Around January 1, 1993, Barbra Streisand signed a contract with Sony Corporation for $2 million a year for 10 years. Suppose the first payment was made on the day of signing and that all other payments are made on the first day of the year. Suppose that all payments are made into a bank account earning 4% a year, compounded annually.

- bahrom7893

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- bahrom7893

a) How much money was in the account
1) On the night of December 31, 1999
2) On the day the last payment was made.
b) What was the present value of the contract on the day it was signed?

- bahrom7893

should be a geom series:
First term - $2mln
nth term - $2*(1.04)^(n-1), i think. I'm not too good with these problems.

- anonymous

You need to account for different cash flows here.
First cash flow:
$2M, $2M, $2M.....up to 1999 from 1993.
Second cash flow:
Compounded interest on $2M for 1 years, $4M for 2 years, $6M for 3 years and so on.
Then, to find present value, you should discount them by 4% (assumed interest value).

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## More answers

- bahrom7893

mehh, i think I gotta do this:
Qn = Q1 + Q1(1.04)+Q1(1.04)^2+...+Qn(1.04)^(n-1)
1.04Qn = Q1(1.04) + Q1(1.04)^2+...+Qn(1.04)^n

- bahrom7893

then subtract, etc, and plug in n = 6

- bahrom7893

@amistre64 or @TuringTest am I right?

- bahrom7893

that's for part a) 1)

- bahrom7893

So for a) 1) I got $13.266 mln and a) 2) $21.1656. Can someone check this pls?

- anonymous

a) 1) 14.43 M

- amistre64

Around January 1, 1993,
Barbra Streisand signed a contract with Sony Corporation for $2 million a year for 10 years.
Suppose the first payment was made on the day of signing
and that all other payments are made on the first day of the year.
Suppose that all payments are made into a bank account
earning 4% a year,
compounded annually: A=P(1.04)^t

- bahrom7893

cinar how did u get that?
Amistre my formula was:
Qn = 2*(1-1.04^n)/(1-1.04)

- bahrom7893

I was using geometric series.

- anonymous

1993-4-5-6-7-1998
he has 2 M money at 1993 put into account for 5 years will get 2.43 profit

- amistre64

the geometric simplifies in the end to that A = P(1.04)^t i believe

- amistre64

but the question is saying that 2M is added to the account each year right?

- anonymous

I used this formula P(1.04)^t like amistre did

- bahrom7893

oh ok amistre, and yes it is being added each year.

- amistre64

then the formula I gave is just for 2M sitting around for t years and nothing being added into it ... gonna have to reconfigure :)

- anonymous

and he will get 6*2 =12 M+2.43=14.43

- bahrom7893

Okay here's what I was doing:
Q1 = 2
Q2 = 2+2*(1.04)
Q3 = 2+2*(1.04)+2*(1.04)^2
Qn = 2+2*(1.04)+2*(1.04)^2+...+2*(1.04)^(n-1)

- anonymous

but maybe you need to calculate each year separatly

- bahrom7893

now multiply everything by 1.04
1.04Qn = 2*(1.04)+...+2*(1.04)^n

- anonymous

he will get 2.08+2 M at 1994

- bahrom7893

Subtract 2nd from first:
Q(1-1.04)=2 - 2*(1.04)^n
Q = 2(1-1.04^n)/(1-1.04)

