Chapter 10, Problem 26P

To determine

(a)

The expected present worth for the expansion cost.

Answer to Problem 26P

The expected present worth for the expansion cost is $\$3111963$.

Explanation of Solution

Given:

If optimistic growth is used in a city, then expansion will be needed in $4$ years and it’s probability is $20\%$.

If most likely growth is used, then expansion will be needed in 8 years.

If pessimistic growth is used, then expansion will be needed in 15 years and its probability is 40%.

Interest rate is $8\%$.

Expansion cost is $\$5400000$.

Concept used:

The total probability is $1$. Therefore, the probability of most likely value will be calculated by deducting optimistic and pessimistic value from $1$.

Write the expression for the present worth factor.

$\left(P/F,i,N\right)=\frac{1}{{\left(1+i\right)}^N}$

Here, interest rate is $i$ and the number of years is $N$.

Calculation:

Calculate the present worth for optimistic scenario ${\left(PW\right)}_O$.

${\left(PW\right)}_O={\left(E\right)}_C\times \left(P/F,i,N\right)\times {\left(P\right)}_O$ .... (I)

Here, expansion cost is ${\left(E\right)}_C$ and probability for optimistic is ${\left(P\right)}_O$.

Substitute, $\left(\$5400000\right)$ for ${\left(E\right)}_C$, $\left(\frac{1}{{\left(1+.06\right)}^4}\right)$ for $\left(P/F,i,N\right)$ and $0.20$ for ${\left(P\right)}_O$.

$\begin{array}{c}{\left(PW\right)}_O=\left(\$5400000\right)\times \left(\frac{1}{{\left(1+.06\right)}^4}\right)\times 0.20\\ =\$855461\end{array}$

Calculate the present worth for most likely ${\left(PW\right)}_M$.

${\left(PW\right)}_M={\left(E\right)}_C\times \left(P/F,i,N\right)\times {\left(P\right)}_M$

Here, probability for most likely is ${\left(P\right)}_M$.

Substitute, $\left(\$5400000\right)$ for ${\left(E\right)}_C$, $\left(\frac{1}{{\left(1+.06\right)}^4}\right)$ for $\left(P/F,i,N\right)$ and $\left(1-0.20-0.4\right)$ for ${\left(P\right)}_M$.

$\begin{array}{c}{\left(PM\right)}_M=\left(\$5400000\right)\times \left(\frac{1}{{\left(1+.06\right)}^8}\right)\times \left(1-0.20-0.4\right)\\ =\$1355210\end{array}$

Calculate the present worth for pessimistic ${\left(PW\right)}_P$.

${\left(PW\right)}_P={\left(E\right)}_C\times \left(P/F,i,N\right)\times {\left(P\right)}_P$

Here, probability for pessimistic is ${\left(P\right)}_P$.

Substitute, $\left(\$5400000\right)$ for ${\left(E\right)}_C$, $\left(\frac{1}{{\left(1+.06\right)}^4}\right)$ for $\left(P/F,i,N\right)$ and $0.40$ for ${\left(P\right)}_O$.

$\begin{array}{c}{\left(PW\right)}_P=\left(\$5400000\right)\times \left(\frac{1}{{\left(1+.06\right)}^{15}}\right)\times 0.40\\ =\$901292\end{array}$

The present worth for the expansion cost is sum of present worth of optimistic, most likely and pessimistic, multiplied with their probabilities.

Calculate the present worth for the expansion cost ${\left(PW\right)}_{\exp }$.

${\left(PW\right)}_{\exp }={\left(PW\right)}_O+{\left(PM\right)}_M+{\left(PM\right)}_P$

Substitute $\$855461$ for ${\left(PW\right)}_O$, $\$1355210$ for ${\left(PM\right)}_M$ and $\$901292$ for ${\left(PM\right)}_P$.

$\begin{array}{c}{\left(PW\right)}_{\exp }=\$855461+\$1355210+\$901292\\ =\$3111963\end{array}$.

Conclusion:

Thus, the present worth for the expansion cost is $\$3111963$.

To determine

(b)

The number of years until expansion.

Answer to Problem 26P

The expected number of years until expansion is $10$.

Explanation of Solution

Concept used:

Expected number of years is the sum of product of number of years and their probabilities in optimistic, most likely and pessimistic growth.

Calculation:

Expected number of years until expansion is given by $E_{Yr}$.

Calculate the expected number of years until expansion $\left(E_{Yr}\right)$.

$E_{Yr}=\left(N_o\times P_o\right)+\left(N_m\times P_m\right)+\left(N_p\times P_p\right)$

Here number of years is $N$ and their probability is $P$ and optimistic, most likely and pessimistic is represented by $O,M\&P$ respectively.

Substitute, $4$ for $N_o$, $8$ for $N_m$, $15$ for $N_p$, $0.2$ for $P_o$, $0.4$ for $P_m$ and $0.4$ for $P_p$.

$\begin{array}{c}{E}_{Yr}=\left(4\times 0.2\right)+\left(8\times 0.4\right)+\left(15\times 0.4\right)\\ =10\end{array}$.

Conclusion:

Thus, the expected number of years until expansion is $10$.

To determine

(c)

The present worth of the expansion cost using expected number of years.

Answer to Problem 26P

The present worth of the expansion cost using expected number of years is $\$3015331$.

Explanation of Solution

Concept used:

The present worth calculation of expansion cost using expected number of years is a product of expansion cost and present worth factor using expected number of years.

Calculation:

The present worth of the expansion cost using expected number of years is given by ${\left(PW\right)}_{Eyr}$.

Calculate present worth of the expansion cost using expected number of years.

${\left(PW\right)}_{Eyr}=$ $\left(E_C\right)$ $\times$ $\left(P/F,6\%,10\text{yr}\right)$

Here, ${\left(E\right)}_C$ is expansion cost.

Substitute, $\left(\$5400000\right)$ for $\left(E_C\right)$ and $\left(\frac{1}{{\left(1+0.06\right)}^{10}}\right)$ for $\left(P/F,6\%,10\text{yr}\right)$

$\begin{array}{c}{\left(PW\right)}_{Eyr}=\left(\$5400000\right)\times \left(\frac{1}{{\left(1+0.06\right)}^{10}}\right)\\ =\$3015331\end{array}$.

Conclusion:

The present worth of the expansion cost using expected number of years is $\$3015331$.

To determine

(d)

If the answers in part ($a)$ and part ($c)$ match.

Answer to Problem 26P

Answer in part ($a)$ and part ($c)$ do not match.

Explanation of Solution

The answer in part $\left(a\right)$ is $\$3111963$ and in part $\left(c\right)$ is $\$3015331$.

The answer in part $\left(a\right)$ and part $\left(c\right)$ do not match.

This is because in part $\left(a\right)$, present worth calculation is done by the addition of present worth of optimistic, most likely and pessimistic multiplied by their respective probabilities and in case of part $\left(b\right)$, the present worth is calculated using expected number of years.

Conclusion:

Answer in part ($a)$ and part ($c)$ do not match.