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May 31st, 2015, 08:50 AM  #1 
Newbie Joined: May 2015 From: Italy Posts: 3 Thanks: 0  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! 
May 31st, 2015, 11:36 AM  #2 
Senior Member Joined: May 2008 Posts: 301 Thanks: 81  Warning: Beer soaked ramblings follow. What have you tried so far? 
May 31st, 2015, 07:43 PM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Anna, you are using many "nonstandard" terms, plus the "English" is difficult to follow. Can you clarify these: Quote:
with the translation into English?  
May 31st, 2015, 09:15 PM  #4 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
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 
June 1st, 2015, 05:22 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Well Abe, looks like you're buddybuddy 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... 
June 1st, 2015, 05:29 AM  #6  
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
The title of the problem states to find present value of a perpetuity Quote:
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  
June 1st, 2015, 06:02 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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 
June 1st, 2015, 07:47 AM  #8 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
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. 
June 2nd, 2015, 05:46 AM  #9 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
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. 
June 2nd, 2015, 09:54 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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 (semiannual starts) ... 34.0 200.00 167.98 6,770.80 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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