Operations and Supply Chain Management The Core

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INVENTORY MANAGEMENT chapter 11 367
Example 11.3: Reorder Point
Consider an economic order quantity case where annual demand D = 1,000 units, economic order
quantity Q = 200 units, the desired probability of not stocking out P = 0.95, the standard deviation
of demand during lead time σ L = 25 units, and lead time L = 15 days. Determine the reorder point.
Assume that demand is over a 250-workday year.
SOLUTION
In our example, ¯
1,000
​d ​ = ​ _____ ​ = 4,​ and lead time is 15 days. We use the equation
250
R = ¯ ​d ​L + z​σ​
​
L ​
​ ​ ​
= 4(15) + z(25)
In this case, z is 1.64.
Completing the solution for R, we have
​R = 4(15) + 1.64(25) = 60 + 41 = 101 units​
This says that when the stock on-hand gets down to 101 units, order 200 more.
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Example 11.4: Order Quantity and Reorder Point
Daily demand for a certain product is normally distributed with a mean of 60 and standard deviation
of 7. The source of supply is reliable and maintains a constant lead time of six days. The cost of placing
the order is $10 and annual holding costs are $0.50 per unit. There are no stock-out costs, and
unfilled orders are filled as soon as the order arrives. Assume sales occur over the entire 365 days of
the year. Find the order quantity and reorder point to satisfy a 95 percent probability of not stocking
out during the lead time.
Excel:
Inventory
Control
SOLUTION
In this problem, we need to calculate the order quantity Q as well as the reorder point R.
¯ ​d ​= 60 S = $10
​σ​ d ​= 7 H = $0.50​ ​
D = 60(365) L = 6
The optimal order quantity is
____
​Q​ opt ​ = ​ √
​ ____ 2DS
____________
H ​ = ​ √
​ ___________
2(60)365(10) ​ = ​√ _______
876,000 ​ = 936 units​
0.50
To compute the reorder point, we need to calculate the amount of product used during the lead time
and add this to the safety stock.
The standard deviation of demand during the lead time of six days is calculated from the variance
of the individual days. Because each day’s demand is independent,
______
L _____
​σ​ L ​ = ​ √
​∑​
​ ​σ​ 2 d ​ = ​√ 6​(7)​ 2 ​ = 17.15​
i=1
Once again, z is 1.64.
​R = ¯ ​d ​L + z​σ​ L ​ = 60(6) + 1.64(17.15) = 388 units​
To summarize the policy derived in this example, an order for 936 units is placed whenever the
number of units remaining in inventory drops to 388.
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