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Can anyone provide a mathematical proof that NPV is always negative when the rate of return is less than the discount rate? That is suppose that we have

  • a single initial input, C0
  • a constant known rate of return on the investment, r > 0
  • and some constant discount rate, rd > r
  • and letting Ct = cashflow in period t

The way I see it, the argument would be something like:

NPV = -C0 + summ( Ct / (1+rd)^t ) 
= -C0 + summ( (r^t*C0) / (1+rd)^t )
... then some magic happens, then ...
< 0 

But how to get from beginning to end, I don't know.

Anecdotally, I see it as, since r < rd and both number will shrink exponentially with time, there would never be a periodwhere the returned cashflow would be greater than the return you could have gotten by investing at the discount rate rd, thus NPV < 0 (and if even this is a wrong way to think about it, please let me know).

Basically, asking for a proof that NPV always < 0 whenever the return on the investment is less than the discount rate for all periods.

Would appreciate a proof and explanation (or even a explanation about why this may be trying to prove something that is not necessarily true). Thanks.

  • What is Ct? Cash value at time t? Surely your second line should be -C0 + summ( (r^t*C0) / (1+rd)^t ) - in other words C0, not Ct – Martin Bonner Dec 22 '17 at 10:10
  • @MartinBonner Yes, my mistake. Changed that. – lampShadesDrifter Dec 23 '17 at 4:45
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The internal rate of return is the value of r which satisifies the equation

0 = C0 + C1 / (1 + r) + C2 / (1 + r)^2 + ... + Cn / (1 + r)^n

The net present value (NPV) is the value of the sum of discounted cash flows:

NPV = C0 + C1 / (1 + rd) + C2 / (1 + rd)^2 + ... + Cn / (1 + rd)^n

Since rd > r, for t = 1, ..., n, we have the simple inequalities:

(1 + rd) > (1 + r)

(1 + rd)^t > (1 + r)^t

Taking the inverse switches the direction of inequality:

1 / (1 + rd)^t < 1 / (1 + r)^t

Since infusions of cash are negative cashflow, C0 < 0. But presumably Ct > 0 for all subsequent time periods (i.e., for t = 1, ..., n). So, multiplying the inequality above by a positive quantity, Ct, preserves the direction of the inequality:

Ct / (1 + rd)^t  < Ct / (1 + r)^t

Now, looking at the sum term-by-term,

0 = C0 + C1 / (1 + r) + C2 / (1 + r)^2 + ... + Cn / (1 + r)^n
  > C0 + C1 / (1 + rd) + C2 / (1 + rd)^2 + ... + Cn / (1 + rd)^n = NPV

This shows that if we valued a cash flow (with an initial investment followed by positive cash flows) using a discount rate which is greater than the rate of return, then NPV would be negative.

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Assuming the second line should be corrected to:

NPV = -C0 + summ( Ct / (1+rd)^t ) 
    = -C0 + summ( (r^t*C0) / (1+rd)^t )

We can continue:

      = -C0 + summ( (r/(1+rd)^t *C0 )

So clearly what we are interested in is, is:

summ( (r/(1+rd))^t ) 

necessarily < 1? All of the terms in the sum are positive, so the maximum value of the sum happens if we sum to infinity. Given that r < 1+rd, we know that (r/(1+rd)) < 1, so the sum to infinity has a finite value. Specifically, the sum to infinity is:

      1
------------  - 1
1 - r/(1+rd)

Let us multiply top and bottom of the fraction by 1+rd and expand 1

  1 + rd         1 + rd - r
----------   -   ----------
1 + rd - r       1 + rd - r

Simplify:

    r
----------
1 + rd - r

Which looks to me like it can be greater than 1. If rd == 5, r = 4, then the expression comes to 4 / 2 == 2 - which implies that the NPV is greater than zero.

This looks to me like it is wrong (but with luck, someone can point out my error)

  • @DStanley The sum of 1 + a + a^2 ... is 1/(1-a). I subtracted 1 to get a+a^2... – Martin Bonner Dec 26 '17 at 18:27

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