## 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 • This Question is Closed 1. bahrom7893 Group Title 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 Group Title 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 Group Title 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 Group Title 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 Group Title then subtract, etc, and plug in n = 6 6. bahrom7893 Group Title @amistre64 or @TuringTest am I right? 7. bahrom7893 Group Title that's for part a) 1) 8. bahrom7893 Group Title So for a) 1) I got$13.266 mln and a) 2) $21.1656. Can someone check this pls? 9. cinar Group Title a) 1) 14.43 M 10. amistre64 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: A=P(1.04)^t

11. bahrom7893 Group Title

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

12. bahrom7893 Group Title

I was using geometric series.

13. cinar Group Title

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 Group Title

the geometric simplifies in the end to that A = P(1.04)^t i believe

15. amistre64 Group Title

but the question is saying that 2M is added to the account each year right?

16. cinar Group Title

I used this formula P(1.04)^t like amistre did

17. bahrom7893 Group Title

oh ok amistre, and yes it is being added each year.

18. amistre64 Group Title

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 Group Title

and he will get 6*2 =12 M+2.43=14.43

20. bahrom7893 Group Title

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 Group Title

but maybe you need to calculate each year separatly

22. bahrom7893 Group Title

now multiply everything by 1.04 1.04Qn = 2*(1.04)+...+2*(1.04)^n

23. cinar Group Title

he will get 2.08+2 M at 1994

24. bahrom7893 Group Title

Subtract 2nd from first: Q(1-1.04)=2 - 2*(1.04)^n Q = 2(1-1.04^n)/(1-1.04)

25. bahrom7893 Group Title

and then I just plugged in n = 6

26. cinar Group Title

4.2432+2 M at 1995

27. cinar Group Title

6.493+2M at 1996

28. cinar Group Title

at end of 1999 15.96M she has

29. amistre64 Group Title

$B_n = B_{n-1}(1.04) + 2m$ $B_{n-1} = (B_{n-2}(1.04)+ 2m)$ $B_n = (B_{n-2}(1.04)+ 2m)(1.04) + 2m$ $B_n = B_{n-2}(1.04)^2+ 2m(1.04) + 2m$ $B_{n-2}=B_{n-3}(1.04)+2m$ $B_n = (B_{n-3}(1.04)+2m)(1.04)^2+ 2m(1.04) + 2m$ $B_n = B_{n-3}(1.04)^3+2m(1.04)^2+ 2m(1.04) + 2m$ $B_n = B_{n-r}(1.04)^r+2m((1.04)^{r-1}+ (1.04)^{r-2}+... +(1.04)+ 1)$ 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 Group Title

well that's pretty much what i used amistre.

31. bahrom7893 Group Title

ughh i hate it when they assign even problems.

32. amistre64 Group Title

i know; but its nice to be able to remember how to come to that conclusion in a general sense :)

33. amistre64 Group Title

and the simplification I believe gets us:$B_n=2m\frac{1-1.04^n}{1-1.04}$

34. bahrom7893 Group Title

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 Group Title

i meant n=6 and n=10

36. bahrom7893 Group Title

What about part two? what is present value? i've never heard of the term before.

37. amistre64 Group Title

present value is; what is the value of the contract at the present time i think

38. bahrom7893 Group Title

so it would just be 2mln?

39. amistre64 Group Title

in other words; how much is left to pay off?

40. bahrom7893 Group Title

ohh do the end minus 2 mln

41. bahrom7893 Group Title

*so the final value at n=10 minus 2mln, that has already been paid, i guess?

42. amistre64 Group Title

end value - piad value = whats left to pay off

43. bahrom7893 Group Title

awesome, thanks a lot!!!!!

44. amistre64 Group Title

ill have to google that up to be sure my idea is sound tho :)

45. bahrom7893 Group Title

uhmm i'm pretty sure u're right.

46. amistre64 Group Title

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 Group Title

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 Group Title

se will get 14 M from company from 1993 to 1999 ( 7 years * 2M=14 M)

49. cinar Group Title

she will get 16.05 M

50. cinar Group Title

for a) 1)

51. amistre64 Group Title

16,221,792 or so if we use that formula ...

52. cinar Group Title

53. bahrom7893 Group Title

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 Group Title

we should be plugging in 6 and 10

55. cinar Group Title

who said that that formula is correct for this problem..

56. bahrom7893 Group Title

i did, and amistre confirmed, u're using the formula that got derived from our formula.

57. bahrom7893 Group Title

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 Group Title

no I am not using your formula.. I am just using A=P(1-n/100)^t

59. cinar Group Title

it can't 13.2 M as I said she will already earn from company 14M

60. cinar Group Title

or maybe I misunderstand question..

61. bahrom7893 Group Title

in 6 yrs she will have earned 12 mln, 6*2 is twelve, that's without interest

62. cinar Group Title

she will get 2 M @ 1993 @ 1994-@5 @6 @7 @8 @9 how many @ sign I used 7 right

63. bahrom7893 Group Title

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 Group Title

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 Group Title

Yr n: 2mln + 2*(1.04) + 2*(1.04)^2 + 2*(1.04)^3 + ... + 2*(1.04)^(n-1)

66. cinar Group Title

then use n=7

67. bahrom7893 Group Title

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 Group Title

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 Group Title

So: 1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 U're right!!!

70. bahrom7893 Group Title

OMGGGGGGGG!!! I'm about to rage quit math.... NOW EVERYTHING MAKES SENSE!

71. bahrom7893 Group Title

1993 - yr1 1994 - yr2 1995 - 3 1996 - 4 1997 - 5 1998 - 6 1999 - 7 2000 - 8 2001 - 9 2002 - 10

72. cinar Group Title

I am happy for you solved it..

73. GT Group Title

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 Group Title

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 Group Title

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 Group Title

ty dumb lol where were u earlier?

77. dumbcow Group Title

away from the computer :) haha

78. bahrom7893 Group Title

haha, i hate calling u dumb, i should start callin u cow lol

79. dumbcow Group Title

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 Group Title

thats how they do it for lottery winners or cash settlements...you want cash now or annual payments ?

81. cinar Group Title

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 Group Title

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 Group Title