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 60 cars per month. The cars cost $70 each, and ordering costs are approximately $15 per order, regardless of the order size. The annual holding cost rate is 20%. Required:
a. Determine the economic order quantity and total annual cost under the assumption that no backorders are permitted.
b. Using a $45 per-unit per-year backorder cost, determine the minimum cost inventory policy and total annual cost for the model racing cars.
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.
d. Would you recommend a no-backorder or a backorder inventory policy for this product? Explain.

Step-by-step explanation:

A) Demand per month= 40 cars

Annual Demand (D)= 12*40 = 480

Fixed Cost per order (K)= 15

Holding Cost= 20% of cost= 60 *0.2 = 12

a. Economic Order Quantity=

Q^{*}={\sqrt {{\frac {2DK}{h

= √(2*480*15)/12

=34.64 ~ 35

Total Cost =P*D+K(D/EOQ)+h(EOQ/2) P= Cost per unit

= 60*480+ 15(480/35) + 12(35/2)

= 28800+ 205.71+ 210

=$29215.71

B). Backorder Cost (b)= $45

Qbo= Q* × √( b+h/ h)

= 35*√(12+45/ 45)

= 35* 1.12

=39.28 ~ 39

Shortage (S)= Qbo * (K/K+b)

= 39* (15/15+45)

= 39* 0.25

= 9.75

Total Cost Minimum=( bS2/ 2Qbo) + P (Qbo- S)2/2Qbo + K(D/Qbo)

=45* 9.752 / 2* 392 + 60 (39-9.75)2/ 2* 392 + 15 ( 480/39)

= 1.40+ 21.9.+ 184.61

=$207.91

C)Length of backorder days (d) = Demand ÷ amount of working days

d = 480 ÷ 300

d = 1.6

Calculate the backorders as the maximum number of backorders divided by the demand per day

s/d = 9.75/1.6 = 6.09 days (answer)

D) Calculate the difference in total between not using backorder:

$207.85 + $207.85 - 207.91 = $207.79

The saving in using backorder is $207.79.

Therefore I would recommend using a backorder

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1