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bahrom7893 Group Title

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.

  • 2 years ago
  • 2 years ago

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  1. bahrom7893 Group Title
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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 years ago
  2. bahrom7893 Group Title
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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.

    • 2 years ago
  3. GT Group Title
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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).

    • 2 years ago
  4. bahrom7893 Group Title
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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

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

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

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

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

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

    • 2 years ago
  10. amistre64 Group Title
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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

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

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

    • 2 years ago
  13. cinar Group Title
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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

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

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

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

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

    • 2 years ago
  18. amistre64 Group Title
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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 :)

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

    • 2 years ago
  20. bahrom7893 Group Title
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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)

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

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

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

    • 2 years ago
  24. bahrom7893 Group Title
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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)

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

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

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

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

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

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

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

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

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

    • 2 years ago
  34. bahrom7893 Group Title
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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.

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

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

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

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

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

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

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

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

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

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

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

    • 2 years ago
  46. amistre64 Group Title
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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 :)

    • 2 years ago
  47. bahrom7893 Group Title
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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?

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

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

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

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

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

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

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

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

    • 2 years ago
  57. bahrom7893 Group Title
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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)

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

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

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

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

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

    • 2 years ago
  63. bahrom7893 Group Title
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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

    • 2 years ago
  64. bahrom7893 Group Title
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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

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

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

    • 2 years ago
  67. bahrom7893 Group Title
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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)

    • 2 years ago
  68. bahrom7893 Group Title
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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.

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

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

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

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

    • 2 years ago
  73. GT Group Title
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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

    • 2 years ago
  74. GT Group Title
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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.

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

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

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

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

    • 2 years ago
  79. dumbcow Group Title
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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

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

    • 2 years ago
  81. cinar Group Title
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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..

    • 2 years ago
  82. cinar Group Title
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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

    • 2 years ago
  83. cinar Group Title
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    • 2 years ago
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