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May 31st, 2015, 08:50 AM   #1
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Present Value Perpetuity

Consider a perpetuity deferred of 1 month and whose payments are: annual and variable as a geometrical progression (first term 120 euros and ratio 1.2) in the first 20 years, no payments in the next 12 years, semestral and constant payments of 200 euros in the next 2 years and annual and variable as an arithmetical progression (ratio 7 years) in the following period. Compute the present value of the perpetuity if the annual interest rate is 2% in the first 30 years and 5% in the following period.

Thank you!
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May 31st, 2015, 11:36 AM   #2
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Warning: Beer soaked ramblings follow.

What have you tried so far?
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May 31st, 2015, 07:43 PM   #3
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Anna, you are using many "non-standard" terms,
plus the "English" is difficult to follow.

Can you clarify these:
Quote:
Originally Posted by Anna View Post
> Consider a perpetuity deferred of 1 month
What does that mean?


> (first term 120 euros and ratio 1.2)
Does that mean the payments increase by 20%

> arithmetical progression (ratio 7 years)
What does "ratio 7 years" mean?

> in the following period.
How long? No information given.
Can you get someone perhaps to help you
with the translation into English?
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May 31st, 2015, 09:15 PM   #4
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The input values

A_1 = 120
i_1 = 2%
g_1 = 1.2%
n_1 = 20
n_db_1 = 1/12

A_2 = 0
i_1 = 2%
i_2 = 5%
n_2_a = 10
n_2_b = 2
n_db_1 = 1/12
n_db_2 = 20

A_3 = 200
i_1 = 2%
i_2 = 5%
n_3 = 4
n_db_1 = 1/12
n_db_2 = 20
n_db_3 = 10
n_db_4 = 2

A_4 = 200
i_1 = 2%
i_2 = 5%
g_2 = 7
n_4 = ∞
n_db_1 = 1/12
n_db_2 = 20
n_db_3 = 10
n_db_4 = 2
n_db_5 = 4
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June 1st, 2015, 05:22 AM   #5
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Well Abe, looks like you're buddy-buddy with the holy spirit...

I simply can't "see" what d'hell is suppose to be happening
after year 34.

Anyway, the FV at end of year 34 is 37,232.073...

[what d'lord is an arithmetical progression (ratio 7 years)]
he wondered...
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June 1st, 2015, 05:29 AM   #6
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The title of the problem states to find present value of a perpetuity

Quote:
and variable as an arithmetical progression (ratio 7 years) in the following period
The word years rhymes with euros. The machine translator forged the Italian word for Euro to a rough equivalent of years in English.

It's an arithmetic progression of an annuity in amount of 200 euros thus highly plausible that the linear gradient is in amount of 7 euros

The following period refers to time remaining after the 34th year till eternity which rhymes with trinity
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June 1st, 2015, 06:02 AM   #7
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Frankly my dear, I don't give a damn (forgot who said that famous line!).
After this point:
Code:
YEAR    PAYMENT    INTEREST      BALANCE
  0                                                      .00
  1          120.00               .00         120.00
  2          144.00             2.40         266.40
...     
 20      3,833.76          406.55    24,567.77
 21              .00          491.36    25,059.13
...
 30              .00          587.22    29,947.97
 31              .00       1,497.40    31,445.37
 32              .00       1,572.27    33,017.64
 32.5      200.00         815.37    34,033.01
...
 34.0      200.00         892.47    37,232.07
...our dear Anna can buy all the pizza she wants...
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June 1st, 2015, 07:47 AM   #8
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Anna would like many of her next generations to be able to buy all the pizza they want.

You had only offered her enough money for herself leaving out the progeny to come.

Your's is Future Value of three annuities at time period 34.

The question is about present value of three annuities and a terminal perpetuity at time period 0.
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June 2nd, 2015, 05:46 AM   #9
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PV = PV_1 + PV_2 + PV_3 + PV_4
PV = 2,182 + 0 + 376.14 + 3079.73
PV = 5,637.87

A_1 = 120
i_1 = 2%
g_1 = 1.2%
n_1 = 20
n_db_1 = 1/12
c_1 = 1
p_1 = 1

aey(i_1,c_1) = (1 + i_1 * c_1)^(1/c_1) - 1
aey(2%,1) = (1 + 2% * 1)^(1/1) - 1
aey(2%,1) = (1 + 0.02)^(1) - 1
aey(2%,1) = (1.02)^(1) - 1
aey(2%,1) = 1.02 - 1
aey(2%,1) = 0.02
aey(2%,1) = 2%

x_1 = [1 + aey(i_1,c_1)]^[p_1]
x_1 = (1 + 2%)^(1)
x_1 = (1.02)^(1)
x_1 = 1.02

aey(g_1,c_1) = (1 + g_1 * c_1)^(1/c_1) - 1
aey(1.2%,1) = (1 + 1.2% * 1)^(1/1) - 1
aey(1.2%,1) = (1 + 0.012)^(1) - 1
aey(1.2%,1) = (1.012)^(1) - 1
aey(1.2%,1) = 1.012 - 1
aey(1.2%,1) = 0.012
aey(1.2%,1) = 1.2%

x_g_1 = [1 + aey(g_1,c_1)]^[p_1]
x_g_1 = (1 + 1.2%)^(1)
x_g_1 = (1.012)^(1)
x_g_1 = 1.012

