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Help with annuities: present value?

Grade 11 stuff.
 
1. Suzie buys a new computer for $2500. She pays $700 and finances the rest at $75.84 per month for 2 and a half years. What annual interest rate, compounded monthly, is Suzie being charged?
 
How do I get 19% for the answer?

Are you supposed to get this : 1800= 75.84 ((1+i/12)^360-1)/(i/12))
If so, how do i isolate for i?
 
2. A lottery has two options for winners collecting their prize: Option A: $1000 each week for life or Option B: $660 000 in one lump sum. The current interest rate is 6.76%/a compounded weekly.
a) Which option would you suggest to a winner who expects to live for another 25 years?
b) When is option A better than option B?

what is the equation for part a?
i have PV=1000((1-(1+0.001298077)^1300)/0.0013)

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1 Answer

Hi Edward,

For information, here are a couple of links that have good descriptions of the annuity equations:

   (1) http://www.investopedia.com/articles/03/101503.asp

   (2) http://www.frickcpa.com/tvom/tvom_fv_annuity.asp

The second site specifically discusses the issue of solving for interest rate -- see section 3, part c.

For your questions, first let me say that annuity calculations are slightly different for the case of payment at the start of period versus payment at the end of period. For example, does Suzie make her payment at the beginning of the month or end of the month ? For the following, I will assume "payment at end of period", also known as "ordinary annuity". ("Payment at beginning" is called "annuity due".)

Question 1: assuming Suzie pays at the end of the month, our equation is

   Pv = C * (1 - (1 + B) ^ -n) / B

where Pv = present value of series of payments, C = Cash flow per period, B = interest rate (as a decimal) per period, and n = number of payments.

Suzie is paying every month, so B = interest rate per month = interest rate per year / 12. Let's use the symbol "i" to represent yearly interest rate. Thus, B = i/12, and

   Pv = C * (1 - (1 + i/12) ^ -n) / (i/12)

For Suzie's case, we have Pv=1800, C=75.84, and n=30 (12*2.5=30 monthly payments), so

   [Equation 1]  1800 = 75.84 * (1 - (1 + i/12) ^ -30) / (i/12)

You asked: are you supposed to get this :
   1800 = 75.84 ((1 + i/12)^360 - 1) / (i/12)) ? ---> No; you should get the equation above.

You also asked: how do I isolate for i ? ---> The answer to this is: you can't isolate for i. The only way to solve the equation for i is an iterative method. In other words, we "guess" at the answer, using some elegant guessing scheme. One such method is called the "bisection method", also known as "interval halving". For reference, see

   http://en.wikipedia.org/wiki/Bisection_method

To use the method, we start by guessing some value for i that is too small, and some value that is too large. For example I think i=0.01 (1% yearly interest) is much too small. If I insert i=0.01 into Equation 1, I get Pv = 75.84 * (1 - (1 + (0.01/12)) ^ -30) / (0.01/12) = 2246.07. So i=0.01 is too small; for such a small interest rate, the present value is much higher than we want. In other words, if the finance company has a low interest rate, Suzie's monthly payments would be smaller than 75.84. For the "too large" value, let's try 50% interest (i=0.50). Inserting into Equation 1, I get Pv = 75.84 * (1 - (1 + (0.50/12)) ^ -30) / (0.50/12) = 1285.29. Thus, i=0.50 (50%) is too large; if the finance company charged that much, Suzie's payments would need to be larger than 75.84. (If it had turned out that 50% were too small, we could try 100% or 200% - just something that gets us a Pv < 1800). To summarize what we have so far:

   i=0.01 ---> Pv=2246.07
   i=0.50 ---> Pv=1285.29

We have bracketed a value for i; that is 0.01 < i < 0.50. We use the bisection method by next trying i=(0.01 + 0.50)/2 = 0.255 (25.5%). That is, we take the mid-point of the lower and upper guesses. Inserting i=0.255 into Equation 1, I get Pv=1669.71. Now we have

   i=0.01 ----> Pv=2246.07
   i=0.255 ---> Pv=1669.71
   i=0.50 ----> Pv=1285.29

So now we know that our desired answer of 1800 is bracketed by 0.01 < i < 0.255. Using the bisection method, we next try i=(0.01 + 0.255)/2 = 0.1325 (13.25%). Again we have taken the mid-point of the lower and upper guesses. Inserting i=0.1325 into Equation 1 I get Pv=1927.79. Now we have

   i=0.01 ------> Pv=2246.07
   i=0.1325 ---> Pv=1927.79
   i=0.255 -----> Pv=1669.71

So now we know that our desired answer of 1800 is bracketed by 0.1325 < i < 0.255. Using the bisection method, we next try i=(0.1325+0.255)/2 = 0.19375 (19.375%). Again we have taken the mid-point of the lower and upper guesses. Inserting i=0.19375 into Equation 1 I get Pv=1792.12.

As you can see, the bisection method is converging very quickly to the desired value of 1800. I'll leave it to you to do a couple of more iterations, and see if you get pretty close to i=0.19 (19%).

Question 2: assuming the $1000 is paid at the end of the week, our equation is again

   Pv = C * (1 - (1 + B) ^ -n) / B

where Pv = present value of series of payments, C = Cash flow per period, B = interest rate (as a decimal) per period, and n = number of payments.

For weekly payments, let's use weeks per year = 365.25 days per year / 7 days per week = 52.18 weeks per year (the 365.25 includes the extra day every 4 years for leap year, and ignores the fact that at the end of the century there might not be that extra day). Therefore, letting B = interest rate per week and i = interest rate per year, we have B = i/52.18, with i=0.0676 (6.76%) and

   Pv = C * (1 - (1 + i/52.18) ^ -n) / (i/52.18)

For the lottery case, we have C=1000, and n=25*52.18=1304.5 weekly payments, so

Pv = 1000 * (1 - (1 + 0.0676/52.18) ^ -1304.5) / (0.0676/52.18) = 629308.20, which is about $30,000 less than the lump sum payment.

Your equation, 1000 * ((1 - (1 + 0.001298077)^1300) / 0.0013), is close except the 1300 should be -1300.

This has been a long answer, and I hope it has helped you.