R PROGRAM for calculating accumulated balances
============================================== 2/13/06

> BCalc.D                                                                    
function (Amts, Times, Rates, Cum=T) 
{
### Discrete payment stream Balance Calculation
### Deposit & withdrawals made in amounts  Amts
###   at times  Times , the last of which is 
###   the final time at which the balance is to 
###   be evaluated. Interest rates are in the 
###   form of instantaneous-compounded (m=infty)
###   annualized rates, or forces of mortality,
###   and are taken to be piecewise constant 
###   between times in vector Times.
### NB. Amts and Times are vectors of same length,
###   and Rates has one fewer component.
###   Cum is Boolean optional parameter, which, 
###   if True, indicates that output Balance is
###   to be given at times Times[-1].
### FUTURE version may give continuous-time 
###   calculation, but for now, finely-discretized
###   Times vector can serve the same purpose.
          M <- length(Rates)
          Accum <- c(1,cumprod(exp(Rates*diff(Times))))
          Depos.Disc <- Amts/Accum
          Balance <- Accum*cumsum(Depos.Disc)
          list(Balance=if(Cum) 
            Balance else Balance[length(Times)], 
               Time = if(Cum) 
            Times else Times[length(Times)])
}

## Here Amts is a vector of dimension the same as Times,
and one component longer than the vector Rates.

> BCalc.D(c(0,rep(.25,40)), seq(0,10,.25), rep(log(1.05),40))
$Balance
 [1]  0.0000000  0.2500000  0.5030681  0.7592418  1.0185594  1.2810594
 [7]  1.5467809  1.8157633  2.0880468  2.3636718  2.6426793  2.9251109
[13]  3.2110086  3.5004148  3.7933727  4.0899259  4.3901184  4.6939950
[19]  5.0016008  5.3129816  5.6281838  5.9472542  6.2702403  6.5971901
[25]  6.9281524  7.2631763  7.6023117  7.9456090  8.2931194  8.6448945
[31]  9.0009867  9.3614489  9.7263348 10.0956987 10.4695955 10.8480808
[37] 11.2312110 11.6190430 12.0116346 12.4090442 12.8113309

$Time
 [1]  0.00  0.25  0.50  0.75  1.00  1.25  1.50  1.75  2.00  2.25  2.50  2.75
[13]  3.00  3.25  3.50  3.75  4.00  4.25  4.50  4.75  5.00  5.25  5.50  5.75
[25]  6.00  6.25  6.50  6.75  7.00  7.25  7.50  7.75  8.00  8.25  8.50  8.75
[37]  9.00  9.25  9.50  9.75 10.00


### This illustration shows that the accumulated balance for an
    immediate annuity (quarterly payments, $1/yr for 10 years,  
    at 5%APR), valued at time 10, is 12.81133, which should 
    agree with our formula:

> (1.05^10)*(1-1.05^(-10))/(4*(1.05^.25-1))                                  
[1] 12.81133

### The virtue of this little program is that it implements the
    calculation for time-varying interest rates. Thus, if we 
    want to know the accumulated balance for r(t) = .04*(1+.01*t)
    of continuous deposits at rate D(t) = 1000*(1-.02*(t-5)^2)
    for 10 years is (to not very high accuracy,
    using 1e3, 1e4, or 1e5 points):

> BCalc.D(1000*(1-.02*((0:1000)/100-5)^2)/100, seq(0,10,.01), 
    .04*(1+.01*seq(0,10-.01,.01)),F)
$Balance
[1] 10387.67
$Time
[1] 10
> BCalc.D(1000*(1-.02*((0:1e4)/1e3-5)^2)/1e3, seq(0,10,1e-3), 
+     .04*(1+.01*seq(0,10-1e-3,1e-3)),F)$Balance                             
[1] 10382.10
> BCalc.D(1000*(1-.02*((0:1e5)/1e4-5)^2)/1e4, seq(0,10,1e-4), 
+     .04*(1+.01*seq(0,10-1e-4,1e-4)),F)$Balance                              
[1] 10381.54

### The exact value, to high accuracy, from the formula, is 
> integrate(function(t) 1000*(1-.02*(t-5)^2)*exp(-.04*t-.0002*t^2),
    0,10)$val*exp(.04*10 + .0002*100)
[1] 10381.48