week 4 solution

Linear Programming Applications

Chapter 4Linear Programming Applications

2. a. Let x1 = units of product 1 produced

x2 = units of product 2 produced

Max 30x1 + 15x2s.t.

x1 + 0.35x2 100 Dept. A

0.30x1 + 0.20x2 36 Dept. B

0.20x1 + 0.50x2 50 Dept. C

x1, x2 0

Solution: x1 = 77.89, x2 = 63.16 Profit = 3284.21

b. The dual price for Dept. A is $15.79, for Dept. B it is $47.37, and for Dept. C it is $0.00. Therefore we would attempt to schedule overtime in Departments A and B. Assuming the current labor available is a sunk cost, we should be willing to pay up to $15.79 per hour in Department A and up to $47.37 in Department B.

c. Let xA = hours of overtime in Dept. A

xB = hours of overtime in Dept. B

xC = hours of overtime in Dept. C

Max 30x1 + 15x2 - 18xA - 22.5xB - 12xCs.t.

x1 + 0.35x2 - xA 100

0.30x1 + 0.20x2 - xB 36

0.20x1 + 0.50x2 - xC 50

xA 10

xB 6

xC 8

x1, x2, xA, xB, xC 0

x1 = 87.21

x2 = 65.12

Profit = $3341.34Overtime

Chapter 4

Dept. A 10 hrs.Dept. B 3.186 hrsDept. C 0 hours

Increase in Profit from overtime = $3341.34 - 3284.21 = $57.13

6. Let x1 = units of product 1

x2 = units of product 2

b1 = labor-hours Dept. A

b2 = labor-hours Dept. B

Max 25x1 + 20x2 + 0b1 + 0b2s.t.

6x1 + 8x2 - 1b1 = 0

12x1 + 10x2 - 1b2 = 0

1b1 + 1b2 900

x1, x2, b1, b2 0

Solution: x1 = 50, x2 = 0, b1 = 300, b2 = 600 Profit: $1,250

8. Let x1 = the number of officers scheduled to begin at 8:00 a.m.

x2 = the number of officers scheduled to begin at noon

x3 = the number of officers scheduled to begin at 4:00 p.m.

x4 = the number of officers scheduled to begin at 8:00 p.m.

x5 = the number of officers scheduled to begin at midnight

x6 = the number of officers scheduled to begin at 4:00 a.m.

The objective function to minimize the number of officers required is as follows:

Min x1 + x2 + x3 + x4 + x5 + x6The constraints require the total number of officers of duty each of the six four-hour periods to be at least equal to the minimum officer requirements. The constraints for the six four-hour periods are as follows:

Time of Day

8:00 a.m. - noon x1 + x6 5

noon to 4:00 p.m. x1 + x2 6

4:00 p.m. - 8:00 p.m. x2 + x3 10

8:00 p.m. - midnight x3 + x4 7

midnight - 4:00 a.m. x4 + x5 4

4:00 a.m. - 8:00 a.m. x5 + x6 6


x1, x2, x3, x4, x5, x6 0

Schedule 19 officers as follows:

x1 = 3 begin at 8:00 a.m.

x2 = 3 begin at noon

x3 = 7 begin at 4:00 p.m.

x4 = 0 begin at 8:00 p.m.

x5 = 4 begin at midnight

x6 = 2 begin at 4:00 a.m.

11. Let xij = units of component i purchased from supplier j

Min 12x11 + 13x1

2

+ 14x13 + 10x21 + 11x22 + 10x23

s.t. x11 + x1

2

+ x13 = 1000

x21 + x22 + x23 = 800

x11 + x21 600

x1

2

+ x22 1000

x13 + x23 800

x11, x12, x13, x21, x22, x23 0

Solution:Supplier

1 2 3

Component 1 600 400 0Component 2 0 0 800

Purchase Cost = $20,400

17. a. Let FM = number of frames manufacturedFP = number of frames purchasedSM = number of supports manufacturedSP = number of supports purchasedTM = number of straps manufacturedTP = number of straps purchased

Min 38FM + 51FP + 11.5SM + 15SP + 6.5TM + 7.5TP

s.t.

3.5FM + 1.3SM + 0.8TM 21,000

2.2FM + 1.7SM 25,200

3.1FM + 2.6SM + 1.7TM 40,800

Chapter 4

FM + FP 5,000

SM + SP 10,000

TM + TP 5,000

FM, FP, SM, SP, TM, TP 0.

