Question

Daily demand for a product is 50 units, with a standard deviation of 30 units. The review period is 5 days and the lead time is 10 days. At the time of review there are 60 units in stock. If 99 percent service probability is desired, how many units should be ordered? (Use Excel's NORMSINV() function to find the correct critical value for the given α-level. Do not round intermediate calculations. Round "z" value to 2 decimal places and final answer to the nearest whole number.) Ordered quantity units

Answer 1

Answer:

The number of units to be ordered is 961.

Step-by-step explanation:

The optimum demand is given as

$q=\bar{d}(L+R)+z\sigma_{L+R}-I$

From the data

Daily demand is given as $\bar{d}$=50

Standard deviation is given as $\sigma=30$

The Lead time is given as L=10 days

The Revier time is given as R=5 days

Initial values given as I=60 units

The confidence level is given as 99% so z is calculated using the NORMSINV() function as NORMSINV(0.99). The value of z is 2.33

Now the standard deviation in terms of the Lead and Review time is given as

$\sigma_{L+R}=\sqrt{L+R}\sigma\\\sigma_{L+R}=\sqrt{10+5}\times 30\\\sigma_{L+R}=\sqrt{15}\times 30\\\sigma_{L+R}=116.18$

Substituting the values in the formula as

$q=\bar{d}(L+R)+z\sigma_{L+R}-I\\q=50(10+5)+2.33\times 116.18-60\\q=50(15)+2.33\times 116.18-60\\q=960.69 \approx 961$

So the number of units to be ordered is 961.