1. LP Extra Issues and Examples

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

3. A Problem with No Feasible Solution 3 X2X2 X1X1 8642086420 24682468 Region Satisfying 3rd Constraint Region Satisfying First 2 Constraints

4. A Solution Region That is Unbounded to the Right 4 X2X2 X1X1 15 10 5 0 5 10 15 Feasible Region X 1 > 5 X 2 < 10 X 1 + 2X 2 > 10

5. A Problem with a Redundant Constraint 5 X2X2 X1X1 30 25 20 15 10 5 0 510 15 20 25 30 Feasible Region 2X 1 + X 2 < 30 X 1 < 25 X 1 + X 2 < 20 Redundant Constraint

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

7. 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

8. Minimization Example Table B.9 60 60 – 50 – 40 40 – 30 – 20 20 – 10 – – ||||||| 0102030405060 X1X1X1X1 X2X2X2X2 Feasible region X 1 = 30 X 2 = 20 X 1 + X 2 = 60 b a

9. 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

10. LP Applications Production-Mix Example Department ProductWiringDrillingAssemblyInspectionUnit Profit XJ201.532.5$ 9 XM8971.5141.0$12 TR291.521.5$15 BR7881.032.5$11 CapacityMinimum Department(in hours)ProductProduction Level Wiring1,500XJ201150 Drilling2,350XM897100 Assembly2,600TR29300 Inspection1,200BR788400

11. 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 1 + 12X 2 + 15X 3 + 11X 4 subject to.5X 1 +1.5X 2 +1.5X 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

12. 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

13. 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 1 +.04X 2 +.025X 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

14. LP Applications Labor Scheduling Example TimeNumber ofTimeNumber of PeriodTellers RequiredPeriodTellers Required 9 AM - 10 AM 101 PM - 2 PM 18 10 AM - 11 AM 122 PM - 3 PM 17 11 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 )

15. 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(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10)

16. 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

17. 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