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The storekeeper of a tyre manufacturing company has to plan the firm's inventory requirements for the valves used in the tyres

Accounting Nov 29, 2020

The storekeeper of a tyre manufacturing company has to plan the firm's inventory requirements for the valves used in the tyres. The monthly usage is 500 valves. The storekeeper has reported that the cost to keep one valve in the storeroom for one year is N$5. The valves are supplied by a firm in Angola which normally delivers the valves between one month and 2 months after the order was placed. The cost to place an order is N$150. The storekeeper has asked you to assist him in calculating some of the inventory levels. Required: Calculate the following inventory levels: a) Economic order quantity b) Re-order level (Re-order point) c) Minimum inventory level (Safety stock) d) Average inventory level e) Maximum inventory level f) Annual ordering cost B) Annual carrying cost

Expert Solution

Economic order quantity : Economic order quantity is order quantity which helps to  minimize the holding cost and ordering cost for your business.It is optimum amount of item that should be ordered at any given point in time , such that total annual cost of carrying and ordering that item is minimized. it sometime also know as optimum lot size.

Formula= Square root of ( 2*D*S/H)

where D=Demand in units( means how much is demand for that item typically annualy , which is to be order for given period of time.)

S= ordering cost ( means incremental cost to process and order other than purchasing amount)

H= Holding cost( means incremental cost to hold one unit in inventory)

Multiply the demand by 2 ,then multiply the result with order cost , then divide the result by holding cost.

for the above question EOQ(Economic order quantity ) is as follows

EOQ= Square root of (2 * D * S/ H)

D= 500*12=6000 units ( Deamand is 500 units monthly , so monthly would be 500*12 = 6000 units)

S =N$150

H=N$5

By putting the values in formula, we get

EOQ= Square root of 2*6000*150/5

= square root of 360000

= 600 valves

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