LP Extra Issues and Examples

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

A Problem with No Feasible Solution 3 X2X2 X1X Region Satisfying 3rd Constraint Region Satisfying First 2 Constraints

A Solution Region That is Unbounded to the Right 4 X2X2 X1X Feasible Region X 1 > 5 X 2 < 10 X 1 + 2X 2 > 10

A Problem with a Redundant Constraint 5 X2X2 X1X Feasible Region 2X 1 + X 2 < 30 X 1 < 25 X 1 + X 2 < 20 Redundant Constraint

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

Minimization Example X 1 =number of tons of black-and-white picture chemical produced X 2 =number of tons of color picture chemical produced Minimize total cost =2,500X 1 +3,000X 2 Subject to: X 1 ≥ 30tons of black-and-white chemical X 2 ≥ 20tons of color chemical X 1 + X 2 ≥ 60tons total X 1, X 2 ≥ $0nonnegativity requirements

Minimization Example Table B – 50 – – 30 – – 10 – – ||||||| X1X1X1X1 X2X2X2X2 Feasible region X 1 = 30 X 2 = 20 X 1 + X 2 = 60 b a

Minimization Example Total cost at a=2,500X 1 +3,000X 2 =2,500 (40)+3,000(20) =$160,000 Total cost at b=2,500X 1 +3,000X 2 =2,500 (30)+3,000(30) =$165,000 Lowest total cost is at point a

LP Applications Production-Mix Example Department ProductWiringDrillingAssemblyInspectionUnit Profit XJ $ 9 XM $12 TR $15 BR $11 CapacityMinimum Department(in hours)ProductProduction Level Wiring1,500XJ Drilling2,350XM Assembly2,600TR29300 Inspection1,200BR788400

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

LP Applications Diet Problem Example A3 oz2 oz 4 oz B2 oz3 oz 1 oz C1 oz0 oz 2 oz D6 oz8 oz 4 oz Feed ProductStock XStock YStock Z

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

LP Applications Labor Scheduling Example TimeNumber ofTimeNumber of PeriodTellers RequiredPeriodTellers Required 9 AM - 10 AM 101 PM - 2 PM AM - 11 AM 122 PM - 3 PM AM - Noon143 PM - 4 PM 15 Noon - 1 PM 164 PM - 5 PM 10 F = Full-time tellers P 1 = Part-time tellers starting at 9 AM (leaving at 1 PM ) P 2 = Part-time tellers starting at 10 AM (leaving at 2 PM ) P 3 = Part-time tellers starting at 11 AM (leaving at 3 PM ) P 4 = Part-time tellers starting at noon (leaving at 4 PM ) P 5 = Part-time tellers starting at 1 PM (leaving at 5 PM )

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

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

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