Solution:Manufacture Purchase

Frames 5000 0Supports 2692 7308

Straps 0 5000

b. Total Cost = $368,076.91

c. Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used.

Constraint

1 Cutting: Slack = 0 350 hours used2 Milling: (25200 - 9623) / 60 = 259.62 hours3 Shaping: (40800 - 18300) / 60 = 375 hours

d. Nothing, there are already more hours available than are being used.

e. Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows that 2714 frames should be purchased.

The optimal solution is as follows:

OPTIMAL SOLUTION

Objective Function Value = 361500.000

Variable Value Reduced Costs -------------- --------------- ------------------ FM 2285.714 0.000 FP 2714.286 0.000 SM 10000.000 0.000 SP 0.000 0.900 TM 0.000 0.600 TP 5000.000 0.000

Constraint Slack/Surplus Dual Prices -------------- --------------- ------------------ 1 0.000 2.000 2 3171.429 0.000 3 7714.286 0.000 4 0.000 -45.000


5 0.000 -14.100 6 0.000 -7.500

19. a. Let x11 = amount of men's model in month 1

x21 = amount of women's model in month 1

x12 = amount of men's model in month 2

x22 = amount of women's model in month 2

s11 = inventory of men's model at end of month 1

s21 = inventory of women's model at end of month 1

s12 = inventory of men's model at end of month 2

s22 = inventory of women's model at end of month

The model formulation for part (a) is given.

Min 120x11 + 90x21 + 120x12 + 90x22 + 2.4s11 + 1.8s21 + 2.4s12 + 1.8s22s.t.

20 + x11 - s11 = 150

or x11 - s11 = 130 Satisfy Demand

[1]

30 + x21 - s21 = 125

or x21 - s21 = 95 Satisfy Demand

[2]

s11 + x12 - s12 = 200 Satisfy Demand [3]

s21 + x22 - s22 = 150 Satisfy Demand [4]

s12 25 Ending Inventory [5]

s22 25 Ending Inventory [6]

Labor Hours: Men’s = 2.0 + 1.5 = 3.5Women’s = 1.6 + 1.0 = 2.6

3.5 x11 + 2.6 x21 900 Labor Smoothingfor [7]

3.5 x11 + 2.6 x21 1100 Month 1

[8]

3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 100 Labor Smoothingfor

[9]-3.5 x11 - 2.6 x21 + 3.5 x12 + 2.6 x22 100 Month 2 [10]

x11, x12, x21, x22, s11, s12, s21, s22 0

The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2. Total Cost = $67,156

Chapter 4

Inventory Schedule

Month 1 63 Men's 0 Women'sMonth 2 25 Men's 25 Women's

Labor Levels

Previous month 1000.00 hoursMonth 1 922.25 hoursMonth 2 1022.25 hours

b. To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to 950, 1050, 50, and 50 respectively. The revised optimal solution is given.

x11 = 201

x21 = 95

x12 = 154

x22 = 175 Total Cost = $67,175

We produce more men's models in the first month and carry a larger men's model inventory; the added cost however is only $19. This seems to be a small expense to have less drastic labor force fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month. Since the added cost is only $19, management might want to experiment with the labor force smoothing restrictions to enforce even less fluctuations. You may want to experiment yourself to see what happens.

21. Decision variables : Regular

Model Month 1 Month 2Bookshelf B1R B2R

Floor F1R F2R

Decision variables : Overtime

Model Month 1 Month 2Bookshelf B1O B2O

Floor F1O F2OLabor costs per unit

Model Regular OvertimeBookshelf .7 (22) = 15.40 .7 (33) = 23.10

Floor 1 (22) = 22 1 (33) = 33

IB = Month 1 ending inventory for bookshelf unitsIF = Month 1 ending inventory for floor model

Objective function


Min 15.40 B1R + 15.40 B2R + 22 F1R + 22 F2R + 23.10 B1O + 23.10 B2O + 33 F1O + 33 F2O + 10 B1R + 10 B2R + 12 F1R + 12 F2R + 10 B1O + 10 B2O + 12 F1O + 12 F2O + 5 IB + 5 IF

or

Min 25.40 B1R + 25.40 B2R + 34 F1R + 34 F2R + 33.10 B1O + 33.10 B2O + 45 F1O + 45 F2O + 5 IB + 5 IFs.t.

