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bahrom7893

  • 2 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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  1. bahrom7893
    • 2 years ago
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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?

  2. bahrom7893
    • 2 years ago
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    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.

  3. GT
    • 2 years ago
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    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).

  4. bahrom7893
    • 2 years ago
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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

  5. bahrom7893
    • 2 years ago
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    then subtract, etc, and plug in n = 6

  6. bahrom7893
    • 2 years ago
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    @amistre64 or @TuringTest am I right?

  7. bahrom7893
    • 2 years ago
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    that's for part a) 1)

  8. bahrom7893
    • 2 years ago
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    So for a) 1) I got $13.266 mln and a) 2) $21.1656. Can someone check this pls?

  9. cinar
    • 2 years ago
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    a) 1) 14.43 M

  10. amistre64
    • 2 years ago
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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: A=P(1.04)^t

  11. bahrom7893
    • 2 years ago
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    cinar how did u get that? Amistre my formula was: Qn = 2*(1-1.04^n)/(1-1.04)

  12. bahrom7893
    • 2 years ago
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    I was using geometric series.

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

  14. amistre64
    • 2 years ago
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    the geometric simplifies in the end to that A = P(1.04)^t i believe

  15. amistre64
    • 2 years ago
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    but the question is saying that 2M is added to the account each year right?

  16. cinar
    • 2 years ago
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    I used this formula P(1.04)^t like amistre did

  17. bahrom7893
    • 2 years ago
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    oh ok amistre, and yes it is being added each year.

  18. amistre64
    • 2 years ago
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    then the formula I gave is just for 2M sitting around for t years and nothing being added into it ... gonna have to reconfigure :)

  19. cinar
    • 2 years ago
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    and he will get 6*2 =12 M+2.43=14.43

  20. bahrom7893
    • 2 years ago
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    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)

  21. cinar
    • 2 years ago
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    but maybe you need to calculate each year separatly

  22. bahrom7893
    • 2 years ago
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    now multiply everything by 1.04 1.04Qn = 2*(1.04)+...+2*(1.04)^n

  23. cinar
    • 2 years ago
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    he will get 2.08+2 M at 1994

  24. bahrom7893
    • 2 years ago
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    Subtract 2nd from first: Q(1-1.04)=2 - 2*(1.04)^n Q = 2(1-1.04^n)/(1-1.04)

  25. bahrom7893
    • 2 years ago
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    and then I just plugged in n = 6

  26. cinar
    • 2 years ago
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    4.2432+2 M at 1995

  27. cinar
    • 2 years ago
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    6.493+2M at 1996

  28. cinar
    • 2 years ago
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    at end of 1999 15.96M she has

  29. amistre64
    • 2 years ago
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    \[B_n = B_{n-1}(1.04) + 2m\] \[B_{n-1} = (B_{n-2}(1.04)+ 2m)\] \[B_n = (B_{n-2}(1.04)+ 2m)(1.04) + 2m\] \[B_n = B_{n-2}(1.04)^2+ 2m(1.04) + 2m\] \[B_{n-2}=B_{n-3}(1.04)+2m\] \[B_n = (B_{n-3}(1.04)+2m)(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{n-3}(1.04)^3+2m(1.04)^2+ 2m(1.04) + 2m\] \[B_n = B_{n-r}(1.04)^r+2m((1.04)^{r-1}+ (1.04)^{r-2}+... +(1.04)+ 1) \] 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 \[B_n = 2m(1.04)^n+2m((1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \] \[B_n = 2m\ ((1.04)^n+(1.04)^{n-1}+ (1.04)^{n-2}+...+(1.04)+ 1) \]

  30. bahrom7893
    • 2 years ago
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    well that's pretty much what i used amistre.

  31. bahrom7893
    • 2 years ago
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    ughh i hate it when they assign even problems.

  32. amistre64
    • 2 years ago
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    i know; but its nice to be able to remember how to come to that conclusion in a general sense :)

  33. amistre64
    • 2 years ago
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    and the simplification I believe gets us:\[B_n=2m\frac{1-1.04^n}{1-1.04}\]

  34. bahrom7893
    • 2 years ago
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    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.

