Ross White's machine shop uses 2,500 brackets during the course of a year, and this usage is relatively constant throughtout the year. These brackets are purchased from a supplier 100 miles away for $15 each, and the lead tie is 2 days. The holding cost per bracket per is 1,50 ( or 10% of the unit cost) adn the ordering cost per order is $18.75.
There are 250 working days per year.
A) What is EOQ?
B) Given the EOQ,what is the average inventory?
What is the annual inventory holding cost?
C) In minimizing cost,how many orders would be made each year? What would be the annual ordring cost?
D) Given the EOQ, What is the total annual inventory cost (including purchase cost)?
E) What is the time between orders?
F) What is ROP?

Answer:

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