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 sq. ft min. $ $11.35
CP sq. ft min. $ $12.30
JR sq. ft min. $ $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
X3 = # JR 8’s
$ – 2(.40) – 4(.20) =
$9.00
$ – 2(.40) – 3(.20) =
$10.00
$ – 2(.40) – 2.5(.20) =
$12.00
MAX Total Expected Weekly Return
MAX Total Expected Weekly Return
MAX 9X1 + 10X2 + 12X3

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
4X1 + 3X X3
7500 ≤
Slide Assemblies Used
Cannot
Exceed
4500
2X1 + 2X2 + 2X3
4500 ≤
Minutes Used
Cannot
Exceed
14400
4.8X X X3 ≤
14400

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

7. =SUMPRODUCT($C$3:$E$3,C10:E10)
=C5-C6-C7-C8
Drag across
=.2*C11
Drag across
=SUMPRODUCT($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?
Shadow price = 1.5
150 is within the Allowable Increase
Slide assemblies are included costs.
Value Per Unit = Original Price + Shadow Price
= = 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.
½ time worker works (4)(5)(60) =1200 minutes per week
Shadow price = 1.25
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
why would the shadow prices change?
Full time worker works (8)(5)(60) =2400 minutes per week
2400 is outside the Allowable Increase
Thus the shadow prices will change.

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 $ = $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 
$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.