Demand for rug-cleaning machines at Clyde’s U-Rent-It is shown in the following

Question

Demand for rug-cleaning machines at Clyde’s U-Rent-It is shown in the following table. Machines are rented by the day only. Profit on the rug cleaners is $11 per day. Clyde has four rug-cleaning machines.

a. Assuming that Clyde’s stocking decision is optimal, what is the implied range of excess cost per machine?

Implied range of excess cost per machine from $________ to $________

b. Suppose now that the $11 mentioned as profit is instead the excess cost per day for each machine and that the shortage cost is unknown. Assuming that the optimal number of machines is four, what is the implied range of shortage cost per machine?

Implied range of shortage cost per machine from $________to $________

Demand Frequency 0 .30           1 .20           2 .20           3 .15           4 .10           5 .05           1.00          

Explanation / Answer

Demand

Frequency

Cummulative Frequency

0

0.3

0.30

1

0.2

0.50

2

0.2

0.70

3

0.15

0.85

4

0.1

0.95

5

0.05

1.00

a)

Determine the implied range of excess cost per machine: Cs = $10 Ce = unknown

So=4

Cs=$11

Ce=unknown

Solution

For 4 machines to be optimal, the SL ratio must be .85 and .95.

Step 1: Set SL = .85 and solve for Ce :

SL=Cs/Cs+Ce

11/11+Ce=

0.85

Ce=1.94(Round to Two decimal places)

Step 2: Set SL = .95 and solve for Ce :

SL=Cs/Cs+Ce

11/11+Ce=

0.95

Ce=0.57(Round to Two decimal places)

So the Implied range of cost per machine from $ 1.94 to 0.57

b)

Suppose now that excess cost per day = $10 and the shortage cost per day is unknown.

Assuming that the optimal number of machines is 4, what is the implied range of shortage cost?

Cs = unknown

Ce = $11

So=4

Solution

For 4 machines to be optimal, the SL ratio must be .85 and .95.

Step 1

Set SL = .85 and solve for Cs :

SL=Cs/Cs+Ce

Cs=0.85(Cs+11)

Cs=0.85Cs+9.35

Cs(0.15)=9.35

Cs=62.32

Step 2

Set SL = .95 and solve for Cs :

SL=Cs/Cs+Ce

Cs=0.95(Cs+11)

Cs=0.95Cs+10.45

Cs(0.05)=10.45

Cs=209

Conclusion: Implied range of shortage cost: $62.32 Cs $209.00

Demand

Frequency

Cummulative Frequency

0

0.3

0.30

1

0.2

0.50

2

0.2

0.70

3

0.15

0.85

4

0.1

0.95

5

0.05

1.00

a)

Determine the implied range of excess cost per machine: Cs = $10 Ce = unknown

So=4

Cs=$11

Ce=unknown

Solution

For 4 machines to be optimal, the SL ratio must be .85 and .95.

Step 1: Set SL = .85 and solve for Ce :

SL=Cs/Cs+Ce

11/11+Ce=

0.85

Ce=1.94(Round to Two decimal places)

Step 2: Set SL = .95 and solve for Ce :

SL=Cs/Cs+Ce

11/11+Ce=

0.95

Ce=0.57(Round to Two decimal places)

So the Implied range of cost per machine from $ 1.94 to 0.57

b)

Suppose now that excess cost per day = $10 and the shortage cost per day is unknown.

Assuming that the optimal number of machines is 4, what is the implied range of shortage cost?

Cs = unknown

Ce = $11

So=4

Solution

For 4 machines to be optimal, the SL ratio must be .85 and .95.

Step 1

Set SL = .85 and solve for Cs :

SL=Cs/Cs+Ce

Cs=0.85(Cs+11)

Cs=0.85Cs+9.35

Cs(0.15)=9.35

Cs=62.32

Step 2

Set SL = .95 and solve for Cs :

SL=Cs/Cs+Ce

Cs=0.95(Cs+11)

Cs=0.95Cs+10.45

Cs(0.05)=10.45

Cs=209

Conclusion: Implied range of shortage cost: $62.32 Cs $209.00

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