Question

Joe​ Henry's machine shop uses 2 comma 500 brackets during the course of a year. These brackets are purchased from a supplier 90 miles away. The following information is known about the​ brackets:
Annual demand: 2,500
Holding cost per bracket per year: $ 1.50
Order cost per order: $ 18.75
Lead time: 2 days
Working days per year: 250 ​
a) What is the​ EOQ? ​(round your response to two decimal​ places). ​
b) What is the average inventory if the EOQ is​ used? ​(round your response to two decimal​ places). What would be the annual inventory holding​ cost? ​​(round your response to two decimal​ places). ​
c) Given the​ EOQ, how many orders will be made​ annually? ​(round your response to two decimal​ places). What would be the annual order​ cost? ​​(round your response to two decimal​ places).
​d) Given the​ EOQ, what is the total annual cost of managing​ (ordering and​ holding) the​ inventory? ​​(round your response to two decimal​ places). ​
e) What is the time between​ orders? ​(round your response to two decimal​ places).
​f) What is the reorder point​ (ROP)? ​(round your response to two decimal​ places).

Answer 1

Answer:

a) 250 brackets;

b) Average inventory: 125 brackets; Annual inventory holding cost: $187.50

c) 10 orders;

d) $375

e) 25 days

f) 20 brackets

Explanation:

a) EOQ = square root of [(2* Order Cost per one order * annual demand] / Holding Cost per bracket per year ] = square root of [ 2* 18.75 * 2,500 / 1.5] = 250 brackets.

b) Average inventory =  EOQ/2 = 125 brackets; Annual inventory holding cost = 125 x 1.5 = $187.5

c) Orders made annually give EOQ = Annual demand / EOQ = 2,500/250 = 10 orders;

d) Total annual cost of managing​ (ordering and​ holding) the​ inventory = 10 x 18.75 + 187.5 = $375

e) Time between​ orders = Total annual working days/ orders made per year = 250/10 = 25 days.

f) The reorder point​ (ROP) = Demand of bracket per working day * lead time = Annual demand * Lead time / total annual working days = 2,500*2/250 = 20 brackets.