Supplement LP formulation Let x11,x22,….x34(12 variables)=job A assigned to machine 1 Min: Z=6X11+7X12+5X13+9X14+…+6X34 Sub to: X11+X12+X13+X14=1 X21+X22+X23+X24=1 X31+X32+X33+X34=1 X11+X21+X31=1 X12+X22+X32=1 X13+X23+X33=1 X13+X23+X33=1 all Xij=1 or 0 Written Assignment-7 is due on next – Sup. P.10-22: Northeastern blood bank Case A) identify demand/supply Total cost If buy all Type: A: (6+2)=8(Pints) 8*30=240 6*36=216 B: (2+4)=6(Pints) 5*35=175 A: (2+3)=5(Pints) 5*40=200 $831 A: (3+2)=5(Pints) Case I: Northeastern blood bank Supply: A : 4 (pints) B : 10 (pints) O : 11 (pints) AB : 7 (pints) Case I: Northeastern blood bank Identify cost matrix: 1) allowable blood transfusion pattern (supply) B $1000 A While A A O $30 AB $1000 (Assign a large case to prevent A $1000 Unallowable transfusion) B B While B AB O $1000 $36 A $1000 AB B $1000 AB AB O $1000 A B O AB dummy supply A 30 1000 1000 30 0 4 B 1000 36 1000 36 0 10 O 35 35 35 35 0 11 AB 1000 1000 1000 40 0 7 demand 8 6 5 5 (8) (32) 24

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posted: | 7/18/2010 |

language: | English |

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