Thursday, June 30, 2005
compount interest and refinancing problem.
i came across a math website explaining the component interest and adv on refinance...
Example: If you invest $2500 at 9% annual interest, compounded quarterly (four times a year), how much money will you have after 10 years?
Solution: We use the compound interest formula:
C = P(1 + r/n)^nt = 2500(1 + 0.09/4)40 = $6087.97.
Example: You borrow $9634.67 at 9% interest, compounded monthly and plan to make the minimum payments of $200 at the end of each month for five years. After two years, what is the payoff amount on the loan?
Solution: After 2 years, there are 5 - 2 = 3 years remaining on the loan, so that payoff amount is:
L = 200(1 - (1 + 0.09/12)-12*3)/(0.09/12) = 6289.36105
or $6289.36
Example: You borrow $200,000 at an annual interest rate of 6.5%, compounded monthly, to purchase a house. You plan on repaying the loan over the course of 30 years and make the minimum payments each month for 10 years, at which time you have the opportunity to refinance the loan at an interest rate of 5.75%. What are the minimum payments on the new loan (assuming you still wish to pay off the loan 30 years after you borrowed the $200,000) and how much money will you save by refinancing?
Solution: We first use the loan formula to find the minimum payments for the original loan:
200000 = P(1 - (1 + 0.065/12)-12*30)/(0.065/12)
which we solve to get P = 1264.136047, or $1264.14.
After 10 years of making monthly payments of $1264.14, there will be 20 years remaining on the original loan. To find the payoff amount we use t = 20 in the loan formula with P = 1264.14, r = 0.065 and n = 12:
L = 20000(1 - (1 + 0.065/12)-12*20)/(0.065/12) = 169552.7829
so the payoff amount is $169,552.78.
So we want to take out a new loan for $169,552.78 at an annual interest rate of 5.75%, compounded monthly, and we wish to repay this new loan over the course of 20 years. We use the loan equation for a third time to find the minimum payments required under the new loan:
169552.78 = P(1 - (1 + 0.0575/12)-12*20)/(0.0575/12)
to get P = 1190.402106, so the new monthly payments will be $1190.40.
This is a remarkably better situation, since for the final 20 years of the loan (or 240 payments) we will only need to pay $1190.40 instead of $1264.14, a savings of $1264.14 - $1190.40 = $73.74 a month. Over the course of 20 years of monthly payments this amounts to a savings of $73.74*12*20 = $17,697.60.
There are often fees associated with refinancing, which can range from a few hundred dollars to several thousand dollars. In this case an investment of $1000, say, would be worthwhile in order to save over 17 times that amount over time.
It should also be noted that when refinancing you will likely have the option of taking out a new 30-year loan, but it makes sense to plan to repay the new loan over the span of time remaining on the original loan. First of all, it will save you more money; secondly, who wants to be making house payments for 40 years?
Example: If you invest $2500 at 9% annual interest, compounded quarterly (four times a year), how much money will you have after 10 years?
Solution: We use the compound interest formula:
C = P(1 + r/n)^nt = 2500(1 + 0.09/4)40 = $6087.97.
Example: You borrow $9634.67 at 9% interest, compounded monthly and plan to make the minimum payments of $200 at the end of each month for five years. After two years, what is the payoff amount on the loan?
Solution: After 2 years, there are 5 - 2 = 3 years remaining on the loan, so that payoff amount is:
L = 200(1 - (1 + 0.09/12)-12*3)/(0.09/12) = 6289.36105
or $6289.36
Example: You borrow $200,000 at an annual interest rate of 6.5%, compounded monthly, to purchase a house. You plan on repaying the loan over the course of 30 years and make the minimum payments each month for 10 years, at which time you have the opportunity to refinance the loan at an interest rate of 5.75%. What are the minimum payments on the new loan (assuming you still wish to pay off the loan 30 years after you borrowed the $200,000) and how much money will you save by refinancing?
Solution: We first use the loan formula to find the minimum payments for the original loan:
200000 = P(1 - (1 + 0.065/12)-12*30)/(0.065/12)
which we solve to get P = 1264.136047, or $1264.14.
After 10 years of making monthly payments of $1264.14, there will be 20 years remaining on the original loan. To find the payoff amount we use t = 20 in the loan formula with P = 1264.14, r = 0.065 and n = 12:
L = 20000(1 - (1 + 0.065/12)-12*20)/(0.065/12) = 169552.7829
so the payoff amount is $169,552.78.
