Problem 14-19 (Algorithmic) The A&M Hobby Shop carries a line of radio-controlle
ID: 364413 • Letter: P
Question
Problem 14-19 (Algorithmic)
The A&M Hobby Shop carries a line of radio-controlled model racing cars. Demand for the cars is
assumed to be constant at a rate of 47 cars per month. The cars cost $60 each, and ordering costs are
approximately $11 per order, regardless of the order size. The annual holding cost rate is 25%.
a. Determine the economic order quantity and total annual cost under the assumption that no
backorders are permitted. If required, round your answers to two decimal places.
Q* = ________
Total Cost = $ ________
b. Using a $41 per-unit per-year backorder cost, determine the minimum cost inventory policy and
total annual cost for the model racing cars. If required, round your answers to two decimal
places.
S* = _________
Total Cost = $ __________
c. What is the maximum number of days a customer would have to wait for a backorder under the
policy in part (b)? Assume that the Hobby Shop is open for business 300 days per year. If
required, round your answer to two decimal places.
Length of backorder period = ________ days
d. Would you recommend a no-backorder or a backorder inventory policy for this product?
Explain. If required, round your answers to two decimal places.
Recommendation would be ____________ inventory policy, since the maximum wait is only _______ days and
the cost savings is $ ________.
e. If the lead time is six days, what is the reorder point for both the no-backorder and backorder
inventory policies? If required, round your answers to two decimal places.
Reorder point for no-backorder inventory policy is _________.
Reorder point for backorder inventory policy is __________.
Explanation / Answer
Given :
Demand For Cars : 40 per month
Annual demand (a): 40*12 = 480
Cost of Car ; $60 per car
ordering Cost (b) : $15 per order
Holding Cost (c): 20% p.a i.e $60*20%= $12 per annum
(a) Economic Order Quantity = [sqrt{(2ab)/c}]
= [sqrt{(2ab)/c} = sqrt{(2*480*15)/12}]
= [sqrt{(2ab)/c} = sqrt{(2*480*15)sqrt{}/12} =sqrt{14400/12}]
= [sqrt{(2ab)/c} = sqrt{(2*480*15)sqrt{}/12} =sqrt{14400/12} =sqrt{1200}]
= 34.64102 or 35 number.
Total Annual cost = Ordering cost + holding Cost
Ordering cost = 480/35 = 13.71 or 14 time = 14*$15 = $210
Holding Cost = 1/2* 35*$60*20% = $420/2 = $210
Total Annual cost = $210+$210 = $420
(b) Minimum ordering quantity : Backorder cost (B) =$45 per car
= [=sqrt{(2ab)/c)} *((c+B)/B)]
= [=sqrt{(2*480*15)/12} *((12*45/45))]
= 34.64102*1.2667 =43.87 or 44 Quantity
(c) No.of. days - Maximum No.of. Days :
number of order = 480/44 = 10.9090 or 11 order
Maximum No.of. days = 300/11 = 27.27 or 27 days.
(d) Annual cost with backorder cost = ordering cost + holding cost
ordering cost = 11*15 = $165
Holding cost = 1/2*44*12 = $264
Total annual cost = $165+$264 = $ 429
since the total anual cost higher in case of back order cost , its better to be with no back order cost .
e) Days with Lead time - 6 days
in case of No backorder cost = 300/14 = 21.42 +6 = 27 days
in case of backordr cost = 300/11= 27+6 = 33 days