How many annual payments of ___(iii)___can be made from this account?

Question

 

A deposit of ___(i)___is made into an account paying an interest rate of 5% compounded annually. How many annual payments of ___(iii)___can be made from this account?

a)  (i)$100,000          (ii)$10,000

b)  (i)$200,000          (ii)$20,000

c)  (i) 10x          (ii) x

d)  (i) 10x          (ii) x                    at r% compounded continuously

Answer 1

To what does (iii) refer?

If P₀ is the initial amount paid then after 1 year the amount plus interest is P₀(1+r/100).

This is the starting amount for the second year P₁=P₀(1+r/100).

So, if P₀=$100,000 and r=5%, then P₁=100000+5000=$105000.

P₂ the amount after 2 years is:

P₂=P₁(1+r/100)=105000(1+1/20)=105000+5250=$110250.

In general then: P_n=P_n-1(1+r/100)=P_n-2(1+r/100)(1+r/100), where n is the number of years, and P_n the amount after this time.

P_n=P_n-2(1+r/100)²=P_n-3(1+r/100)³=...=P₀(1+r/100)ⁿ. (1+r/100)ⁿ can be called the growth factor.

Using this formula P₃=100000×1.05³=$115762.50. 

If n=1 (one year), P₁=$105,000, which is 10.5 payments of $10000.

The growth factor G can also be written G=P_n/P₀=P₀(1+r/100)ⁿ/P₀=(1+r/100)ⁿ. If G=10P₀ then we need to find n.

So (1+r/100)ⁿ=10; when r=5%, 1.05ⁿ=10, nlog₁₀(1.05)=log₁₀(10)=1; n=1/log₁₀(1.05)=47.19 years approx.

So after about 47 years at annual compound interest of 5% fixed annual rate, the amount would have grown to 10 times its original value.

However, if payments of $x are made each year to reduce the outstanding amount then the starting value at the beginning of each successive year will be different.

P₁=P₀(1+r/100)-x; P₂=P₁(1+r/100)-x=(P₀(1+r/100)-x)(1+r/100)-x,

P₂=P₀(1+r/100)²-x(1+r/100)-x. Generalising:

P_n=P₀(1+r/100)ⁿ-x(1+(1+r/100)+(1+r/100)²+...+(1+r/100)^n-1). When P_n=0 the annual payments would have cancelled the amount which has accumulated. We are given a fixed rate 5% of compound interest, but we have different possible values for the annual payment.

Let's first look at the geometric progression 1+(1+r/100)+(1+r/100)²+...+(1+r/100)^n-1.

If we let p=1+r/100 then the progression is 1+p+p²+...+p^n-1 which sums to:

(pⁿ-1)/(p-1).

r is fixed at 5% so p=1.05 and this sum S=(1.05ⁿ-1)/(5/100)=20(1.05ⁿ-1).

So we have P₀pⁿ=xS, or P₀×1.05ⁿ=20x(1.05ⁿ-1). This relates the initial amount P₀ and the annual payment x.

It can also be written: P₀=20x(1-1.05^-n) or P₀/x=20(1-1.05^-n).

So P₀=$100000 or $200000 and x=$10000 or $20000.

This gives us some values for (P₀,x,P₀/x):

(100000,10000,10), (200000,10000,20), (100000,20000,5), (200000,20000,10).

Therefore we only have to consider three values for P₀/x: 5, 10 and 20. Let y=P₀/x.

So we have 20(1-1.05^-n)=5, 10 or 20, for which we can find three different values for n. However, when y=20, 1.05^-n=0 which has no solution (n→∞). That leaves y=5 or 10.

1-1.05^-n=y/20, 1-y/20=1.05^-n, log₁₀(1-y/20)=-nlog₁₀(1.05), n=-log₁₀(1-y/20)/log₁₀(1.05).

When y=5, n=-log(0.75)/log(1.05)=5.89 years. When y=10, n=-log(0.5)/log(1.05)=14.21 years.

When the amount is $100,000 and the annual payment $20,000, the period is 5.89 years.

When the amount is $100,000 and the annual payment $10,000, the period is 14.21 years.

When the amount is $200,000 and the annual payment $20,000, the period is 14.21 years.