Internal Rate of return
In case you forgot the formula:
w^n + w^{n-1} + w^{n-2} + \dots + w = \frac{w^{n+1} - w}{w - 1}
Let P0, P1, P2,..., Pk be a future payment flow. This means that at the time instant 0 we pay P0, at the time instant 1, we pay P1 and so on. If a payment is negative, usually P0, this means that we receive money instead of paying. The Net Present Value of this flow, denoted by NPV(r), is defined as the sum of all present values of the above payments, for a given interest rate r. The rate r is the rate which is applied at each time instant. To find the Internal Rate of Return (IRR) of this flow, we follow the next steps:
1. We define a set of interest rates r0, r1, r2,... rn.
2. We calculate the quantities: NPV (r0), NPV (r1), NPV (r2),..., NPV (rn).
3. We find successive ra, rb such that NPV(ra) > 0 and NPV(rb) < 0.
4. We find a straight line joining the points: (ra, NPV(ra)), (rb, NPV(rb)).
5. We find the value r* where this straight line meets the horizontal r-axis.
6. IRR = r*
A company takes, at the "zero" time instant, a loan equal to 71 money units and plans to pay the first month 15 money units, the second 28 and the third 30, that is P1 = 15, P2 = 28, P3 = 30
(a) (10 rp) Using the above procedure calculate the IRR for this payment flow. Use r0 = 0.14, applied annually, and increase it with step h = 0.01 up to the value r = 0.16.
For r0=0,14 we have:
NPV(0{,}14) = P_0 \cdot \left(1 + \frac{0{,}14}{12}\right)^0 + P_1 \cdot \left(1 + \frac{0{,}14}{12}\right)^{-1} + P_2 \cdot \left(1 + \frac{0{,}14}{12}\right)^{-2} + P_3 \cdot \left(1 + \frac{0{,}14}{12}\right)^{-3} = 0{,}158975
For r0=0,15 we have:
NPV(0{,}15) = P_0 \cdot \left(1 + \frac{0{,}15}{12}\right)^0 + P_1 \cdot \left(1 + \frac{0{,}15}{12}\right)^{-1} + P_2 \cdot \left(1 + \frac{0{,}15}{12}\right)^{-2} + P_3 \cdot \left(1 + \frac{0{,}15}{12}\right)^{-3} = 0{,}0302743
For r0=0,16 we have:
NPV(0{,}16) = P_0 \cdot \left(1 + \frac{0{,}16}{12}\right)^0 + P_1 \cdot \left(1 + \frac{0{,}16}{12}\right)^{-1} + P_2 \cdot \left(1 + \frac{0{,}16}{12}\right)^{-2} + P_3 \cdot \left(1 + \frac{0{,}16}{12}\right)^{-3} = -0{,}09806
Therefore, we seek for a straight line y=Ar+B, passing through the points (0,15, 0,0302743), (0,16, 0,09806). By substitution we find:
А = −12,8334; B=1,95529
So, this straight line is vanished at the point
-12{,}8334 \cdot r + 1{,}95529 = 0 \Rightarrow r^* = 0{,}152359
IRR \approx 0{,}15
(b) (10 rp) If the company, instead of following this payment policy, decides to give 25 money units on the third month and some amount of money the fifth month, what should this amount be, in order the whole loan to be repaid? As an annual interest rate, use the IRR we found before.
The present value of the 25 money units, paid at the third month is:
\frac{25}{\left(1 + \frac{0{,}15}{12}\right)^3} = 24{,}0855
where, we have taken the interest rate equal to 0,15. So, the rest of the loan is:
71 − 24,0855 = 46,9145
In order to repay it the fifth month, we must pay:
46{,}9145 \cdot \left(1 + \frac{0{,}15}{12}\right)^5 = 49{,}9209
money units. This is the requested P5 value.
(c) (10 rp) The company has started to repay the loan by 5 equal monthly payments, covering both principal and interest and with an interest rate equal to the IRR, found before. Suddenly after the second payment, the interest rate is increased by 3%. The government, in order to help, decides to cover the 25% of the loan which has not been paid yet and to extend the number of total payments from 5 to 8. Find the amount of each equal monthly payment, the company must pay from now on, to repay the loan.
First, we have to calculate how much the original equal-payment was. Let us denote it by R, we have:
R \cdot \left(1 + \frac{0{,}15}{12}\right)^{-1} + R \cdot \left(1 + \frac{0{,}15}{12}\right)^{-2} + R \cdot \left(1 + \frac{0{,}15}{12}\right)^{-3} + R \cdot \left(1 + \frac{0{,}15}{12}\right)^{-4} + R \cdot \left(1 + \frac{0{,}15}{12}\right)^{-5} = 71
or
R \cdot \left(w^5 + w^4 + w^3 + w^2 + w\right) = 71
where w = \left(1 + \frac{0{,}15}{12}\right)^{-1} = 0{,}987654 Using the formula for adding terms of geometric progression, we get:
R \cdot \frac{w^6 - w}{w - 1} = 71 \quad \Rightarrow \quad R = 71 \cdot \frac{w - 1}{w^6 - w} = 14{,}7369
Therefore, the present value of the two already-done payments is:
R \cdot w + R \cdot w^2 = 14{,}7369 \cdot \left(0{,}987654 + (0{,}987654)^2\right) = 28{,}9302
So, the rest of the loan is 71-28,9302=42,0698. This amount is reduced by 25%, due to the government intervention. So, the new rest of the loan is:
42{,}0698 - 0{,}25 \cdot 42{,}0698 = 31{,}5523
This quantity must be repaid with six equal payments (denoted by H), starting the third month, with new annually interest rate equal to \tilde{r} = 18\% This means that:
H \cdot \left( s^3 + s^4 + s^5 + s^6 + s^7 + s^8 \right) = 31{,}5523
with:
s = \left(1 + \frac{0{,}18}{12}\right)^{-1} = 0{,}985222
Working as before, we get:
H \cdot \left( s^3 + s^4 + s^5 + s^6 + s^7 + s^8 \right) = H \cdot s^3 \cdot \left( 1 + s + s^2 + s^3 + s^4 + s^5 \right)
H \cdot s^3 \cdot \frac{s^6 - 1}{s - 1} = 31{,}5523 \quad \Rightarrow \quad H = 31{,}5523 \cdot \frac{s - 1}{s^9 - s^3} = 5{,}70561
This is the new monthly payment, upon request.