Question 18.9: Assume everything is identical to Example 18.6 except that t......

Based on equation 18.4,

$P_o(1- f)=\sum\limits_{t=1}^{n}{\frac{DP_t}{(1+k_p)^t} +\frac{P_n}{(1+k_p)^n} }$      (18.4)

$k_p$ =
$ 61.2632 million = $\frac{\mathrm{\$ \>7.3220\> million}}{(1+k_p)^1}+\frac{\mathrm{\$ \>7.1784\> million}}{(1+k_p)^2}+…+\frac{\mathrm{\$ \>62.7142\> million}}{(1+k_p)^6}$

Like $k_d$, the approximate formula to determine kp is provided by Equation 18.5.
$\,$           $= DP (1\, –\, d)\, (1 + f) \,–\, d$               (18.5)
$\,$          $=$ 12% (1 – 0.02) (1 + 0.0195) – 2%
$\,$          $=$ 11.99% – 2% = 9.99% or 10 per cent.
The value of $k_p$ is likely to be between 10 and 11 per cent.

$\,$          $k_p$ = 10% + $\left\lgroup\frac{\mathrm{\$ \>62.1545\> million − \$\> 61.2632 \>million =\$ \>0.8913\> million}}{\mathrm{\$ \>62.1545\> million − \$\> 59.6809  =\$ \>2.4736\> million}} \right\rgroup$

$\,$                    =10% + 0.36% = 10.36 per cent.
$\,$   In the event of the currency in which borrowings are made appreciating vis-à-vis the base currency (US dollar in the present context), Equation 18.6 provides the basis of determining the approximate value of kp.

$\qquad \qquad k_p = DP\, (1 + r) \,(1 + f) + r$                    (18.6)