- bahrom7893

and then I just plugged in n = 6

- anonymous

4.2432+2 M at 1995

- anonymous

6.493+2M at 1996

- anonymous

at end of 1999 15.96M she has

- amistre64

\[B_n = B_{n-1}(1.04) + 2m\]
\[B_{n-1} = (B_{n-2}(1.04)+ 2m)\]
\[B_n = (B_{n-2}(1.04)+ 2m)(1.04) + 2m\]
\[B_n = B_{n-2}(1.04)^2+ 2m(1.04) + 2m\]
\[B_{n-2}=B_{n-3}(1.04)+2m\]
\[B_n = (B_{n-3}(1.04)+2m)(1.04)^2+ 2m(1.04) + 2m\]
\[B_n = B_{n-3}(1.04)^3+2m(1.04)^2+ 2m(1.04) + 2m\]
\[B_n = B_{n-r}(1.04)^r+2m((1.04)^{r-1}+ (1.04)^{r-2}+... +(1.04)+ 1) \]
when r=n we get B0
\[B_n = B_{n-n}(1.04)^n+2m((1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]
\[B_n = B_{0}(1.04)^n+2m((1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]
B0 = 2m
\[B_n = 2m(1.04)^n+2m((1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]
\[B_n = 2m\ ((1.04)^n+(1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]

- bahrom7893

well that's pretty much what i used amistre.

- bahrom7893

ughh i hate it when they assign even problems.

- amistre64

i know; but its nice to be able to remember how to come to that conclusion in a general sense :)

- amistre64

and the simplification I believe gets us:\[B_n=2m\frac{1-1.04^n}{1-1.04}\]

- bahrom7893

lol kk awesome, then i was right. Now all we gotta do is plug in x = 6 for 1999 and x=10 for the last payment.

- bahrom7893

i meant n=6 and n=10

- bahrom7893

What about part two? what is present value? i've never heard of the term before.

- amistre64

present value is; what is the value of the contract at the present time i think

- bahrom7893

so it would just be 2mln?

- amistre64

in other words; how much is left to pay off?

- bahrom7893

ohh do the end minus 2 mln

- bahrom7893

*so the final value at n=10 minus 2mln, that has already been paid, i guess?

- amistre64

end value - piad value = whats left to pay off

- bahrom7893

awesome, thanks a lot!!!!!

- amistre64

ill have to google that up to be sure my idea is sound tho :)

- bahrom7893

uhmm i'm pretty sure u're right.

- amistre64

http://www.frickcpa.com/tvom/TVOM_PV_Annuity.asp
try out my idea and then use this to see how close or how far off we are :)

- bahrom7893

hmm i got 19mlns using ur idea, and 16mlns using that formula
http://www.wolframalpha.com/input/?i=2*%28%281-%281%2B0.04%29^%28-10%29%29%2F0.04%29
could my previous answers be wrong?

- anonymous

se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)

- anonymous

she will get 16.05 M

- anonymous

for a) 1)

- amistre64

16,221,792 or so if we use that formula ...

- anonymous

##### 1 Attachment

- bahrom7893

Ughh Cinar u gotta use this formula:
Qn = 2*(1-1.04^n)/(1-1.04)
I'm just wondering, for 6 years, do we plug in n=6, or 5?
same for ten years, do we plug in x = 10, or 9?

- bahrom7893

we should be plugging in 6 and 10

- anonymous

who said that that formula is correct for this problem..

- bahrom7893

i did, and amistre confirmed, u're using the formula that got derived from our formula.

- bahrom7893

Which is the accounting approach, or whatever. This is a geometric series:
Qn = 2 + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3+...+2*(1.04)^(n-1)

- anonymous

no I am not using your formula..
I am just using A=P(1-n/100)^t

- anonymous

it can't 13.2 M as I said she will already earn from company 14M

- anonymous

or maybe I misunderstand question..

- bahrom7893

in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest

- anonymous

she will get 2 M @ 1993 @ 1994-@5 @6 @7 @8 @9 how many @ sign I used 7 right

- bahrom7893

Now with interest:
Year 1: 2mln
Year 2: 2*(1.04) + 2 (the interest on the existing plus 2 more million)
Year 3: (2+2*(1.04))*(1.04) + 2 (the interest on existing amount plus 2 more million)
Year 3 simplifies to: 2*(1.04)+2*(1.04)^2+2
Year 4: Year3 * 1.04 + 2, which becomes:
Year 4: (2*(1.04)+2*(1.04)^2+2)*1.04) + 2

