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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.

Mathematics
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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?
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.
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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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
then subtract, etc, and plug in n = 6
@amistre64 or @TuringTest am I right?
that's for part a) 1)
So for a) 1) I got $13.266 mln and a) 2) $21.1656. Can someone check this pls?
a) 1) 14.43 M
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
cinar how did u get that? Amistre my formula was: Qn = 2*(1-1.04^n)/(1-1.04)
I was using geometric series.
1993-4-5-6-7-1998 he has 2 M money at 1993 put into account for 5 years will get 2.43 profit
the geometric simplifies in the end to that A = P(1.04)^t i believe
but the question is saying that 2M is added to the account each year right?
I used this formula P(1.04)^t like amistre did
oh ok amistre, and yes it is being added each year.
then the formula I gave is just for 2M sitting around for t years and nothing being added into it ... gonna have to reconfigure :)
and he will get 6*2 =12 M+2.43=14.43
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)
but maybe you need to calculate each year separatly
now multiply everything by 1.04 1.04Qn = 2*(1.04)+...+2*(1.04)^n
he will get 2.08+2 M at 1994
Subtract 2nd from first: Q(1-1.04)=2 - 2*(1.04)^n Q = 2(1-1.04^n)/(1-1.04)
and then I just plugged in n = 6
4.2432+2 M at 1995
6.493+2M at 1996
at end of 1999 15.96M she has
\[B_n = B_{n-1}(1.04) + 2m\] \[B_{n-1} = (B_{n-2}(1.04)+ 2m)\] \[B_n = (B_{n-2}(1.04)+ 2m)(1.04) + 2m\] \[B_n = B_{n-2}(1.04)^2+ 2m(1.04) + 2m\] \[B_{n-2}=B_{n-3}(1.04)+2m\] \[B_n = (B_{n-3}(1.04)+2m)(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{n-3}(1.04)^3+2m(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{n-r}(1.04)^r+2m((1.04)^{r-1}+ (1.04)^{r-2}+... +(1.04)+ 1) \] 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 \[B_n = 2m(1.04)^n+2m((1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \] \[B_n = 2m\ ((1.04)^n+(1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]
well that's pretty much what i used amistre.
ughh i hate it when they assign even problems.
i know; but its nice to be able to remember how to come to that conclusion in a general sense :)
and the simplification I believe gets us:\[B_n=2m\frac{1-1.04^n}{1-1.04}\]
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.
i meant n=6 and n=10
What about part two? what is present value? i've never heard of the term before.
present value is; what is the value of the contract at the present time i think
so it would just be 2mln?
in other words; how much is left to pay off?
ohh do the end minus 2 mln
*so the final value at n=10 minus 2mln, that has already been paid, i guess?
end value - piad value = whats left to pay off
awesome, thanks a lot!!!!!
ill have to google that up to be sure my idea is sound tho :)
uhmm i'm pretty sure u're right.
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 :)
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?
se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)
she will get 16.05 M
for a) 1)
16,221,792 or so if we use that formula ...
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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?
we should be plugging in 6 and 10
who said that that formula is correct for this problem..
i did, and amistre confirmed, u're using the formula that got derived from our formula.
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)
no I am not using your formula.. I am just using A=P(1-n/100)^t
it can't 13.2 M as I said she will already earn from company 14M
or maybe I misunderstand question..
in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest
she will get 2 M @ 1993 @ 1994-@5 @6 @7 @8 @9 how many @ sign I used 7 right
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
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
Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n-1)
then use n=7
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)
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.
So: 1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 U're right!!!
OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!
1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 2000 - 8 2001 - 9 2002 - 10
I am happy for you solved it..
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
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.
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 \[PV = 2(1+1.04^{-1}+1.04^{-2} +...1.04^{-9})\] \[PV=\frac{2(1-1.04^{-10})}{1-1.04^{-1}}=16.871\]
ty dumb lol where were u earlier?
away from the computer :) haha
haha, i hate calling u dumb, i should start callin u cow lol
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
thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?
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..
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
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