VIPSolutionsGive step-by-step solution with explanation and final answer: In a replacement analysis for a small Silicon Valley start-up data server, the following data are known about the challenger: the initial investment is $11,000; there is no annual maintenance cost for the first three years, however, it will be $3,000 in each of years four and five, and then $5,000 in the sixth year and increasing by $1,000 each year until year 10. The salvage value is always $0, and MARR is 9% per year. What is the minimum EUAC of this challenger?
Question:
Give step-by-step solution with explanation and final answer: In a replacement analysis for a small Silicon Valley start-up data server, the following data are known about the challenger: the initial investment is $11,000; there is no annual maintenance cost for the first three years, however, it will be $3,000 in each of years four and five, and then $5,000 in the sixth year and increasing by $1,000 each year until year 10. The salvage value is always $0, and MARR is 9% per year. What is the minimum EUAC of this challenger?
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Asked by: mayank
Created at: 2025-09-15 19:23:25
Asked by: mayank
Created at: 2025-09-15 19:23:25
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# Step-by-Step Solution: Minimum EUAC of the Challenger
Let's solve the problem of finding the **minimum Equivalent Uniform Annual Cost (EUAC)** for the data server (challenger) given the cash flow and MARR (Minimum Attractive Rate of Return) of 9%.
---
## 1. **Understanding the Cash Flows**
- **Initial Investment (Year 0):** $11,000
- **Maintenance Costs:**
- Years 1–3: $0
- Years 4–5: $3,000 per year
- Year 6: $5,000
- Year 7: $6,000
- Year 8: $7,000
- Year 9: $8,000
- Year 10: $9,000
- **Salvage Value (Year 10):** $0
---
## 2. **Tabular Representation of Cash Flows**
| Year | Cash Flow |
|------|-----------|
| 0 | -11,000 |
| 1 | 0 |
| 2 | 0 |
| 3 | 0 |
| 4 | -3,000 |
| 5 | -3,000 |
| 6 | -5,000 |
| 7 | -6,000 |
| 8 | -7,000 |
| 9 | -8,000 |
| 10 | -9,000 |
---
## 3. **EUAC Calculation for Different Lives**
Since the maintenance cost keeps increasing, the optimal replacement period is not obviously at 10 years. We need to compute EUAC for various possible lives (from 4 up to 10 years) and select the minimum.
We'll use the formula for EUAC:
\[
EUAC = \frac{PW_{\text{costs}}}{P/A, i, n}
\]
Where:
- \( PW_{\text{costs}} \) = Present Worth of costs for a life of \( n \) years
- \( P/A, i, n \) = Present Worth Factor for interest rate \( i \) and \( n \) years
Let’s proceed for each possible life.
---
### **Life = 4 Years**
- **Cash Flows:** Year 0: -11,000; Year 4: -3,000
\[
PW = -11,000 + \frac{-3,000}{(1+0.09)^4}
\]
\[
(1+0.09)^4 = 1.4116
\]
\[
PW = -11,000 + \frac{-3,000}{1.4116} = -11,000 - 2,126.5 = -13,126.5
\]
\[
