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LP Extra Issues and Examples

LP Extra Issues and Examples. Special Cases in LP. Infeasibility Unbounded Solutions Redundancy Degeneracy More Than One Optimal Solution. X 2. 8 6 4 2 0. Region Satisfying 3rd Constraint. 2 4 6 8. X 1. Region Satisfying First 2 Constraints. A Problem with No Feasible Solution.

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LP Extra Issues and Examples

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  1. LP Extra Issues and Examples

  2. Special Cases in LP • Infeasibility • Unbounded Solutions • Redundancy • Degeneracy • More Than One Optimal Solution

  3. X2 8 6 4 2 0 Region Satisfying 3rd Constraint 2 4 6 8 X1 Region Satisfying First 2 Constraints A Problem with No Feasible Solution

  4. X2 15 10 5 0 X1 > 5 X2 < 10 Feasible Region X1 +2X2 > 10 5 10 15 X1 A Solution Region That is Unbounded to the Right

  5. X2 30 25 20 15 10 5 0 Redundant Constraint 2X1 + X2 < 30 X1 < 25 X1 +X2 < 20 Feasible Region X1 5 10 15 20 25 30 A Problem with a Redundant Constraint

  6. Sensitivity Analysis • Changes in the Objective Function Coefficient • Changes in Resources (RHS) • Changes in Technological (LHS) Coefficients

  7. Minimization Example X1 = number of tons of black-and-white picture chemical produced X2 = number of tons of color picture chemical produced Minimize total cost = 2,500X1 + 3,000X2 Subject to: X1 ≥ 30 tons of black-and-white chemical X2 ≥ 20 tons of color chemical X1 + X2≥ 60 tons total X1, X2≥ $0 nonnegativity requirements

  8. X2 60 – 50 – 40 – 30 – 20 – 10 – – X1 + X2= 60 X1= 30 X2= 20 | | | | | | | 0 10 20 30 40 50 60 X1 Minimization Example Table B.9 Feasible region b a

  9. Minimization Example Total cost at a = 2,500X1 + 3,000X2 = 2,500 (40) + 3,000(20) = $160,000 Total cost at b = 2,500X1 + 3,000X2 = 2,500 (30) + 3,000(30) = $165,000 Lowest total cost is at point a

  10. Department Product Wiring Drilling Assembly Inspection Unit Profit XJ201 .5 3 2 .5 $ 9 XM897 1.5 1 4 1.0 $12 TR29 1.5 2 1 .5 $15 BR788 1.0 3 2 .5 $11 Capacity Minimum Department (in hours) Product Production Level Wiring 1,500 XJ201 150 Drilling 2,350 XM897 100 Assembly 2,600 TR29 300 Inspection 1,200 BR788 400 LP Applications Production-Mix Example

  11. LP Applications X1 = number of units of XJ201 produced X2 = number of units of XM897 produced X3 = number of units of TR29 produced X4 = number of units of BR788 produced Maximize profit = 9X1 + 12X2 + 15X3 + 11X4 subject to .5X1 + 1.5X2 + 1.5X3 + 1X4 ≤ 1,500 hours of wiring 3X1 + 1X2 + 2X3 + 3X4 ≤ 2,350 hours of drilling 2X1 + 4X2 + 1X3 + 2X4 ≤ 2,600 hours of assembly .5X1 + 1X2 + .5X3 + .5X4 ≤ 1,200 hours of inspection X1 ≥ 150 units of XJ201 X2 ≥ 100 units of XM897 X3 ≥ 300 units of TR29 X4 ≥ 400 units of BR788

  12. Feed Product Stock X Stock Y Stock Z A 3 oz 2 oz 4 oz B 2 oz 3 oz 1 oz C 1 oz 0 oz 2 oz D 6 oz 8 oz 4 oz LP Applications Diet Problem Example

