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Production Mix Problem Graphical Solution 0 10 20 30 40 50 60 med lrg 40 30 20 10 0 Electronics Cabinetry Profit (10,20) (Optimal Product Mix!) Profit = 80*10 + 120*20 = $3200/mo 27-inch sets 20-inch sets Feasible Region

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Product Mix Problem Primal Problem Formulation Production Capacity Constraints: 6*med + 15*lrg < = 360 4*med + 5*lrg < = 140 med > = 0; lrg > = 0 Market Constraints: med < = 15 lrg < = 40 Objective Function: Maximize: Profit = 80*med + 120*lrg (hours/set)*sets = hours (sets) ($/set)*sets = $

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Primal Problem Primal Problem (Interpretation of Dual Prices) Row Dual Price (Marginal value of an 1 1.000000 additional unit of 2 2.666667 the resource) 3 16.00000 4 0.0000000 5 0.0000000

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Work-Scheduling Problem Formulate as Linear Programming Problem Let xi = number of employees beginning work on day i, i = 1,…,7 Define variables: Write objective function: Min z = x1 + x2 + x3 + x4 + x5 + x6 + x7 Impose constraints: x1 + x4 + x5 + x6 + x7 > 17 (Monday) x1+ x2 + x5 + x6 + x7 > 13 (Tuesday) x1+ x2 + x3 + x6 + x7 > 15 (Wednesday) x1+ x2 + x3 + x4 + x7 > 19 (Thursday) x1+ x2 + x3 + x4 + x5 > 14 (Friday) x2 + x3 + x4 + x5 + x6 + x7 > 16 (Saturday) x3 + x4 + x5 + x6 + x7 > 11 (Sunday) xi > 0 (i= 1, …, 7) (Non-negativity)

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Work-Scheduling Problem (LINGO Model: with Integer Constraint) MIN = @SUM( EMPLOYEES: NHIRED); @FOR( NEEDS( I): @SUM( EMPLOYEES( J): CONSTRAINTS( I, J) * NHIRED( J)) >= NREQUIRED( I)); ! We want NHIRED to be integer; @FOR(EMPLOYEES(I): @GIN(NHIRED(I))); END

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Work-Scheduling Problem (LINGO Solution: with Integer Constraint) Optimal solution found at step: 8 Objective value: 23.00000 Branch count: 1 Variable Value NHIRED( MONDAY) 7.000000 1.000000 NHIRED( TUESDAY) 3.000000 1.000000 NHIRED( WEDNESDAY) 2.000000 1.000000 NHIRED( THURSDAY) 7.000000 1.000000 NHIRED( FRIDAY) 1.000000 1.000000 NHIRED( SATURDAY) 3.000000 1.000000 NHIRED( SUNDAY) 0.0000000 1.000000

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Allocation of Scarce Resources, II Transportation Problems Powerco, Electrical Power Company Power Transmission Costs: ($/million kwh) To Supply FromCity 1 City 2 City 3 City 4 (million kwh) Plant 1 $8 $6 $10 $9 35 Plant 2 $9 $12 $13 $7 50 Plant 3$14 $9 $16 $5 40 Demand 45 20 30 30

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Transportation Problem Powerco Power Plant: Formulation Define Variables: Let Xij = number of (million kwh) produced at plant i and sent to city j. Objective Function: Min z = 8*X11 + 6*X12 + 10*X13 + 9*X14 + 9*X21 + 12*X22 + 13*X23 + 7*X24 + 14*X31 + 9*X32 + 16*X33 + 5*X34 Supply Constraints: X11 + X12 + X13 + X14 < = 35 (Plant 1) X21 + X22 + X23 + X24 < = 50 (Plant 2) X31 + X32 + X33 + X34 < = 40 (Plant 3)

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Transportation Problem Powerco Power Plant: Formulation (Cont’d.) Demand Constraints: X11 + X21 +X31 > = 45 (City 1) X12 + X22 +X32 > = 20 (City 2) X13 + X23 +X33 > = 30 (City 3) X14 + X24 +X34 > = 30 (City 4) Nonnegativity Constraints: Xij > = 0 (i=1,..,3; j=1,..,4) Balanced? Total Demand = 45 + 20 + 30 + 30 = 125 Total Supply = 35 + 50 + 40 = 125 Yes, Balanced Transportation Problem! (If problem is unbalanced, add dummy supply or demand as required.)