(P/A, 9\%, 4) = \frac{1 - (1+0.09)^{-4}}{0.09} = \frac{1 - 0.7107}{0.09} = 3.209
\]
\[
EUAC = \frac{-13,126.5}{3.2397} = -4,052
\]
---
### **Life = 5 Years**
- **Cash Flows:** Year 0: -11,000; Years 4–5: -3,000 each
\[
PW = -11,000 + \frac{-3,000}{(1+0.09)^4} + \frac{-3,000}{(1+0.09)^5}
\]
\[
(1+0.09)^5 = 1.5386
\]
\[
PW = -11,000 + \frac{-3,000}{1.4116} + \frac{-3,000}{1.5386}
= -11,000 - 2,126.5 - 1,950.0 = -15,076.5
\]
\[
(P/A, 9\%, 5) = \frac{1 - (1+0.09)^{-5}}{0.09} = \frac{1 - 0.6499}{0.09} = 3.889
\]
\[
EUAC = \frac{-15,076.5}{3.8897} = -3,876
\]
---
### **Life = 6 Years**
- Years 4–5: -3,000
- Year 6: -5,000
\[
PW = -11,000 + \frac{-3,000}{1.4116} + \frac{-3,000}{1.5386} + \frac{-5,000}{(1+0.09)^6}
\]
\[
(1+0.09)^6 = 1.6761
\]
\[
PW = -11,000 - 2,126.5 - 1,950.0 - 2,984.6 = -18,061.1
\]
\[
(P/A, 9\%, 6) = \frac{1 - (1+0.09)^{-6}}{0.09} = \frac{1 - 0.5962}{0.09} = 4.487
\]
\[
EUAC = \frac{-18,061.1}{4.4869} = -4,027
\]
---
### **Life = 7 Years**
Add Year 7: -6,000
\[
(1+0.09)^7 = 1.8279
\]
\[
PW = -11,000 - 2,126.5 - 1,950.0 - 2,984.6 - \frac{6,000}{1.8279}
\]
\[
6,000/1.8279 = 3,282.4
\]
\[
PW = -11,000 - 2,126.5 - 1,950.0 - 2,984.6 - 3,282.4 = -21,343.5
\]
\[
(P/A, 9\%, 7) = \frac{1 - (1+0.09)^{-7}}{0.09} = \frac{1 - 0.5471}{0.09} = 5.032
\]
\[
EUAC = \frac{-21,343.5}{5.0321} = -4,242
\]
---
### **Life = 8 Years**
Add Year 8: -7,000
\[
(1+0.09)^8 = 1.9924
\]
\[
7,000/1.9924 = 3,514.5
\]
\[
PW = -21,343.5 - 3,514.5 = -24,858.0
\]
\[
(P/A, 9\%, 8) = \frac{1 - (1+0.09)^{-8}}{0.09} = \frac{1 - 0.5022}{0.09} = 5.534
\]
\[
EUAC = \frac{-24,858.0}{5.5343} = -4,493
\]
---
### **Life = 9 Years**
Add Year 9: -8,000
\[
(1+0.09)^9 = 2.1707
\]
\[
8,000/2.1707 = 3,685.8
\]
\[
PW = -24,858.0 - 3,685.8 = -28,543.8
\]
\[
(P/A, 9\%, 9) = \frac{1 - (1+0.09)^{-9}}{0.09} = \frac{1 - 0.4607}{0.09} = 5.999
\]
\[
EUAC = \frac{-28,543.8}{5.9993} = -4,759
\]
---
### **Life = 10 Years**
Add Year 10: -9,000
\[
(1+0.09)^{10} = 2.3641
\]
\[
9,000 / 2.3641 = 3,806.4
\]
\[
PW = -28,543.8 - 3,806.4 = -32,350.2
\]
\[
(P/A, 9\%, 10) = \frac{1 - (1+0.09)^{-10}}{0.09} = \frac{1 - 0.4224}{0.09} = 6.437
\]
\[
EUAC = \frac{-32,350.2}{6.4370} = -5,028
\]
---
## 4. **Summary Table**
| Life (years) | Present Worth (\$) | EUAC (\$) |
|--------------|-------------------|-----------|
| 4 | -13,126.5 | -4,052 |
| 5 | -15,076.5 | -3,876 |
| 6 | -18,061.1 | -4,027 |
| 7 | -21,343.5 | -4,242 |
| 8 | -24,858.0 | -4,493 |
| 9 | -28,543.8 | -4,759 |
| 10 | -32,350.2 | -5,028 |
---
## 5. **Find the Minimum EUAC**
From the table, the **minimum (least negative) EUAC is at 5 years**:
\[
\boxed{\$3,876}
\]
(per year, rounded to the nearest dollar; negative sign indicates cost)
---
## **Final Answer**
### The minimum EUAC of this challenger is **\$3,876 per year**, occurring at a 5-year replacement cycle.
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