So we want to take out a new loan for $169,552.78 at an annual interest rate of 5.75%, compounded monthly, and we wish to repay this new loan over the course of 20 years. We use the loan equation for a third time to find the minimum payments required under the new loan:
169552.78 = P(1 - (1 + 0.0575/12)-12*20)/(0.0575/12)
to get P = 1190.402106, so the new monthly payments will be $1190.40.
This is a remarkably better situation, since for the final 20 years of the loan (or 240 payments) we will only need to pay $1190.40 instead of $1264.14, a savings of $1264.14 - $1190.40 = $73.74 a month. Over the course of 20 years of monthly payments this amounts to a savings of $73.74*12*20 = $17,697.60.
There are often fees associated with refinancing, which can range from a few hundred dollars to several thousand dollars. In this case an investment of $1000, say, would be worthwhile in order to save over 17 times that amount over time.
It should also be noted that when refinancing you will likely have the option of taking out a new 30-year loan, but it makes sense to plan to repay the new loan over the span of time remaining on the original loan. First of all, it will save you more money; secondly, who wants to be making house payments for 40 years?
Thursday, June 16, 2005
engine tuneup
today my car had the check engine light stay on. i realized there comes the problem again.. it ran 75,000 miles all ready, looks like the car is also like a human having trouble when it gets older... anyway, i ended up in tuning up the engine for $150+tax ($162) in c.a.r.s international, guy called aurel fixed it in a matter of an hour.
another interesting thing was, my car has a chip in build, which allows any hand-held auto-scanner to get information about the car if in trouble, it was able to send diagnostic information also, so it is cool.. no need to work on your gut-feeling any more... it had listed 4 errors, misfiring.. all of them.. but he ended up in replacing the spark plug ($6 /each) and the wire and fuel filter... after that i fly like a bird to the office.
Wednesday, June 15, 2005
Marketing Strategy concepts
Thursday, June 09, 2005
cable deal
found that sbc is giving more deals on phone and internet. as cox is much expensive after 3 months, thinking of moving to it.
right now, with 3 months deal on cox, this is my monthly expenses on the tech...
cox (internet) - 24.99
cox (cable) - 29.99
vonage (phone) - 24.99
t-mobile (cell)- 39.99 (1-year commitment)
pretty expensive...
but an alternative i have been thinking is:
sbc(internet) - 25.00 (1-year commitment)
sbc(phone) - 40.00 (no canada)
wondering whether i can some bucks..
right now, with 3 months deal on cox, this is my monthly expenses on the tech...
cox (internet) - 24.99
cox (cable) - 29.99
vonage (phone) - 24.99
t-mobile (cell)- 39.99 (1-year commitment)
pretty expensive...
but an alternative i have been thinking is:
sbc(internet) - 25.00 (1-year commitment)
sbc(phone) - 40.00 (no canada)
wondering whether i can some bucks..
Wednesday, June 08, 2005
did an interview.
the candidate was supposedly expert in .NET but did not seem to. I gave a score of 3.5 in the scale of 1 to 5
- the resume claimed lots of work done on .NET Remoting and .NET Webservices.
- atleast the interview gave me a chance to refresh my memory on those two.
.NET REMOTING
o client, server(listener), remote object (marshalbyrefobject, byvalue through seriablizable)
o soapsuds.exe to create (server metadata)
o IIS can act as a listener
o server.exe.config can provide registration mechanism
o activation mode (singlecall - singletone)
o system.runtime.remoting (.channels.tcp)
o server-activated object, client-activated object
.NET Web services
o wsdl to create client proxy
o [webmethod]
o Async?
o based on .NET remoting with soap formatter
General questions I had asked:
o GAC
o Boxing/unboxing
o abstract and interface
o vb class /.net class
o vb and vb.net
o delegates and events
- the resume claimed lots of work done on .NET Remoting and .NET Webservices.
- atleast the interview gave me a chance to refresh my memory on those two.
.NET REMOTING
o client, server(listener), remote object (marshalbyrefobject, byvalue through seriablizable)
o soapsuds.exe to create (server metadata)
o IIS can act as a listener
o server.exe.config can provide registration mechanism
o activation mode (singlecall - singletone)
o system.runtime.remoting (.channels.tcp)
o server-activated object, client-activated object
.NET Web services
o wsdl to create client proxy
o [webmethod]
o Async?
o based on .NET remoting with soap formatter
General questions I had asked:
o GAC
o Boxing/unboxing
o abstract and interface
o vb class /.net class
o vb and vb.net
o delegates and events