PV_1 = A_1 * (x_1)^(-n_db_1) * [1 - (x_1)^(-n_1) * (x_g_1)^(n_1)] / [x_1 - x_g_1]

PV_1 = 120 * (1.02)^(-1/12) * [1 - (1.02)^(-20) * (1.012)^(20)] / [1.02 - 1.012]

PV_1 = 120 * 0.99835114192125220508028728270302 * [1 - (0.67297133310805768745821793902561) * (1.2694343623807925926447777212967)] / 0.008

PV_1 = 120 * 0.99835114192125220508028728270302 * [1 - 0.8542929351445791862475479166361] / 0.008

PV_1 = 120 * 0.99835114192125220508028728270302 * 0.1457070648554208137524520833639 / 0.008

PV_1 = 120 * 0.99835114192125220508028728270302 * 18.213383106927601719056510420488

PV_1 = 120 * 18.183351823050415529842735519299

PV_1 = 2,182.00

A_2 = 0
i_1 = 2%
i_2 = 5%
n_2_a = 10
n_2_b = 2
n_db_1 = 1/12
n_db_2 = 20
n_db_3 = 10
c_2 = 1
p_2 = 1
c_3 = 1
p_3 = 1

aey(i_1,c_2) = (1 + i_1 * c_2)^(1/c_2) - 1
aey(2%,1) = (1 + 2% * 1)^(1/1) - 1
aey(2%,1) = (1 + 0.02)^(1) - 1
aey(2%,1) = (1.02)^(1) - 1
aey(2%,1) = 1.02 - 1
aey(2%,1) = 0.02
aey(2%,1) = 2%

x_2 = [1 + aey(i_1,c_2)]^[p_2]
x_2 = (1 + 2%)^(1)
x_2 = (1.02)^(1)
x_2 = 1.02

aey(i_2,c_3) = (1 + i_2 * c_3)^(1/c_3) - 1
aey(5%,1) = (1 + 5% * 1)^(1/1) - 1
aey(5%,1) = (1 + 0.05)^(1) - 1
aey(5%,1) = (1.05)^(1) - 1
aey(5%,1) = 1.05 - 1
aey(5%,1) = 0.05
aey(5%,1) = 5%

x_3 = [1 + aey(i_2,c_3)]^[p_3]
x_3 = (1 + 5%)^(1)
x_3 = (1.05)^(1)
x_3 = 1.05

PV_2 = A_2 * (x_1)^(-n_db_1) * (x_1)^(-n_db_2) * { [1 - (x_2)^(-n_2_a)] / [x_2 - 1] + (x_1)^(-n_db_3) * [1 - (x_3)^(-n_2_b)] / [x_3 - 1] }
PV_2 = 0 * (1.02)^(-1/12) * (1.02)^(-20) * { [1 - (1.02)^(-10)] / [1.02 - 1] + (1.02)^(-10) * [1 - (1.05)^(-2)] / [1.05 - 1] }
PV_2 = 0 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * { [1 - 0.82034829987515527699797252835421] / 0.02 + 0.82034829987515527699797252835421 * [1 - 0.90702947845804988662131519274376] / 0.05 }
PV_2 = 0 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * { 0.17965170012484472300202747164579 / 0.02 + 0.82034829987515527699797252835421 * 0.09297052154195011337868480725624 / 0.05 }
PV_2 = 0 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * { 8.9825850062422361501013735822895 + 0.82034829987515527699797252835421 * 1.8594104308390022675736961451248 }
PV_2 = 0 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * { 8.9825850062422361501013735822895 + 1.5253641857089055037150509597517 }
PV_2 = 0 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 10.507949191951141653816424542041
PV_2 = 0 * 0.99835114192125220508028728270302 * 7.0715485759390973592513539675937
PV_2 = 0 * 7.0598885959404027140349349469266
PV_2 = 0

