The annual demand for a product is 17,400 units. The weekly demand is 335 units

Question

The annual demand for a product is 17,400 units. The weekly demand is 335 units with a standard deviation of 80 units. The cost to place an order is $29.00, and the time from ordering to receipt is eight weeks. The annual inventory carrying cost is $0.20 per unit. a. Find the reorder point necessary to provide a 99 percent service probability. (Use Excel's NORMSINVO function to find the correct critical value for the given a-level. Round "z" value to 2 decimal places.) Reorder point b. Suppose the production manager is asked to reduce the safety stock of this item by 60 percent. If she does so, what will the new service probability be? (Use Excel's NORMSDISTO function to find the correct probability for your computed Z-value. Round "z" value to 2 decimal places and final answer to 1 decimal place.) Service probability

Explanation / Answer

Answer to question a :

Standard deviation of weekly demand = 80 weeks

Lead time ( i.e time from ordering to receipt ) = 8 weeks

Hence, standard deviation of demand during lead time

= Standard deviation of weekly demand x Square root ( Lead time)

= 80 x square root ( 8)

= 80 x 2.828

= 226.24

Z value for 98 percent service probability = NORMSINV ( 0.98) = 2.053

Therefore ,

Safety stock

= Z value x Standard deviation of demand during lead time

= 2.053 x 226.24

= 464.47 ( 465 rounding to next higher whole number)

Thus , Safety stock = 465 units

Hence,

Reorder point

= Weekly demand x Lead time ( weeks ) + safety stock

= 335 x 8 + 465

=2680 + 465

= 3145 units

REORDER POINT = 3145 UNITS

Answer to question b:

If the safety stock reduces by 60 percent, new safety stock becomes 40 percent of initial safety stock

Therefore ,

New safety stock = 40% of 465 units = 186

Let the Z value for the corresponding Service probability = Z1

Hence,

Z1 x Standard deviation of demand during lead time = New safety stock

Or, 226.24.Z1 = 186

Or, Z1 = 186/226.24 = 0.82 ( rounded to 2 decimal places )

Corresponding probability as derived from standard normal distribution table for Z1 = 0.82 is 0.79389

Therefore , Service probability will be = 0.79389 x 100 = 79.389 % ( or 79.4 % rounded to 1 decimal place )

NEW SERVICE PROBABILITY WILL BE = 79.4%

REORDER POINT = 3145 UNITS

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