Chapter 3

Q. 3.PP.38

A company is considering an investment in a project that requires an initial net investment of Rs 3,000 with an expected cash flow (CFAT) generated over three years as follows:
(a) What is the expected NPV of this project? (Assume that the probability distributions are
\,     independent and the risk-free rate of interest in the market is 0.05).
(b) Calculate the standard deviation about the expected value.
(c) Find the probability that the NPV will be less than zero (Assume that the distribution is normal
\,     and continuous).
(d) What is the probability that the NPV will be greater than zero?
(e) What is the probability that NPV will be (i) between the range of Rs 500 and Rs 750, (ii) between
\,     the range of Rs 400 and Rs 600, (iii) at least Rs 300 and (iv) at least Rs 1,000.

Probability CFAT Probability CFAT Probability CFAT
0.2 Rs 800 0.1 Rs 800 0.1 Rs 800
0.5    1,000 0.3    1,000 0.2    1,000
0.2    1,500 0.4    1,500 0.4    1,500
0.1    2,000 0.2    2,000 0.3   2,000

Step-by-Step

Verified Solution

Standard\> deviation \> about \> the \> expected \> value:

\qquad\quad\>\sigma =\frac{417}{(1 +0.05)^2}+\frac{400}{(1 +0.05)^4}+\frac{316}{(1 +0.05)^6}
=\sqrt{3,86,639}=622

(c) The \>probability\> that \>the\> NPV\> will\> be\> zero\> or \>less:

Z=\frac{0-NPV}{\sigma}=\frac{0- 663}{622} =-.1.0659 = – 1.07

\, According to Table Z, the probability of the NPV being zero is 0.3577; therefore, the probability of the NPV being less than zero would be 0.5 – 0.3577 = 0.1423 = 14.23 per cent.
(d) The\> probability \>of \>the \>NPV\> being\> greater\> than \>zero \>would \>be:

1 – 0.1423 = 0.8577 = 85.77 per cent

(e) (i) The \>probability \>of \> the \> NPV \> being \> within \> the \> range \> of \>Rs \> 500
and \> Rs \> 750:

Z_1=\frac{\mathrm{Rs \>500- Rs\> 663}}{\mathrm{Rs\> 622}}=(0.262)\qquad\qquad \,Z_2=\frac{\mathrm{Rs \>750- Rs\> 663}}{\mathrm{Rs\> 622}}=(0.141)

\, According to Table Z, the probability corresponding to the respective values of Z_1 and Z_2 is 0.1026 and 0.0557. Summing up the values, we have, 0.1583, that is 15.83 per cent.
(ii) Between\> the \>range\> of\> Rs\>400 >and \>Rs\> 600

Z_1=\frac{\mathrm{Rs \>400- Rs\> 663}}{\mathrm{Rs\> 622}}=(0.42)\qquad\qquad \,Z_2=\frac{\mathrm{Rs \>600- Rs\> 663}}{\mathrm{Rs\> 622}}=(0.10)

\, According to Table Z, the probability corresponding to the respective values of Z are 0.1628 and 0.0398 respectively.
\, To put it explicitly: The probability of NPV having a value between Rs 400 and Rs 663 = 0.1628. The probability of it having a value between Rs 600 and Rs 663 = 0.0398. Therefore, the probability of having its value between Rs 400 and Rs 600 would be = 0.1628 – 0.0398 = – 0.1230 = 12.3 per cent.
(iii) At \>least \>Rs \>300

Z=\frac{\mathrm{Rs \>300- Rs\> 663}}{\mathrm{Rs\> 622}}=(0.5836)

\, According to Table Z, probability of the NPV being 300 = 0.2190. The probability of having NPV at least equal to Rs 300 would be more by 0.50 (area to the right side of mean), that is, 0.7190 or 71.9 per cent.
(iv) At\> least \>Rs \>1,000

Z=\frac{\mathrm{Rs \>1,000- Rs\> 663}}{\mathrm{Rs\> 622}}=0.5418

\, According to Table Z the probability of having the NPV value Rs 1,000 is 0.254. The probability of having NPV Rs 1,000 or more would be 0.5 – 0.2054 = 0.2946 = 29.46.

(a)\> Determination \>of\> expected \>NPV:

Period\> 3 Period\>2 Period\> 1
Cash\>flow
(CF × Pj)
P_j CF Cash\>flow
(CF × Pj)
P_j CF Cash\>flow
(CF × Pj)
P_j CF
Rs 160 0.2 Rs 800 Rs 80 0.1 Rs 800 Rs 80 0.1 Rs 800
     500 0.5   1,000    300 0.3   1,000   200 0.2   1,000
     300 0.2   1,500    600 0.4   1,500   600 0.4   1,500
      200 0.1   2,000    400 0.2   2,000   600 0.3   2,000
\overline{(CF_3)}\>\overline{1,160} \overline{(CF_2)}\>\overline{1,380} Mean \overline{(CF_1)}\>\overline{1,480}
Determination \>of \>NPV
Total\>PV PV\>factor\>(0.05) CF
Rs 1,409 0.952 Rs 1,480
      1,252 0.907       1,380
     1,002


0.864       1,160
     3,663 Total present value
     3,000


Less cash outflows
        663 NPV

(b) \>Determination \>of \>standard\> deviation \>for\> each \>period:

Period \>1
(CF – \overline{CF_1})^2Pj_1 P_\mathrm{j1} (×) (CF_\mathrm{j1}\overline{CF_1})^2
Rs 46,240 0.1 × Rs 4,62,400
    46,080 0.2 ×     2,30,400
         160 0.4 ×              400
      81,120


0.3 ×      2,70,400
  1,73,600
\sigma_1=\sqrt{1,73,600}=417
Period \>2
(CFj_2 –  \overline{CF_2})^2Pj_2 Pj_2 (×) (CFj_2 –  \overline{CF_2})^2
Rs 33,640 0.1 × Rs 3,36,400
     43,320 0.3 ×      1,44,400
        5,760 0.4 ×         14,400
      76,880


0.2 ×     3,84,400
   1,59,600
\sigma_2=\sqrt{1,59,600}=400
Period \>3
(CF_{j3} –  \overline{CF_3})^2P_{j3} P_{j3} (×) (CF_{j3} –  \overline{CF_3})^2
Rs 25,920 0.2 × Rs 1,29,600
      12,530 0.5 ×          25,600
      22,120 0.2 ×       1,15,600
      70,560


0.1 ×      7,05,600
     1,31,130
\sigma_3=\sqrt{1,31,130}=362