Question

Demand /yearly = 12,000, S = $16, and H =10% of the Price

Offer. Discount Quantity, Discount (%), Discount Price (P)

1. 1 to 999, no discount, $5.00

2. 1,000 to 1,999, 4%, $4.80

3. 2,000 and over, 5%, $4.75

(a) Evaluate the number of computers that the store manager should order in each replacement lot.

(b) Maximum Inventory

(c) Average Inventory

(d) The total annual number of order cycles

(e) Length of the order cycle

(f) Average flow time

(g) Annual inventory-related cost (Yearly total cost)

Answer 1

a)

For order quantity 1 to 999,

EOQ = sqrt(2DS/h) = sqrt((2*12000*16)/(10%*5)) = 876.356092

Total inventory related cost = order cost+holding cost+purchase cost = (12000/876.356092)*16+(876.356092/2)*(10%*5)+12000*5
=60438.17805

For order quantity 1000 to 1999,

EOQ = sqrt(2DS/h) = sqrt((2*12000*16)/(10%*4.8)) = 894.427191

Ordering quantity is adjusted upwards to 1000 to avail discount

Total inventory related cost = order cost+holding cost+purchase cost = (12000/1000)*16+(1000/2)*(10%*4.8)+12000*4.8
=658032

For order quantity 2000 and over,

EOQ = sqrt(2DS/h) = sqrt((2*12000*16)/(10%*4.75)) = 899.1223791

Ordering quantity is adjusted upwards to 2000 to avail discount

Total inventory related cost = order cost+holding cost+purchase cost = (12000/2000)*16+(2000/2)*(10%*4.75)+12000*4.75
=57571

As Total inventory related cost is minimum with order quantity = 2000, the number of computers that the store manager should order in each replacement lot = 2000

b) Maximum inventory = EOQ = 2000
c) Average inventory = EOQ/2 = 1000
d) total number of order cycles = 12000/2000 = 6
e) length of order cycle = total number of months in an year/number of order cycles = 12/6 = 2 month
f) Average flow time = optimal ordering quantity/(2*demand per month) = 2000/(2*(12000/12)) = 1 month
g) Total Annual inventory related cost = order cost+holding cost+purchase cost = (12000/2000)*16+(2000/2)*(10%*4.75)+12000*4.75
=57571