Actuarial Outpost Loan question
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#1
11-10-2017, 05:59 PM
 Futon Member SOA Join Date: Jul 2016 Studying for FM Posts: 128
Loan question

A borrower took out a loan of 100,000 and promised to repay it with a payment at the end of each year for 30 years.
The amount of each of the first ten payments equals the amount of interest due. The amount of each of the next ten payments equals 150% of the amount of interest due. The amount of each of the last ten payments is X.
The lender charges interest at an annual effective rate of 10%.
Calculate X.
(A) 3,204
(B) 5,675
(C) 7,073
(D) 9,744
(E) 11,746

So I'm kind of lost. But here is my intuition:

$100000=10000a_{\overline{10|}10}+v^{10}15000a_{\ov erline{10|}10}+v^{20}Xa_{\overline{10|}10}$

Obviously I got the wrong answer but any guidance would be greatly appreciated.

__________________________________________________ _____________________________________________
Edit: So Mimetex is really poorly optimized here. That strange group of letters after 15000a is supposed to be overline. I literally copy pasted the annuity code for each coefficient:

Code:
100000=10000a_{\overline{10|}10}+v^{10}15000a_{\overline{10|}10}+v^{20}Xa_{\overline{10|}10}

Last edited by Futon; 11-10-2017 at 06:04 PM..
#2
11-10-2017, 06:18 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,848

[quote=Futon;9162273]Your problem starts in year 12. Once 15000 is paid in year 11, the balance at the start of year 12 is less than 100,000, so 15000 is more than 150% of the interest due.
#3
11-10-2017, 06:50 PM
 Futon Member SOA Join Date: Jul 2016 Studying for FM Posts: 128

[quote=Gandalf;9162286]
Quote:
 Originally Posted by Futon Your problem starts in year 12. Once 15000 is paid in year 11, the balance at the start of year 12 is less than 100,000, so 15000 is more than 150% of the interest due.
Right so... I get that interest is paid for the first 10 years so the PV after 10 payments is still 100000. So what should I do after?

$100000=15000a_{\overline{10|}10}+v^{10}Xa_{\overli ne{10|}10}$

doesn't yield me the right answer either.

Edit: again, it supposed to be a overline 10 at the second coefficient.
#4
11-10-2017, 06:59 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,650

In a given year if you pay (1+k) x i x Loan Balance at the end of the year, k x i x Loan Balance will reduce principal so the remain proportion is (1- ki). If you continued to do this the Balance would decline by that proportion each year.
#5
11-10-2017, 07:46 PM
 Futon Member SOA Join Date: Jul 2016 Studying for FM Posts: 128

Sorry it's not clicking. I finally get that 150% decreases the principal (sorry didn't read Gandalf's comment clearly enough). But I have a hard time finding the pattern and making an arithmetic annuity for it.

1000000(.1)(1.5)-1000000(.1)= 5000

95000(.1)(1.5)-95000(.1)=4750

90250(.1)(1.5)-90250(.1)=4512.5

Thus no pattern. Am I not supposed to use annuities for this question?

Edit: Nevermind I did find a pattern. 5000/4750 = 4750/4512.5 = 1.05263

$100000= [5000v+5000v^{2}(1.05263^{-1})+...+5000v^{10}(1.05263^{-9})] + v^{10}Xa_{\overline{10|}10}$

$100000= [5000v\frac{1-\frac{1}{(1.1)(1.05263)}^{10}}{1-\frac{1}{(1.1)(1.05263)}}] + v^{10}Xa_{\overline{10|}10}$

I got X=31389.31128 which is very wrong.

Last edited by Futon; 11-10-2017 at 08:01 PM..
#6
11-10-2017, 09:47 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,650

Don't discount the amounts of principal repayment. It's simpler to look at the pattern for the outstanding loan balance than the pattern for the principal repayments.
#7
11-10-2017, 09:50 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,848

The problem with you latest attempt is that it says the total paid in year 11 is 5000. No, 5000 is the amount that reduces principle.
Quote:
 Originally Posted by Academic Actuary In a given year if you pay (1+k) x i x Loan Balance at the end of the year, k x i x Loan Balance will reduce principal so the remain proportion is (1- ki). If you continued to do this the Balance would decline by that proportion each year.
Use that bolded statement. What is the balance at the end of year 11? At the end of year 12? ... At the end of year 20?

