THE EQUATION IS NOT WRONG, I’m going to use mohammad.bellal’s demonstration but using the correct expression to V, similar to the one used to derive the unlevered beta V = Dx(1-t) + E
Let’s derive this formula using V = Dx(1-t) + E
Ro = (D/V)xRd(1-t) + (E/V)xRe
Ro - (DxRd(1-t)) / V = (E/V)xRe
Taking LCM and cutting both denominators (V)
VRo - DRd(1-t) = ERe
(VxRo)/E - (DxRd(1-t))/E= Re
Here we go with V = D(1.t)+E, Substituting V with D(1-t)+E
((D(1-t)+E)Ro)/E - (DxRd(1-t))/E = Re
Rewrite it
(D(1-t)/E x Ro) + ERo/E - D/E x Rd(1-t) = Re
(D(1-t)/E x Ro) + ERo/E - D/E x Rd(1-t) = Re
(D(1-t)/E)Ro + Ro - (D/E)Rd(1-t) = Re
Ro + (D(1-t)/E)Ro - (D/E)Rd(1-t) = Re
Taking D/E as common
Ro + (D/E) (Ro - Rd)(1-t) = Re
Rewrite it
Re = Ro + (Ro - Rd)(1-t)D/E
It’s derived in the correct way as Modiggliani Miller proposed first!
The issue with the problem is weather the capital structure is D / (D+E) = 50% or D / [D(1-t)+E] = 50%
I believe that they gave us the expression for D / (D(1-t)xE) but I have not found the correct answer, so please help me!!!
My thoughts in order to answer the problem assuiming that re = 16% is correct, are below:
In order to have an answer equal to C: re=16, we need that D/E = 1.5 (16% = 10% + (10%-5%)*(1-0.2)*1.5)
So with D/E = 1.5 and t = 0,2 , D/[D*(1-t) +E] will be equal to 0.681818, see how I found it:
If you divide the expresion D/[D*(1-t) +E] by E in both the numerator and denominator you will have (D/E) / [(D/E)*(1-t) +1], and by substituting with D/E = 1.5 and t = 0.2 I arrived to D/ D(1-T)+E = 0.681818
The thing that freaks me out is that there should be a mistake in my calculation or in the way I propose the problem as the solution doesn’t seem to be ok:
WACC = [D / (D(1-t)+E)] x rd x (1-t) + [E / (D(1-t)+E) x re]
10% = 0.681818 x 5% x (1-0.2) + [50% x re]
reordering
re = (10% - [0.681818 x 5% x (1-0.2)] ) / 50%
re = 14.5455%
Would someone help me find a solution without telling me that the formula proposed by MM is wrong and we should change 50 years of papers (and the cfa curriculum too)?
tks