1. Linear Programming Example 4 Determining Objective Function Coefficients Interpretation of Shadow Prices Interpretation of Reduced Costs Ranges of Optimality/Feasibility

2. The Problem A manufacturer of docking stations for computers can make three different styles from laminated wood. –Each docking station requires 2 slide assemblies. –Screws, braces and other hardware required to produce the docking stations are in abundant supply and will not affect production. –Each week it can assign up to 6 workers working 8 hours per day, 5 days a week for production – sunk cost. –Each week it can purchase up to 7500 sq. ft. of the laminated wood for $0.20 per sq. ft. 4500 slide assemblies for $0.40 each Station Wood Labor Cost of Selling Model Required Required Hardware Price SL 1 4 sq. ft 4.8 min.$0.75 $11.35 CP 6 3 sq. ft. 6.6 min.$0.90 $12.30 JR 8 2.5 sq. ft. 7.2 min.$1.10 $14.40

3. Questions What is the optimal production schedule and weekly profit? If 150 extra slide assemblies became available, what is the most you would be willing to pay for them? If a half-time worker could be added to the labor force, what is the most we would be willing to pay him. If an additional full-time worker were added, what is the most we would be willing to pay him? What is the minimum selling price for the CP 6 model that would justify its production? Within what range of values for the net profit of JR 8’s will the optimal solution remains the same?

4. Decision Variables/Objective X 1 = # SL 1’s produced weekly X 2 = # CP 6’s produced weekly X 3 = # JR 8’s produced weekly MAX Total Expected Weekly Return Net Weekly Unit Profit (Selling Price) – (Hardware Cost) -(Slide Cost) – (Wood Cost) $11.35 -.75 – 2(.40) – 4(.20) =$9.00 $12.30 -.90 – 2(.40) – 3(.20) =$10.00 $14.40 - 1.10 – 2(.40) – 2.5(.20) =$12.00 MAX Total Expected Weekly Return MAX 9X 1 + 10X 2 + 12X 3

5. Constraints Cannot use more than 7500 feet of wood Cannot use more than 4500 slide assem. Cannot use more than (6 workers)x(8hr/day) x(5 days/week)x(60min/hr) = 14,400 min Feet of wood used Cannot Exceed 7500 4X 1 + 3X 2 + 2.5X 3 ≤ 7500 Slide Assemblies Used Cannot Exceed 4500 2X 1 + 2X 2 + 2X 3 ≤ 4500 Minutes Used Cannot Exceed 14400 4.8X 1 + 6.6X 2 + 7.2X 3 ≤ 14400

6. Complete Model MAX 9 X 1 + 10 X 2 + 12X 3 s.t. 4 X 1 + 3 X 2 + 2.5X 3 ≤ 7500 2 X 1 + 2 X 2 + 2X 3 ≤ 4500 4.8X 1 + 6.6X 2 + 7.2X 3 ≤ 14400 All X’s ≥ 0

7. =.2*C11 Drag across =C5-C6-C7-C8 Drag across =SUMPRODUCT($C$3:$E$3,C10:E10) Drag down

8. What is the optimal production schedule and weekly profit? 750 Sl 1’s 0 CP 6’s 1500 JR 8’5 $24,750 Weekly Profit

9. If 150 extra slide assemblies became available, what is the most you would be willing to pay for them? Shadow price = 1.5 150 is within the Allowable Increase Slide assemblies are included costs. Value Per Unit = Original Price + Shadow Price =.40 +1.50 = 1.90 150 units Worth 150(1.90) = $285

10. If a half-time worker could be added to the labor force, what is the most we would be willing to pay him. Shadow price = 1.25 ½ time worker works (4)(5)(60) =1200 minutes per week 1200 is within the Allowable Increase Labor is a sunk cost. 1200 minutes Worth 1200(1.25) = $1500

11. If an additional full-time worker were added What is the most you are willing to pay? Full time worker works (8)(5)(60) =2400 minutes per week 2400 is outside the Allowable Increase Thus the shadow prices will change. Re-solve the problem: total profit = $27,000. The most you are willing to pay = $27,000 – 24,750 = $2,250

12. What is the minimum selling price for the CP 6 model that would justify its production? Reduced Cost = -1.25 Profit and hence the selling price would have to improve by $1.25 Thus selling price must rise to $12.30 + 1.25 = $13.55

13. Within what range of values for the net profit of JR 8’s will the optimal solution remains the same? Range of Optimality 12 – 1.67  12 + 1.50 $10.33  $13.50

14. Review How to calculate “net” objective function coefficients. How to interpret the shadow price of an included cost. How to interpret a shadow price of a sunk cost. How to interpret a reduced cost. How to use a range of feasibility. How to use a range of optimality.