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Allocation of Scarce Resources, I

Labor Type 20-inch`(med)’ 27-inch`(lrg)’ Hrs Available/Month. Electronics 6 15 360 Cabinetry 4 5 140. Allocation of Scarce Resources, I. Television Manufacturing Company. Market:

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Allocation of Scarce Resources, I

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  1. Labor Type 20-inch`(med)’ 27-inch`(lrg)’ Hrs Available/Month Electronics 6 15 360 Cabinetry 4 5 140 Allocation of Scarce Resources, I Television Manufacturing Company Market: 20-inch sets < 15 sets per month 27-inch sets < 40 sets per month Example: Production Mix Problem Production Coefficients:(hours of labor per TV set) Profit per TV set:20-inch --- $80;27-inch --- $120

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

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

  4. Linear Programming Formulation (LINGO Model: Sets & Data) MODEL: !Product Mix (Television Manufacturing); SETS: PRODUCTS/ MEDIUM, LARGE /: PROFIT, QUANTITY, SALESMAX; RESOURCES /ELECTRONICS, CABINETRY/: AVAILABLE; PRODUCTION (RESOURCES, PRODUCTS): CONSUMPTION; ENDSETS DATA: PROFIT = 80 120 ; SALESMAX = 15 40 ; AVAILABLE = 360 140; CONSUMPTION = 6 15 4 5; ENDDATA

  5. Linear Programming Formulation (LINGO Model: Objective & Constraints) MAX = @SUM( PRODUCTS: PROFIT * QUANTITY ); @FOR( RESOURCES( I): @SUM( PRODUCTS( J): CONSUMPTION( I, J) * QUANTITY( J)) <= AVAILABLE( I)); @FOR( PRODUCTS: QUANTITY <= SALESMAX); END

  6. Linear Programming Formulation (LINGO Model: Solution) Optimal solution found at step: 0 Objective value: 3200.000 Variable Value QUANTITY( MEDIUM) 10.0000 QUANTITY( LARGE) 20.0000

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

  8. Product Mix Problem Production Pricing Constraints: 6*eprice + 4*cprice +mprice> = 80 15*eprice+ 5*cprice +lprice> = 120 eprice > = 0; cprice > = 0 (hours/set)*($/hr) = $/set Dual Problem Formulation Objective Function: Minimize: Value = 360*eprice + 140*cprice +15*mprice + 40*lprice Hrs*($/hr) = $ Sets*($/set)=$

  9. Dual Problem MODEL: !Product Mix (Television Manufacturing); !Dual Problem; SETS: RESOURCES/ ELECTRONICS, CABINETRY, MEDSALES, LRGSALES /: PRICE, AVAILABLE; PRODUCTS /MEDIUM, LARGE/: PROFIT; RETURN (PRODUCTS, RESOURCES): EARNINGS; ENDSETS DATA: PROFIT = 80 120 ; AVAILABLE = 360 140 15 40; EARNINGS = 6 4 1 0 15 5 0 1; ENDDATA LINGO Model: Sets & Data

  10. Dual Problem MIN = @SUM( RESOURCES: AVAILABLE * PRICE ); @FOR( PRODUCTS( I): @SUM( RESOURCES( J): EARNINGS( I, J) * PRICE( J)) >= PROFIT( I)); END LINGO Model: Objective & Constraints

  11. Dual Problem Optimal solution found at step: 3 Objective value: 3200.000 Variable Value PRICE( ELECTRONICS) 2.66667 PRICE( CABINETRY) 16.00000 PRICE( MEDSALES) 0.00000 PRICE( LRGSALES) 0.00000 LINGO Model: Solution Row Slack or Surplus Dual Price 1 3200.000 1.000000 2 0.0000000 -10.00000 3 0.0000000 -20.00000

  12. Primal Problem (LINGO Model: Solution) Optimal solution found at step: 0 Objective value: 3200.000 Variable Value QUANTITY( MEDIUM) 10.0000 QUANTITY( LARGE) 20.0000 Row Slack or Surplus Dual Price 1 3200.000 1.000000 2 0.0000000 2.666667 3 0.0000000 16.00000 4 0.0000000 0.0000000 5 0.0000000 0.0000000

  13. 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

  14. Work-Scheduling Problem Post-Office Hiring of Full-Time Employees Number of full-time employees required Employee Requirements: (a) Day 1 = Monday 17 Day 2 = Tuesday 13 Day 3 = Wednesday 15 Day 4 = Thursday 19 Day 5 = Friday 14 Day 6 = Saturday 16 Day 7 = Sunday 11 (b) Each employee must work five consecutive days and then receive two days off.

  15. Work-Scheduling Problem Define variables: Let xi = number of employees beginning work on day i, i = 1,…,7 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) Formulate as Linear Programming Problem

  16. Work-Scheduling Problem MODEL: !Work-Schedule (Post-Office); SETS: EMPLOYEES/ MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY, SUNDAY/: NHIRED; NEEDS /MON, TU, WED, THUR, FRI, SAT, SUN/: NREQUIRED; MATRIX (NEEDS, EMPLOYEES): CONSTRAINTS; ENDSETS DATA: NREQUIRED = 17 13 15 19 14 16 11; CONSTRAINTS = 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1; ENDDATA (LINGO Model: Sets & Data)

  17. Work-Scheduling Problem MIN = @SUM( EMPLOYEES: NHIRED); @FOR( NEEDS( I): @SUM( EMPLOYEES( J): CONSTRAINTS( I, J) * NHIRED( J)) > NREQUIRED( I)); END (LINGO Model: Objective & Constraints)

  18. Work-Scheduling Problem Optimal solution found at step: 7 Objective value: 22.33333 Variable Value NHIRED( MONDAY) 1.333333 NHIRED( TUESDAY) 3.333333 NHIRED( WEDNESDAY) 2.000000 NHIRED( THURSDAY) 7.333333 NHIRED( FRIDAY) 0.0000000 NHIRED( SATURDAY) 3.333333 NHIRED( SUNDAY) 5.000000 (LINGO Model: Solution)

  19. Work-Scheduling Problem Row Slack or Surplus Dual Price 1 22.33333 1.000000 2 0.0000000 -0.3333333 3 0.0000000 0.0000000 4 0.0000000 -0.3333333 5 0.0000000 -0.3333333 6 0.0000000 -0.5551115E-16 7 0.0000000 -0.3333333 8 6.666667 0.0000000 (LINGO Model: Dual Prices)

  20. Work-Scheduling Problem 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 (LINGO Model: with Integer Constraint)

  21. Work-Scheduling Problem 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 (LINGO Solution: with Integer Constraint)

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