.7 B1R + 1 F1R 2400 Regular time: month 1

.7 B2R + 1 F2R 2400 Regular time: month 2

.7B1O + 1 F1O 1000 Overtime: month 1

.7B2O + 1 F2O 1000 Overtime: month 2B1R + B1O - IB = 2100 Bookshelf: month 1IB + B2R + B2O = 1200 Bookshelf: month 2F1R + F1O - IF = 1500 Floor: month 1IF + F2R + F2O = 2600 Floor: month 2

OPTIMAL SOLUTION


Variable Value Reduced Costs -------------- --------------- ------------------ B1R 2100.000 0.000 B2R 1200.000 0.000 F1R 930.000 0.000 F2R 1560.000 0.000 B1O 0.000 0.000 B2O 0.000 0.000 F1O 610.000 0.000 F2O 1000.000 0.000 IB 0.000 1.500 IF 40.000 0.000

Constraint Slack/Surplus Dual Prices -------------- --------------- ------------------ 1 0.000 11.000 2 0.000 16.000 3 390.000 0.000 4 0.000 5.000 5 0.000 -33.100 6 0.000 -36.600 7 0.000 -45.000 8 0.000 -50.000OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper Limit ------------ --------------- --------------- --------------- B1R 23.900 25.400 25.400

Chapter 4

B2R No Lower Limit 25.400 25.400 F1R 34.000 34.000 36.143 F2R 34.000 34.000 50.000 B1O 33.100 33.100 No Upper Limit B2O 33.100 33.100 No Upper Limit F1O 40.000 45.000 45.000 F2O No Lower Limit 45.000 45.000 IB 3.500 5.000 No Upper Limit IF 0.000 5.000 7.143

RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit ------------ --------------- --------------- --------------- 1 2010.000 2400.000 3010.000 2 2010.000 2400.000 2440.000 3 610.000 1000.000 No Upper Limit 4 610.000 1000.000 1040.000 5 1228.571 2100.000 2657.143 6 1142.857 1200.000 1757.143 7 890.000 1500.000 1890.000 8 2560.000 2600.000 2990.000

23. Let F = number of windows manufactured in FebruaryM = number of windows manufactured in MarchA = number of windows manufactured in AprilIm = increase in production level necessary during month mDm = decrease in production level necessary during month msm = ending inventory in month m

Min 1I1 + 1I2 + 1I3 + 0.65D1 + 0.65D2 + 0.65D3

s.t.9000 + F - s1 = 15,000 February Demand

or

(1) F1 - s1 = 6000

(2) s1 + M - s2 = 16,500 March Demand

(3) s2 + A - s3 = 20,000 April Demand

F - 15,000 = I1 - D1 Change in February Productionor

(4) F - I1 + D1 = 15,000

M - F = I2 - D2 Change in March Productionor

(5) M - F - I2 + D2 = 0


A - M = I3 - D3 Change in April Productionor

(6) A - M - I3 + D3 = 0

(7) F 14,000 February Production Capacity

(8) M 14,000 March Production Capacity

(9) A 18,000 April Production Capacity

(10) s1 6,000 February Storage Capacity

(11) s2 6,000 March Storage Capacity

(12) s3 6,000 April Storage Capacity

Optimal Solution: Cost = $6,450

February March April

Production Level 12,000 14,000 16,500Increase in Production 0 2,000 2,500Decrease in Production 3,000 0 0Ending Inventory 6,000 3,500 0

Case Problem 3: Textile Mill Scheduling

Let X3R = Yards of fabric 3 on regular loomsX4R = Yards of fabric 4 on regular loomsX5R = Yards of fabric 5 on regular loomsX1D = Yards of fabric 1 on dobbie loomsX2D = Yards of fabric 2 on dobbie loomsX3D = Yards of fabric 3 on dobbie loomsX4D = Yards of fabric 4 on dobbie loomsX5D = Yards of fabric 5 on dobbie loomsY1 = Yards of fabric 1 purchasedY2 = Yards of fabric 2 purchasedY3 = Yards of fabric 3 purchasedY4 = Yards of fabric 4 purchasedY5 = Yards of fabric 5 purchased

Profit Contribution per YardManufactured Purchased

1 0.33 0.19

2 0.31 0.16

Fabric 3 0.61 0.504 0.73 0.54

5 0.20 0.00

Production Times in Hours per Yard

Chapter 4

Regular Dobbie

1 — 0.21598

2 — 0.21598

Fabric 3 0.1912 0.1912 4 0.1912 0.1912

5 0.2398 0.2398

Model may use a Max Profit or Min Cost objective function.

