A 20-year loan of 20,000 may be repaid under one of two methods: a. Amortization method with equal annual payments at an effective rate of
6.5%
b. Sinking fund method in which the lender receives an effective rate of
8%
and the sinking fund earns an annual effective rate of
j
. Both methods require a payment of
X
at the end of the each year for 20 years. Calculate j.
Question:
A 20-year loan of 20,000 may be repaid under one of two methods: a. Amortization method with equal annual payments at an effective rate of
6.5%
b. Sinking fund method in which the lender receives an effective rate of
8%
and the sinking fund earns an annual effective rate of
j
. Both methods require a payment of
X
at the end of the each year for 20 years. Calculate j.
This is AI generated Response.
Asked by: Test User 100128
Created at: 2025-05-21 10:29:33
Final Answer:
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Let's break down the problem step by step.
**Given:**
- Principal: \( P = 20,000 \)
- Loan term: \( n = 20 \) years
- **Method a:** Amortization, **effective rate:** \( i_a = 6.5\% \)
- **Method b:** Sinking fund, lender receives **effective rate:** \( i_l = 8\% \), sinking fund earns **rate:** \( j \)
- Both methods require **equal annual payment:** \( X \)
We are to **find \( j \)** so that the annual payment \( X \) is the same under both methods.
---
### 1. **Amortization Method**
Annual payment for a fully amortized loan:
\[
X = P \cdot \frac{i_a}{1 - (1+i_a)^{-n}}
\]
Plug in values:
\[
X = 20,000 \cdot \frac{0.065}{1 - (1.065)^{-20}}
\]
First, calculate \( (1.065)^{-20} \):
\[
(1.065)^{20} \approx 3.5075 \implies (1.065)^{-20} \approx 0.2852
\]
So,
\[
X = 20,000 \cdot \frac{0.065}{1 - 0.2852}
= 20,000 \cdot \frac{0.065}{0.7148}
= 20,000 \cdot 0.09093
= \$1,818.60
\]
---
### 2. **Sinking Fund Method**
For each year:
- **Interest payment to lender:** \( P \cdot i_l = 20,000 \times 0.08 = \$1,600 \)
- **Sinking fund deposit:** \( S \) (to accumulate to $20,000 in 20 years at rate \( j \))
Sinking fund deposit:
\[
S = \frac{P}{s_{\overline{n}|j}}
\]
Where:
\[
s_{\overline{n}|j} = \frac{(1+j)^n - 1}{j}
\]
So,
\[
S = 20,000 \cdot \frac{j}{(1+j)^{20} - 1}
\]
**Total annual payment:** \( X = \text{Interest} + \text{Sinking Fund Deposit} \)
\[
X = 1,600 + 20,000 \cdot \frac{j}{(1+j)^{20} - 1}
\]
---
### 3. **Equate the Payments**
Set the two expressions for \( X \) equal:
\[
1,818.60 = 1,600 + 20,000 \cdot \frac{j}{(1+j)^{20} - 1}
\]
\[
1,818.60 - 1,600 = 20,000 \cdot \frac{j}{(1+j)^{20} - 1}
\]
\[
218.60 = 20,000 \cdot \frac{j}{(1+j)^{20} - 1}
\]
\[
\frac{218.60}{20,000} = \frac{j}{(1+j)^{20} - 1}
\]
\[
0.01093 = \frac{j}{(1+j)^{20} - 1}
\]
\[
0.01093 \times ((1+j)^{20} - 1) = j
\]
\[
0.01093 (1+j)^{20} - 0.01093 = j
\]
\[
0.01093 (1+j)^{20} - j = 0.01093
\]
\[
0.01093 (1+j)^{20} - j = 0.01093
\]
---
### 4. **Solve for \( j \):**
Let \( x = 1+j \):
\[
0.01093 x^{20} - (x-1) = 0.01093
\]
\[
0.01093 x^{20} - x + 1 = 0.01093
\]
\[
0.01093 x^{20} - x + 1 - 0.01093 = 0
\]
\[
0.01093 x^{20} - x + 0.98907 = 0
\]
Now solve for \( x \) (and thus \( j \)). This is a transcendental equation and must be solved numerically.
---
#### **Numerical solution:**
Start with an estimate. The amortization rate is 6.5%, but in the sinking fund method, the fund must earn less than this since the lender is being paid 8%. Try \( j \approx 4\% \) to \( 6\% \).
Let’s try \( j = 5\% \):
\[
x = 1.05
\]
\[
x^{20} = 1.05^{20} \approx 2.6533
\]
\[
0.01093 \times 2.6533 = 0.02901
\]
\[
0.02901 - 1.05 + 0.98907 = 0.02901 - 0.06093 = -0.03192
\]
That’s negative. Try \( j = 6\% \):
\[
x = 1.06, \quad x^{20} = 3.2071
\]
\[
0.01093 \times 3.2071 = 0.03504
\]
\[
0.03504 - 1.06 + 0.98907 = 0.03504 - 0.07093 = -0.03589
\]
Still negative, but slightly more negative.
Try \( j = 4\% \):
\[
x = 1.04, \quad x^{20} = 2.1911
\]
\[
0.01093 \times 2.1911 = 0.02394
\]
\[
0.02394 - 1.04 + 0.98907 = 0.02394 - 0.05093 = -0.02699
\]
Still negative, but less so. Let's try \( j = 7\% \):
\[
x = 1.07, \quad x^{20} = 3.8697
\]
\[
0.01093 \times 3.8697 = 0.04232
\]
\[
0.04232 - 1.07 + 0.98907 = 0.04232 - 0.08093 = -0.03861
\]
All values are negative. Let's check our setup.
Alternatively, let's use the version from earlier:
\[
0.01093 (1+j)^{20} - j = 0.01093
\]
\[
0.01093 (1+j)^{20} = 0.01093 + j
\]
\[
(1+j)^{20} = 1 + \frac{j}{0.01093}
\]
Try \( j = 0.05 \):
\[
1 + \frac{0.05}{0.01093} = 1 + 4.574 = 5.574
\]
\[
(1.05)^{20} = 2.6533
\]
2.65 is much less than 5.57, so \( j \) needs to be higher.
Try \( j = 0.15 \):
\[
1 + \frac{0.15}{0.01093} = 1 + 13.73 = 14.73
\]
\[
(1.15)^{20} = 16.37
\]
Now, 16.37 is more than 14.73, so the correct value is between 5% and 15%.
Try \( j = 0.12 \):
\[
1 + \frac{0.12}{0.01093} = 1 + 10.98 = 11.98
\]
\[
(1.12)^{20} = 9.646
\]
Still less. Try \( j = 0.135 \):
\[
1 + \frac{0.135}{0.01093} = 1 + 12.35 = 13.35
\]
\[
(1.135)^{20} = 13.36
\]
Now, \( (1.135)^{20} \approx 13.36 \) and the right hand side is 13.35. This is **very close**!
---
### **Final Answer**
\[
\boxed{j \approx 13.5\%}
\]
**So, the sinking fund must earn an annual effective rate of approximately 13.5% for the payments to be equal under both methods.**
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