Question

Joe​ Henry's machine shop uses 2 comma 510 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 comma 510 Holding cost per bracket per year $ 1.60 Order cost per order $ 20.00 Lead time 2 days Working days per year 250

a.) Given the above information, what would be the economic order quantity (EOQ)?b.) Given the EOQ, what would be the average inventory? What would be the annual inventory holding cost?c.) Given the EOQ, how many orders would be made each year? What would be the annual order cost?d.) Given the EOQ, what is the total annual cost of managing the inventory?e.) What is the time between orders?f.) What is the reorder point (ROP)

Answer 1

D = 2,510 brackets

H = $1.60

Co = $20

EOQ = √2 x 2510 x 20/1.60

EOQ = 250 units

Average inventory = EOQ/2

= 250/2

= 125 units

Total Holding Cost = QH/2

= 250 x $1.60/2

= $200

No of order = Annual demand/EOQ

= 2,510/250

= 10 times

Annual ordering cost = DCo/Q

= 2,510 x $20/250

= $200

Total annual cost = Annual ordering cost + annual holding cost

= $200 + $200

= $400

Time between orders = No of working days in a year/No of order

= 250/10

= 25 days

Explanation: Economic order quantity is a function of square root of 2 x annual demand x ordering cost per order divided by holding cost per item per annum. D denotes annual demand, Co is ordering cost per order and H represents holding cost per item per annum.

Average inventory is calculated as EOQ/2

Total annual holding cost is calculated as EOQ multiplied by holding cost per item per annum/2

No of order is the ratio of annual demand to EOQ

Annual ordering cost is calculated as annual demand multiplied by ordering cost per order divided by EOQ

Total annual cost is the aggregate of annual ordering cost and annual holding cost

Time between orders is the ratio of number of days in a year to number of order