A_3 = 200
i_1 = 2%
i_2 = 5%
n_3 = 4
n_db_1 = 1/12
n_db_2 = 20
n_db_3 = 10
n_db_4 = 2
c_4 = 1/2 = 0.5
p_4 = 1/2 = 0.5

aey(i_2,c_4) = (1 + i_2 * c_4)^(1/c_4) - 1
aey(5%,1/2) = (1 + 5% * 0.5)^(1/0.5) - 1
aey(5%,1/2) = (1 + 0.025)^(2) - 1
aey(5%,1/2) = (1.025)^(2) - 1
aey(5%,1/2) = 1.050625 - 1
aey(5%,1/2) = 0.050625
aey(5%,1/2) = 5.0625%

x_4 = [1 + aey(i_2,c_4)]^[p_4]
x_4 = (1 + 5.0625%)^(0.5)
x_4 = (1.050625)^(0.5)
x_4 = 1.025


PV_3 = A_3 * (x_1)^(-n_db_1) * (x_1)^(-n_db_2) * (x_1)^(-n_db_3) * (x_3)^(-n_db_4) * [1 - (x_4)^(-n_3)] / [x_4 - 1]

PV_3 = 200 * (1.02)^(-1/12) * (1.02)^(-20) * (1.02)^(-10) * (1.05)^(-2) * [1 - (1.025)^(-4)] / [1.025 - 1]

PV_3 = 200 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * [1 - 0.90595064479975482710675106634991] / 0.025

PV_3 = 200 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 0.09404935520024517289324893365009 / 0.025

PV_3 = 200 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 3.7619742080098069157299573460034

PV_3 = 200 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 3.4122215038637704451065372752865

PV_3 = 200 * 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 2.7992101094920896677619571393648

PV_3 = 200 * 0.99835114192125220508028728270302 * 1.8837881590344437078106716643333

PV_3 = 200 * 1.8806820597097703294076708373094

PV_3 = 376.14


A_4 = 200
i_1 = 2%
i_2 = 5%
g_2 = 7
n_4 = 8
n_db_1 = 1/12
n_db_2 = 20
n_db_3 = 10
n_db_4 = 2
n_db_5 = 4

PV_4 = (x_1)^(-n_db_1) * (x_1)^(-n_db_2) * (x_1)^(-n_db_3) * (x_3)^(-n_db_4) * (x_4)^(-n_db_5) * [A_4 / (x_3 - 1) + g_2 / (x_3 - 1)^2]

PV_4 = (1.02)^(-1/12) * (1.02)^(-20) * (1.02)^(-10) * (1.05)^(-2) * (1.025)^(-4) * [200/(1.05 - 1) + 7 / (1.05 - 1)^2]

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 0.90595064479975482710675106634991 * [200/ 0.05 + 7 / (0.05)^2]

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 0.90595064479975482710675106634991 * [200/0.05 + / 0.0025]

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 0.90595064479975482710675106634991 * [4000 + 2800]

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 0.90595064479975482710675106634991 * 6800

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 0.90702947845804988662131519274376 * 6160.4643846383328243259072511794

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 0.82034829987515527699797252835421 * 5587.7227978578982533568319738588

PV_4 = 0.99835114192125220508028728270302 * 0.67297133310805768745821793902561 * 4583.878897396372768803563152801

PV_4 = 0.99835114192125220508028728270302 * 3084.8190923867305648278126147848

PV_4 = 3079.73

Last edited by AbrahamA; June 2nd, 2015 at 05:59 AM.
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June 2nd, 2015, 09:54 AM   #10
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Abe, did you trip and hit your head on one of the 4 corners of the Earth?
How do you expect a poor student like Anna to understand all that junk?

I can see your result for PV1 is correct...too scared to do more checking.

I see you're using 1.012 (not 1.2 as I did) as the payment increase during
1st 20 years, then some iffy assumption that payment will increase by a
constant 7 euros beginning in 35th year.

Anyway, using your result of 5,637.87 as correct, this is what the
story is for the 1st 34 years:
Code:
Year      Payment        Interest        Balance
  0                                              5,637.87
  1         -120.00         112.76       5,630.63
  2         -121.44         112.61       5,621.80
...     
 20        -150.53         103.54       5,129.88
 21              .00          102.60       5,232.48
...
 30              .00          122.61       6,253.30
 31              .00          312.67       6,565.97 (5% starts)
 32              .00          328.30       6,894.27
 32.5     -200.00         170.25       6,864.52 (semi-annual starts)
...
 34.0     -200.00         167.98       6,770.80
So we have 6,770.80 to accommodate whatever is to happen during
year35 and after...and only the Holy Spirit knows what that is.

To end: I wouldn't be surprised if Anna is playing jokes on us.
She's posted a completely unclear/ridiculous scenario...
like, how the hell is 7 euros per year suppose to help buy groceries!!

Last edited by Denis; June 2nd, 2015 at 10:21 AM.
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