Week 4 Solution

Linear Programming Applications

Chapter 4 Linear Programming Applications 2.

a.

Let

x1 = units of product 1 produced x2 = units of product 2 produced Max

30x1

+

15x2

x1 0.30x1

+

≤

100

Dept. A

+

0.35x2 0.20x2

≤

36

Dept. B

0.20x1

+

0.50x2

≤

50

Dept. C

s.t.

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

x1 0.30x1

+

0.35x2

-

xA

+

0.20x2

0.20x1

+

0.50x2

-

22.5xB

-

12xC

s.t. -

xB -

xC

xA xB xC x1, x2, xA, xB, xC ≥ 0 x1 = 87.21

≤

100

≤

36

≤

50

≤

10

≤

6

≤

8

Chapter 4

x2 = 65.12 Profit = $3341.34 Overtime Dept. A Dept. B Dept. C

10 hrs. 3.186 hrs 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

6x1 12x1

+

-

1b1

+

0b2

s.t. 8x2 + 10x2

1b1

=

0

-

1b2

=

0

+

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 + x6

The 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 noon to 4:00 p.m.

x1 x1

+ x6 + x2

≥

5

≥

6

Linear Programming Applications

4:00 p.m. - 8:00 p.m.

x2

8:00 p.m. - midnight

+ x3 x3

midnight - 4:00 a.m.

≥ 10 + x4 x4

+ x5

4:00 a.m. - 8:00 a.m.

x5

+ x6

≥

7

≥

4

≥

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

+ 14x13

+ 10x21

+

11x22

+ 10x23

2 s.t. x11

+

x1

+

x13

= 1000

2 x21 x11

+ x1

+

x22

+

x23

x21 +

=

800

≤

600

≤ 1000

x22

2 x13

+

x23

x11, x12, x13, x21, x22, x23 ≥ 0 Solution: 1 Component 1 Component 2

17. a.

Let

FM FP SM SP TM TP

= = = = = =

600 0

Supplier 2

3

400 0 0 800 Purchase Cost = $20,400

number of frames manufactured number of frames purchased number of supports manufactured number of supports purchased number of straps manufactured number of straps purchased

≤

800

Chapter 4

Min s.t.

38FM 3.5FM 2.2FM 3.1FM FM

+ 51FP

+ 11.5SM + + +

+

+ 15SP

1.3SM 1.7SM 2.6SM

+ 6.5TM

+ 7.5TP

+ 0.8TM + 1.7TM

FP SM

+

SP

TM FM, FP, SM, SP, TM, TP ≥ 0.

+

TP

≤ ≤ ≤ ≥ ≥ ≥

21,000 25,200 40,800 5,000 10,000 5,000

Solution: Frames Supports Straps

Manufacture 5000 2692 0

Purchase 0 7308 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 2 3

Cutting: Milling: Shaping:

Slack = 0 350 hours used (25200 - 9623) / 60 = 259.62 hours (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 = Variable -------------FM FP SM SP TM TP

361500.000

Value --------------2285.714 2714.286 10000.000 0.000 0.000 5000.000

Reduced Costs -----------------0.000 0.000 0.000 0.900 0.600 0.000

Linear Programming Applications

Constraint -------------1 2 3 4 5 6

19. a.

Let

Slack/Surplus --------------0.000 3171.429 7714.286 0.000 0.000 0.000

Dual Prices -----------------2.000 0.000 0.000 -45.000 -14.100 -7.500

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 + 1.8s22

s.t. 20 + x11 - s11 = 150 or x11 - s11 = 130

Satisfy Demand

[1]

Satisfy Demand

[2]

s11 + x12 - s12 = 200

Satisfy Demand

[3]

s21 + x22 - s22 = 150

Satisfy Demand

[4]

30 + x21 - s21 = 125 or x21 - s21 = 95

Labor Hours:

s12

≥ 25

Ending Inventory

[5]

s22

≥ 25

Ending Inventory

[6]

Labor Smoothing for

[7]

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

3.5 x11 + 2.6 x21 ≥ 900 3.5 x11 + 2.6 x21 ≤ 1100

Month 1

[8] 3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 ≤ 100 [9]

Labor Smoothing for

Chapter 4 -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 Inventory Schedule 63 Men's 25 Men's

Month 1 Month 2

0 Women's 25 Women's

Labor Levels 1000.00 hours 922.25 hours 1022.25 hours

Previous month Month 1 Month 2 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 Bookshelf Floor

Month 1 B1R F1R

Month 2 B2R F2R

Month 1 B1O F1O

Month 2 B2O F2O

Regular .7 (22) = 15.40

Overtime .7 (33) = 23.10

Decision variables : Overtime Model Bookshelf Floor Labor costs per unit Model Bookshelf

Linear Programming Applications

Floor

1 (22) = 22

1 (33) = 33

IB = Month 1 ending inventory for bookshelf units IF = 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 IF

s.t. .7 B1R + 1 F1R .7 B2R + 1 F2R .7B1O + 1 F1O .7B2O + 1 F2O B1R + B1O - IB IB + B2R + B2O F1R + F1O - IF IF + F2R + F2O

