force.ijs

Script: ~addons/finexec/finexec/basicfinance/force.ijs
Contributor: William Szuch
Updated: 2022 6 25
Depend: nil
Definitions: loaded to locale base
Status: done
Script source: force.ijs

Definitions for solving force of interest (continous compounding) problems.

Equations:

Definitions

E	ez
F	f, fc, fct, fct1, fct1_E, fct_E, fn, ft
G	g, gc, gct, gct1, gct1_E, gct_E, gn, gt
Z	ze

fct1_E	d	Explicit form of fct1
fct_E	d	Explicit form of fct
gct1_E	d	Explicit form of gct1
gct_E	d	Explicit form of gct

ez (monad)

Form: tacit
Equivalent effective interest rate for a constant force
of interest rate for one time unit.

Syntax

ez(Z)
Z = constant force of interest rate for one time unit.

Example

   ez(0.05)
0.0512711

   ez(_0.05 0 0 0.05 0.1)
_0.0487706 0 0 0.0512711 0.105171

f (monad)

Form: tacit
Present value of an amount of 1 payable in one time unit
for a force of interest rate Z.

Syntax

f(Z)
Z = constant force of interest rate for one time unit.

Example

   f(0.05)
0.951229

   f(0.01 0.05 0.1)
0.99005 0.951229 0.904837

fc (monad)

Form: tacit
Present value of an amount of 1 paid continuosly over one time unit
for a force of interest rates Z.

Syntax

fc(Z)
Z = constant force of interest rate for one time unit.

Example

   fc(0.05)
0.975412

   fc(0.01 0.05 0.1)
0.995017 0.975412 0.951626

   fc(0 0.1)
1 0.951626

fct (dyad)

Form: tacit
Present value of continous payments over the period of
of T time units where the rate of payment per time unit
is one for a constant force of interest Z.

Syntax

(Z)fct(T)
Z = constant force of interest rate per time unit over the period.
T = period in time units of continous payments

Example

   (0.0)fct(10)
10

   (0.05)fct(10)
7.86939

   (0.05)fct(0 5 10)
0 4.42398 7.86939

   (0 0.05 0.10)fct(10)
10 7.86939 6.32121

   (0 0.05 0.10 0.15)fct(0 5 10)
 0       0       0       0
 5 4.42398 3.93469 3.51756
10 7.86939 6.32121 5.17913

fct1 (dyad)

Form: tacit
Restricted form of fct: only single values for Z and T.
Present value of continous payments over the period of
of T time units where the rate of payment per time unit
is one for a constant force of interest Z.
Equation: (Z)fct1(T)

(Z)fct(T)
Z = constant force of interest rate per time unit over the period.
T = period in time units of continous payments

Example

   (0)fct1(0)
0

   (0)fct1(10)
10

   (0.05)fct1(0)
0
   (0.05)fct1(10)
7.86939

fn (dyad)

Form: tacit
Present value of N payments of 1 paid at the beginning of each time unit
for a constant force of interest rate Z over the period.
ie. Payments in advance

Syntax

(Z)fn(N)
Z = constant force of interest rate per time unit over the period.
N = integer number of payments over the period

Example

   (0.0)fn(10)
10

   (0.05)fn(10)
8.06776

   (0.05)fn(0 5 10)
0 4.53551 8.06776

   (0 0.05 0.10)fn(10)
10 8.06776 6.64253

   (0 0.05 0.10 0.15)fn(0 5 10)
 0       0       0       0
 5 4.53551 4.13471 3.78797
10 8.06776 6.64253 5.57727

ft (dyad)

Form: tacit
Present value of an amount of 1 payable in T time units
for a constant force of interest Z over the period.

Syntax

(Z)ft(T)
Z = constant force of interest rate per time unit over a period
T = number of time units in the period

Example

   (0.05)ft(5)
0.778801

   (0.05)ft(0 5 10)
1 0.778801 0.606531

   (0.0 0.05 0.10)ft(5)
1 0.778801 0.606531

   (0.0 0.05 0.1 0.15)ft(0 5 10)
 1        1        1
1 0.778801 0.606531
1 0.606531 0.367879
1 0.472367  0.22313

g (monad)

Form: tacit
Future value of an amount of 1 in one time unit
for a force of interest rates Z.

Syntax

g(Z)
Z = constant force of interest rate for one time unit.

Example

   g(0.05)
1.05127

   g(0 0.01 0.05 0.1)
1 1.01005 1.05127 1.10517

gc (monad)

Form: tacit
Future value of an amount of 1 paid continuosly over one time unit
for a force of interest rates Z.

Syntax

gc(Z)
Z = constant force of interest rate for one time unit.

Example

   gc(0.05)
1.02542

   gc(0.01 0.05 0.1)
1.00502 1.02542 1.05171

   gc(0 0.1)
1 1.05171

gct (dyad)

Form: tacit
Present value of continous payments over the period of
of T time units where the rate of payment per time unit
is one for a constant force of interest Z.

Syntax

(Z)gct(T)
Z = constant force of interest rate per time unit over the period.
T = period in time units of continous payments

Example

   (0.0)gct(10)
10

   (0.05)gct(10)
12.9744

   (0.05)gct(0 5 10)
0 5.68051 12.9744

   (0 0.05 0.10)gct(10)
10 12.9744 17.1828

   (0 0.05 0.10 0.15)gct(0 5 10)
 0       0       0       0
 5 5.68051 6.48721 7.44667
10 12.9744 17.1828 23.2113

gct1 (dyad)

Syntax

(Z)gct(T)
Z = constant force of interest rate per time unit over the period.
T = period in time units of continous payments

Example

   (0)gct1(0)
0

   (0)gct1(10)
10

   (0.05)gct1(0)
0
   (0.05)gct1(10)
12.9744

gn (dyad)

Form: tacit
Future value of N payments of 1 payable at intervals for a time unit.
for a constant force of interest Z.
accumulated to the time of the last payment.
(ie: payments in arrears)

Syntax

(Z)gn(N)
Z = constant force of interest rate per time unit over the period.
N = number of payments over the period

Example

   (0.0)gn(10)
10

   (0.05)gn(10)
12.6528

   (0.05)gn(0 5 10)
0 5.53968 12.6528

   (0 0.05 0.10)gn(10)
10 12.6528 16.338

   (0 0.05 0.10 0.15)gn(0 5 10)
 0       0       0       0
 5 5.53968 6.16826 6.90212
10 12.6528  16.338 21.5139

gt (dyad)

Form: tacit
Future value of an amount of 1 payable in T time units
for a constant force of interest Z over the period.

Syntax

(Z)gt(T)
Z = constant force of interest rate per time unit over a period
T = number of time units in the period

Example

   (0.05)gt(5)
1.28403

   (0.05)gt(0 5 10)
1 1.28403 1.64872

   (0.0 0.05 0.10)gt(5)
1 1.28403 1.64872

   (0.0 0.05 0.1 0.15)gt(0 5 10)
1       1       1       1
1 1.28403 1.64872   2.117
1 1.64872 2.71828 4.48169

ze (monad)

Form: tacit
Equivalent constant force of interest rate for a
constant effective interest rate for one time unit.

Syntax

ez(E)
E = constant effective interest rate for one time unit.

Example

   ze(0.05)