bahrom7893
  • bahrom7893
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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chestercat
  • chestercat
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bahrom7893
  • 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
  • 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
  • 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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bahrom7893
  • 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
  • bahrom7893
then subtract, etc, and plug in n = 6
bahrom7893
  • bahrom7893
@amistre64 or @TuringTest am I right?
bahrom7893
  • bahrom7893
that's for part a) 1)
bahrom7893
  • bahrom7893
So for a) 1) I got $13.266 mln and a) 2) $21.1656. Can someone check this pls?
anonymous
  • anonymous
a) 1) 14.43 M
amistre64
  • 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
  • bahrom7893
cinar how did u get that? Amistre my formula was: Qn = 2*(1-1.04^n)/(1-1.04)
bahrom7893
  • bahrom7893
I was using geometric series.
anonymous
  • 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
  • amistre64
the geometric simplifies in the end to that A = P(1.04)^t i believe
amistre64
  • amistre64
but the question is saying that 2M is added to the account each year right?
anonymous
  • anonymous
I used this formula P(1.04)^t like amistre did
bahrom7893
  • bahrom7893
oh ok amistre, and yes it is being added each year.
amistre64
  • 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
  • anonymous
and he will get 6*2 =12 M+2.43=14.43
bahrom7893
  • 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
  • anonymous
but maybe you need to calculate each year separatly
bahrom7893
  • bahrom7893
now multiply everything by 1.04 1.04Qn = 2*(1.04)+...+2*(1.04)^n
anonymous
  • anonymous
he will get 2.08+2 M at 1994
bahrom7893
  • bahrom7893
Subtract 2nd from first: Q(1-1.04)=2 - 2*(1.04)^n Q = 2(1-1.04^n)/(1-1.04)
bahrom7893
  • bahrom7893
and then I just plugged in n = 6
anonymous
  • anonymous
4.2432+2 M at 1995
anonymous
  • anonymous
6.493+2M at 1996
anonymous
  • anonymous
at end of 1999 15.96M she has
amistre64
  • 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
  • bahrom7893
well that's pretty much what i used amistre.
bahrom7893
  • bahrom7893
ughh i hate it when they assign even problems.
amistre64
  • amistre64
i know; but its nice to be able to remember how to come to that conclusion in a general sense :)
amistre64
  • amistre64
and the simplification I believe gets us:\[B_n=2m\frac{1-1.04^n}{1-1.04}\]
bahrom7893
  • 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
  • bahrom7893
i meant n=6 and n=10
bahrom7893
  • bahrom7893
What about part two? what is present value? i've never heard of the term before.
amistre64
  • amistre64
present value is; what is the value of the contract at the present time i think
bahrom7893
  • bahrom7893
so it would just be 2mln?
amistre64
  • amistre64
in other words; how much is left to pay off?
bahrom7893
  • bahrom7893
ohh do the end minus 2 mln
bahrom7893
  • bahrom7893
*so the final value at n=10 minus 2mln, that has already been paid, i guess?
amistre64
  • amistre64
end value - piad value = whats left to pay off
bahrom7893
  • bahrom7893
awesome, thanks a lot!!!!!
amistre64
  • amistre64
ill have to google that up to be sure my idea is sound tho :)
bahrom7893
  • bahrom7893
uhmm i'm pretty sure u're right.
amistre64
  • 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
  • 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
  • anonymous
se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)
anonymous
  • anonymous
she will get 16.05 M
anonymous
  • anonymous
for a) 1)
amistre64
  • amistre64
16,221,792 or so if we use that formula ...
anonymous
  • anonymous
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bahrom7893
  • 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
  • bahrom7893
we should be plugging in 6 and 10
anonymous
  • anonymous
who said that that formula is correct for this problem..
bahrom7893
  • bahrom7893
i did, and amistre confirmed, u're using the formula that got derived from our formula.
bahrom7893
  • 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
  • anonymous
no I am not using your formula.. I am just using A=P(1-n/100)^t
anonymous
  • anonymous
it can't 13.2 M as I said she will already earn from company 14M
anonymous
  • anonymous
or maybe I misunderstand question..
bahrom7893
  • bahrom7893
in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest
anonymous
  • anonymous
she will get 2 M @ 1993 @ 1994-@5 @6 @7 @8 @9 how many @ sign I used 7 right
bahrom7893
  • 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
  • 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
  • bahrom7893
Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n-1)
anonymous
  • anonymous
then use n=7
bahrom7893
  • 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
  • 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
  • bahrom7893
So: 1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 U're right!!!
bahrom7893
  • bahrom7893
OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!
bahrom7893
  • bahrom7893
1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 2000 - 8 2001 - 9 2002 - 10
anonymous
  • anonymous
I am happy for you solved it..
anonymous
  • 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
  • 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
  • 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
  • bahrom7893
ty dumb lol where were u earlier?
dumbcow
  • dumbcow
away from the computer :) haha
bahrom7893
  • bahrom7893
haha, i hate calling u dumb, i should start callin u cow lol
dumbcow
  • 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
  • dumbcow
thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?
anonymous
  • 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
  • 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
  • anonymous
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