## Very difficult question (need help)

Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance, Engineering Mathematics ...

### Re: Very difficult question (need help)

Erimess wrote:
Denis wrote:
Erimess wrote:I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that.

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`          j = 10%       i = 12%Year      Deposit      Interest       Balance0               .00          .00           .001           1000.00          .00       1000.002           1100.00       120.00       2220.003           1210.00       266.40       3696.40`

Li'l "picture", \$1000 deposit increasing by 10% annually, rate 12% annually.

F = Future value (?)
D = initial deposit (1000)
j = deposit increase (.10)
i = interest rate (.12)
n = number of years (3)

F = D*[(1 + i)^n - (1 + j)^n] / (i - j)

F = 1000(1.12^3 - 1.10^3) / (.12 - .10) = 3696.40

Basically all that simple...google will give ya lotsa sites, like:
http://www.financeformulas.net/Growing- ... Value.html

This isn't the same thing. I could've done that in Excel easily enough. (Though it never hurts to add an equation to the repertoire.)

This is a present value relative to the withdrawals, payments growing at 5% annually, but earning (we assume) 10% semi-annually. That equation is growing to a future value, not shrinking from a present value.

YES, it's the same thing...though hard to see, I'll admit.
I used it that way because (being lazy) it was easier to "do the typing"!

Take the ending balance of 3696.40 and get its Present Value: 3696.40 / 1.12^3 = ~2631.02.
So this means that if a payment of \$1000 increasing by 10% yearly is made for 3 years,
then the amount that can be "borrowed" is \$2631.02
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`           j = 10%       i = 12%Year       Payment      Interest       Balance0               .00          .00       2631.021          -1000.00       315.72       1946.742          -1100.00       233.61       1080.353          -1210.00       129.65           .00`

In the actual problem, the balance required in the account after 35 years, to accomodate a withdrawal
of \$132,691.91 yearly increasing by 5% yearly would basically be obtained the same way.

So you don't scold me again(!), here's the direct formula:
P = Payment (?)
A = Amount borrowed (2631.02)
n = number of periods (3)
i = interest rate (.12)
j = payment increase (.10)

P = A(i - j) / [1 - (1 + j)^n / (1 + i)^n]

P = 2631.02(.12 - .10) / (1 - 1.10^3 / 1.12^3) = 1000
I'm just an imagination of your figment...
Denis
Super User

Posts: 2921
Joined: Mon Oct 31, 2011 7:29 pm

### Re: Very difficult question (need help)

More on "the formula"!
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`       Payments accumulation      Loan, no payments0                         .00               2631.021  1000.00      .00   1000.00     315.72    2946.742  1100.00   120.00   2220.00     353.61    3300.353  1210.00   266.40   3696.40     396.05    3696.40`

As shown, we can look at the payments as accumulating in a separate account,
the loan in a separate account; both accounts end up the same, of course.

The "payments account" is looked at this way:
1000(1.12)^2 = 1254.40
1100(1.12)^1 = 1232.00
1210(1.12)^0 = 1210.00
====================
total.......... = 3696.40

So the requirement is made up of a SUM; this is shown as such in Mark's formula.
The formula I've shown produces same results, in a direct way.
I'm just an imagination of your figment...
Denis
Super User

Posts: 2921
Joined: Mon Oct 31, 2011 7:29 pm

### Re: Very difficult question (need help)

Code: Select all
`#      DATE         DEP/-WD      INTEREST         BALANCE1     Jan.1/1        300.00           .00          300.002     Feb.1/1        300.00          2.45          602.453     Mar.1/1        300.00          4.92          907.37....419  Nov.1/35        300.00      8,680.68    1,072,157.77420  Dec.1/35        300.00      8,754.00    1,081,211.771    Jan.1/36   -133,978.13      8,827.93      956,061.572    Jan.1/37   -140,677.04     97,996.31      913,380.84....9    Jan.1/44   -197,946.72     36,694.91      196,747.7110   Jan.1/45   -207,844.05     20,166.63        9,070.29KB   Jan.1/46                      929.71       10,000.00 : KB = Kicks Bucket!`

Ok, since we're in agreement that 1st deposit is at beginning, plus
there is a 1 year period between last withdrawal and \$10,000 donation,
the 1st withdrawal will be \$133,978.13.
Quite close to Mark's \$133,930.49, where \$10,000 donation is not 1 year later.
I'm just an imagination of your figment...
Denis
Super User

Posts: 2921
Joined: Mon Oct 31, 2011 7:29 pm

### Re: Very difficult question (need help)

Denis wrote:This was the rate per the problem: "in an account that earns an average of 10% compounded semi-annually".
10% cpd semi-annually means 10.25% annually ; 1.05^2 - 1 = .1025

Yeah. But that would also mean 5% at the semi-annual mark, which would turn into 10.25% annually once compounded. But I first figured out what all the \$300 payments would have earned at the semi-annual point, which would be at 5%, or .8333% per month. It doesn't compound til you tack the interest on. So the second half of the year, you're earning 5% on the original plus the interest, equating to 10.25% as a yield. But it's still 5% per semi-annual period. Way I learned it - way I've always done it.

Oh, actually, you know what. We're both right. .8333% per month is still correct, as long as you treat it like simple interest within the six month period, which is actually what I learned and should have done. (i.e. 300*.8333*6 + 300*.8333*5, etc) I just looked back at my work and realized I compounded the .8333% per month! I didn't mean to do that. (Probably sounds like the hard way, but most of my classes were using charts, not equations.)
So this is how liberty dies... with thunderous applause.

Erimess