Max 0.61X3R + 0.73X4R + 0.20X5R + 0.33X1D + 0.31X2D + 0.61X3D + 0.73X4D + 0.20X5D+ 0.19Y1 + 0.16Y2 + 0.50Y3 + 0.54Y4

or

Min 0.49X3R + 0.51X4R + 0.50X5R+ 0.66X1D + 0.55X2D + 0.49X3D + 0.51X4D + 0.50X5D+ 0.80Y1 + 0.70Y2 + 0.60Y3 + 0.70Y4 + 0.70Y5

Regular Hours Available

30 Looms x 30 days x 24 hours/day = 21600

Dobbie Hours Available

8 Looms x 30 days x 24 hours/day = 5760

Constraints:

Regular Looms:0.192X3R + 0.1912X4R + 0.2398X5R 21600

Dobbie Looms:0.21598X1D + 0.21598X2D + 0.1912X3D + 0.1912X4D + 0.2398X5D 5760

Demand Constraints

X1D + Y1 = 16500 X2D + Y2 = 22000

X3R + X3D + Y3 = 62000X4R + X4D + Y4 = 7500 X5R + X5D + Y5 = 62000

OPTIMAL SOLUTION


Variable Value Reduced Costs -------------- --------------- ------------------ X3R 27711.29297 0.00000 X4R 7500.00000 0.00000 X5R 62000.00000 0.00000 X1D 4669.13672 0.00000


X2D 22000.00000 0.00000 X3D 0.00000 0.01394 X4D 0.00000 0.01394 X5D 0.00000 0.01748 Y1 11830.86328 0.00000 Y2 0.00000 0.01000 Y3 34288.70703 0.00000 Y4 0.00000 0.08000 Y5 0.00000 0.06204 Constraint Slack/Surplus Dual Prices -------------- --------------- ------------------ 1 0.00000 0.57531 2 0.00000 0.64821 3 0.00000 0.19000 4 0.00000 0.17000 5 0.00000 0.50000 6 0.00000 0.62000 7 0.00000 0.06204

OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper Limit ------------ --------------- --------------- --------------- X3R 0.50000 0.61000 0.62394 X4R 0.71606 0.73000 No Upper Limit X5R 0.18252 0.20000 No Upper Limit X1D 0.31426 0.33000 0.34000 X2D 0.30000 0.31000 No Upper Limit X3D No Lower Limit 0.61000 0.62394 X4D No Lower Limit 0.73000 0.74394 X5D No Lower Limit 0.20000 0.21748 Y1 0.18000 0.19000 0.20574 Y2 No Lower Limit 0.16000 0.17000 Y3 0.48606 0.50000 0.61000 Y4 No Lower Limit 0.54000 0.62000 Y5 No Lower Limit 0.00000 0.06204

RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit ------------ --------------- --------------- --------------- 1 16301.60059 21600.00000 28156.00000 2 4751.55957 5760.00000 8315.23047 3 4669.13672 16500.00000 No Upper Limit 4 10169.13672 22000.00000 26669.13672 5 27711.29297 62000.00000 No Upper Limit 6 0.00000 7500.00000 35211.29297 7 34660.54688 62000.00000 84095.07813

Production/Purchase Schedule (Yards)

Regular Looms

Dobbie Looms Purchased

Chapter 4

1 4669 11831

2 22000

Fabric 3 27711 34289

4 7500

5 62000

Projected Profit: $62,531.92

Value of 9th Dobbie Loom

Dual Price (Constraint 2) = 0.64821 per hour dobbie

Monthly Value of 1 Dobbie Loom

(30 days)(24 hours/day)($0.64821) = $466.71

Note: This change is within the Right-Hand Side Ranges for Constraint 2.

Discussion of Objective Coefficient Ranges

For example, fabric one on the dobbie loom shares ranges of 0.31426 to 0.34 for the profit maximization model or 0.64426 to 0.67 for the cost minimization model.

