91Ó°ÊÓ

The standard deviation of a collection of datais a measure of its dispersion from the mean. It quantifies the absolute variability of a distribution; the larger the dispersion or variability, the greater thestandard deviation and the degree of the value's divergence from its mean.

Given,

Daily demand for a product = 100 units

the standard deviation = 25 units

Review period = 10 days

lead time = 6 days

percent service probability = 98%

$\begin{array}{rcl}{\mathrm{δ}}_{\mathrm{L}+\mathrm{T}}& =& \sqrt{\mathrm{leadtime}}×\mathrm{Standard}\mathrm{deviation}\\ {\mathrm{δ}}_{\mathrm{L}+\mathrm{T}}& =& \sqrt{16}\left(25\right)\\ & =& 100\mathrm{units}\\ & & \end{array}$

So, by considering 98% service probability, apply the formula NORMSINV (098). The value of z was obtained as 2.05.

Substitute the values in equation (1).

$\begin{array}{rcl}\mathrm{q}& =& \mathrm{d}\left(\mathrm{T}+\mathrm{L}\right)+{\mathrm{³úδ}}_{\mathrm{L}+\mathrm{T}}-\mathrm{I}\\ \mathrm{q}& =& 100\left(6+9\right)+\left(2.05\right)×\left(100\right)-50\\ \mathrm{q}& =& 1,755\mathrm{units}\end{array}$

So, 1755 units should be ordered to meet the requirement.