Question

Daily demand for a product is 170 units, with a standard deviation of 30 units. The review period is 20 days and the lead time is 6 days. At the time of review there are 40 units in stock.
If 98 percent service probability is desired, how many units should be ordered?

Answer 1

Answer:

4,693.58 units

Explanation:

The computation of the number of unit to be ordered is shown below:

But before that we need to find out the standard deviation of review period and the lead time which is shown below:

$\sigma (T+L) = \sqrt{(T+L) \sigma^{2} }$

$= \sqrt{(20+6)30^{2} }$

= 152.97

Now the number of units to be ordered is

$Q = D(T+L) + Z (\sigma ) - I$

where,

T = Review period

L = Lead time

D = daily demand

I = Inventory

Z = service probability

Now placing these values to the above formula

So,

= 170 units ×  (20 + 6) + 2.05 × 152.97 - 40 units

= 4,420 units + 313.58 - 40  units

= 4,693.58 units

The 98% service probability is 2.05 and the same is to be considered