Chapter 10, Problem 38P

To determine

i.

Expected savings per year, expected life and rate of return for expected values.

Answer to Problem 38P

Expected savings per year is $1980, expected life is 6 years and rate of return for the expected values is 12.18%.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Calculation:

Expected Savings (EAS):

$\text{EAS}\text{=}\left(\begin{array}{l}\left(\begin{array}{l}{\text{Saving}}_{\text{Optimistic}}\\ {\text{×SavingProbability}}_{\text{Optimistic}}\end{array}\right)\text{+}\left(\begin{array}{l}{\text{Saving}}_{\text{Pessimistic}}\\ \text{×Savin}{\text{Probability}}_{\text{Pessimistic}}\end{array}\right)\\ \text{+}\left({\text{Saving}}_{\text{Most}\text{Likely}}\text{×}\text{Saving}{\text{Probability}}_{\text{Most}\text{likely}}\right)\end{array}\right)$

= $\left(\left(22000\times 0.1\right)+\left(18000\times 0.2\right)+\left(20000\times 0.7\right)\right)$

$=(2200+3600+14000)$

$=19800$

Expected Saving - $19800

Expected Useful Life:

$\begin{array}{l}\text{Joint}\text{probability}\\ \left(\begin{array}{l}\left(\text{Useful}{\text{life}}_{\text{Optimistic}}\text{×Useful}\text{life}{\text{probability}}_{\text{Optimistic}}\right)\\ \text{+}\left(\text{Useful}{\text{life}}_{\text{Pessimistic}}\text{×Useful}\text{life}{\text{Probability}}_{\text{Pessimistic}}\right)\\ \text{+}\left(\text{Useful}\text{life}{\text{probability}}_{\text{Most}\text{likely}}\text{×Useful}\text{life}{\text{probability}}_{\text{Most}\text{likely}}\right)\end{array}\right)\end{array}$

= $\left(\left(4\times \frac{1}{6}\right)+\left(12\times \frac{1}{6}\right)+\left(5\times \frac{2}{3}\right)\right)$

$=\left(\left(4\times 0.17\right)+\left(12\times 0.17\right)+\left(5\times 0.67\right)\right)$

$=\left(0.68+2.04+3.35\right)$

$=6.07$

Hence, expected useful life is 6 years

Rate of return (RR) for optimistic:

Present value can be calculated as follows:

$\text{Cost=}\text{Joint}\text{Probability}\text{of}\text{savings}\left(\frac{{\text{(1+Interest)}}^{\text{Joint}\text{probability}}}{{\text{Interest×(1+Interest)}}^{\text{Joint}\text{Probability}}}\right)$

$81000=19800\left(\frac{{\left(1+Interest\right)}^6-1}{Interest\times {(1+Interest)}^6}\right)$

$\frac{81000}{19800}=\left(\frac{{\left(1+Interest\right)}^6-1}{Interest\times {(1+Interest)}^6}\right)$

$4.091=\left(\frac{{\left(1+Interest\right)}^6-1}{Interest\times {(1+Interest)}^6}\right)$

Interest rate = 12%

$4.091=\left(\frac{{\left(1+0.12\right)}^6-1}{0.12\times {(1+0.12)}^6}\right)$

$4.091=\left(\frac{1.9929-1}{0.1218\times (1.9929)}\right)$

$4.091=4.091$

As calculated value is equal to actual Value, thus rate of return is 12.18%.

Conclusion:

Expected savings per year is $1980, expected life is 6 years and rate of return for the expected values is 12.18%.

To determine

ii.

Expected rate of return for combination of savings per year and life.

Answer to Problem 38P

Expected rate of return for combination of savings per year and life is 10.48%.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Calculation:

Expected rate of return for combination of savings per year and life:

$\mathrm{Cos}t=\left(\begin{array}{l}ExpectedSavings\times \mathrm{Pr}obabilit{y}_{4years}\left(\frac{{(1+Interest)}^4-1}{Interest{(1+Interest)}^4}\right)\\ +ExpectedSaving\times \mathrm{Pr}obabilit{y}_{5years}\left(\frac{{(1+Interest)}^5-1}{Interest{(1+Interest)}^5}\right)\\ +ExpectedSaving\times \mathrm{Pr}obabilit{y}_{12years}\left(\frac{{(1+Interest)}^{12}-1}{Interest{(1+Interest)}^{12}}\right)\end{array}\right)$

$81000=\left(\begin{array}{l}19800\times \frac{1}{6}\left(\frac{{\left(1+Interest\right)}^4-1}{Interest{\left(1+Interest\right)}^4}\right)\\ +19800\times \frac{2}{3}\left(\frac{{\left(1+Interest\right)}^5-1}{Interest{\left(1+Interest\right)}^5}\right)\\ +19800\times \frac{1}{6}\left(\frac{{\left(1+Interest\right)}^{12}-1}{Interest{\left(1+Interest\right)}^{12}}\right)\end{array}\right)$

Interest rate = 11%

$81000=\left(\begin{array}{l}19800\times 0.1667\left(\frac{1+0.11{)}^4-1}{0.11{(1+0.11)}^4}\right)\\ +19800\times 0.6667\left(\frac{1+0.11{)}^5-1}{0.11{(1+0.11)}^5}\right)\\ +19800\times 0.1667\left(\frac{1+0.11{)}^{12}-1}{0.11{(1+0.11)}^{12}}\right)\end{array}\right)$

$81000=\left(\begin{array}{l}3300.66\left(\frac{1.5181-1}{0.11(1.5181}\right)\\ +1320.66\left(\frac{1.6851-1}{0.11(1.6851)}\right)\\ +3300.66\left(\frac{(3.4985)-1}{0.11(3.4985)}\right)\end{array}\right)$

$81000=\left(\begin{array}{l}3300.66\left(\frac{0.5181}{0.167}\right)\\ +1320.66\left(\frac{0.6851}{0.1854}\right)\\ +3300.66\left(\frac{(2.4985)}{0.3848}\right)\end{array}\right)$

$81000=\left(\begin{array}{l}3300.66(3.1024)\\ +1320.66(3.6959)\\ +3300.66(6.4924)\end{array}\right)$

$\begin{array}{l}81000=10240.12+48788.28+21429.06\\ 81000\succ 80457.46\end{array}$

When interest rate is substituted as 11%, then calculated value is less than the actual value. Thus decrease interest rate further. If 10.84% is substituted, then actual value is equivalent to calculated value.

Conclusion:

Thus, 10.84% is rate of return for combination of savings per year.

To determine

iii.

Answers for part a and b match or not.

Answer to Problem 38P

Answers donot match.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Concept used:

Since time period varies in part a and part b, rate of return for part a and part b are different.

Conclusion:

Answers donot match.