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  #11  
Old 05-14-2019, 08:19 PM
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There's a quicker way to get the answer. Calculate the level payment that would have to be made at the end of every second year.

0 FV
-5000 PV
10 N
18.147 I/Y

CPT PMT


Set this equal to X(1+i) + 2X.
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Old 05-14-2019, 08:41 PM
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Originally Posted by Academic Actuary View Post
There's a quicker way to get the answer. Calculate the level payment that would have to be made at the end of every second year.

0 FV
-5000 PV
10 N
18.147 I/Y

CPT PMT


Set this equal to X(1+i) + 2X.
Thanks. That's a much simpler way.

But what I don't get it in this method why the payment of 2009 (X) is moved to 2008 (PMT is assumed on even years) since it started in 2007.

So, 2007 to 2008 is only 1 year but payment period is 2 years. So, shouldn't the first level payment be on 2009? If so, 2X payment earns (1+i) from 2008 to 2009 and then X is added on 2009,

giving it 2X(1+i) + X = PMT

Why is that not?
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Old 05-14-2019, 10:21 PM
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I messed up. I thought they went X, 2X,..... and not 2X,X.....
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Old 05-14-2019, 10:26 PM
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Originally Posted by Academic Actuary View Post
There's a quicker way to get the answer. Calculate the level payment that would have to be made at the end of every second year.

0 FV
-5000 PV
10 N
18.147 I/Y

CPT PMT


Set this equal to X(1+i) + 2X.
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  #15  
Old 05-15-2019, 05:46 AM
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I messed up. I thought they went X, 2X,..... and not 2X,X.....
Thanks! It works. I get the correct answer 10,571 using 2X(1+i) + X = PMT = 1,118.39
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  #16  
Old 05-16-2019, 08:19 PM
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Q: A fund earning 8% effective is being accumulated with payments of 500 at the beginning of each year for 20 years. Find the maximum number of withdrawals of 1,000 which can be made at the end of each year under the
condition that once withdrawals start they must continue through the end of the 20-year period.


I solved the problem by letting the fund accumulate with payments of 500 for 20 years and then withdrawing from beginning of 20th period/end of 19th period. It gave me answer of 14 which is what's on the book.

My question is what does they must continue through the end of the 20-year period means?

Does it mean the withdrawal has to start before the end of 20th period or the withdrawal has to start in such time so as the 14 withdrawals of 1000 must finish within the end of 20th period (hence starting at the beginning of 7th period max)?

Or just withdrawals are allowed to go over 20th period and hence up to the end of 33rd period in my case?
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Old 05-16-2019, 09:47 PM
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Does it mean the withdrawal has to start in such time so as the 14 withdrawals of 1000 must finish within the end of 20th period (hence starting at the beginning of 7th period max)?
This
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Old 05-17-2019, 10:25 AM
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This
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Old 05-20-2019, 10:54 AM
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Given that δ(t) = 2/(10+t); t ≥ 0; find a4.

My approach: 1+i = e^δ(t)

So, a4= v + v^2 + v^3 + v^4

= 1/e^(2/11) + 1/[e^(2/11*2/12)] + 1/[e^(2/11*2/12*2/13)] + 1/[e^(2/11*2/12*2/13*2/14)]

But answer on the book : a4 = summation of(k=1 to 4) [10/(10+ k)]^2

Where did I go wrong?
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Old 05-20-2019, 11:11 AM
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Quote:
Originally Posted by jubair07 View Post
Given that δ(t) = 2/(10+t); t ≥ 0; find a4.

My approach: 1+i = e^δ(t)

So, a4= v + v^2 + v^3 + v^4

= 1/e^(2/11) + 1/[e^(2/11*2/12)] + 1/[e^(2/11*2/12*2/13)] + 1/[e^(2/11*2/12*2/13*2/14)]

But answer on the book : a4 = summation of(k=1 to 4) [10/(10+ k)]^2

Where did I go wrong?
1+i = e^δ(t) Is an expression valid only for the value of 1+i at the the moment of time t. Since delta and hence i are changing continuously, it is not the case that the discount factor for year 1 is equal to e^(2/11). You have to solve for the discount factor by integrating. Similarly for later periods.
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