Question

Lieutenant Commander Data is planning to make his monthly (every 30 days) trek to Gamma Hydra City to pick up a supply of isolinear chips. The trip will take Data about four days. Before he leaves, he calls in the order to the GHC Supply Store. He uses chips at an average rate of seven per day (seven days per week) with a standard deviation of demand of three per day. He needs a 95 percent service probability.

a. If he currently has 40 chips in inventory, how many should he order? (Use Excel's NORMSINV() function to find the correct critical value for the given a-level.)

Number of chips _____

b. What is the most he will ever have to order? (Use Excel's NORMSINV() function to find the correct critical value for the given a-level. )

Maximum order quantity _____

Answer 1

Answer:

a

 $N  =  209 \ unites$

b

 $N_{max}  =  249$  

Step-by-step explanation:

From the question we are told that

   The daily demand is  $d =  7/day$

   The  standard deviation is  $\sigma  =  3 /day$

   The  service probability is  $s =  95\%  =  0.95$

    The number of chips  in the inventory is $k  =  40$

   The number of days in his monthly is  $T =  30 \  days$

    The lead time is  $L  =  4$

 

Generally the number he should order (optimal order ) is mathematically represented as

      $N  =  d(L  +  T)+(\sigma_{T + L}  * z)  -k$

Here $\sigma__{T+L }}$ is the monthly standard deviation which is mathematically evaluated as

            $\sigma__{T  + L}} =  \sqrt{(T + L) * 1^2}$

=>         $\sigma__{T  + L}}  =  \sqrt{34}$

=>          $\sigma__{T+L}} =  5.83$

Also  z-value  for the  95% service probability  from the z-table  is  1.96

So

      $N  =  7 *  (30 +4 ) + (1.96 *  5.83) - 40$

=>     $N  =  209 \ unites$

For the maximum order quantity  k =  0

So

    $N_{max} = d(L  +  T)+(\sigma_{T + L}  * z)  -0$

=> $N_{max}  =  7 *  (30 +4 ) + (1.96 *  5.83) - 0$

=>  $N_{max}  =  249$