≤ ≤ ≤ ≤ = = = =

2400 2400 1000 1000 2100 1200 1500 2600

Regular time: month 1 Regular time: month 2 Overtime: month 1 Overtime: month 2 Bookshelf: month 1 Bookshelf: month 2 Floor: month 1 Floor: month 2

OPTIMAL SOLUTION Objective Function Value =

241130.000

Variable -------------B1R B2R F1R F2R B1O B2O F1O F2O IB IF

Value --------------2100.000 1200.000 930.000 1560.000 0.000 0.000 610.000 1000.000 0.000 40.000

Reduced Costs -----------------0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.500 0.000

Constraint -------------1 2 3 4

Slack/Surplus --------------0.000 0.000 390.000 0.000

Dual Prices -----------------11.000 16.000 0.000 5.000

Chapter 4

5 6 7 8

0.000 0.000 0.000 0.000

-33.100 -36.600 -45.000 -50.000

OBJECTIVE COEFFICIENT RANGES Variable -----------B1R B2R F1R F2R B1O B2O F1O F2O IB IF

Lower Limit --------------23.900 No Lower Limit 34.000 34.000 33.100 33.100 40.000 No Lower Limit 3.500 0.000

Current Value --------------25.400 25.400 34.000 34.000 33.100 33.100 45.000 45.000 5.000 5.000

Upper Limit --------------25.400 25.400 36.143 50.000 No Upper Limit No Upper Limit 45.000 45.000 No Upper Limit 7.143

Current Value --------------2400.000 2400.000 1000.000 1000.000 2100.000 1200.000 1500.000 2600.000

Upper Limit --------------3010.000 2440.000 No Upper Limit 1040.000 2657.143 1757.143 1890.000 2990.000

RIGHT HAND SIDE RANGES Constraint -----------1 2 3 4 5 6 7 8

23.

Lower Limit --------------2010.000 2010.000 610.000 610.000 1228.571 1142.857 890.000 2560.000

Let

F M A Im Dm sm

= = = = = =

number of windows manufactured in February number of windows manufactured in March number of windows manufactured in April increase in production level necessary during month m decrease in production level necessary during month m 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

Linear Programming Applications

F - 15,000 = I1 - D1

Change in February Production

or (4)

F - I1 + D1 = 15,000 M - F = I2 - D2

Change in March Production

or (5)

M - F - I2 + D2 = 0 A - M = I3 - D3

Change in April Production

or (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

Production Level Increase in Production Decrease in Production Ending Inventory

February 12,000 0 3,000 6,000

Case Problem 3: Textile Mill Scheduling Let X3R = Yards of fabric 3 on regular looms X4R = Yards of fabric 4 on regular looms X5R = Yards of fabric 5 on regular looms X1D = Yards of fabric 1 on dobbie looms X2D = Yards of fabric 2 on dobbie looms X3D = Yards of fabric 3 on dobbie looms X4D = Yards of fabric 4 on dobbie looms X5D = Yards of fabric 5 on dobbie looms Y1 = Yards of fabric 1 purchased Y2 = Yards of fabric 2 purchased Y3 = Yards of fabric 3 purchased Y4 = Yards of fabric 4 purchased Y5 = Yards of fabric 5 purchased

March 14,000 2,000 0 3,500

April 16,500 2,500 0 0

Chapter 4

Profit Contribution per Yard

Fabric

1 2 3 4 5

Manufactured 0.33 0.31 0.61 0.73 0.20

Purchased 0.19 0.16 0.50 0.54 0.00

1 2 3 4 5

Regular — — 0.1912 0.1912 0.2398

Dobbie 0.21598 0.21598 0.1912 0.1912 0.2398

Production Times in Hours per Yard

Fabric

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 X2D + Y2 X3R + X3D + Y3 X4R + X4D + Y4 X5R + X5D + Y5

= 16500 = 22000 = 62000 = 7500 = 62000

Linear Programming Applications

OPTIMAL SOLUTION Objective Function Value = Variable -------------X3R X4R X5R X1D X2D X3D X4D X5D Y1 Y2 Y3 Y4 Y5 Constraint -------------1 2 3 4 5 6 7

62531.91797

Value --------------27711.29297 7500.00000 62000.00000 4669.13672 22000.00000 0.00000 0.00000 0.00000 11830.86328 0.00000 34288.70703 0.00000 0.00000 Slack/Surplus --------------0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000

Reduced Costs -----------------0.00000 0.00000 0.00000 0.00000 0.00000 0.01394 0.01394 0.01748 0.00000 0.01000 0.00000 0.08000 0.06204 Dual Prices -----------------0.57531 0.64821 0.19000 0.17000 0.50000 0.62000 0.06204

OBJECTIVE COEFFICIENT RANGES Variable -----------X3R X4R X5R X1D X2D X3D X4D X5D Y1 Y2 Y3 Y4 Y5

Lower Limit --------------0.50000 0.71606 0.18252 0.31426 0.30000 No Lower Limit No Lower Limit No Lower Limit 0.18000 No Lower Limit 0.48606 No Lower Limit No Lower Limit

