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

ID: 349272 • Letter: D

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 $17 per day. Clyde has 3 rug-cleaning machines.

  
a. Assuming that Clyde’s stocking decision is optimal, what is the implied range of excess cost per machine? (Enter smaller value in first box and larger value in second box. Do not round intermediate calculations. Round your answers to 2 decimal places. Omit the "$" sign in your response.)

Implied range of excess cost per machine from $  to $  

b.
Your answer from part a has been presented to Clyde, who protests that the amount is too low. Does this suggest an increase or a decrease in the number of rug machines he stocks?

Increase

Decrease



c. Suppose now that the $17 mentioned as profit is instead the excess cost per day for each machine and that the shortage cost is unknown. Assuming that the optimal number of machines is four, what is the implied range of shortage cost per machine? (Enter smaller value in first box and larger value in second box. Do not round intermediate calculations. Round your answers to 2 decimal places. Omit the "$" sign in your response.)

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

Given:

Profit on cleaners = $17/day. Clyde has 3 machines.

Demand

Frequency

cumulative 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

1

           

a.Implied range of excess cost per machine:

So=3

If the machine is in shortage we lose a day’s profit for one machine

Shortage cost, Cs = $17

Excess cost, Ce = unknown

As we see from cumulative frequency - For 3 machines to be optimal, the Service level must be between 0..70 to 0.85.

Step 1

Set SL = .70 and solve for Ce:

SL=Cs/(Cs+Ce )=17/(17+Ce )=.70

17=0.70(17+Ce)

17=11.9+0.70Ce

17-11.9=.70Ce

5.1=0.70Ce

Ce=$7.29

Step 2

Set SL ratio = .85 and solve for Ce:

SL=Cs/(Cs+Ce )=17/(17+Ce )=.85

17=.85(17+Ce)

Ce=$3.00

Conclusion

Implied range of excess cost: $3.00. C_e $7.29

b.         If the excess cost is higher (which shows that there are excess machines compared to service level required), then he should decrease the number of machines should be decreased.

If excess cost increases, SL, and optimum stocking level decrease

c.         So=3

Cs = unknown

Ce = $17

For 3 machines to be optimal, the SL ratio must be 0.70 and 0.85

Step 1

Set SL = .70 and solve for Cs:

SL=Cs/(Cs+Ce )=Cs/(Cs+17)=.70

Cs=$31.67

Step 2

Set SL = .85 and solve for Cs:

SL=Cs/(Cs+Ce )=Cs/(Cs+17)=.85

Cs=.85(C_s+17)

Cs=.85Cs+14.45

Cs - .85Cs=14.45

Cs=$96.33

Conclusion

Implied range of shortage cost: $31.67 C_s $96.33.

Demand

Frequency

cumulative 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

1

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