SOLUTION: What is the present worth of a P250 annuity starting at the end of the third year and continuing to the end of the fourth year, if the annual interest rate is 5%? With cash flow di

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Question 1178719: What is the present worth of a P250 annuity starting at the end of the third year and continuing to the end of the fourth year, if the annual interest rate is 5%? With cash flow diagram.
Found 3 solutions by CPhill, ikleyn, Edwin McCravy:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let's break down this problem to find the present worth of the annuity.
**Understanding the Annuity:**
* **Annuity Amount (PMT):** P250
* **Start Time:** End of the third year
* **End Time:** End of the fourth year
* **Annual Interest Rate (r):** 5% or 0.05
**Cash Flow Diagram:**
```
Time (Years): 0 1 2 3 4
Cash Flow: 0 0 0 250 250
```
**Calculations:**
1. **Present Value at the End of Year 2:**
* We need to calculate the present value of the annuity payments at the end of the second year (one year before the first payment).
* This is a simple two-payment annuity.
* Use the present value of an ordinary annuity formula:
* PV = PMT * [(1 - (1 + r)^-n) / r]
* PV = 250 * [(1 - (1.05)^-2) / 0.05]
* PV = 250 * [(1 - 0.907029478) / 0.05]
* PV = 250 * [0.092970522 / 0.05]
* PV = 250 * 1.85941044
* PV ≈ 464.85261
* This means that at the end of year 2, the payments are worth about P464.85
2. **Present Value at Time 0 (Now):**
* Now, we need to discount the present value at the end of year 2 back to the present (time 0).
* PV(at t=0) = PV(at t=2) / (1 + r)^2
* PV(at t=0) = 464.85261 / (1.05)^2
* PV(at t=0) = 464.85261 / 1.1025
* PV(at t=0) ≈ 421.63502
**Therefore, the present worth of the P250 annuity is approximately P421.64.**

Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
What is the present worth of a P250 annuity starting at the end of the third year
and continuing to the end of the fourth year, if the annual interest rate is 5%?
With cash flow diagram.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


I will use slightly different logic comparing with @CPhill.


The cash flow diagram


Years     1      2      3       4   

Deposits  0      0     250     250

Amount    0      0     250     250 + 0.05*250 + 250 = 512.50.

                       end     end        
                       of the  of the
                       year    year


In this problem, the amount at the account is P250 at the end of the 3rd year 
(after first depositing), and it is not a subject of compounding at the end of the year 3.


First compounding will happen at the end of the 4th year, and immediately after that 
the next amount of P250 will be deposited, giving the total at the end of year 4 as 
250 + 0.05*250 + 250 = 512.50.


So, at the end of the 4th year, after first compounding and the second deposit, 
the amount at the account is 512.50.


To get the present value for "now", which is the beginning of the year 1, we should rewind 
this amount of P512.50 4 (four) years back.


So, the present value is   =  = P421.64  (rounded).


ANSWER.  The present value is  P421.64  (rounded).

Solved.


/////////////////////////////////


Below is my comment to Edwin's post.

Edwin, I did not get, to which address did you directed/targeted your post - to mine or
to the address of @CPhill, who solves here old problems one after another in numbers of hundreds ?

Edwin, you see students everywhere.

Stop, your students are a myth, they are mythical.

Just starting from the year 2021 we all at this forum work almost exclusively
for AI (= Artificial Intelligence), without knowing it, without understanding it
and without even suspecting it.

In the school, nobody will give such problem to a student, because it is one level above the school level.

Also, after posting a problem and responding a solution, this solution is viewed by others 15 times
in the archive, according to the statistic.

So, your notice is interesting, but irrelevant to the process of functioning of this forum.

Have a nice day !



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
This problem was posted almost 4 years ago! The student is probably no longer in
school.

Edwin

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