The Warren W. Fisher Computer Corporation purchases 8,000 transistors each year as components in minicomputers. The unit cost of each transistor is $10, and the cost of carrying one transistor in inventory for a year is $3. Ordering cost is $30 per order. Assume that Fisher operates on a 200-day working year. a. What is the optimal order quantity? b. What is the expected number of orders placed each year? c. What is the expected time between orders? d. What is the total cost?
The Warren W. Fisher Computer Corporation purchases 8,000 transistors each year as components in minicomputers. The unit cost of each transistor is $10, and the cost of carrying one transistor in inventory for a year is $3. Ordering cost is $30 per order. Assume that Fisher operates on a 200-day working year. a. What is the optimal order quantity? b. What is the expected number of orders placed each year? c. What is the expected time between orders? d. What is the total cost?
Chapter1: Financial Statements And Business Decisions
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![### Inventory Management Problem: Warren W. Fisher Computer Corporation
The Warren W. Fisher Computer Corporation purchases 8,000 transistors each year as components in minicomputers. The unit cost of each transistor is $10, and the cost of carrying one transistor in inventory for a year is $3. Ordering cost is $30 per order. Assume that Fisher operates on a 200-day working year.
**a. What is the optimal order quantity?**
**b. What is the expected number of orders placed each year?**
**c. What is the expected time between orders?**
**d. What is the total cost?**
**Solution:**
To solve the above questions, we will employ the Economic Order Quantity (EOQ) model, which is a fundamental principle in inventory management. The EOQ formula is given by:
\[ EOQ = \sqrt{\frac{2DS}{H}} \]
Where:
- \( D \) is the annual demand (8,000 transistors).
- \( S \) is the ordering cost per order ($30).
- \( H \) is the holding cost per unit per year ($3).
By plugging the values into the EOQ formula, we can find the optimal order quantity.
1. **Optimal Order Quantity (Q\*)**
\[ Q\* = \sqrt{\frac{2 \times 8000 \times 30}{3}} = \sqrt{160000} = 400 \text{ transistors} \]
2. **Expected Number of Orders**
\[ \text{Number of Orders} = \frac{D}{Q\*} = \frac{8000}{400} = 20 \text{ orders/year} \]
3. **Expected Time Between Orders**
Since the company operates 200 days a year:
\[ \text{Time Between Orders} = \frac{200}{\text{Number of Orders}} = \frac{200}{20} = 10 \text{ days} \]
4. **Total Cost (TC)**
The total cost comprises the ordering cost, holding cost, and purchasing cost. The total cost formula is:
\[ TC = \left(\frac{D}{Q\*} \times S\right) + \left(\frac{Q\*}{2} \times H\right) + (D \times \text{Unit Cost}) \]
Let's calculate each part:
- Ordering Cost = \(\frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe23436e2-3710-4584-9afd-a6782603d0bb%2F5c507b49-119b-4b1e-9bf5-d8b689ebdb79%2Fdm1zol3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Inventory Management Problem: Warren W. Fisher Computer Corporation
The Warren W. Fisher Computer Corporation purchases 8,000 transistors each year as components in minicomputers. The unit cost of each transistor is $10, and the cost of carrying one transistor in inventory for a year is $3. Ordering cost is $30 per order. Assume that Fisher operates on a 200-day working year.
**a. What is the optimal order quantity?**
**b. What is the expected number of orders placed each year?**
**c. What is the expected time between orders?**
**d. What is the total cost?**
**Solution:**
To solve the above questions, we will employ the Economic Order Quantity (EOQ) model, which is a fundamental principle in inventory management. The EOQ formula is given by:
\[ EOQ = \sqrt{\frac{2DS}{H}} \]
Where:
- \( D \) is the annual demand (8,000 transistors).
- \( S \) is the ordering cost per order ($30).
- \( H \) is the holding cost per unit per year ($3).
By plugging the values into the EOQ formula, we can find the optimal order quantity.
1. **Optimal Order Quantity (Q\*)**
\[ Q\* = \sqrt{\frac{2 \times 8000 \times 30}{3}} = \sqrt{160000} = 400 \text{ transistors} \]
2. **Expected Number of Orders**
\[ \text{Number of Orders} = \frac{D}{Q\*} = \frac{8000}{400} = 20 \text{ orders/year} \]
3. **Expected Time Between Orders**
Since the company operates 200 days a year:
\[ \text{Time Between Orders} = \frac{200}{\text{Number of Orders}} = \frac{200}{20} = 10 \text{ days} \]
4. **Total Cost (TC)**
The total cost comprises the ordering cost, holding cost, and purchasing cost. The total cost formula is:
\[ TC = \left(\frac{D}{Q\*} \times S\right) + \left(\frac{Q\*}{2} \times H\right) + (D \times \text{Unit Cost}) \]
Let's calculate each part:
- Ordering Cost = \(\frac
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