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Transportation Problem Powerco Power Plant: LINGO Formulation model: ! A 3 Plant, 4 Customer Transportation Problem; SETS: PLANT / P1, P2, P3/ : CAPACITY; CUSTOMER / C1, C2, C3, C4/ : DEMAND; ROUTES( PLANT, CUSTOMER) : COST, QUANTITY; ENDSETS ! The objective; [OBJ] MIN = @SUM( ROUTES: COST * QUANTITY);

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Transportation Problem Powerco Power Plant: LINGO Formulation (Cont’d.) ! The demand constraints; @FOR( CUSTOMER( J): [DEM] @SUM( PLANT( I): QUANTITY( I, J))>= DEMAND( J)); ! The supply constraints; @FOR( PLANT( I): [SUP] @SUM( CUSTOMER(J): QUANTITY(I,J))<=CAPACITY( I)); ! Here are the parameters; DATA: CAPACITY = 35, 50, 40 ; DEMAND = 45, 20, 30, 30; COST = 8, 6, 10, 9, 9, 12, 13, 7, 14, 9, 16, 5; ENDDATA end

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Transportation Problem Powerco Power Plant: LINGO Solution Optimal solution found at step: 7 Objective value: 1020.000 QUANTITY( P1, C1) 0.0000000 QUANTITY( P1, C2) 10.00000 QUANTITY( P1, C3) 25.00000 QUANTITY( P1, C4) 0.0000000 QUANTITY( P2, C1) 45.00000 QUANTITY( P2, C2) 0.0000000 QUANTITY( P2, C3) 5.000000 QUANTITY( P2, C4) 0.0000000 QUANTITY( P3, C1) 0.0000000 QUANTITY( P3, C2) 10.00000 QUANTITY( P3, C3) 0.0000000 QUANTITY( P3, C4) 30.00000 All Quantities are Integers!

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Transportation Problem Powerco Power Plant: LINGO Solution (Cont’d.) Row Slack or Surplus Dual Price OBJ 1020.000 1.000000 DEM( C1) 0.0000000 -9.000000 DEM( C2) 0.0000000 -9.000000 DEM( C3) 0.0000000 -13.00000 DEM( C4) 0.0000000 -5.000000 SUP( P1) 0.0000000 3.000000 SUP( P2) 0.0000000 0.0000000 SUP( P3) 0.0000000 0.0000000 All constraints are binding! (Typical of Balanced Transportation Problems; results in simple algorithms)

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Transportation Problem Powerco Power Plant: Sensitivity Analysis Row Dual Price OBJ 1.00000 DEM( C1) -9.00000(-v1) DEM( C2) -9.00000 (-v2) DEM( C3) -13.00000(-v3) DEM( C4) -5.00000(-v4) SUP( P1) 3.00000(-u1) SUP( P2) 0.00000(-u2) SUP( P3) 0.00000(-u3) Changes in total cost due to changes in demand and supply: z = v1* C1 + v2* C2 + v3* C3 + v4* C4 + u1* P1 + u2* P2 + u3* P3 e.g. C2 = 1, P1 = 1 z = 9*1 + (-3)*1 = $6

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Transportation Problem Powerco Power Plant: Sensitivity Analysis Row Dual Price OBJ 1.00000 DEM( C1) -9.00000(-v1) DEM( C2) -9.00000 (-v2) DEM( C3) -13.00000(-v3) DEM( C4) -5.00000(-v4) SUP( P1) 3.00000(-u1) SUP( P2) 0.00000(-u2) SUP( P3) 0.00000(-u3) Changes in total cost due to changes in demand and supply: z = v1* C1 + v2* C2 + v3* C3 + v4* C4 + u1* P1 + u2* P2 + u3* P3 e.g. C2 = 1, P1 = 1 z = 9*1 + (-3)*1 = $6

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Powerco Power Plant: Sensitivity Analysis (Cont’d.) Variable Value Reduced Cost QUANTITY( P1, C1) 0.00 2.00 (c11) QUANTITY( P1, C2) 10.00 0.00 (c12) QUANTITY( P1, C3) 25.00 0.00 (c13) QUANTITY( P1, C4) 0.00 7.00 (c14) QUANTITY( P2, C1) 45.00 0.00 (c21) QUANTITY( P2, C2) 0.00 3.00 (c22) QUANTITY( P2, C3) 5.00 0.00 (c23) QUANTITY( P2, C4) 0.00 2.00 (c24) QUANTITY( P3, C1) 0.00 5.00 (c31) QUANTITY( P3, C2) 10.00 0.00 (c32) QUANTITY( P3, C3) 0.00 3.00 (c33) QUANTITY( P3, C4) 30.00 0.00 (c34) z = c11* Q11 + c12* Q12 + c13* Q13 + c14* Q14 + c21* Q21 + c22* Q22 + c23* Q23 + c24* Q24 + c31* Q31 + c32* Q32 + c33* Q33 + c34* Q34 Nonbasic Basic Nonbasic Basic Nonbasic Basic Nonbasic Basic Nonbasic Basic

