Chapter 7, Problem 36P

(a):

To determine

Identify the number of i* values.

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Year	Cash flow (NC)
0	-5,000
1	-10,100
2	4,500
3	6,500
4	8,500
5	10,500
6	12,500

Cumulative cash (CC) flow can be calculated using the following formula:

${\text{CC}}_{\text{i}}={\text{CC}}_{\text{i}-\text{1}}+{\text{NC}}_{\text{i}}$ (1)

In Equation (1), the term i-1 indicates the previous time period.

Cumulative cash flow of in Year 1 can be calculated by substituting the respective values in Equation (1).

$\begin{array}{c}{\text{CC}}_{\text{1}}=-5,000+\left(-10,100\right)\\ =-15,100\end{array}$

Cumulative cash flow in Year 1 is -$15,100.

Table-2 shows the net cash flow and cumulative cash flow that is obtained using Equation (1).

Table -2

Year	Cash flow (NC)	CC
0	-5,000	-5,000
1	-10,100	-15,100
2	4,500	-10,600
3	6,500	-4,100
4	8,500	4,400
5	10,500	14,900
6	12,500	27,400

In Descartes’ rule, net cash flow is used, whereas in Norstrom’s criterion, the cumulative cash flow is used.

Norstrom’s criterion states that there is one root (i*) that is available when the cash flow starts with a negative sign and only one direction change.

Descartes’ rule states that the number of i* is equal to the number of direction change in the cash flow.

Table-2 shows that the NC cash flow has one change in direction. Thus, it would have one non-negative i* value.

Table-2 shows that the CC cash flow starts with a negative sign, and the cash flow has one change in direction. Thus, it would have one non-negative i* value.

(b):

To determine

Calculation of rate of return.

Explanation of Solution

Table-1 reveals that the incremental cash flow (I) is $2,000. Time period (n) is 5 years.

Rate of return (i) can be calculated as follows:

$\begin{array}{c}\text{NC0}=\frac{\text{NC1}}{{\left(1+\text{i}\right)}^{\text{1}}}+\left(\text{NC2}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{I}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\right)\frac{\text{1}}{{\left(1+\text{i}\right)}^{\text{1}}}\\ \text{5,000}=\frac{-10,100}{{\left(1+\text{i}\right)}^{\text{1}}}+\left(\text{4,500}\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\text{I}\frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}-\frac{\text{5}}{{\left(1+\text{i}\right)}^{\text{5}}}\right)\right)\frac{\text{1}}{{\left(1+\text{i}\right)}^{\text{1}}}\end{array}$

Substitute the rate of return as 2% in the above calculation by trial-and-error method.

$\begin{array}{c}\text{5,000}=\frac{-10,100}{{\left(1+\text{0}\text{.33}\right)}^{\text{1}}}+\left(\begin{array}{l}\text{4,500}\left(\frac{{\left(1+\text{0}\text{.33}\right)}^{\text{5}}-1}{\text{0}\text{.33}{\left(1+\text{0}\text{.33}\right)}^{\text{5}}}\right)+\text{2,000}\left(\frac{1}{0.33}\right)\\ \times \left(\frac{{\left(1+\text{0}\text{.33}\right)}^{\text{5}}-1}{\text{0}\text{.33}{\left(1+\text{0}\text{.33}\right)}^{\text{5}}}-\frac{\text{5}}{{\left(1+\text{0}\text{.33}\right)}^{\text{5}}}\right)\end{array}\right)\frac{\text{1}}{{\left(1+\text{0}\text{.33}\right)}^{\text{1}}}\\ \text{5,000}=\frac{-10,100}{1.33}+\left(\begin{array}{l}\text{4,500}\left(\frac{4.1616-1}{\text{0}\text{.33}\left(4.1616\right)}\right)+\text{2,000}\left(3.0303\right)\\ \times \left(\frac{4.1616-1}{\text{0}\text{.33}\left(4.1616\right)}-\frac{\text{5}}{4.1616}\right)\end{array}\right)\frac{\text{1}}{1.33}\\ \text{5,000}=-7,593.98+\left(\text{4,500}\left(\frac{3.1616}{1.3733}\right)+\text{6,060}\text{.6}\left(\frac{3.1616}{1.3733}-1.2015\right)\right)\left(0.7519\right)\\ \text{5,000}=-7,593.98+\left(\text{4,500}\left(2.3022\right)+\text{6,060}\text{.6}\left(2.3022-1.2015\right)\right)\left(0.7519\right)\\ \text{5,000}=-7,593.98+\left(10,359.9+\text{6,060}\text{.6}\left(1.1007\right)\right)\left(0.7519\right)\\ \text{5,000}=-7,593.98+\left(10,359.9+6,670.9\right)\left(0.7519\right)\\ \text{5,000}=-7,593.98+12,805.46\\ \text{5,000}<52,11.48\end{array}$

The calculated value is greater than the initial year net cash flow. Thus, increase the rate of return to 33.7%

