The company Q-Tec is specialised on Computers. Currently, the model C453X is ver
ID: 363804 • Letter: T
Question
The company Q-Tec is specialised on Computers. Currently, the model C453X is very popular. It is sold to the final customer in 45 retail stores. These stores are only open on 4 days per week (for historical reasons, Thursday – Sunday). During these days, the daily demand of C453X in each of the stores is normally distributed with mean demand 50 units and standard deviation 25 units. The company uses an s,Q policy to manage the local warehouse at each of the stores individually. Q-Tec orders directly from the supplier and receives the orders after 9 weeks.
a) In order to have a CSL of at least 95% in each of their local storages, how must the reorder point (s) be set? How large is the total safety stock in the network?
b) The last year was not very successful for Q-Tec. Therefore they consider to close all their stores and to offer their products only online. The customers would then receive the goods from a central warehouse. If in this warehouse the CSL should be 99.76%, how large is the safety stock? How much can be saved in terms of safety stock compared to the situation in a)
c) Finally, Q-Tec decides on a different alternative. There will be one central new warehouse (CSL = 99.76%) but the 45 stores remain. The central warehouse will get the products from the supplier and will then be delivered to the stores with a lead time of only 2 days. In the stores, again a CSL of 95% should be reached. How does this affect the inventory in the network if you assume that the products belong to Q-Tec when they arrive in the central warehouse? Compare your results with a) and b) and give a recommendation and justification.
Explanation / Answer
Answer to question a :
Z value corresponding to Customer service level of 95 % ( i.e ins tock probability of 0.95 ) will be = NORMSINV ( 0.95 ) = 1.6448
Mean daily demand of each store = m = 50 units
Standard deviation of daily demand of each unit = Sd = 25 units
Lead time of delivery after placement of order = 9 weeks x 4 days/ week = 36 days
Therefore ,
Standard deviation of demand during lead time
= Standard deviation of daily demand x Square root ( Lead time )
= 25 x Square root ( 36 )
= 25 x 6
= 150
Therefore ,
Safety stock requirement
= Zvalue x Standard deviation of demand during lead time
= 1.6448 x 150
= 246.72 ( 247 rounded to nearest whole number )
Therefore, Safety stock requirement = 247
Therefore ,
Reorder point for each store
= Mean daily demand x Lead time + Safety stock
= 50 / day x 36 days + 247
= 1800 + 247
= 2047
Safety stock in the system = Safety stock/ store x Number of stores ( 45 stores)
= 247 x 45
= 11115
REORDER POINT FOR EACH STORE = 2047
SAFETY STOCK IN THE SYSTEM = 11115
Answer to question B :
Average daily demand at central warehouse
= Average daily demand of each store x 45 stores
= 50 x 45
= 2250
Standard deviation of daily demand at central warehouse
= Standard deviation of daily demand at each store x Square root ( number of stores)
= 25 x Square root ( 45 )
= 25 x 6.708
= 167.705
Since delivery lead time will be 9 weeks or 36 days, standard deviation of demand
during lead time
= Standard deviation of daily demand x Square root ( 36)
= 167.705 x 6
= 1006.23
Z value corresponding to customer service level of 99.76% ( probability 0.9976) = 2.82
Therefore ,
Safety stock requirement at central warehouse
= Zvalue x standard deviation of daily demand at central warehouse
= 2.82 x 1006.23
= 2837.56 ( 2838 rounded to nearest whole number )
Amount of safety stock at 45 warehouses will be 2838 which gets reduced from 11115 as in scenario #a . Thus amount that can be saved in terms of safety stock = 11115 – 2838 = 8277 numbers
SAFETY STOCK = 2838
AMOUNT THAT CAN BE SAVED IN TERMS OF SAFETY STOCK = 8277 UNITS
Answer to question # c:
Lead time of delivery from central warehouse to individual store = 2 days
Daily demand at individual store = 50 units
Standard deviation of daily demand = 25 units
Hence , Standard deviation of demand during lead time of 2 days
= 25 x Square root ( 2 )
= 25 x 1.414
= 35.35
Z value corresponding to service level of 95% ( probability o.95) = 1.6448
Therefore safety stock requirement at store for supply from central warehouse
= Z value x Standard deviation of demand during lead time
= 1.6448 x 35.35
= 57.568 ( 58 rounded to next higher whole number )
Hence reorder point for supply from new warehouse
= Daily demand x Lead time + Safety stock
= 50 x 2 + 58
= 158
The comparative picture of system inventory for all these options as follows :
OPTION
IN PROCESS INVENTORY ( DAILY DEMAND X LEAD TIME )
SAFETY STOCK
TOTAL INVENTORY IN THE SYSTEM ( IN PROCESS + SAFETY STOK)
OPTION #a
( 50/day x 36 days) FOR EACH STORE X 45 STORES =81000
11115
81000 + 11115 = 92115
OPTION #b
( 50/day-store x 45 Stores) x 36 days = 81000
2838
81000+ 2838 = 83838
OPTION #c
81000 ( as above) + 50/day for each store x Lead time of 2 days between central warehouse and each store x 45 store = 81000 + 4500 = 85500
2838 ( as above) + 58 for each store x 45 = 2838 + 2610 = 5448
85500 + 5448 = 90948
REORDER POINT FOR EACH STORE = 2047
SAFETY STOCK IN THE SYSTEM = 11115