# OM groupassignment emporium Essay

Submitted By Bklyn77
Words: 1081
Pages: 5

GSM 5113: OPERATIONS MANAGEMENT

CASE STUDY 6 WRITE UP: PARTS EMPORIUM CASE STUDY

GROUP 4 MEMBERS:

NAME
BUVANESWAREN NAIR
VIVEKSARATI A/L SANDRASIGARAN
MOHD FARIZ SHAHRAN
DR. RAIMOND SELKE

MATRIC ID
PBSxxxxx
PBSxxxxx
GMxxxxx
GMxxxxx
GMxxxxx

1

A.

Synopsis
This case describes the problems facing Sue McCaskey, the new materials manager of a wholesale distributor of auto parts. She seeks ways to cut the bloated inventories while improving customer service. Back orders with excessive lost sales are all too frequent.
Inventories were much higher than expected when the new facility was built, even though sales have not increased. Summary data on inventory statistics, such as inventory turns, are not available. McCaskey decides to begin with a sample of two products to uncover the nature of the problems—the EG151 exhaust gasket and the DB032 drive belt. B.

Analysis
We now find appropriate policies for a Q system, beginning with the exhaust gasket. Shown here are the calculations of the EOQ and R, followed by a cost comparison between this continuous review system and the one now being used. The difference is what can be realized by a better inventory control system. Reducing lost sales due to back orders is surely the biggest benefit.

Question 1: Put yourself in Sue McCaskey’s position and prepare a detailed report to Dan
Block and Ed Spriggs on managing the inventory of the EG151 and DB032.

a. New plan
Begin by estimating annual demand and the variability in the demand during the lead time for this first item. Working with the weekly demands for the first 21 weeks of this year and assuming 52 business weeks per year, we find the EOQ as follows: Weekly demand average = 102 gaskets/week
Annual demand (D) = 102(52) = 5304 gaskets
Holding cost = \$1.85 per gasket per year (or 0.21 × 0.68 × \$12.99)
Ordering cost = \$20 per order

2

Turning to R, the Normal Distribution appendix shows that a 95% cycle-service level corresponds to a z = 1.65. We then use the EG151 data to find the standard deviation of demand.
Standard deviation in weekly demand = 2.86 gaskets
Standard deviation in demand during lead time σ dLT = 2.86 2 = 4.04
R = Average demand during the lead time + Safety stock
= 2(102) + 1.65(4.04) = 210.66, or 211 gaskets
b. Cost comparison
After developing their plan, students can compare its annual cost with what would be experienced with current policies.
Cost Category
Ordering cost
Holding cost (cycle inventory)
TOTAL

Current Plan

Proposed Plan

\$707
139
\$846

\$313
314
\$627

The total of these two costs for the gasket is reduced by 26 percent (from \$846 to
\$627) per year. The safety stock with the proposed plan may be higher than the current plan, if the reason for the excess back orders is that no safety stock is now being held (inaccurate inventory records or a faulty replenishment system are other explanations). We cannot determine the safety stock level (if any) in the current system. The extra cost of safety stock for the proposed system is minimal, however. Only seven gaskets are being proposed as safety stock, and their annual holding cost is just another \$1.85(7) = \$12.95.
Surely the lost sales due to back orders are substantial with the current plan and will be much less with the proposed plan. One symptom of such losses is that 11 units are on back order in week 21. A lost sale costs a minimum of \$4.16 per gasket (0.32. × \$12.99). If 10 percent of annual sales were lost with the current policy, this cost would be \$4.16(0.10)(5,304) = \$2,206 per year. Such a loss would be much reduced with the 95% cycle-service level implemented with the proposed plan.
2. DB032 Drive Belt
a. New plan
The following demand estimates are based on weeks 13 through 21. Weeks 11 and 12 are excluded from the analysis because the new product’s start-up makes them unrepresentative. We find the EOQ as follows:
Weekly…