  13. LP Applications X1 = number of pounds of stock X purchased per cow each month X2 = number of pounds of stock Y purchased per cow each month X3 = number of pounds of stock Z purchased per cow each month Minimize cost = .02X1 + .04X2 + .025X3 Ingredient A requirement: 3X1 + 2X2 + 4X3 ≥ 64 Ingredient B requirement: 2X1 + 3X2 + 1X3 ≥ 80 Ingredient C requirement: 1X1 + 0X2 + 2X3 ≥ 16 Ingredient D requirement: 6X1 + 8X2 + 4X3 ≥ 128 Stock Z limitation:X3 ≤ 80 X1, X2, X3 ≥ 0 Cheapest solution is to purchase 40 pounds of grain X at a cost of $0.80 per cow

  14. Time Number of Time Number of Period Tellers Required Period Tellers Required 9 AM - 10 AM 10 1 PM - 2 PM 18 10 AM - 11 AM 12 2 PM - 3 PM 17 11 AM - Noon 14 3 PM - 4 PM 15 Noon - 1 PM 16 4 PM - 5 PM 10 LP Applications Labor Scheduling Example F = Full-time tellers P1 = Part-time tellers starting at 9 AM (leaving at 1 PM) P2 = Part-time tellers starting at 10 AM (leaving at 2 PM) P3 = Part-time tellers starting at 11 AM (leaving at 3 PM) P4 = Part-time tellers starting at noon (leaving at 4 PM) P5 = Part-time tellers starting at 1 PM (leaving at 5 PM)

  15. Minimize total daily manpower cost = $75F + $24(P1 + P2 + P3 + P4 + P5) LP Applications F + P1≥ 10 (9 AM - 10 AM needs) F + P1 + P2≥ 12(10 AM - 11 AM needs) 1/2 F + P1 + P2 + P3≥ 14(11 AM - 11 AM needs) 1/2 F + P1 + P2 + P3 + P4≥ 16(noon - 1 PM needs) F + P2 + P3 + P4+ P5≥ 18(1 PM - 2 PM needs) F + P3 + P4+ P5≥ 17 (2 PM - 3 PM needs) F + P4+ P5≥ 15(3 PM - 7 PM needs) F + P5≥ 10(4 PM - 5 PM needs) F ≤ 12 4(P1 + P2 + P3 + P4 + P5) ≤ .50(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10)

  16. Minimize total daily manpower cost = $75F + $24(P1 + P2 + P3 + P4 + P5) LP Applications F + P1≥ 10 (9 AM - 10 AM needs) F + P1 + P2≥ 12(10 AM - 11 AM needs) 1/2 F + P1 + P2 + P3≥ 14(11 AM - 11 AM needs) 1/2 F + P1 + P2 + P3 + P4≥ 16(noon - 1 PM needs) F + P2 + P3 + P4+ P5≥ 18(1 PM - 2 PM needs) F + P3 + P4+ P5≥ 17 (2 PM - 3 PM needs) F + P4+ P5≥ 15(3 PM - 7 PM needs) F + P5≥ 10(4 PM - 5 PM needs) F ≤ 12 4(P1 + P2 + P3 + P4 + P5 ) ≤ .50(112) F, P1, P2 , P3, P4, P5 ≥ 0

  17. Minimize total daily manpower cost = $75F + $24(P1 + P2 + P3 + P4 + P5) First Second Solution Solution F = 10 F = 10 P1 = 0 P1 = 6 P2 = 7 P2 = 1 P3 = 2 P3 = 2 P4 = 2 P4 = 2 P5 = 3 P5 = 3 LP Applications There are two alternate optimal solutions to this problem but both will cost $1,086 per day F + P1≥ 10 (9 AM - 10 AM needs) F + P1 + P2≥ 12(10 AM - 11 AM needs) 1/2 F + P1 + P2 + P3≥ 14(11 AM - 11 AM needs) 1/2 F + P1 + P2 + P3 + P4≥ 16(noon - 1 PM needs) F + P2 + P3 + P4+ P5≥ 18(1 PM - 2 PM needs) F + P3 + P4+ P5≥ 17 (2 PM - 3 PM needs) F + P4+ P5≥ 15(3 PM - 7 PM needs) F + P5≥ 10(4 PM - 5 PM needs) F ≤ 12 4(P1 + P2 + P3 + P4 + P5 ) ≤ .50(112) F, P1, P2 , P3, P4, P5 ≥ 0

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