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

4,693.58 units

Step-by-step 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