(Note: If the lead time is given as a continuous time distribution take the mid point.)
Example: Demand for a product during an order period is assumed to be normally distributed with mean of 1000
units and standard deviation of 40 units. What % service can a company expect to provide (i) if it satisfies the
average demand only (ii) if it carries a safety stock of 60 units.
Solution:
1. If the company provides only average demand, we can expect only 50% service level.
2. The standard normal variate Z is computed with the following formula.
Z = (Safety stock - 0)/ Standard deviation
= (60 - 0)/40 = 1.5
The area under normal curve for Z = 1.5 is 0.4332.
Service level = 0.50 + 0.4332 = 0.9332 04 93.32%
Example: A manufacturer of water filters purchases components in EOQ's of 850 units/order. Total demand
averages 12000 components per year and MAD = 32 units per month. If the manufacturer carries a safety stock of
80 units, what service level does the this give the firm?
Solution:
Standard deviation
2/ 0.8 32 / 0.8 40MAD MAD
Z = (S S - 0)/S.D = (80 - 0)/40 = 2
The area under normal curve for Z = 2 = 0.4772.
Service level = 0.9772 or 97.72%
Example: A firm has normally distributed forecast of usage with MAD = 60 units. It desires a service level, which
limits the stock, outs to one order cycle per year.
(1) How much safety stock should be carried if the order quantity is normally a week's supply?
(2) How much safety stock should be carried if the order quantity is weeks supply.
Solution:
No. of orders = 52/year.
1 stock out in 52 weeks means a 98 % = (51/52) values
For 98% area, the value of
Z = 2.05 (from tables)
S D = MAD/0.8 = 60/08
Z = (S S - 0 )/S D= (S S - 0)/60/0.8
2.05 = (S S - 0 ) x 0.8/60
SS = 2.05 x 60/0.8 = 154 units.
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