Then write the equation, looking forward, at the end of year 20. The present value of a level annuity of X must equal that balance.

ETA: ninja'd by AA's response, which he posted as I was composing mine.
#8
11-11-2017, 12:28 AM
 NattyMo Member Non-Actuary Join Date: Nov 2016 Posts: 606

Quote:
 Originally Posted by Futon Right so... I get that interest is paid for the first 10 years so the PV after 10 payments is still 100000. So what should I do after? $100000=15000a_{\overline{10|}10}+v^{10}Xa_{\overli ne{10|}10}$ doesn't yield me the right answer either. Edit: again, it supposed to be a overline 10 at the second coefficient.
Good.

Quote:
 Originally Posted by Futon Sorry it's not clicking. I finally get that 150% decreases the principal (sorry didn't read Gandalf's comment clearly enough). But I have a hard time finding the pattern and making an arithmetic annuity for it. 1000000(.1)(1.5)-1000000(.1)= 5000 95000(.1)(1.5)-95000(.1)=4750 90250(.1)(1.5)-90250(.1)=4512.5 Thus no pattern. Am I not supposed to use annuities for this question? Edit: Nevermind I did find a pattern. 5000/4750 = 4750/4512.5 = 1.05263 $100000= [5000v+5000v^{2}(1.05263^{-1})+...+5000v^{10}(1.05263^{-9})] + v^{10}Xa_{\overline{10|}10}$ $100000= [5000v\frac{1-\frac{1}{(1.1)(1.05263)}^{10}}{1-\frac{1}{(1.1)(1.05263)}}] + v^{10}Xa_{\overline{10|}10}$ I got X=31389.31128 which is very wrong.
\begin{align}
\text{LET: }
B_t &= \text{ "balance at time t"}
\text{THEN: }
B_{10} &= 100,000
B_{11} &= B_{10} \, (1+i) - B_{10} \, (1.5i)
&= B_{10} \, [1+i-1.5i]
&= B_{10} \, [1-0.5i]
B_{12} &= B_{11} \, (1+i)-B_{11} \, (1.5i)
&= B_{11} \, [1+i-1.5i]
&= B_{11} \, [1-0.5i]
&= (B_{10} \, [1-0.5i]) \, [1-0.5i]
&= B_{10} \, [1-0.5i]^2
&\text{ . }
&\text{ . }
&\text{ . }
B_{20} &= B_{19} \, (1+i)-B_{19} \, (1.5i)
&= B_{19} \, [1+i-1.5i]
&= B_{19} \, [1-0.5i]
&= (B_{10} \, [1-0.5i]^9) \, [1-0.5i]
&= B_{10} \, [1-0.5i]^{10}
\end{align}
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Last edited by NattyMo; 11-11-2017 at 12:32 AM..
#9
11-11-2017, 02:41 AM
 NattyMo Member Non-Actuary Join Date: Nov 2016 Posts: 606

Quote:
 Originally Posted by Futon Right so... I get that interest is paid for the first 10 years so the PV after 10 payments is still 100000. So what should I do after? $100000=15000a_{\overline{10|}10}+v^{10}Xa_{\overli ne{10|}10}$ doesn't yield me the right answer either. Edit: again, it supposed to be a overline 10 at the second coefficient.
(Don't mind me -- I'm just tinkering with your code to see why your second a-angle-10 is wonky.)

