chap5.pdf

Math 618 Notes Chapter 5 ? Cash Flows At time tk, Ak in, Bk out, so the net amount is Ck = Ak ?Bk (0 = t0 < t1 < ··· < tn). ? i > ?1 is an IRR for the cash flow Ck if nsummationdisplay k=0 Ck(1+i)tn?tk = 0, or nsummationdisplay k=0 Ckvtk = 0. ? Uniqueness of IRR (Example 5.2) Excample (#5.1.4) Transaction A: ?5(1+i)3 +3.72(1+i)2+4 = 0 BA II: CF0 = ?5, CO1 = 3.72, FO1 = 1, CO2 = 0, FO2 = 1, CO3 = 4, FO3 = 1 =? 25.330403%. TI-83: irr(?5,{3.72,0,4},{1,1,1}) = 25.330403% Transaction B: ?5(1+i)3 +3(1+i)2+1.7(1+i)+3 = 0 BA II: CF0 = ?5, CO1 = 3, FO1 = 1, CO2 = 1.7, FO2 = 1, CO3 = 3, FO3 = 1 =? 25.328044%. TI-83: irr(?5,{3,1.7,3},{1,1,1}) = 25.328044% Let PA and PB be the present value of transaction A and B respectively. Then PA = ?5+3.72v +4v3, PB = ?5+3v +1.7v2 +3v3 A is preferable to B ?? PA > PB ?? PA ?PB > 0. We have PA ?PB = 0.72v ?1.7v2 +v3 > 0 ?? 0.72?1.7v +v2 > 0 ?? v < 0.8 or v > 0.9 ?? i > 0.25 or i < 0.111111... ? Dollar-Weighted Rate (similar to IRR, but uses simple rate) Future value of all the deposits ?Future value of all the withdrawals = = End value of the balance. ? Time-Weighted Rate (1+i1)(1+i2)···(1+in) = 1+i Example (#5.2.3) Dates 1/1/05 7/1/05 1/1/06 Unit Price 1 0.8 1 # Units +100,000 +125,000 225,000 Cash Fllow (100,000) (100,000) 225,000 ? Dollar-Weighted: 100000(1+i)+100000(1+0.5i) = 225000 =? i = .1¯6. ? Time-Weighted: 1/1/05 to 7/1/05: $100000 goes down to $80000, i1 = ?20% = ?0.2 or 1+i1 = 80000100000 = 0.8 7/1/05 to 1/1/06: $180000 goes up to $225000, i2 = 25% = 0.25 or 1+i2 = 225000180000 = 1.25 We have (1?0.2)(1+0.25) = 1 = 1+0 or 0.8·1.25 = 1 = 1+0. Therefore, i = 0.