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Optimal Production Schedule and Profit Analysis for Docking Stations

Explore a linear programming example to determine the optimal production schedule and weekly profit for a manufacturer producing three styles of docking stations. Learn about shadow prices, reduced costs, and ranges of optimality and feasibility.

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Optimal Production Schedule and Profit Analysis for Docking Stations

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  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 why would the shadow prices change? • 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 • Net Weekly Unit Profit • (Selling Price) – (Hardware Cost) • (Slide Cost) – (Wood Cost) X1 = # SL 1’s produced weekly X2 = # CP 6’s produced weekly X3 = # JR 8’s produced weekly $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 Total Expected Weekly Return MAX 9X1 + 10X2 + 12X3

  5. Constraints Feet of wood used Cannot Exceed 7500 • 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 7500 4X1 + 3X2 + 2.5X3 ≤ Slide Assemblies Used Cannot Exceed 4500 4500 2X1 + 2X2 + 2X3 ≤ Minutes Used Cannot Exceed 14400 14400 4.8X1 + 6.6X2 + 7.2X3 ≤

  6. Complete Model MAX 9 X1 + 10 X2 + 12X3 s.t. 4 X1 + 3 X2 + 2.5X3≤ 7500 2 X1 + 2 X2 + 2X3 ≤ 4500 4.8X1 + 6.6X2 + 7.2X3 ≤ 14400 All X’s ≥ 0

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

  8. What is the optimal production schedule and weekly profit? • 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? 150 is within the Allowable Increase Shadow price = 1.5 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 1200 is within the Allowable Increase ½ time worker works (4)(5)(60) =1200 minutes per week 1200 minutes Worth 1200(1.25) = $1500 Labor is a sunk cost.

  11. 2400 is outside the Allowable Increase If an additional full-time worker were added why would the shadow prices change? Full time worker works (8)(5)(60) =2400 minutes per week Thus the shadow prices will change.

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

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

  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.

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