My Math Forum Compund Interest

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 May 31st, 2013, 05:48 PM #1 Newbie   Joined: May 2013 Posts: 2 Thanks: 0 Compund Interest I have no idea where to begin! A two-payment stream consisting of $1750 due today and$2900 due in 18 months is to be replaced by an economically equivalent stream comprised of an undetermined payment due in 9 months and a payment of \$3000 due in 19 months. Calculate the unknown replacement payment if money is worth 9% compounded monthly.
 June 1st, 2013, 08:04 AM #2 Newbie   Joined: May 2013 Posts: 23 Thanks: 0 Re: Compund Interest The data are: $r=0.09$ $C_1=1750\qquad t_1=0$ $C_2=2900\qquad t_2=18$ $C_1^*=?\qquad t_1^*=9$ $C_2^*=3000\qquad t_2^*=19$ In order that the two operations be equivalent it is required that the financial situation be the same for the creditor at the most distant time. It is not clear from context if the capitals due must be capitalized on maturity or are already the effective amounts. In the first case we have: $C_1(1+r)^{t_1+t_2^*}+C_2(1+r)^{t_2+t_2^*-t_2}=C_1^*(1+r)^{t_1^*+t_2^*-t_1^*}+C_2^*(1+r)^{t_2^*}$ that is $C_1(1+r)^{t_2^*}+C_2(1+r)^{t_2^*}=C_1^*(1+r)^{t_2^ *}+C_2^*(1+r)^{t_2^*}$ In the second case we have: $C_1(1+r)^{t_2^*}+C_2(1+r)^{t_2^*-t_2}=C_1^*(1+r)^{t_2^*-t_1^*}+C_2^*$
 June 1st, 2013, 04:57 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 Re: Compund Interest f = 1 + .09/12 = 1.0075 x = f^9(1750 + 2900/f^18 - 3000/f^19) = 1799.11522182... OR x = (1750f^19 + 2900f - 3000) / f^10 = 1799.11522182...
 June 2nd, 2013, 08:20 AM #4 Newbie   Joined: May 2013 Posts: 23 Thanks: 0 Re: Compund Interest My memories in financial mathematics are perhaps a bit far. I forgot to divide by 12, which is the interest rate per annum (http://math.about.com/library/weekly/aa042002a.htm), as Davis noted in its solution.
 June 2nd, 2013, 04:40 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 Re: Compund Interest Who is Davis? Didn't see that name at your suggested site...

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