Answer:
219.57 units
Explanation:
Given :
Daily demand, d = 7 per day
Standard deviation, = 2 per day
Service probability = 98%
Total number of days per week = 7
Lead time , L = 3 days
On hand inventory, I = 35
Now calculating the optimal order quantity by using the given formula,
[tex]$q=d(T+L)+ z \ \sigma_{r+L}-I$[/tex] .............(i)
First, we will find out the value of [tex]$\sigma_{r+L}$[/tex] and z.
Therefore,
[tex]$\sigma_{r+L}=\sqrt{(30+3)(2)^2}$[/tex]
[tex]$=\sqrt{132}$[/tex]
= 11.48
Now the value of z can be found out from the z-table,
Z value for 98% service level = 2.054
Now putting the value of [tex]$\sigma_{r+L}$[/tex] and z in equation (i), we get,
[tex]$q=d(T+L)+ z \ \sigma_{r+L}-I$[/tex]
= (7)(30+3)+(2.054)(11.48) - 35
= 231 + 23.57 - 35
= 219.57 units
So the optimal number of the units required to be order = 219.57 units