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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.

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Presentation on theme: "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."— Presentation transcript:

1 Production Mix Problem Graphical Solution med lrg Electronics Cabinetry Profit (10,20) (Optimal Product Mix!) Profit = 80* *20 = $3200/mo 27-inch sets 20-inch sets Feasible Region

2 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 = $

3 Primal Problem Primal Problem (Interpretation of Dual Prices) Row Dual Price (Marginal value of an additional unit of the resource)

4 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)

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

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

7 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

8 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*X *X13 + 9*X14 + 9*X *X *X23 + 7*X *X31 + 9*X *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)

9 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 = = 125 Total Supply = = 125 Yes, Balanced Transportation Problem! (If problem is unbalanced, add dummy supply or demand as required.)

10 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 ROUTES: COST * QUANTITY);

11 Transportation Problem Powerco Power Plant: LINGO Formulation (Cont’d.) ! The demand CUSTOMER( J): PLANT( I): QUANTITY( I, J))>= DEMAND( J)); ! The supply PLANT( I): 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

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

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

14 Transportation Problem Powerco Power Plant: Sensitivity Analysis Row Dual Price OBJ DEM( C1) (-v1) DEM( C2) (-v2) DEM( C3) (-v3) DEM( C4) (-v4) SUP( P1) (-u1) SUP( P2) (-u2) SUP( P3) (-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

15 Transportation Problem Powerco Power Plant: Sensitivity Analysis Row Dual Price OBJ DEM( C1) (-v1) DEM( C2) (-v2) DEM( C3) (-v3) DEM( C4) (-v4) SUP( P1) (-u1) SUP( P2) (-u2) SUP( P3) (-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

16 Powerco Power Plant: Sensitivity Analysis (Cont’d.) Variable Value Reduced Cost QUANTITY( P1, C1) (c11) QUANTITY( P1, C2) (c12) QUANTITY( P1, C3) (c13) QUANTITY( P1, C4) (c14) QUANTITY( P2, C1) (c21) QUANTITY( P2, C2) (c22) QUANTITY( P2, C3) (c23) QUANTITY( P2, C4) (c24) QUANTITY( P3, C1) (c31) QUANTITY( P3, C2) (c32) QUANTITY( P3, C3) (c33) QUANTITY( P3, C4) (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

17 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: $20/sailboat/quarter) Quarter Regular Overtime

18 Inventory Problems as Transportation Problems Sailco Corporation: Formulation Supply Nodes: Node Description Capacity, Sailboats 1 Initial Inventory st quarter, regular st quarter, overtime nd quarter, regular nd quarter, overtime rd quarter, regular rd quarter, overtime th quarter, regular th quarter, overtime 150 Total 770

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

20 Inventory Problems as Transportation Problems Sailco Corporation: Formulation (Costs) Cost Coefficients: ($/Sailboat) Supply Demand Dummy I R OT R 4 M OT 5 M R 6 M M OT 7 M M R 8 M M M OT 9 M M M M is a very large, arbitrary, positive number.

21 Assignment Problems Personnel Assignment: Time (hours) Job 1Job 2Job 3Job 4 Person Person Person Person

22 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

23 Assignment Problem Algorithm Row Minimum Column Minimum Subtract Row Minimum from Each Row:

24 Assignment Problem Algorithm Subtract Column Minimum from Each Column: 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:

25 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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