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bahrom7893
 4 years ago
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
 4 years ago
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.

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bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0a) 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
 4 years ago
Best ResponseYou've already chosen the best response.0should be a geom series: First term  $2mln nth term  $2*(1.04)^(n1), i think. I'm not too good with these problems.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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).

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0mehh, i think I gotta do this: Qn = Q1 + Q1(1.04)+Q1(1.04)^2+...+Qn(1.04)^(n1) 1.04Qn = Q1(1.04) + Q1(1.04)^2+...+Qn(1.04)^n

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0then subtract, etc, and plug in n = 6

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 or @TuringTest am I right?

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0that's for part a) 1)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0So for a) 1) I got $13.266 mln and a) 2) $21.1656. Can someone check this pls?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1Around 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
 4 years ago
Best ResponseYou've already chosen the best response.0cinar how did u get that? Amistre my formula was: Qn = 2*(11.04^n)/(11.04)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0I was using geometric series.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0199345671998 he has 2 M money at 1993 put into account for 5 years will get 2.43 profit

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the geometric simplifies in the end to that A = P(1.04)^t i believe

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1but the question is saying that 2M is added to the account each year right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I used this formula P(1.04)^t like amistre did

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok amistre, and yes it is being added each year.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1then the formula I gave is just for 2M sitting around for t years and nothing being added into it ... gonna have to reconfigure :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and he will get 6*2 =12 M+2.43=14.43

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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)^(n1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but maybe you need to calculate each year separatly

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0now multiply everything by 1.04 1.04Qn = 2*(1.04)+...+2*(1.04)^n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0he will get 2.08+2 M at 1994

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Subtract 2nd from first: Q(11.04)=2  2*(1.04)^n Q = 2(11.04^n)/(11.04)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0and then I just plugged in n = 6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0at end of 1999 15.96M she has

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[B_n = B_{n1}(1.04) + 2m\] \[B_{n1} = (B_{n2}(1.04)+ 2m)\] \[B_n = (B_{n2}(1.04)+ 2m)(1.04) + 2m\] \[B_n = B_{n2}(1.04)^2+ 2m(1.04) + 2m\] \[B_{n2}=B_{n3}(1.04)+2m\] \[B_n = (B_{n3}(1.04)+2m)(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{n3}(1.04)^3+2m(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{nr}(1.04)^r+2m((1.04)^{r1}+ (1.04)^{r2}+... +(1.04)+ 1) \] when r=n we get B0 \[B_n = B_{nn}(1.04)^n+2m((1.04)^{n1}+ (1.04)^{n2}+...+(1.04)+ 1) \] \[B_n = B_{0}(1.04)^n+2m((1.04)^{n1}+ (1.04)^{n2}+...+(1.04)+ 1) \] B0 = 2m \[B_n = 2m(1.04)^n+2m((1.04)^{n1}+ (1.04)^{n2}+...+(1.04)+ 1) \] \[B_n = 2m\ ((1.04)^n+(1.04)^{n1}+ (1.04)^{n2}+...+(1.04)+ 1) \]

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0well that's pretty much what i used amistre.

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0ughh i hate it when they assign even problems.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i know; but its nice to be able to remember how to come to that conclusion in a general sense :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1and the simplification I believe gets us:\[B_n=2m\frac{11.04^n}{11.04}\]

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0lol 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
 4 years ago
Best ResponseYou've already chosen the best response.0i meant n=6 and n=10

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0What about part two? what is present value? i've never heard of the term before.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1present value is; what is the value of the contract at the present time i think

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0so it would just be 2mln?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1in other words; how much is left to pay off?

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0ohh do the end minus 2 mln

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0*so the final value at n=10 minus 2mln, that has already been paid, i guess?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1end value  piad value = whats left to pay off

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0awesome, thanks a lot!!!!!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ill have to google that up to be sure my idea is sound tho :)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0uhmm i'm pretty sure u're right.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://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
 4 years ago
Best ResponseYou've already chosen the best response.0hmm i got 19mlns using ur idea, and 16mlns using that formula http://www.wolframalpha.com/input/?i=2*%28%281%281%2B0.04%29^%2810%29%29%2F0.04%29 could my previous answers be wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.116,221,792 or so if we use that formula ...

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Ughh Cinar u gotta use this formula: Qn = 2*(11.04^n)/(11.04) 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
 4 years ago
Best ResponseYou've already chosen the best response.0we should be plugging in 6 and 10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0who said that that formula is correct for this problem..

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0i did, and amistre confirmed, u're using the formula that got derived from our formula.

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Which 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)^(n1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no I am not using your formula.. I am just using A=P(1n/100)^t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it can't 13.2 M as I said she will already earn from company 14M

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or maybe I misunderstand question..

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0she will get 2 M @ 1993 @ 1994@5 @6 @7 @8 @9 how many @ sign I used 7 right

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Now 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
 4 years ago
Best ResponseYou've already chosen the best response.0See 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n1)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0no, 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: Yrn1.04Yrn = 2  2*(1.04)^n (all terms in between will cancel)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Yrn(11.04) = 2(11.04^n) Yrn = 2(11.04^n)/(11.04), that's the amount she has in her bank in year n.

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0So: 1993  yr1 1994  yr2 1995  3 1996  4 1997  5 1998  6 1999  7 U're right!!!

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.01993  yr1 1994  yr2 1995  3 1996  4 1997  5 1998  6 1999  7 2000  8 2001  9 2002  10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am happy for you solved it..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Present Value is the discounted value of the cash flows. It will be: $2 + $2/1.04 + $2/(1.04)^2 + ...... + ...... + $2/(1.04)^n1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just 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 \[FV = 2(1+1.04 +1.04^{2}+...1.04^{9})\] \[FV = \frac{2(1.04^{10}1)}{1.041} = 24.012\] 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 \[PV = 2(1+1.04^{1}+1.04^{2} +...1.04^{9})\] \[PV=\frac{2(11.04^{10})}{11.04^{1}}=16.871\]

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0ty dumb lol where were u earlier?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0away from the computer :) haha

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0haha, i hate calling u dumb, i should start callin u cow lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok.. you are sayin that \[Q_n=2*\left( \frac{1x^n}{1x}\right)\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0your 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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