1. Hewlett Packard (HP) produces ink-jet printers in China for sales in Europe.
ID: 344698 • Letter: 1
Question
1. Hewlett Packard (HP) produces ink-jet printers in China for sales in Europe. Printers sold in different European countries differ in terms of the power supply unit and packaging that is used with the printer. Currently, HP assembles printers for sale in each of the individual countries. The distribution of weekly demands in these different countries are all normally distributed with the means and standard deviations shown below. Assume that demands in each country are independent of demands in all other countries. The lead time from the production facility in China to Europe is 12 weeks. Countr France Germany Spain Italy Portugal UK Mean Demand 6,000 8,000 7,500 3,200 1,750 11,000 Standard Deviation 3,000 3,400 2,800 1,600 800 4,250 How much safety inventory does HP require to satisfy all demand in the European countries in order to satisfy a CSL of 0.98? HP decides to build a centralized distribution center (DC) in Europe. HP will ship the base printer unit (without the power supply unit and final packaging) from China to the DC. When the DC receives an order, the DC will assemble the power supplies, package the unit for final sale and ship the completed printer package to the appropriate country. Given that the lead time will remain 12 weeks and HP sll wants to satisfy a CSL of 0.98, how much savings in safety inventory for printers can HP expect by following this policy? a) b)Explanation / Answer
Answer to question a:
Z value of CSL of 0.98 = NORMSINV ( 0.98 ) = 2.053
Lead time of delivery = 12 weeks for all countries in Europe
Safety stock for a particular country = Z value x Square root ( Lead time in weeks ) x Weekly standard deviation
= 2.053 x Square root ( 12) x weekly standard deviation
= 2.053 x 3.464 x weekly standard deviation
= 7.11 x weekly standard deviation
Therefore,
Total safety stock requirement at all 6 locations in Europe
= 7.11 x Sum of standard deviations of weekly demands at 6 locations
= 7.11 x ( 3000 + 3400 + 2800 + 1600 + 800 + 4250 )
= 7.11 x 15850
= 112693
HP REQUIRES SAFETY INVENTORY OF 112693 [RINTERS TO SATISFY ALL DEMANDS IN EUROPEAN COUNTRIES
Answer to question b :
When demand of all warehouses are aggregated in centralised distribution center , ,
Weekly standard deviation of cumulative demand
= Square root of ( sum of variances of demand for all locations )
= Square root ( 3000^2 + 3400^2 + 2800^2 + 1600^2 + 800^2 + 4250^2 )
= Square root ( 49662500)
= 7047.16
Z value of CSL 0.98 = NORMSINV ( 0.98 ) = 2.053
Therefore , safety inventory = Z value x weekly standard deviation of cumulative demand x Square root ( Lead time in weeks ) = 2.053 x 7047.16 x Square root ( 12) = 2.053 x 7047.16 x 3.464 =50116.52 ( 50117 rounded to nearest whole number)
Thus savings in safety inventory HP can expect = 112693 – 50117 = 62576
SAVINGS IN SAFETY INVENTORY FOR PRINTERS HP CAN EXPECT = 62576 NUMBERS
HP REQUIRES SAFETY INVENTORY OF 112693 [RINTERS TO SATISFY ALL DEMANDS IN EUROPEAN COUNTRIES
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