ENGINEERING ECONOMY DIGITAL ACCESS
ENGINEERING ECONOMY DIGITAL ACCESS
8th Edition
ISBN: 2810022611683
Author: Blank
Publisher: MCG
Question
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Chapter 7, Problem 36P

(a):

To determine

Identify the number of i* values.

(a):

Expert Solution
Check Mark

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Year

Cash flow

(NC)

0-5,000
1-10,100
24,500
36,500
48,500
510,500
612,500

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

CCi=CCi1+NCi (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).

CC1=5,000+(10,100)=15,100

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
24,500-10,600
36,500-4,100
48,5004,400
510,50014,900
612,50027,400

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.

(b):

Expert Solution
Check Mark

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:

NC0=NC1(1+i)1+(NC2((1+i)n1i(1+i)n)+I((1+i)n1i(1+i)nn(1+i)n))1(1+i)15,000=10,100(1+i)1+(4,500((1+i)51i(1+i)5)+I1i((1+i)51i(1+i)55(1+i)5))1(1+i)1

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

5,000=10,100(1+0.33)1+(4,500((1+0.33)510.33(1+0.33)5)+2,000(10.33)×((1+0.33)510.33(1+0.33)55(1+0.33)5))1(1+0.33)15,000=10,1001.33+(4,500(4.161610.33(4.1616))+2,000(3.0303)×(4.161610.33(4.1616)54.1616))11.335,000=7,593.98+(4,500(3.16161.3733)+6,060.6(3.16161.37331.2015))(0.7519)5,000=7,593.98+(4,500(2.3022)+6,060.6(2.30221.2015))(0.7519)5,000=7,593.98+(10,359.9+6,060.6(1.1007))(0.7519)5,000=7,593.98+(10,359.9+6,670.9)(0.7519)5,000=7,593.98+12,805.465,000<52,11.48

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

5,000=10,100(1+0.337)1+(4,500((1+0.337)510.337(1+0.337)5)+2,000(10.337)×((1+0.337)510.337(1+0.337)55(1+0.337)5))1(1+0.337)15,000=10,1001.337+(4,500(4.272310.337(4.2723))+2,000(2.9676)×(4.272310.337(4.2723)54.2723))11.3375,000=7,554.23+(4,500(3.27231.4398)+2,000(2.9676)×(3.27231.43981.1703))(0.7479)5,000=7,554.23+(4,500(2.2727)+5,935.2×(1.1024))(0.7479)5,000=7,554.23+(10,227.15+6,542.96)(0.7479)5,000=7,554.23+(10,227.15+6,542.96)(0.7479)5,000=7,554.23+12452.375,0004,988.14

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.

(c):

Expert Solution
Check Mark

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:

NC=NC1(1+i)1+(NC2((1+i)n1i(1+i)n)+I((1+i)n1i(1+i)nn(1+i)n))1(1+i)15,000=10,100(1+0.15)1+(4,500((1+0.15)510.15(1+0.15)5)+I(10.15)((1+0.15)510.15(1+0.15)55(1+0.15)5))1(1+0.15)15,000=10,1001.15+(4,500(2.011410.15(2.0114))+I(6.6667)(2.011410.15(2.0114)52.0114))(11.15)5,000=8,782.61+(4,500(1.01140.3017)+I(6.6667)(1.01140.30172.4858))(0.8696)5,000=8,782.61+(4,500(3.3523)+I(6.6667)(3.35232.4858))(0.8696)5,000=8,782.61+(15,085.35+I(5.7767))(0.8696)5,000=8,782.61+13,118.22+I(5.0234)I(5.0234)=5,000+8,782.6113,118.22I(5.0234)=5,000+8,782.6113,118.22I=664.395.0234=132.26

Thus, the incremental value is $132.26.

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