Example 2015.1: Time to refinance?

In the US, it’s typical to borrow a fairly substantial portion of the cost of a new house from a bank. The cost of these loans, the mortgage rate, varies over time depending on what the financial wizards see in their crystal balls. What this means over time is that when the mortgage rates go down, the cost of living in your own house magically decreases–you take a new loan at the lower rate and pay off your old loan with it– then you only have to pay off the new loan at the lower rate. You can find mortgage rate calculators on the web very easily– if you don’t mind their collecting your data and being bombarded with ads if you let their cookies trace you.

Instead, you can use SAS or R to calculate what you might pay for a new loan with various posted rates. There are some sophisticated tools available for either package if you’re interested in the remaining principal or the proportion of each payment that’s principal. Here, we just want to check the monthly payment.

R

We’ll begin by writing a little function to calculate the monthly payment from the principal, interest rate (in per cent), and term (in years) of the loan. This is basic stuff, but the code here is adapted from a function
written by Thomas Girke
of UC Riverside
.

mortgage <- function(principal=300000, rate=3.875, term=30) { 
  J <- rate/(12 * 100)
  N <- 12 * term
  M <- principal*J/(1-(1+J)^(-N))
  monthPay <<- M
  return(monthPay)
}

To compare the monthly costs for a series of loans offered by a local bank, we’ll input the bank’s loans as a data frame. To save typing, we’ll use the rep()

function to generate the term of the loan and the points.

offers = data.frame(
  principal = rep(275000, times=9),
  term = rep(c(30,20,15), each=3), 
  points = rep(c(0,1,2), times=3),
  rate = c(3.875, 3.75, 3.5, 3.625, 3.5, 3.375, 3, 2.875, 2.75))

> offers

  principal term points  rate
1    275000   30      0 3.875
2    275000   30      1 3.750
3    275000   30      2 3.500
4    275000   20      0 3.625
5    275000   20      1 3.500
6    275000   20      2 3.375
7    275000   15      0 3.000
8    275000   15      1 2.875
9    275000   15      2 2.750

(Points are an up-front cost a borrower can pay to lower the mortgage rate for the loan.) With the data and function in hand, it’s easy to add the monthly cost to the data frame:

offers$monthly = with(offers, mortgage(rate=rate, term=term, principal=principal))

> offers

  principal term points  rate  monthly
1    275000   30      0 3.875 1293.152
2    275000   30      1 3.750 1273.568
3    275000   30      2 3.500 1234.873
4    275000   20      0 3.625 1612.610
5    275000   20      1 3.500 1594.889
6    275000   20      2 3.375 1577.282
7    275000   15      0 3.000 1899.100
8    275000   15      1 2.875 1882.611
9    275000   15      2 2.750 1866.210

In theory, each of these costs are fair, and the borrower should choose based on monthly costs they can afford, as well as whether they see a better value in having money in hand to spend on a better quality of life or to invest it in savings or in paying off their house sooner. Financial professionals often discuss things like the total dollars spent or the total spent on interest vs. principal, as well.

SAS

The SAS/ETS package provides the LOAN procedure, which can calculate the detailed analyses mentioned above. For simple calculations like this one, we can use the mort
function in the data step. It will find and return the missing one of the four parameters– principal, payment, rate, and term. To enter the data in a manner similar to R, we’ll use array
statements and do
loops.

data t;
principal = 275000; 
array te [3] (30,20,15);
array po [3] (0,1,2); 
array ra [9] (.03875, .0375, .035, .03625, .035, 
              .03375, .03, .02875, .0275);
do i = 1 to 3;
  do j = 1 to 3;
    term = te[i];
 points = po[j];
 rate = ra[ 3 * (i-1) +j];
 monthly = mort(principal,.,rate/12, term*12);
    output;
  end;
end;
run;

proc print noobs data = t; 
var principal term points rate monthly; run;

principal    term    points      rate     monthly

  275000      30        0      0.03875    1293.15
  275000      30        1      0.03750    1273.57
  275000      30        2      0.03500    1234.87
  275000      20        0      0.03625    1612.61
  275000      20        1      0.03500    1594.89
  275000      20        2      0.03375    1577.28
  275000      15        0      0.03000    1899.10
  275000      15        1      0.02875    1882.61
  275000      15        2      0.02750    1866.21

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, and statsblogs
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