To summarize:
a. In scenario #1, both loans run full term.
b. In scenario #2, the shorter loan runs full term, and the longer loan is
paid off early at the same time.
c. Empirically, we find that when the loan interest rates are equal, the
IRRs for scenarios #1 and 2 are equal.
d. Empirically, we find that when the loan interest rates are different, the
IRRs for scenarios #1 and 2 are different.
Previously, I offered mathematical proofs to explain of #c and #d. But they
depended on assertions that I left unproven. To clean up the sloppiness
in the previous proofs....
In the scenario #1 formula, if rA = rB (same interest rates), then
r1 = rB.
This seems to be the most intuitive assertion, yet oddly I am having trouble
with the proof.
0 = -prin + SUM{(pmtA+pmtB)/(1+r1)^i, i=1,nA}
+ SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -(prinA + prinB) + SUM{pmtA/(1+r1)^i, i=1,nA}
+ SUM{pmtB/(1+r1)^i, i=1,nA} + SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -prinA + SUM{pmtA/(1+r1)^i, i=1,nA} +
-prinB + SUM{pmtB/(1+r1)^i, i=1,nB}
Note the following formulas, which are fundamental to loan arithmetic (the
loan amount is the NPV of the payments discounted at the interest rate):
[1] 0 = -prinA + SUM{pmtA/(1+rA)^i, i=1,nA}
[2] 0 = -prinB + SUM{pmtB/(1+rB)^i, i=1,nB}
Since rA = rB, then r1 = rB = rA is at least one solution for the scenario
#1 formula, since 0 = 0 + 0.
Arguably, that is only a "sufficient" proof, not a "necessary and
sufficient" proof. Theoretically, some IRR formulas can have more than one
solution. But I believe we should have only one IRR solution in each of
these scenarios since the sign of the cash flows changes only once.
BTW, I am talking about mathematical equality. There can always be
infinitesimal inequalities that are artifacts of floating-point
computations.
when rA <> rB, we expect r1 <> rB in scenario #1.
Proof by contradiction; that is, assume r1 = rB, and prove that contradicts
the requirement rA <> rB.
0 = -prin + SUM{(pmtA+pmtB)/(1+r1)^i, i=1,nA}
+ SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -(prinA + prinB) + SUM{pmtA/(1+r1)^i, i=1,nA}
+ SUM{pmtB/(1+r1)^i, i=1,nA} + SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -prinA + SUM{pmtA/(1+r1)^i, i=1,nA} +
-prinB + SUM{pmtB/(1+r1)^i, i=1,nB}
= -prinA + SUM{pmtA/(1+r1)^i, i=1,nA} +
-prinB + SUM{pmtB/(1+rB)^i, i=1,nB}
Based on [2] above, the 2nd line is zero. So:
0 = -prinA + SUM{pmtA/(1+r1)^i, i=1,nA}
Since that has the same form as [1] above, then r1 = rA; and since r1 = rB,
then rA = rB. That contradicts our requirement (rA <> rB). Ergo, r1 <> rB
when rA <> rB [....]
It should be clear that r1 <> r2
Proof by contradiction; that is, assume r1 = r2, and prove that contracdicts
the requirement rA <> rB.
The scenario #1 and 2 formulas are both equal to zero. So:
-prin + SUM{(pmtA+pmtB)/(1+r1)^i, i=1,nA} + SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -prin + SUM{(pmtA+pmtB)/(1+r2)^i, i=1,nA} + fvB/(1+r2)^nA
-prin + SUM{(pmtA+pmtB)/(1+r1)^i, i=1,nA} + SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= -prin + SUM{(pmtA+pmtB)/(1+r1)^i, i=1,nA} + fvB/(1+r1)^nA
SUM{pmtB/(1+r1)^i, i=nA+1,nB}
= fvB/(1+r1)^nA
SUM{pmtB/(1+r1)^i, i=1,nB-nA}/(1+r1)^nA
= SUM{pmtB/(1+rB)^i, i=1,nB-nA}/(1+r1)^nA
That is true only if r1 = rB. But that contradicts our requirement
(rA <> rB), since we proved above that if rA <> rB, then r1 <> rB
(and the contrapositive: if r1 = rB, then rA = rB). Ergo, r1 <> r2
when rA <> rB.
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