Lot size-reorder point systems: Example 5a A maintenance manager has collected h
ID: 428915 • Letter: L
Question
Lot size-reorder point systems: Example 5a A maintenance manager has collected historical data and found the following for the repair parts needed: Annual demand is Normally distributed with mean 14 units per year and standard deviation 3.73 Unit cost of the part is $150 per year Annual interest rate 20% i mean demand rate (unitslyear) t lead time D Random variable representing F() aumulative probability distribution of D o It takes 45 days to receive an order demand during the lead time Assume 365 days per year Cost of placing a purchase order is $10 foed or setup cost to place an order (S/order e unit cost (Sunitlyear) h holding cost (Suit . P penalty cost per unit of unsatisfied demand o Backorder cost is estimated at $40 per unit per year How many units should he order and at what frequency? This means, we want to find Q* and R*. Note that information on demand during lead time (D) was not given.We need to scale the annual demand to the lead time! - (see a discussion on page 277)Explanation / Answer
D = average annual demand = 14
K = ordering cost = $10
h = holding cost per unit per annum = 14% x $150 = $21
p = cost of backordering = $40
Optimal order quantity, Q* = {(2.D.K / h).(1+ h/p)}1/2 = SQRT((2*14*10/21)*(1+21/40)) = 4.51 units or 5 units (rounded off)
t = average lead time = 45/365 = 0.123 years
S = std. dev. of annual demand = 3.73
Sltd = std. dev. of lead time demand = S * SQRT(t) = 3.73 * SQRT(0.123) = 1.308
Optimal service level = p / (p+h) = 40/(40+21) = 0.6557
So, correspondingly, Z = NORMSINV(0.6557) = 0.40
Safety stock = Z.Sltd = 0.40 x 1.308 = 0.52
Average lead time demand = 14 x 0.123 = 1.722
So, reorder point, R* = Average lead time demand + Safety stock = 1.722+0.52 = 2.24 or 2 units
So, the policy should be to order for 5 units when the inventory level falls to 2 units.
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