I think the book just had an error, your answer is correct by using the Annuity PV formula for years 13 and 5 and taking one away from the other
I need help answering this problem:
A sequence of 8 annual payments of $750 each, with the first payment due at the beginning of the 6th year; money is worth 6%. (The beginning of year 6 is the end of year 5.) Find the present value.
This is my answer: 750 [ (1-1.06^-13) / .06 ] - 750 [ (1-1.06^-5) / .06 ] = $3,480.24
and this is the book's: $3,689 .
The book didn't provide a solution, so I would really appreciate it if someone can help me find where my solution went wrong. Thank you!
the book has correct answer.
Firstly why did u take 750 when the payment is 575.82 ?
Secondly if you are taking 13 and 5 it says that your payments are made in advance, but the formula you used is for payments being made at the end. If you wanna use this formula only then you must take 12 and 4.
here i attached what i did did gave an answer 2,834.31€
Why would you use 575.82? Since its an annuity problem you use the actual value of the payments, not discounted value since the formula discounts the payment for you. The payments are made in years 6 and 14 so we use the end of year 5 and 13. This is also correct.
The answer you gave at the end is about 800 too low.... so it is not correct.
I'm not getting you, the problem clearly says that the payments are 575.82 so, why would you take 750?
And the first payment is at the beginning of 6th year which is end of 5th year. If you are considering 5 and 13 that says you have 5 years which have no payment at the beginning of each year and 8 years with a payment of 575.82 at the beginning of each year.
you are using this right , but this will work when your payments are made at the end of each year (i.e., by taking 4 and 12).
Use , when payments are made in advance (i.e., by taking 5 and 13).
Either of them will give you correct answer with payment 575.82 not 750.
check this in wolframalpha:
575.82 [ (1-1.06^(-13)) /( .06/1.06) ] - 575.82 [ (1-1.06^(-5)) / (.06/1.06) ] - Wolfram|Alpha
I'm not getting you either, where did you get the 575.82? The OP clearly stated the payments are 750. Also the reason we use years 5 and 13 is because beginning of the 6th year = end of 5th year, why would you have to go back 1 year more?
Also since the answer you have given is clearly not the same as that of the text book, your answer is obvious not correct either. Stop trolling
Ok my bad.
But just change the payment from 575.82 to 750 gives me the answer as 3689 which is the exact. You can check.
NB: in my browser it shows me the payments in euros, that's why i have taken 575.82 yesterday.
Why don't you look at it this way:
at end of year 4, a loan is set up at 8 payments of $750, rate 6%; 1st payment end of year 5:
that'll give you a present value (loan proceeds) of $4,657.
Now just get present value of that today: 4657 / 1.06^4 = 3689
It looks like you're using the formula for an annuity immediate (when interest is applied at the end of the year), rather than for an annuity due (when interest is applied at the beginning of the year).
The formula should be
So, what you typed should look like this:
750 [[ (1-1.06^-13) / .06 ]/1.06] - 750 [[ (1-1.06^-5) / .06 ]/1.06]=3,689.0544
I hope that helps.
I took a different approach to a solution:
I used Excel and created a simple timeline of 12 periods, populating zero into the first four, and 750 in periods five through twelve, thus representing four time-periods with zero payments, followed by eight with $750.
I then used the NPV function with a 6% rate, and the stream of cash flows (Note: with the first 750 entered in period 5, this is assuming payment at the END of the 5th period {or the beginning of #6})
The NPV evaluates to $3,689.05; the same as the text book.
To Amul28's comment about Euros vs Dollars, it wouldn't matter as long as the same units are used throughout...
Present value formulaI need help answering this problem:
A sequence of 8 annual payments of 750 each, with the first payment due at the beginning of the 6th year; money is worth 6%.
(The beginning of year 6 is the end of year 5.) Find the present value.
This is my answer: 750 [ (1-1.06^-13) / .06 ] - 750 [ (1-1.06^-5) / .06 ] = 3,480.24
and this is the book's: 3,689 .
The book didn't provide a solution, so I would really appreciate it if someone can help me find where my solution went
wrong. Thank you!
whereCode:PV = R . (1+i*type) . (1+i)^-d [ 1 - (1+i)^-n ] / i
R = annuity payment
i = interest rate
type = 0 for end of period payments, 1 for start of period payments
d = time period by which annuity is deferred
n = total number of payments
Your problem has the following data
Code:PV = R . (1+i*type) . (1+i)^-d [ 1 - (1+i)^-n ] / i R = 750 i = 6% = 0.06 type = 1 for start of period annuity d = 5 as the first payment is deferred by 5 periods n = 8 as there are 8 annuity payments 750 (1+6%)^-5 (1+6%) [ 1-(1+6%)^-8 ] / 6% 750 (1+0.06)^-5 (1+0.06) [ 1-(1+0.06)^-8 ] / 0.06 750 (1.06)^-5 (1.06) [ 1-(1.06)^-8 ] / 0.06 750 (0.74725817286605716719189988974845) (1.06) [ 1-0.62741237134182678250493686881491 ] / 0.06 750 (0.74725817286605716719189988974845) (1.06) [ 0.37258762865817321749506313118509 ] / 0.06 750 (0.74725817286605716719189988974845) (1.06) (6.2097938109695536249177188530833) 750 (0.74725817286605716719189988974845) (6.5823814396277268424127819842683) 750 (4.9187383276836621437572364513326) PV = 3689.05
Hello Wilmer
I only found the thread on the second page of the forum, thus overlooked the date and assumed it must be from recent months
BTW, is there a restriction of invoking older threads?
Do they just bury the dead. I thought Thy Lord had the powers to resurrect the dead to life.
I had assumed the formula I presented will help out those seeking present value of a deferred annuity
If you think I had erred on judgement, then let Thy Lord be a witness that I shall not make the same mistake twice
But then I tend to forget things