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March 8th, 2017, 10:34 AM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800  Interesting Loan case...
A loan is obtained, at monthly payments, over 12 months. The amount that went to principal at 3rd pay't = 804.34 The amount that went to principal at 7th pay't = 837.00 How much was borrowed, at what rate, at what monthly pay't? This'll probably wake up Sir Jonah 
March 12th, 2017, 06:33 AM  #2 
Senior Member Joined: May 2008 Posts: 267 Thanks: 55  Beer soaked ramblings follow. Insuficient data to resolve problem. Suppose P3=804.34 & I3=amount that went to interest at 3rd pay't, we then have (P3+I3)(1+i)^[(123+1)]=P3 OR (P3+I3){1(1+i)^[(123+1)]}=I3. Solving for I3 in both equations leads to an identity of I3=P3[(1+i)^101] for both. Methinks I3 should be given to solve amount borrowed and rate. I would be very interested indeed as to how you propose to solve this one Sir D with the given info. Last edited by jonah; March 12th, 2017 at 06:46 AM. 
March 12th, 2017, 10:18 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800 
To eliminate "back and forth"'s, here's the resulting loan: (the givens are earmarked with *) Code: MTH [PAYMENT] PRINCIPAL INTEREST BALANCE 0 10,000.00 1 888.48 788.48 100.00 9,211.52 ... 3* 888.48 804.34* 84.14 7,610.81 ... 7* 888.48 837.00* 51.48 4,312.21 ... 11 888.48 870.98 17.50 879.69 12* 888.48 879.69 8.79 .00 (i being the monthly loan rate), then: 804.34*(1 + i)^(73) = 837.00 : solve to get i = .01 The rest is fairly straightforward. OK Sir Jonah? 
March 14th, 2017, 06:16 AM  #4  
Senior Member Joined: May 2008 Posts: 267 Thanks: 55  Brandy soaked revelation follows. Quote:
I was about to correct myself but I see that you beat me to it. Turns out I be wrong. Took a day to abstain from me anti allergy meds and gave myself me favorite brandy and it all seemed to be clear somehow. Don't know why I did't notice it but it should have been apparent that with R as the monthly payment, we should have the following system of equations: R(1+i)^[(123+1)]=804.34 & R(1+i)^[(127+1)]=837. A little algebra and we get something like i=.01000017213 or .01and the rest is history.  
March 14th, 2017, 06:35 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800  
March 18th, 2017, 10:33 AM  #6 
Senior Member Joined: May 2008 Posts: 267 Thanks: 55  Beer soaked ramblings follow. Somebody just asked me how he or she can solve for the amount borrowed given Sir D's insight. Sorry for the delay, been traveling a lot. Given my system of equations, you solve for either i (or R), then R and then A, the amount borrowed using the the present value for the ordinary annuity. Using Sir D's equation, as you have seen, you simply solve for i, then A (if you're not aware of the formula I used to come up with them system of equations), and then R using the aforementioned present value formula for the ordinary annuity. Accordingly, for A, we should have A=804.34(1.01)^(13)+ ...+804.34+...+804.34 (1.01)^(123) A=804.34(1.01)^(13)*[(1.01)^121]/.01 Last edited by jonah; March 18th, 2017 at 10:37 AM. 
March 18th, 2017, 11:01 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800  
March 18th, 2017, 08:42 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800  Code: MTH [PAYMENT] PRINCIPAL INTEREST BALANCE 0 10,000.00 1 888.48 788.48 100.00 9,211.52 ... 3(e) 888.48 804.34(u) 84.14 7,610.81 ... 7(f) 888.48 837.00(v) 51.48 4,312.21 ... 11 888.48 870.98 17.50 879.69 12(n) 888.48 879.69 8.79 .00 These are my formulas: i = (v/u)^[1 / (fe)]  1 : monthly interest rate = .01 p = u(1+i)^(ne+1) : monthly payment = 888.48 a = p[1  (1+i)^(n)] / i : loan amount = 10000.00 Agree? Last edited by Denis; March 18th, 2017 at 08:48 PM. 
March 19th, 2017, 06:12 AM  #9 
Senior Member Joined: May 2008 Posts: 267 Thanks: 55  
March 19th, 2017, 06:14 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,121 Thanks: 800 
You "absolutely" has no booze this weekend?


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