Replacing Equivalent Payments with a Single Payment

Angel198

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Aug 5, 2019
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Hey, so I know how to find a single payment but only with 2 payments. There seems to be 3 to calculate here and I'm not sure what n should be for PV3.
Here is the question:
What single payment in 3 years would replace payments of...
A payment due 2 years ago of $2000
A payment due today of $3000
A payment of $5000 in 7 years
The interest rate is 3% compounded semi-annually

The equation is FV1+FV2+PV3=

I've been at it for hours and I know it will be obvious but I've tried every alternative. Any help is greatly appreciated. Thanks!!
 
"I've been at it for hours..." - Never do that.

I'll give you one for free.

i = 0.03 compounded semi-annually
j = 0.03 / 2 = 0.015 = Semi-annual interest rate.
r = 1+j = 1.015 = One, semi-monthly accumulation factor.
s = 1/r = 0.985221675 = One, semi-monthly discount factor.

We're going 3 years into the future.

"A payment due 2 years ago of $2000"

2000 * r^(2*2+3*2)

"A payment due today of $3000"

3000 * r^(3*2)

"A payment of $5000 in 7 years"

5000 * s^(7*2-3*2)

Once you have all the pieces, just build it!

Also, this is not a unique way to go about it. Build it any accurate way that makes sense to you. Most importantly, get the time periods and the interest compounding to match up.
 
I am getting $10400.61.... that appears to be wrong! I have tried every possible combination. How do I get n for the pv of 5000?
 
I am getting $10400.61.... that appears to be wrong! I have tried every possible combination. How do I get n for the pv of 5000?
"tried every possible combination" - Don't do that. Just try the right combination.

Please show intermediate values:

2000 * r^(4+6) = 2000 * r^10 = ??

3000 * r^6 = 3000 * r^6 = ??

5000 * s^(14-6) = 5000 * s^8 = ??
 
2000(1+.015)^10=2321.08165
3000(1+.015)^6=3280.329792
5000(1+.015)^-8=5632.462933
=10039.96706

I just submitted it and it actually accepted it. Thank you so so so so much.
 
I used FV+FV+PV and got it but I do usually try to think of it in PV. I came out with 10039.96706, which the app accepted (online course).
 
I get the concept and am now stuck on equivalent payment streams. It's not urgent, I can probably figure it out, but don't have the time.
Here it is:
A payment 2 years ago of 6000 and a payment due in 3 years of 4000
is being replaced by
1000 due today, 3000 in 30 months, an unknown payment in 5 years.
Interest is 4% quarterly

I know the equation is FV1+FV2=FV3+FV4+x
The only thing I am lost at is whether the first payment stream should be *5 or 2?
6000(1+.01)^ ....(10?)
4000(1+.01)^......(8?)

1000(1+.01)^.....(10?)
3000(1+.01)^
The visual timeline doesn't seem to help when comparing
 
I like to do these all in present value. Somehow it is easier for me to think about things relative to today. And we need to convert everything to quarters because that is our compounding period.

I am guessing that what is intended here is 4% per annum compounded quarterly. So the quarterly rate is 4% / 4 = 1%.

6000 due 2 years ago, so minus 8 quarters.
4000 due 3 years from now, so plus 12 quarters.
1000 due today, so 0 quarters.
3000 due 30 months from now, so 2.5 years, or plus 10 quarters.
x due 5 years from now so plus 20 quarters.

Notice that if you treat the present as zero, past periods are negative, and future periods are positive. This keeps me from getting my brains scrambled.

You have now done the preliminary work required before setting up the equation.

[MATH]\therefore \dfrac{6000}{1.01^{-8}} + \dfrac{4000}{1.01^{12}} = \dfrac{1000}{1.01^0} + \dfrac{3000}{1.01^{10}} + \dfrac{x}{1.01^{20}}.[/MATH]
Now you can proceed in a number of ways, but I hate fractions so I would clear fractions first.

[MATH]1.01^{20} * \left ( \dfrac{6000}{1.01^{-8}} + \dfrac{4000}{1.01^{12}} \right ) = 1.01^{20} * \left ( \dfrac{1000}{1.01^0} + \dfrac{3000}{1.01^{10}} + \dfrac{x}{1.01^{20}} \right ) \implies .[/MATH]
[MATH]6000(1.01)^{\{20-(-8)\}} + 4000(1.01)^{(20-12)} = 1000(1.01)^{(20-0)} + 3000(1.01)^{(20 -10)} + x(1.01)^{(20-20)} \implies [/MATH]
[MATH]6000(1.01)^{28} + 4000(1.01)^8 = 1000(1.01)^{20} + 3000(1.01)^{10} + x(1.01)^0.[/MATH]
Which gets you to the equation that you were looking for with no muss no fuss. Furthermore it simplifies nicely to

[MATH]6000(1.01)^{28} + 4000(1.01)^8 = 1000(1.01)^{20} + 3000(1.01)^{10} + x \implies[/MATH]
[MATH]x = 1000\{6(1.01)^{28} + 4(1.01)^8 - (1.01)^{20} - 3(1.01)^{10}\}.[/MATH]
Now it is just arithmetic.

If you wanted to define x in terms of thousands of dollars, you could eliminate all those thousands from the get go.

EDIT: My original equation was done in terms of present value, but when I cleared fractions, that transformed it into a future value equation. Probably you were taught in terms of future value to avoid fractions, but, as I said, it gets tricky thinking that way when some numbers are in the past relative to the present. But obviously the present value and future value equations are just different ways to get to the same result. You just need to insure that you are consistent with your temporal reference point.
 
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I wouldn't get too hung up on "Present Value" or "Future Value". Those are important, but you really might want to concentrate on WHEN the payment exists. You can let the calculations worry about the direction.

In my first post, I defined r (accumulation factor) and s (discount factor). That really was unnecessary.

Accumulate Two Periods
r^2 = s^(-2)

Discount Two Periods
r^(-2) = s^2

They are mostly the same thing.
 
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