## 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

## 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 $\sigma_3=\sqrt{1,31,130}=362$