Note here that since demand for the fabrics is fixed, both the profit maximization and cost minimization models will provide the same optimal solution. However, the interpretation of the ranges for the objective function coefficients differ for the two models. In the profit maximization case, the coefficients are profit contributions. Thus, the range information indicates how price per unit and cost per unit may vary simultaneously. That is, as long as the net changes in price per unit and cost per unit keep the profit contributions within the ranges, the solution will remain optimal. In the cost minimization model, the coefficients are costs per unit. Thus, the range information indicates that assuming price per unit remains fixed how much the cost per unit may vary and still maintain the same optimal solution.

Case Problem 4: Workforce Scheduling

1. Let tij = number of temporary employees hired under option i (i = 1, 2, 3) in month j (j = 1 for January, j = 2 for February and so on)

The following table depicts the decision variables used in this case problem.

Jan. Feb. Mar. Apr. May JuneOption 1 t11 t12 t13 t14 t15 t16

Option 2 t21 t22 t23 t24 t25

Option 3 t31 t32 t33 t34

Costs: Contract cost plus training cost

Option Contract Cost Training Cost Total Cost1 $2000 $875 $28752 $4800 $875 $56753 $7500 $875 $8375


Min. 2875(t11 + t12 + t13 + t14 + t15 + t16)+ 5675(t21 + t22 + t23 + t24 + t25)+ 8375(t31 + t32 + t33 + t34)

One constraint is required for each of the six months.

Constraint 1: Need 10 additional employees in January

t11 = number of temporary employees hired under Option 1 (one-month contract) in Januaryt21 = number of temporary employees hired under Option 2 (two-month contract) in Januaryt31 = number of temporary employees hired under Option 3 (three-month contract) in January

t11 + t21 + t31 = 10

Constraint 2: Need 23 additional employees in February

t12 , t22 and t32 are the number of temporary employees hired under Options 1, 2 and 3 in February.

But, temporary employees hired under Option 2 or Option 3 in January will also be available to satisfy February needs.

t21 + t31 + t12 + t22 + t32 = 23

Note: The following table shows the decision variables used in this constraint

Jan. Feb. Mar. Apr. May JuneOption 1 t12

Option 2 t21 t22

Option 3 t31 t32

Constraint 3: Need 19 additional employees in March


Option 2 t22 t23

Option 3 t31 t32 t33

t31 + t22 + t32 + t13 + t23 + t33 = 19

Constraint 4: Need 26 additional employees in May


Option 2 t23 t24

Option 3 t32 t33 t34

t32 + t23 + t33 + t14 + t24 + t34 = 26

Constraint 5: Need 20 additional employees in May

Chapter 4


Option 2 t24 t25

Option 3 t33 t34

t33 + t24 + t34 + t15 + t25 = 20

Constraint 6: Need 14 additional employees in June


Option 2 t25

Option 3 t34

t34 + t25 + t16 = 14

Optimal Solution: Total Cost = $313,525

Jan. Feb. Mar. Apr. May JuneOption 1 0 1 0 0 6 0Option 2 3 0 0 0 0Option 3 7 12 0 14

2.Option Number Hired Contract Cost Training Cost Total Cost

1 7 $14,000 $6,125 $20,1252 3 $14,400 $2,625 $17,0253 33 $247,500 $28,875 $276,375

Total: $275,900 $37,625 $313,525

3. Hiring 10 full-time employees at the beginning of January will reduce the number of temporary employees needed each month by 10. Using the same linear programming model with the right-hand sides of 0, 13, 9, 16, 10 and 4, provides the following schedule for temporary employees:

Jan. Feb. Mar. Apr. May JuneOption 1 0 4 0 0 3 0Option 2 0 0 0 3 0Option 3 0 9 0 4

Option Number Hired Contract Cost Training Cost Total Cost1 7 $14,000 $6,125 $20,1252 3 $14,400 $2,625 $17,0253 13 $97,500 $11,375 $108,875Total: 23 $146,025

Full-time employees cost:

Training cost: 10($875) = $8,750Salary: 10(6)(168)($16.50) = $166,320Total Cost = $146,025 + $8750 + $166,320 = $321,095


Hiring 10 full-time employees is $321,095 - $313,525 = $7,570 more expensive than using temporary employees. Do not hire the 10 full-time employees. Davis should continue to contract with WorkForce to obtain temporary employees.