- bahrom7893

See the pattern:
Yr 1: 2mln + 0*(1.04)
Yr 2: 2mln + 2*(1.04)
Yr 3: 2mln + 2*(1.04) + 2*(1.04)^2
Yr 4: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3

- bahrom7893

Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n-1)

- anonymous

then use n=7

- bahrom7893

no, now multiply everything by 1.04
1.04 * Yrn = 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + 2*(1.04)^4 + ... + 2*(1.04)^(n)
Subtract this from the previous equation:
Yrn-1.04Yrn = 2 - 2*(1.04)^n (all terms in between will cancel)

- bahrom7893

Yrn(1-1.04) = 2(1-1.04^n)
Yrn = 2(1-1.04^n)/(1-1.04), that's the amount she has in her bank in year n.

- bahrom7893

So: 1993 - yr1
1994 - yr2
1995 - 3
1996 - 4
1997 - 5
1998 - 6
1999 - 7
U're right!!!

- bahrom7893

OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!

- bahrom7893

1993 - yr1
1994 - yr2
1995 - 3
1996 - 4
1997 - 5
1998 - 6
1999 - 7
2000 - 8
2001 - 9
2002 - 10

- anonymous

I am happy for you solved it..

- anonymous

Present Value is the discounted value of the cash flows. It will be:
$2 + $2/1.04 + $2/(1.04)^2 + ...... + ...... + $2/(1.04)^n-1

- anonymous

If someone says they will pay you $2 million a year from now, and you can earn an interest of 4% on that money, you would have them pay 2/1.04 today. It is the same as $2 million a year from now at 4% interest.

- dumbcow

just to put it all together: units are in millions
Part1 , since it wants the balance at end of year 1999, then it includes 7 payments plus the interest on last payment
\[FV = 2(1.04 +1.04^{2}+...1.04^{7})\]
using formula for sum of geometric sequence
\[FV = \frac{2(1.04)(1.04^{7}-1)}{1.04 -1} = 16.428\]
Part 2: balance on jan 1, 2002, includes 10 payments with no interest on last payment
\[FV = 2(1+1.04 +1.04^{2}+...1.04^{9})\]
\[FV = \frac{2(1.04^{10}-1)}{1.04-1} = 24.012\]
Part 3: present value asks what is the present value of 24.012 M future dollars
use similar geometric sequence except the common ratio is 1/1.04 =1.04^-1= .9615
\[PV = 2(1+1.04^{-1}+1.04^{-2} +...1.04^{-9})\]
\[PV=\frac{2(1-1.04^{-10})}{1-1.04^{-1}}=16.871\]

- bahrom7893

ty dumb lol where were u earlier?

- dumbcow

away from the computer :)
haha

- bahrom7893

haha, i hate calling u dumb, i should start callin u cow lol

- dumbcow

just a side note on present value
if you take that 16.871 and multiply it by 1.04^9 you get the future value of 24.012
so 10 years of 2M payments at 4% is worth 16.871M cash right now

- dumbcow

thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?

- anonymous

ok.. you are sayin that
\[Q_n=2*\left( \frac{1-x^n}{1-x}\right)\]
I put here x for 1.04
have you tried to find first year?
n=1 right
you will get 2M
But I am saying that she has 2M at 1st Jan 1993 and 2.08M at 31 Dec 1993
and she has 4.08M at 1st Jan 1994 and 4.2443 at 31 Dec 1994
and so on..
therefore we know that for n=1
there are two different value
first one for 1st Jan second one for 31st Dec.
so you need to consider two point in a year..

- anonymous

your formula only works to find value of 1st of year.. (1st Jan each year)
for example for Dec 31 1999 you will get ( you will put n=7)
15.797M
but in fact you found the money which is on the 1st Jan 1999
as I said before money which she got on 31st Dec 1999 is 16.5M

- anonymous

##### 1 Attachment

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