VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:Compare the alternatives shown on the basis of their capitalized costs using a MARR of 10% per year. i — — 75000 500 ; nat Operating cost, Sper | 4g 000 alvage value, § 000,000 I The capitalized cost for alternative M is $~ [6230843968] @ , and the capitalized cost for alternative Nis $= |. The best alternative selected on the basis of their capitalized costs is alternative [ N_| @-
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give answer in 2 step with explanation at the end of each step and final answer at the end:
Compare the alternatives shown on the basis of their capitalized costs using a MARR of 10% per year.
i — —
75000 500
; nat Operating cost, Sper | 4g 000
alvage value, § 000,000
I
The capitalized cost for alternative M is $~ [6230843968] @ , and the capitalized cost for alternative Nis $= |.
The best alternative selected on the basis of their capitalized costs is alternative [ N_| @-
This is AI generated Response.
Asked by: Pagal
Created at: 2025-06-29 06:05:35
Asked by: Pagal
Created at: 2025-06-29 06:05:35
Final Answer:
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Let's solve the problem step by step:
---
### **Step 1: Calculate Capitalized Cost for Alternative M**
**Capitalized Cost Formula (for a finite life project):**
\[
CC = \text{First Cost} + \left(\frac{\text{Annual Operating Cost}}{i}\right) - \left(\frac{\text{Salvage Value}}{(1+i)^n - 1}\right)
\]
Where:
- \( i = 0.10 \) (MARR, or 10%)
- \( n = 5 \) years
Plug in the values for M:
- First Cost = -170,000
- Annual Operating Cost = -45,000
- Salvage Value = 8,000
\[
CC_M = -170,000 + \left(\frac{-45,000}{0.10}\right) - \left(\frac{8,000}{(1.10)^5 - 1}\right)
\]
\[
= -170,000 + (-450,000) - \left(\frac{8,000}{0.61051}\right)
\]
\[
= -170,000 - 450,000 - 13,106.6
\]
\[
= -633,106.6
\]
**Explanation:**
We computed the capitalized cost for M by applying the formula for a finite project, adjusting each component for the provided interest rate and life span.
---
### **Step 2: Calculate Capitalized Cost for Alternative N**
**Capitalized Cost Formula (for infinite life):**
\[
CC = \text{First Cost} + \left(\frac{\text{Annual Operating Cost}}{i}\right) - \left(\frac{\text{Salvage Value}}{i}\right)
\]
Where:
- \( i = 0.10 \)
Plug in the values for N:
- First Cost = -625,000
- Annual Operating Cost = -14,000
- Salvage Value = 1,000,000
\[
CC_N = -625,000 + \left(\frac{-14,000}{0.10}\right) - \left(\frac{1,000,000}{0.10}\right)
\]
\[
= -625,000 - 140,000 - 10,000,000
\]
\[
= -10,765,000
\]
**Explanation:**
For infinite life (alternative N), we use the perpetual cost formula, subtracting the perpetuity value of the salvage.
---
### **Final Answer**
- **Capitalized cost for M:** $-633,106.6
- **Capitalized cost for N:** $-10,765,000
**Best alternative:**
Since we are dealing with costs (negative values), the *least negative* value is better (lower cost).
Therefore, **Alternative M** has the lower capitalized cost and is the better alternative based on capitalized costs.
---
#### **Summary Table:**
| Alternative | Capitalized Cost |
|-------------|------------------|
| M | $-633,106.6 |
| N | $-10,765,000 |
**Final Answer:**
The best alternative based on capitalized costs is **alternative M**.
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