Chapter 5, Problem 72QP

a)

Summary Introduction

To determine: The amount that to accumulate by the time of retirement.

Explanation of Solution

Computation of present value is as follows:

$\begin{array}{c}\text{PV}=\text{C}\times \left[\frac{1}{i}-\frac{1}{i{\left(1+i\right)}^{\text{n}}}\right]\\ =\$30,000\times \left[\frac{1}{0.10}-\frac{1}{0.10{\left(1.10\right)}^{15}}\right]\\ =\$30,000\times 7.6061\\ =\$228,183\end{array}$

Therefore, the present value is $228,183.

b)

Summary Introduction

To determine: The savings amount required per year to finance the retirement consumption.

Explanation of Solution

Computation of savings amount required per year to finance the retirement consumption is as follows:

$\begin{array}{c}\text{PV of the retirement goal}=\frac{\text{PV}}{{\left(1+i\right)}^{\text{n}}}\\ =\frac{\$228,183}{{\left(1.10\right)}^{30}}\\ =\$13,076.8\end{array}$

The present value of the savings of 30 year stream will be same as this present value, therefore:

$\begin{array}{c}C\times \left[\frac{1}{0.10}-\frac{1}{0.10\times {(1.10)}^{30}}\right]=\$13,076.80\\ \text{Therefore, the value of C is}\Rightarrow C=PMT=\$1,387.18\end{array}$

Therefore, the required annual savings is $1,387.18.

c)

Summary Introduction

To determine: The cost of loaf of bread at the time of retirement.

Explanation of Solution

Computation of cost of loaf of bread at the time of retirement is as follows:

$\begin{array}{c}\text{Future value }=\text{ pv}\times {\left(1+r\right)}^{\text{n}}\\ =\$1\times {\left(1.04\right)}^{30}\\ =\$3.24\end{array}$

Therefore, the cost of loaf of bread at the time of retirement is $3.24.

d)

Summary Introduction

To determine: The real amount of saving.

Explanation of Solution

Computation of real amount of saving is as follows:

$\begin{array}{c}\text{Real interest rate }=\frac{\left(1+\text{nominal interest rate}\right)}{\left(\text{1}+\text{inflation rate}\right)}-1\\ =\frac{1.10}{1.04}-1\\ =0.0577\text{or 5}\text{.77\%}\end{array}$

$\begin{array}{c}\text{Retirement goal in real terms}=\text{ C}\times \left[\frac{1}{i}-\frac{1}{i\left(1+i\right)}\right]\\ =\$30,000\times \left[\frac{1}{0.0577}-\frac{1}{0.0577\times {(1.0577)}^{15}}\right]\\ =\$295,796.61\end{array}$

Therefore, the real amount of saving is $295,796.61.

e)

Summary Introduction

To determine: The predetermined real annual savings.

Explanation of Solution

Computation of predetermined real annual savings is as follows:

The future value of your 30-years savings stream must equal to $295,796.61. Then we solve for (PMT) in the following equation:

$\begin{array}{l}C\times \left[\frac{{1.0577}^{30}-1}{0.0577}\right]=\$295,796.61\\ \text{therefore the value of C}\Rightarrow C=PMT=\$3,895.66\end{array}$

The savings amount is comparatively higher than in the alternate situation because the rate at which purchasing power grows is less than the nominal interest rate, 10%.

f)

Summary Introduction

To determine: The nominal value of saving.

Explanation of Solution

Computation of nominal value of saving is as follows:

If the real amount saved is $3,895.66 and prices rise at 4% per year, then the amount saved at the end of 1 year, in nominal terms it will be $4,051.49$\left(\$3,895.66\times 1.04\right)$.

f)

Summary Introduction

To determine: The nominal value of saving in the 30^th year.

Explanation of Solution

Computation of nominal value of saving in the 30^th year is as follows:

The 30th year will require nominal savings of $12,635.17$\left(\$3,895.66\times {\left(1.04\right)}^{30}\right)$.