$\begin{array}{c}\text{5,000}=\frac{-10,100}{{\left(1+\text{0}\text{.337}\right)}^{\text{1}}}+\left(\begin{array}{l}\text{4,500}\left(\frac{{\left(1+\text{0}\text{.337}\right)}^{\text{5}}-1}{\text{0}\text{.337}{\left(1+\text{0}\text{.337}\right)}^{\text{5}}}\right)+\text{2,000}\left(\frac{1}{0.337}\right)\\ \times \left(\frac{{\left(1+\text{0}\text{.337}\right)}^{\text{5}}-1}{\text{0}\text{.337}{\left(1+\text{0}\text{.337}\right)}^{\text{5}}}-\frac{\text{5}}{{\left(1+\text{0}\text{.337}\right)}^{\text{5}}}\right)\end{array}\right)\frac{\text{1}}{{\left(1+\text{0}\text{.337}\right)}^{\text{1}}}\\ \text{5,000}=\frac{-10,100}{1.337}+\left(\begin{array}{l}\text{4,500}\left(\frac{4.2723-1}{\text{0}\text{.337}\left(4.2723\right)}\right)+\text{2,000}\left(2.9676\right)\\ \times \left(\frac{4.2723-1}{\text{0}\text{.337}\left(4.2723\right)}-\frac{\text{5}}{4.2723}\right)\end{array}\right)\frac{\text{1}}{1.337}\\ \text{5,000}=-7,554.23+\left(\begin{array}{l}\text{4,500}\left(\frac{3.2723}{1.4398}\right)+\text{2,000}\left(2.9676\right)\\ \times \left(\frac{3.2723}{1.4398}-1.1703\right)\end{array}\right)\left(0.7479\right)\\ \text{5,000}=-7,554.23+\left(\text{4,500}\left(2.2727\right)+5,935.2\times \left(1.1024\right)\right)\left(0.7479\right)\\ \text{5,000}=-7,554.23+\left(10,227.15+6,542.96\right)\left(0.7479\right)\\ \text{5,000}=-7,554.23+\left(10,227.15+6,542.96\right)\left(0.7479\right)\\ \text{5,000}=-7,554.23+12452.37\\ 5,000\approx 4,988.14\end{array}$

Since the calculated value is nearly equal to the initial cash flow, it is confirmed that the rate of return is 33.7%.

(c):

To determine

Calculation of incremental value.

Explanation of Solution

The initial cash flow (NC0) is $5,000. The first year cash flow (NC1) is $10,100. The second year cash flow is $4,500. Rate of return (i) is 15%. Time period (n) is 5 year.

Incremental cash flow from Year 3 (I) can be calculated as follows:

$\begin{array}{c}\text{NC}=\frac{\text{NC1}}{{\left(1+\text{i}\right)}^{\text{1}}}+\left(\text{NC2}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{I}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\right)\frac{\text{1}}{{\left(1+\text{i}\right)}^{\text{1}}}\\ \text{5,000}=\frac{-10,100}{{\left(1+\text{0}\text{.15}\right)}^{\text{1}}}+\left(\begin{array}{l}\text{4,500}\left(\frac{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}-1}{\text{0}\text{.15}{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\right)\\ +\text{I}\left(\frac{1}{\text{0}\text{.15}}\right)\left(\frac{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}-1}{\text{0}\text{.15}{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}-\frac{\text{5}}{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\right)\end{array}\right)\frac{\text{1}}{{\left(1+\text{0}\text{.15}\right)}^{\text{1}}}\\ \text{5,000}=\frac{-10,100}{1.15}+\left(\begin{array}{l}\text{4,500}\left(\frac{2.0114-1}{\text{0}\text{.15}\left(2.0114\right)}\right)\\ +\text{I}\left(6.6667\right)\left(\frac{2.0114-1}{\text{0}\text{.15}\left(2.0114\right)}-\frac{\text{5}}{2.0114}\right)\end{array}\right)\left(\frac{\text{1}}{1.15}\right)\\ \text{5,000}=-8,782.61+\left(\begin{array}{l}\text{4,500}\left(\frac{1.0114}{0.3017}\right)\\ +\text{I}\left(6.6667\right)\left(\frac{1.0114}{0.3017}-2.4858\right)\end{array}\right)\left(0.8696\right)\\ \text{5,000}=-8,782.61+\left(\begin{array}{l}\text{4,500}\left(3.3523\right)\\ +\text{I}\left(6.6667\right)\left(3.3523-2.4858\right)\end{array}\right)\left(0.8696\right)\\ \text{5,000}=-8,782.61+\left(15,085.35+\text{I}\left(5.7767\right)\right)\left(0.8696\right)\\ \text{5,000}=-8,782.61+13,118.22+\text{I}\left(5.0234\right)\\ \text{I}\left(5.0234\right)=5,000+8,782.61-13,118.22\\ \text{I}\left(5.0234\right)=5,000+8,782.61-13,118.22\\ \text{I}=\frac{664.39}{5.0234}\\ =132.26\end{array}$

Thus, the incremental value is $132.26.