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Inventory Problems as Transportation Problems Sailco Corporation: Manufacturer of Sailboats Demand: 1st Quarter 40 (Sailboats) 2nd Quarter 60 3rd Quarter 75 4th Quarter 25 Supply: (Initial inventory: 10) Production: (Storage @ $20/sailboat/quarter) Quarter Regular Overtime 1st40@$400150@$440 2nd40@$400150@$440 3rd40@$400150@$440 4th40@$400150@$440

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Inventory Problems as Transportation Problems Sailco Corporation: Formulation Supply Nodes: Node Description Capacity, Sailboats 1 Initial Inventory 10 2 1st quarter, regular 40 3 1st quarter, overtime 150 4 2nd quarter, regular 40 5 2nd quarter, overtime 150 6 3rd quarter, regular 40 7 3rd quarter, overtime 150 8 4th quarter, regular 40 9 4th quarter, overtime 150 Total 770

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Inventory Problems as Transportation Problems Sailco Corporation: Formulation (Cont’d.) Demand Nodes: Node DescriptionDemand, Sailboats 1 1st quarter 40 2 2nd quarter 60 3 3rd quarter 75 4 4th quarter 25 5 Dummy 570 Total 770

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Inventory Problems as Transportation Problems Sailco Corporation: Formulation (Costs) Cost Coefficients: ($/Sailboat) Supply Demand 1 234 Dummy I 1 0 20 40 60 0 R 2 400 420 440 460 0 OT 3 450 470 490 510 0 R 4 M 400 420 440 0 OT 5 M 450 470 490 0 R 6 M M 400 420 0 OT 7 M M 450 470 0 R 8 M M M 400 0 OT 9 M M M 450 0 M is a very large, arbitrary, positive number.

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Assignment Problems Personnel Assignment: Time (hours) Job 1Job 2Job 3Job 4 Person 1 14 5 8 7 Person 2 2 12 6 5 Person 3 7 8 3 9 Person 4 2 4 6 10

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Assignment Problem Formulation Define Variables: Let Xij = 1 if ith person is assigned to jth job Xij = 0 if ith person is not assigned to jth job Objective Function: Min z = 14*X11 + 5*X12 + … + 10*X44 Personnel Constraints: X11 + X12 + X13 + X14 = 1 X21 + X22 + X23 + X24 = 1 X31 + X32 + X33 + X34 = 1 X41 + X42 + X43 + X44 = 1 Demand Constraints: X11 + X21 + X31 + X41 = 1 X12 + X22 + X32 + X42 = 1 X13 + X23 + X33 + X43 = 1 X14 + X24 + X34 + X44 = 1 Binary Constraints: Xij = 0 or Xij = 1

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Assignment Problem Algorithm 14 5 8 7 2 12 6 5 7 8 3 9 2 4 6 10 Row Minimum 5 2 3 2 9 0 3 2 0 10 4 3 4 5 0 6 0 2 4 8 0 0 0 2 Column Minimum Subtract Row Minimum from Each Row:

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Assignment Problem Algorithm 9 0 3 0 0 10 4 1 4 5 0 4 0 2 4 6 Subtract Column Minimum from Each Column: 10 0 3 0 0 9 3 0 5 5 0 4 0 1 3 5 Subtract Minimum uncrossed value from uncrossed values and add to twice-crossed values: Draw lines to cross out zeros and read solution from zeros 0 Solution:

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Hungarian Method Step 1: Find the minimum element in each row of the m x m cost matrix. Construct a new matrix by subtracting from each cost the minimum cost in its row. For this new matrix, find the minimum cost in each column. Construct a new matrix (called the reduced cost matrix) by subtracting from each cost the minimum cost in its column. Step 2: Draw the minimum number of lines (horizontal and/or vertical) that are needed to cover all the zeros in the reduced cost matrix. If m lines are required, an optimal solution is available among the covered zeros in the matrix. If fewer than m lines are needed, proceed to Step 3. Step 3: Find the smallest nonzero element (call its value k) in the reduced cost matrix that is uncovered by the lines drawn in Step 2. Now subtract k from each uncovered element of the reduced cost matrix and add k to each element that is covered by two lines. Return to Step 2.

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