Current Value --------------0.61000 0.73000 0.20000 0.33000 0.31000 0.61000 0.73000 0.20000 0.19000 0.16000 0.50000 0.54000 0.00000

Upper Limit --------------0.62394 No Upper Limit No Upper Limit 0.34000 No Upper Limit 0.62394 0.74394 0.21748 0.20574 0.17000 0.61000 0.62000 0.06204

Current Value ---------------

Upper Limit ---------------

RIGHT HAND SIDE RANGES Constraint ------------

Lower Limit ---------------

Chapter 4

1 2 3 4 5 6 7

16301.60059 4751.55957 4669.13672 10169.13672 27711.29297 0.00000 34660.54688

21600.00000 5760.00000 16500.00000 22000.00000 62000.00000 7500.00000 62000.00000

28156.00000 8315.23047 No Upper Limit 26669.13672 No Upper Limit 35211.29297 84095.07813

Production/Purchase Schedule (Yards) Regular Looms

Fabric

1 2 3 4 5

Dobbie Looms 4669 22000

Purchased 11831

27711 7500 62000

34289

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.

Linear Programming Applications

Option 1 Option 2 Option 3

Jan. t11 t21 t31

Feb. t12 t22 t32

Mar. t13 t23 t33

Apr. t14 t24 t34

May t15 t25

June t16

Costs: Contract cost plus training cost Option 1 2 3

Contract Cost $2000 $4800 $7500

Training Cost $875 $875 $875

Total Cost $2875 $5675 $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 January t21 = number of temporary employees hired under Option 2 (two-month contract) in January t31 = 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. Option 1 Option 2 Option 3

t21 t31

Feb. t12 t22 t32

Mar.

Apr.

May

June

Constraint 3: Need 19 additional employees in March

Option 1 Option 2 Option 3

Jan.

Feb.

t31

t22 t32

Mar. t13 t23 t33

Apr.

May

June

Chapter 4

t31 + t22 + t32 + t13 + t23 + t33 = 19 Constraint 4: Need 26 additional employees in May Jan. Option 1 Option 2 Option 3

Feb.

Mar.

t32

t23 t33

Apr. t14 t24 t34

May

June

t32 + t23 + t33 + t14 + t24 + t34 = 26 Constraint 5: Need 20 additional employees in May Jan.

Feb.

Option 1 Option 2 Option 3

Mar.

Apr.

t33

t24 t34

May t15 t25

June

t33 + t24 + t34 + t15 + t25 = 20

Constraint 6: Need 14 additional employees in June Jan.

Feb.

Mar.

Option 1 Option 2 Option 3

Apr.

May

June t16

t25 t34

t34 + t25 + t16 = 14 Optimal Solution: Total Cost = $313,525

Option 1 Option 2 Option 3

Jan. 0 3 7

Feb. 1 0 12

Mar. 0 0 0

Apr. 0 0 14

May 6 0

June 0

2. Option 1 2 3

3.

Number Hired 7 3 33 Total:

Contract Cost $14,000 $14,400 $247,500 $275,900

Training Cost $6,125 $2,625 $28,875 $37,625

Total Cost $20,125 $17,025 $276,375 $313,525

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:

Option 1 Option 2

Jan. 0 0

Feb. 4 0

Mar. 0 0

Apr. 0 3

May 3 0

June 0

Linear Programming Applications

Option 3

Option 1 2 3 Total:

0

9

Number Hired 7 3 13 23

0

4

Contract Cost $14,000 $14,400 $97,500

Training Cost $6,125 $2,625 $11,375

Total Cost $20,125 $17,025 $108,875 $146,025

Full-time employees cost: Training cost: 10($875) = $8,750 Salary: 10(6)(168)($16.50) = $166,320 Total 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 1 2 3

Cost $2000 $4800 $7500

Training Cost $700 $700 $700

Total Cost $2700 $5500 $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 xij = 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, cij , 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 ∑ ∑ cij xij . 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

Chapter 4 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:

∑x

= Li

for all fixed-tonnage contracts

∑x

≤ Si

for all variable-tonnage contracts

ij

j

ij

j

The demand constraints specify the number of mWh of electricity that must be generated by each generating unit. Let aij = 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 as follows:

∑a

x = Dj

ij ij

i

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 Miami Fort #5 #7 RAG

Beckjord

East Bend Zimmer

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 Addingto n

0

0

0

0

0

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 #5 111.84

3.

Miami Fort #7 136.97

Beckjor d 127.24

East Bend 103.85

Zimme r 114.51

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

Linear Programming Applications

Miami Fort #5 13,300

Miami Fort #7 12,069

Beckjor d 12,354

East Bend 13,000

Zimme r 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 Miami Fort #5 #7 Beckjord East Bend 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

200,000

0

0

0

0

0

0

0

0

0

Addington Waterloo 6.

Zimmer

The shadow prices for the demand constraints are as follows: Miami Fort #5 21

Miami Fort #7 20

Beckjor d 20

East Bend 18

Zimme r 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:

Chapter 4