Inventory Control Modules

Quantative Analysis for Management
Chapter 6 - Inventory Control Modules


Question 7.
Shoe shine is a local retail shoe store on the north side of Centerville. Annual demand for a popular sandal is 500 pairs, and John Dirk, the owner of the Shoe Shine, has been in the habit of ordering 100 pairs at a time. John estimates that the ordering cost is $10 per order. The cost of the sandal is $5 per pair.
(A) For John’s ordering policy to be correct, what would the carrying cost as a percentage of the unit cost have to be?
Q should be square root of 2DCo/Ch, that is,
Q = sqrt(2DCo/Ch)
100= sqrt(2*500*10/Ch)
10000=2*500*10/Ch
Ch=10000/10000=$1 per unit per year
So the carrying cost per year per unit is 1/5=20% of the unit cost.

(B) If the carrying cost were 10% of the cost, what would the optimal ordering quantity be?
Q = sqrt( 2DCo/Ch)= sqrt(2*500*10/0.5)= sqrt(20000)=141.4=141

Question 8.
Ross White's machine shop uses 2,500 brackets during the course of a year, and this usage is relatively constant throughout the year. These brackets are purchased from a supplier 100 miles away for $15 each, and the lead time is 2 days. The holding cost per bracket per year is $1.50 (or 10% of the unit cost) and the ordering cost per order is $18.75.

There are 250 working days per year.

(a) What is the EOQ?
Q = sqrt( 2DCo/Ch)= sqrt( 2*2500*18.75/1.5)=250

(b) Given the EOQ, what is the average inventory? What is the annual inventory holding cost?
Average inventory =Q/2=250/2=125
Annual inventory holding cost=Q*Ch/2=125*1.5=$187.5

(c) In minimizing cost, how many orders would be made each year? What would be the annual ordering cost?
D/Q=2500/250=10, so 10 orders would be made each year.
annual ordering cost = Co*D/Q = 18.75*10=$187.5

(d) Given the EOQ, what is the total annual inventory cost (including purchase cost)?
annual inventory cost= Annual inventory holding cost+ annual ordering cost + purchase cost
=187.5+187.5+2500*15=$37875

(e) What is the time between orders?
Since we need to order 10 times a year, and the lead time is 2 days, the time between orders is
365/10 – 2 =34.5 (days)

(f) What is the ROP?
ROP=LT*D=(2/365)*2500=13.6986=14

Question 9.
Ross White (Question 8) wants to reconsider his decision of buying the brackets and is considering making the brackets in-house. He has determined that set-up costs would be $25 in machinist time and lost production time, and 50 brackets could be produced in a day once the machine has been set-up. Ross estimates that the cost (including labor time and materials) of producing one bracket would be $14.80. The holding costs would be 10% of this cost.
a. What is the daily demand rate?
2500/365=6.85 per day

b. What is the optimal production quantity?
sqrt( 2DCo/Ch)=sqrt(2*2500*25/1.48)= 290.619=291

c. How long will it take to produce the optimal quantity?
291/50=5.82 days
d. How much inventory is sold during the production run time?
6.85*5.82= 40
e. If Ross uses the optimal production quantity, what would be the maximum inventory level?
291-40=251
e. What would be the average inventory level?
291/2=145.5
f. What is the annual inventory cost?
The annual setting up cost =25*1000/291 =85.91
The annual holding cost = 291*1.48/2 =215.34
The annual inventory cost=85.91+215.34=301.25
g. How many production would there be each year?
1000/291=3.44
I. What would be the annual set up cost?
The annual setting up cost=25*1000/291 =85.91
J. Given the optimal production run size, what is the total annual inventory cost?
The annual inventory cost=85.91+215.34=301.25
Remark: Is this question the same as f? Or question f should be the annual holding cost?
k. If the lead time is one-half day, what is the ROP?
ROP=LT*D=(1.5/365)*1000=4.10959=4





Question 10.
North Manufacturing has a demand for 1,000 pumps each year. The cost of a pump is $50. It cost North Manufacturing $40 to place an order, and the carrying cost is 25% of the unit cost. If pumps are ordered in quantities of 200, North Manufacturing can get a 3% discount on the cost of pumps. Should North Manufacturing orders 200 pumps at a time and take the 3% discount?
Let’s first calculate the total cost of ordering 200 pumps.
If ordering 200 pumps at a time, the total cost is
The ordering cost=Co*D/Q =1000*40/200=200
The holding cost = Q*Ch/2 = 200*0.25*50/2=1250
The purchasing cost = 1000*50*(1-0.03)= 48500
The total cost=200 + 1250+48500=$49950

The EOQ i= sqrt( 2DCo/Ch)=sqrt(2*1000*40/12.5)=80, and the total cost is
40*1000/80+ 80*12.5/2+1000*50=$51000, which is higher than the cost of ordering 200 pumps.
So North Manufacturing should order 200 pumps at a time and take the 3% discount

Formulas.

 EOQ minimizes the sum of holding and setup costs
 Q = 2DCo/Ch
D = annual demand
Co = ordering/setup costs
Ch = cost of holding one unit of inventory


 Quantity to which inventory is allowed to drop before replenishment order is made

 Need to order EOQ at the Reorder Point:

ROP = D X LT
D = Demand rate per period
LT = lead time in periods