Jennifer, a senior project manager, estimated net cash flow after taxes (CFAT) for a large project she is working on. The additional CFAT of $2,800,000 in year 10 is the salvage value, estimated at 9% of the $31 million first cost for all capital assets.

Question

The PW value at the current MARR of 7% per year is



PW = ?31,000 + 5400(P?A,7%,6) + 2040(P?A,7%,4)(P?F,7%,6) + 2800(P?F,7%,10) = $767



Jennifer believes the MARR will vary over a relatively narrow range, as will the CFAT, especially during the out years of 7 through 10. She is willing to accept the other estimates as certain. Use the following probabil­ity distribution assumptions for MARR and CFAT to perform a simulation—hand- or spreadsheet-based.

MARR. Uniform distribution over the range 6% to 10%.

CFAT, years 7 through 10. Uniform distribu­tion over the range $1,600,000 to $2,400,000 for each year.

Answer 1

Plot the resulting PW distribution. Should the plan be accepted using decision making under cer­tainty? Under risk?

Using a spreadsheet, the steps in Sec. 19.5 are applied.



1. CFAT given for years 0 through 6.

2. i varies between 6% and 10%.

CFAT for years 7-10 varies between $1,600,000 and $2,400,000, written in $1000.

3. Uniform for both i and CFAT values.

4. Set up a spreadsheet. The example below has the following relations:



Col A: = RAND ( )* 100 to generate random numbers from 0-100.

Col B, cell B13: = INT((.04*A13+6) *100)/10000 converts the RN to i between

0.06 and 0.10. The % designation changes it to an interest rate between

6% and 10%.

Col C: = RAND( )* 100 to generate random numbers from 0-100.

Col D, cell D13: = INT (8*C13+1600) converts RN to a CFAT between $1600

and $2400.



Ten samples of i and CFAT for years 7-10 are below in columns B and D,

respectively (highlighted).



5. Columns F, G and H give 3 CFAT sequences, for example only, using rows 4, 5 and 6 RN generations. The entry for cells F11 through F13 is = D4 and cell F14 is

= D4+2800, where S = $2800. The PW values are obtained using the NPV function.

6. Plot the PW values for as large a sample as desired. Or, following the logic of

Figure 19-14, a spreadsheet relation can count the + and – PW values, with average and standard deviation calculated for the sample.

7. Conclusion:

For certainty, accept the plan since PW = $767 exceeds zero at 7% per year.

For risk, the result depends on the preponderance of positive PW values from the

simulation, and the distribution of PW obtained in step 6.