- #1
mathmari
Gold Member
MHB
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Hey!
A worker wishing to retire for $20$ years would be paid an advance pension of $6000$ euro. The capital required for this is to be paid in the course of $15 $ years by successively paid annual contributions. At the beginning of the accumulation period, the deposit is once $10,000 $ Euro. How much are the annual contributions to be paid if the interest rate is $ 4 \% $ during the entire term?
I have done the following:
Let $X$ be the amount of the annual contributions.
After 15 years the present amount will be $$K_{15}=10000 \cdot 1.04^{15}+X\sum_{i=0}^{14}1.04^i=10000 \cdot 1.04^{15}+X\frac{1.04^{14+1}-1}{1.04-1} \\ =10000 \cdot 1.04^{15}+X\frac{1.04^{15}-1}{0.04}$$
right? (Wondering)
After that, after the 20 years the present value will be $$K_{15} \cdot 1.04^{20}-6000\sum_{i=0}^{19}1.04^i=K_{15} \cdot 1.04^{20}-6000\frac{1.04^{19+1}-1}{1.04-1} \\ =K_{15} \cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}$$
right?
After the 20 years the value must be $0$, or not? (Wondering)
So, we have the following:
$$K_{15} \cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}=0 \\ \Rightarrow \left ( 10000 \cdot 1.04^{15}+X\frac{1.04^{15}-1}{0.04}\right )\cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}=0 \\ \Rightarrow 10000 \cdot 1.04^{35}+X\frac{1.04^{15}-1}{0.04}\cdot 1.04^{20}=6000\frac{1.04^{20}-1}{0.04} \\ \Rightarrow X\frac{1.04^{15}-1}{0.04}\cdot 1.04^{20} = 6000\frac{1.04^{20}-1}{0.04}-10000 \cdot 1.04^{35} \\ \Rightarrow X = \frac{6000(1.04^{20}-1)}{ 1.04^{20}(1.04^{15}-1)}-\frac{400 \cdot 1.04^{35}}{ 1.04^{20}(1.04^{15}-1)} \\ \Rightarrow X = \frac{6000(1.04^{20}-1)}{ 1.04^{20}(1.04^{15}-1)}-\frac{400 \cdot 1.04^{15}}{1.04^{15}-1}$$
Is this correct so far?
According to Wolfram this is about 3172,88.
Can this be correct? (Wondering)
A worker wishing to retire for $20$ years would be paid an advance pension of $6000$ euro. The capital required for this is to be paid in the course of $15 $ years by successively paid annual contributions. At the beginning of the accumulation period, the deposit is once $10,000 $ Euro. How much are the annual contributions to be paid if the interest rate is $ 4 \% $ during the entire term?
I have done the following:
Let $X$ be the amount of the annual contributions.
After 15 years the present amount will be $$K_{15}=10000 \cdot 1.04^{15}+X\sum_{i=0}^{14}1.04^i=10000 \cdot 1.04^{15}+X\frac{1.04^{14+1}-1}{1.04-1} \\ =10000 \cdot 1.04^{15}+X\frac{1.04^{15}-1}{0.04}$$
right? (Wondering)
After that, after the 20 years the present value will be $$K_{15} \cdot 1.04^{20}-6000\sum_{i=0}^{19}1.04^i=K_{15} \cdot 1.04^{20}-6000\frac{1.04^{19+1}-1}{1.04-1} \\ =K_{15} \cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}$$
right?
After the 20 years the value must be $0$, or not? (Wondering)
So, we have the following:
$$K_{15} \cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}=0 \\ \Rightarrow \left ( 10000 \cdot 1.04^{15}+X\frac{1.04^{15}-1}{0.04}\right )\cdot 1.04^{20}-6000\frac{1.04^{20}-1}{0.04}=0 \\ \Rightarrow 10000 \cdot 1.04^{35}+X\frac{1.04^{15}-1}{0.04}\cdot 1.04^{20}=6000\frac{1.04^{20}-1}{0.04} \\ \Rightarrow X\frac{1.04^{15}-1}{0.04}\cdot 1.04^{20} = 6000\frac{1.04^{20}-1}{0.04}-10000 \cdot 1.04^{35} \\ \Rightarrow X = \frac{6000(1.04^{20}-1)}{ 1.04^{20}(1.04^{15}-1)}-\frac{400 \cdot 1.04^{35}}{ 1.04^{20}(1.04^{15}-1)} \\ \Rightarrow X = \frac{6000(1.04^{20}-1)}{ 1.04^{20}(1.04^{15}-1)}-\frac{400 \cdot 1.04^{15}}{1.04^{15}-1}$$
Is this correct so far?
According to Wolfram this is about 3172,88.
Can this be correct? (Wondering)