  35. bahrom7893
    • 2 years ago
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    i meant n=6 and n=10

  36. bahrom7893
    • 2 years ago
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    What about part two? what is present value? i've never heard of the term before.

  37. amistre64
    • 2 years ago
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    present value is; what is the value of the contract at the present time i think

  38. bahrom7893
    • 2 years ago
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    so it would just be 2mln?

  39. amistre64
    • 2 years ago
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    in other words; how much is left to pay off?

  40. bahrom7893
    • 2 years ago
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    ohh do the end minus 2 mln

  41. bahrom7893
    • 2 years ago
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    *so the final value at n=10 minus 2mln, that has already been paid, i guess?

  42. amistre64
    • 2 years ago
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    end value - piad value = whats left to pay off

  43. bahrom7893
    • 2 years ago
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    awesome, thanks a lot!!!!!

  44. amistre64
    • 2 years ago
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    ill have to google that up to be sure my idea is sound tho :)

  45. bahrom7893
    • 2 years ago
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    uhmm i'm pretty sure u're right.

  46. amistre64
    • 2 years ago
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    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 :)

  47. bahrom7893
    • 2 years ago
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    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?

  48. cinar
    • 2 years ago
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    se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)

  49. cinar
    • 2 years ago
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    she will get 16.05 M

  50. cinar
    • 2 years ago
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    for a) 1)

  51. amistre64
    • 2 years ago
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    16,221,792 or so if we use that formula ...

  52. cinar
    • 2 years ago
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  53. bahrom7893
    • 2 years ago
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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?

  54. bahrom7893
    • 2 years ago
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    we should be plugging in 6 and 10

  55. cinar
    • 2 years ago
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    who said that that formula is correct for this problem..

  56. bahrom7893
    • 2 years ago
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    i did, and amistre confirmed, u're using the formula that got derived from our formula.

  57. bahrom7893
    • 2 years ago
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    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)

  58. cinar
    • 2 years ago
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    no I am not using your formula.. I am just using A=P(1-n/100)^t

  59. cinar
    • 2 years ago
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    it can't 13.2 M as I said she will already earn from company 14M

  60. cinar
    • 2 years ago
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    or maybe I misunderstand question..

  61. bahrom7893
    • 2 years ago
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    in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest

  62. cinar
    • 2 years ago
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    she will get 2 M @ 1993 @ 1994-@5 @6 @7 @8 @9 how many @ sign I used 7 right

  63. bahrom7893
    • 2 years ago
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    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

  64. bahrom7893
    • 2 years ago
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    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

  65. bahrom7893
    • 2 years ago
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    Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n-1)

  66. cinar
    • 2 years ago
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    then use n=7

  67. bahrom7893
    • 2 years ago
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    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)

  68. bahrom7893
    • 2 years ago
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    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.

  69. bahrom7893
    • 2 years ago
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    So: 1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 U're right!!!

  70. bahrom7893
    • 2 years ago
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    OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!

  71. bahrom7893
    • 2 years ago
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    1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 2000 - 8 2001 - 9 2002 - 10

  72. cinar
    • 2 years ago
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    I am happy for you solved it..

  73. GT
    • 2 years ago
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    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

  74. GT
    • 2 years ago
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    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.

  75. dumbcow
    • 2 years ago
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    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 \[PV = 2(1+1.04^{-1}+1.04^{-2} +...1.04^{-9})\] \[PV=\frac{2(1-1.04^{-10})}{1-1.04^{-1}}=16.871\]

  76. bahrom7893
    • 2 years ago
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    ty dumb lol where were u earlier?

  77. dumbcow
    • 2 years ago
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    away from the computer :) haha

  78. bahrom7893
    • 2 years ago
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    haha, i hate calling u dumb, i should start callin u cow lol

  79. dumbcow
    • 2 years ago
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    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

  80. dumbcow
    • 2 years ago
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    thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?

  81. cinar
    • 2 years ago
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    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..

  82. cinar
    • 2 years ago
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    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

  83. cinar
    • 2 years ago
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