4. With the lower training costs, the costs per employee for each option are as follows:

Option Cost Training Cost Total Cost1 $2000 $700 $27002 $4800 $700 $55003 $7500 $700 $8200

Resolving the original linear programming model with the above costs indicates that Davis should hire all temporary employees on a one-month contract specifically to meet each month's employee needs. Thus, the monthly temporary hire schedule would be as follows: January - 10; February - 23; March - 19; April - 26; May - 20; and June - 14. The total cost of this strategy is $302,400. Note that if training costs were any lower, this would still be the optimal hiring strategy for Davis.

Case Problem 5: Cinergy Coal Allocation

A linear programming model can be used to determine how much coal to buy from each of the mining companies and where to ship it. Let

= tons of coal purchased from supplier i and used by generating unit j

The objective function minimizes the total cost to buy and burn coal. The objective function coefficients, , are the cost to buy coal at mine i, ship it to generating unit j,

and burn it at generating unit j. Thus, the objective function is . In computing the objective function coefficients three inputs must be added: the cost of the coal, the transportation cost to the generating unit, and the cost of processing the coal at the generating unit.

There are two types of constraints: supply constraints and demand constraints. The supply constraints limit the amount of coal that can be bought under the various contracts. For the fixed-tonnage contracts, the constraints are equalities. For the variable-tonnage contracts, any amount of coal up to a specified maximum may be purchased. Let Li represent the amount that must be purchased under fixed-tonnage contract i and Si represent the maximum amount that can be purchased under variable-tonnage contract i. Then the supply constraints can be written as follows:

for all fixed-tonnage contracts

for all variable-tonnage contracts

The demand constraints specify the number of mWh of electricity that must be generated by each generating unit. Let = mWh hours of electricity generated by a ton of coal purchased from supplier i and used by generating unit j, and Dj = mWh of electricity demand at generating unit j. The demand constraints can then be written

Chapter 4

as follows:

for all generating units

Note: Because of the large number of calculations that must be made to compute the objective function and constraint coefficients, we developed an Excel spreadsheet model for this problem. Copies of the data and model worksheets are included after the discussion of the solution to parts (a) through (f).

1. The number of tons of coal that should be purchased from each of the mining companies and where it should be shipped is shown below:

Miami Fort # 5

Miami Fort # 7 Beckjord East Bend Zimmer

RAG 0 0 61,538 288,462 0

Peabody 217,105 11,278 71,617 0 0

American 0 0 0 0 275,000

Consol 0 0 33,878 0 166,122

Cyprus 0 0 0 0 0Addington 0 200,000 0 0 0

Waterloo 0 0 98,673 0 0

The total cost to purchase, deliver, and process the coal is $53,407,243.

2. The cost of the coal in cents per million BTUs for each generating unit is as follows:

Miami Fort #5

Miami Fort #7

Beckjord

East Bend

Zimmer

111.84 136.97 127.24 103.85 114.51

3. The average number of BTUs per pound of coal received at each generating unit is shown below:

Miami Fort #5

Miami Fort #7

Beckjord

East Bend

Zimmer

13,300 12,069 12,354 13,000 12,468

4. The sensitivity report shows that the shadow price per ton of coal purchased from American Coal Sales is -$13 per ton and the allowable increase is 88,492 tons. This means that every additional ton of coal that Cinergy can purchase at the current price of $22 per ton will decrease cost by $13. So even paying $30 per ton, Cinergy will decrease cost by $5 per ton. Thus, they should buy the additional 80,000 tons; doing so will save them $5(80,000) = $400,000.

5. If the energy content of the Cyprus coal turns out to be 13,000 BTUs per ton the procurement plan changes as shown below:


Miami Fort # 5

Miami Fort # 7 Beckjord East Bend Zimmer

RAG 0 0 61,538 288,462 0

Peabody 36,654 191,729 71,617 0 0

American 0 0 0 0 275,000

Consol 0 0 33,878 0 166,122

Cyprus 0 0 85,769 0 0

Addington 200,000 0 0 0 0

Waterloo 0 0 0 0 0

6. The shadow prices for the demand constraints are as follows:

Miami Fort #5

Miami Fort #7

Beckjord

East Bend

Zimmer

21 20 20 18 19

The East Bend unit is the least cost producer at the margin ($18 per mWh), and the allowable increase is 160,000 mWh. Thus, Cinergy should sell the 50,000 mWh over the grid. The additional electricity should be produced at the East Bend generating unit. Cinergy’s profit will be $12 per mWh.

The Excel data and model worksheets used to solve the Cinergy coal allocation problem are as follows:

week 4 solution

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