\begin{align}
100000=15000a_{\overline{10|}10}+v^{10}Xa_{\overli ne{10|}10}
100000&=15000a_{\overline{10|}10}+v^{10}Xa_{\overl ine{10|}10} &\text{ [1] }
100000&=15000a_{\overline{10|}10}+v^{10} Xa_{\overline{10|}10} &\text{ [2] }
100,000 &= 15,000 a_{\overline{10|}10} + v^{10} X a_{\overline{10|}10} &\text{ [3] }
100,000 &= 15,000 a_{\overline{10|}10} + \nu^{10} X a_{\overline{10|}10} &\text{ [4] }
100,000 &= 15,000 a_{\overline{10|}10} + X a_{\overline{10|}10} v^{10} &\text{ [5] }
100,000 &= 15,000a_{\overline{10|}10}+v^{10}Xa_{\overline{10| }10} &\text{ [6] }
100,000 &= 15,000 a_{\overline{10|}10} + X a_{\overline{10|}10} v^{10} &\text{ [7] }
100,000 &= 15,000 a_{\overline{10|}10} + v^{10} X a_{\overline{10|}10} &\text{ [8] }
100,000 &= 15,000 \, a_{\overline{10|}10} + v^{10} \, X \, a_{\overline{10|}10} &\text{ [9] }
100,000 &= 15,000 a_{\overline{10|}0.10} + v^{10} X a_{\overline{10|}0.10} &\text{ [10] }
100,000 &= 15,000 a_{\overline{10|}0.10} + X a_{\overline{10|}0.10} \nu^{10} &\text{ [11] }

100,000 &= 15,000 \, a_{\overline{10|} \, 0.10} + X \, a_{\overline{10|} \, 0.10} \, \nu^{10} &\text{ [12] }

100,000 &= 15,000 \, a_{\overline{10|} \, 0.10} + X \, a_{\overline{10|} \, 0.10} \, v^{10} &\text{ [13] }
100,000 &= 15,000 \, a_{\overline{10|} \, 0.10} + X \, \left( a_{\overline{10|} \, 0.10} \right) \, \nu^{10} &\text{ [14] }
100,000 &= 15,000 \, a_{\overline{10|} \, 0.10} + X \, \left( a_{\overline{10|} \, 0.10} \right) \, v^{10} &\text{ [15] }
100,000 &= 15,000 \, a_{\lcroof{10} \, 0.10} + \nu^{10} \, X \, a_{\lcroof{10} \, 0.10} &\text{ [16] }
100,000 &= 15,000 \, a_{\bar{10|} \, 0.10} + X \, a_{\bar{10|} \, 0.10} \, v^{10} &\text{ [17] }
100,000 &= 15,000 \, a_{\bar{10|} \, 0.10} \, + \, X \, a_{\bar{10|} \, 0.10} \, \nu^{10} &\text{ [18] }
100,000 &= 15,000 \, a_{\bar{10|}0.10} \, + \, X \, a_{\bar{10|}0.10} \, v^{10} &\text{ [19] }
100,000 &= 15,000 \, a_{\bar{10|}0.10} \, + \, X \, a_{\bar{10|}0.10} \, \nu^{10} &\text{ [20] }
\end{align}

(Fixed in line [2] by putting a space between X and v^10.)
__________________
Favorite Quote(s):

Spoiler:

“Progress isn't made by early risers. It's made by lazy men trying to find easier ways to do something.”

“Don't handicap your children by making their lives easy.”

“There are no dangerous weapons; there are only dangerous men.”

“A desire not to butt into other people's business is at least eighty percent of all human 'wisdom'...and the other twenty percent isn't very important.”

“Belief gets in the way of learning.”

“The first principle of freedom is the right to go to hell in your own handbasket.”

“Does history record any case in which the majority was right?”

#10
11-12-2017, 02:27 PM
 Futon Member SOA Join Date: Jul 2016 Studying for FM Posts: 128

I took a day break from this problem hoping to be able to solve this with a fresh mind.

Year 11: If I pay (1.5)x(.1)x100000=15000, (.5)x(.1)x(100000)=5000 will reduce principle so the remaining proportion is 100000(1-(.5)(.1))=95000

Year 12: 1000000(1-(.5)(.1))(1-(.5)(.1))= 90250

Year 13: 1000000(1-(.5)(.1))(1-(.5)(.1))(1-(.5)(.1))=85737.5

Year 20: 1000000(1-(.5)(.1))^10=59873.69392

59873.69392= R x (a angle 10)

R= 9744

ooooooooooooooooooooooooooooooooooooooh :o

Thanks a lot guys. Much appreciated.

@NattyMo

$100000&=15000a_{\overline{10|}10}+v^{10} Xa_{\overline{10|}10}$

Oh you're right! I wonder why.