Inventory Management and Economic Order Quantity (EOQ) Analysis
Problem #12-1: ABC Inventory Analysis for the Welsh Corporation
The Welsh Corporation utilizes 10 key components in one of its manufacturing plants. The task is to perform an ABC analysis based on the provided data and explain the rationale behind the classification.
Data and Calculations
| SKU | Item Cost ($) | Annual Demand |
|---|---|---|
| WC219 | 0.10 | 12,000 |
| WC008 | 1.20 | 22,500 |
| WC916 | 3.20 | 700 |
| WC887 | 0.41 | 6,200 |
| WC397 | 5.00 | 17,300 |
| WC654 | 2.10 | 350 |
| WC007 | 0.90 | 225 |
| WC419 | 0.45 | 8,500 |
| WC971 | 7.50 | 2,950 |
| WC713 | 10.50 | 1,000 |
ABC Classification and Explanation
Based on the calculated annual dollar usage and cumulative percentages, one possible ABC classification is as follows:
- A Items (30% of items, 86.5% of value): WC397, WC008, WC971
- B Items (20% of items, 9.1% of value): WC713, WC419
- C Items (50% of items, 4.4% of value): WC887, WC916, WC219, WC654, WC007
This classification suggests that a small portion of items (A items) account for a significant portion of the total inventory value, while a large portion of items (C items) contribute to a relatively small portion of the value. B items fall in between these two extremes.
Problem #12-2: A-B-C Classification for Health Care Facility Inventory
The following table presents data on monthly volume and unit costs for a sample of 16 items from a health care facility’s inventory. The objective is to develop an A-B-C classification for these items.
Data and Classification
| Item | Unit Cost | Usage | Dollar Usage | Category |
|---|---|---|---|---|
| F95 | 30 | 800 | 24,000 | A |
| Z45 | 80 | 250 | 16,000 | A |
| K35 | 25 | 600 | 15,000 | A |
| P05 | 16 | 500 | 8,000 | B |
| F14 | 20 | 300 | 6,000 | B |
| D45 | 10 | 550 | 5,500 | B |
| K36 | 36 | 150 | 5,400 | B |
| D57 | 40 | 120 | 4,800 | B |
| K34 | 10 | 200 | 2,000 | C |
| D52 | 15 | 110 | 1,650 | C |
| M20 | 20 | 80 | 1,600 | C |
| F99 | 20 | 60 | 1,200 | C |
| N08 | 30 | 40 | 1,200 | C |
| D48 | 12 | 90 | 1,080 | C |
| M10 | 16 | 25 | 400 | C |
| P09 | 10 | 30 | 300 | C |
Problem #12-3: Economic Order Quantity (EOQ) for Flour
A bakery purchases flour in 25-pound bags at $30 per bag, with an annual usage of 4,860 bags. The cost to prepare and receive an order is $10, and annual holding costs are $75 per bag. The goal is to determine the EOQ and associated costs.
EOQ Calculation
Using the EOQ formula:
EOQ = √(2DS / H) = √(2 * 4,860 * $10 / $75) ≈ 36 bags
Average Inventory, Orders per Year, and Total Costs
- Average number of bags on hand: Q*/2 = 36/2 = 18 bags
- Orders per year: D/Q* = 4,860/36 = 135 orders
- Total ordering cost: S * D/Q* = $10 * 135 = $1,350
- Total holding cost: H * Q*/2 = $75 * 18 = $1,350
- Total cost: Ordering cost + Holding cost = $1,350 + $1,350 = $2,700
Problem #12-4: EOQ Analysis for Garden Variety Flower Shop
Garden Variety Flower Shop uses 750 clay pots per month, purchased at $2 each. Annual carrying costs are 30% of the cost, and ordering costs are $20 per order. The manager currently uses an order size of 1,500 pots. The task is to analyze the cost implications and benefits of using the EOQ.
EOQ Calculation and Cost Comparison
EOQ = √(2DS / H) = √(2 * 9,000 * $20 / $0.60) ≈ 774.6 pots (rounded to 775 pots)
Total cost with EOQ: $464.71
Total cost with order size of 1,500: $570
Additional annual cost with current order size: $570 – $464.71 = $105.29
Benefits of Using EOQ
Besides cost savings, using the EOQ would reduce the required storage space by approximately half.
Problem #12-5: EOQ and Order Frequency for Mail-Order House
A mail-order house uses 18,000 boxes annually, with carrying costs of $0.60 per box per year and ordering costs of $96. The price per box varies based on the order quantity. The objective is to determine the optimal order quantity and number of orders per year.
EOQ Calculation and Price Break Analysis
EOQ = √(2DS / H) = √(2 * 18,000 * $96 / $0.60) ≈ 2,400 boxes
Comparing total costs at different price breaks:
- TC2,400 = $2,880 + $1,440 = $4,320
- TC5,000 = $5,750 + $720 = $6,470
- TC10,000 = $10,400 + $360 = $10,760
The optimal order quantity is 2,400 boxes, resulting in the lowest total cost.
Number of Orders per Year
Orders per year: D/Q* = 18,000/2,400 = 7.5 orders (rounded to 8 orders)
Problem #12-6: EOQ Analysis for the Friendly Sausage Factory
The Friendly Sausage Factory (FSF) can produce hot dogs at a rate of 5,000 per day and supplies them to restaurants at a steady rate of 250 per day. The setup cost is $66, annual holding costs are $0.45 per hot dog, and the factory operates 300 days a year. The goal is to find the optimal run size, number of runs per year, and run length.
EOQ Calculation and Run Parameters
EOQ = √(2DS / H) = √(2 * 75,000 * $66 / $0.45) ≈ 4,812 hot dogs
- Number of runs per year: D/Q* = 75,000/4,812 ≈ 16 runs
- Run length: Q*/p = 4,812/5,000 ≈ 1 day
Problem #12-7: EOQ and Production Run Analysis for New Product Component
A company is starting production of a new product. The manager of the component production department wants to determine the optimal run quantity, production run length, and inventory buildup rate.
EOQ Calculation and Production Run Parameters
EOQ = √(2DS / H) = √(2 * 20,000 * $300 / $10) ≈ 1,095.45 units (rounded to 1,414 units considering daily production rate)
- Production run length: Q*/p = 1,414/200 ≈ 7 days
- Inventory buildup rate during production: 200 – 80 = 120 units per day