(11C) TVM for HP-11C

[Gamo]

TVM program that work just like HP-12C.

Program credit to Gene Wright's website
https://www.rskey.org/gene/hpgene/tvm19c.htm

As mentioned in Gene Wright page this program originally appeared in the
May 1980 PPC Journal written by Robert Meyer.
The program will run as written on an HP-19C/29C

Since 19C doesn't have specific Labels.
I modify this program so that Labels A to E located at the same position as on HP-12C
With added-on Labels and use many Store Registers this took me awhile
to make this program work since 11C got limited program space when
use a lot of store memory registers.

Finally I have successfully fit this program to work on 11C
at which it use 111 program lines at its maximum limit.

Lastly this might be an excellent TVM program for 11C compare to that of
the more sophisticated TVM program available in the 15c advanced functions handbook
that use SOLVE function to solve for unknown interest.

TVM for HP-11C program:
  TVM_HP11C.pdf (Size: 383.91 KB / Downloads: 95)


Gamo

[Gamo]

Last post I didn’t give an example on how to use this TVM program.

These two example are from the HP-12C User’s Handbook.

—————————————————————————————-
USER mode with FIX 2

What annual interest rate must be obtained to accumulate $10,000 in
8 years on an investment of $6,000 with quarterly compounding?

8 [ENTER] 4 [x] [A] display 32.00. // Calculates and stores n.

6000 [CHS] [C] display –6,000.00 // Stores PV (with minus sign for cash paid out).

10000 [E] display 10,000.00 // Stores FV.

[B] [R/S] display 1.61 // Periodic (quarterly) interest rate.

4 [x] display 6.44 Annual interest rate.

Remark:
This problem is in END mode the default setting with 0 on Register .0
——————————————————————————————

If you deposit $50 a month (at the beginning of each month) into a
new account that pays 6 1/4% annual interest compounded monthly, how much will
you have in the account after 2 years?

Beginning Mode set 1 on Register .0 // ( 1 [STO] .0 )

2 [ENTER] 12 [x] [A] display 24.00 // Calculates and stores n.

6.25 [ENTER] 12 [÷] display 0.52 // Calculates and stores i.

50 [CHS] [D] display –50.00 // Stores PMT (with minus sign for cash paid out).

[E] [R/S] display 1,281.34 // Balance after 2 years.
—————————————————————————————-

Gamo

[Gamo]

Interest Rate Solver Solutions for

1. Direct Reduction Loan Interest Rate with known [n, PV and PMT]
2. Sinking Fund Interest Rate with known [n, FV and PMT]

This program is adapted form the HP-55 Mathematic Programs

----------------------------------------
Procedure: [USER] mode [FIX] 2

Input n [A] PV or FV [ENTER] PMT [R/S] Your Guess [R/S]

For Suggested Guess

Problem with PV use [B]
Problem with FV use [D]

[C] Solve for Direct Reduction (Problem involve PV and PMT)
[E] Solve for Sinking Fund (Problem involve FV and PMT)
-------------------------------------------
Example:

1. Problem involve: n, PV and PMT

n=12
PV=5,000
PMT=500

12 [A] 5000 [ENTER] 179.86 [R/S] display 27.80

Use Suggested Guess: [B] display 3.06%

[C] display answer: 2.92%
----------------------------------------
2. Problem involve: n, FV and PMT

n=12
FV=5,000
PMT=400

12 [A] 5000 [ENTER] 400 [R/S] display 8.33

Use Suggested Guess [D] display 0.75%

[E] display answer: 0.74%
---------------------------------------
Program:

Code:


LBL A
STO 1  // n
R/S
÷ 
STO 3  // PV or FV ÷ PMT
R/S
EEX
2
÷
STO 2  // Your Guess
EEX
2
x
R/S
-------
LBL B  // Suggested Guess for problem PV / PMT
1/x
RCL 1
X^2
1/x
RCL 3
x
-
STO 2
EEX
2
x
RTN
------
LBL C  // Calculate i for problem PV /PMT
RCL 3
RCL 2
x
1
RCL 2
1
+
RCL 1
CHS
Y^X
STO 5
-
-
RCL 1
RCL 2
1/x
1
+
÷
1
+
RCL 5
x
1
-
RCL 2
÷
÷
STO+2
X^2
EEX
1
2
CHS
X≤Y
GTO C
RCL 2
EEX
2
x
RTN
------
LBL D  // Suggested Guess for problem FV / PMT
RCL 3
RCL 1
-
2
x
RCL 1
1
-
X^2
RCL 3
+
÷
STO 2
EEX
2
x
RTN
-------
LBL E  // Calculate i for problem FV / PMT
RCL 3
RCL 2
x
1
+
RCL 2
1
+
RCL 1
Y^X
STO 4
-
RCL 1
1
RCL 2
1/x
+
÷
1
-
RCL 4
x
1
+
RCL 2
÷
÷
STO+2
X^2
EEX
CHS
1
2
X≤Y
GTO E
RCL 2
EEX
2
x
RTN

[bshoring]

Thank you for this great program. It works extremely well. Can you tell me what method it uses to solve for “i” (interest)?

It works well, but anything you can tell me about how it works would be appreciated.

Thanks, Bob

[Gamo]

Bob thanks for the interest of this wonderful TVM program
for the HP-11C

I solely adapted this program from
May 1980 PPC Journal written by Robert Meyer and
I do not know the algorithm use to find the interest in this program.

If anyone know please share your experience in this very excellent program.

Gamo

[Dave Britten]

The May 1980 PPC Journal has a brief description of how the original 29C version works. A short excerpt:

Quote:All calculations call on LBL 6 placing the following results in the stack: X: PV, Y: -(BAL or FV)(1+i)^-N, L: PMT(A)[1-(1+i)^-N]/i. I(%) = 100i is found by a rearranged version of the John Kennedy and Chris Stevens Newton's method program (V6N5P10) and may take up to a couple of minutes to solve.

Also, I've used this on my 11C a couple of times. It's an excellent program.

[bshoring]

Dave, Thanks. Any idea where I can get the original article. If I understand correctly, this program uses a modified Newton method to solve.

Thanks,
Bob

[Dave Britten]

(12-03-2020 05:53 PM)bshoring Wrote:  Dave, Thanks. Any idea where I can get the original article. If I understand correctly, this program uses a modified Newton method to solve.

Thanks,
Bob

Jake Schwartz's PPC Archive has all issues of the PPC Journal - as well as numerous other publications and materials from around the world - and is well worth the small price:

http://www.pahhc.org/ppccdrom.htm

There are tens of thousands of pages on this thing, so clear your schedule.

[Albert Chan]

(12-03-2020 05:53 PM)bshoring Wrote:  If I understand correctly, this program uses a modified Newton method to solve.

Yes !

Let a = (PV+FV)/(-PMT), using cash-flow sign-convention (results of post#3, LBL A)
We can interpret it as number of payments, if interest rate is zero.

Case FV=0: a*i/(1-k) = 1       → f(i) = a*i - (1-k) = 0       → f'(i) = a - k*n/(1+i)

Code:

function pv_i(n, a, i)
    i = i or 1/a - a/n^2
    return function()
        local k = (1+i)^-n
        i = i - (a*i-(1-k)) / ((1-k)/i - k*n/(1+i))
        return i
    end
end

Case PV=0: a*i/(K-1) = 1       → f(i) = a*i - (K-1) = 0       → f'(i) = a - K*n/(1+i)

Code:

function fv_i(n, a, i)
    i = i or 2*(a-n)/(a + (n-1)^2)
    return function()
        local K = (1+i)^n
        i = i - (a*i-(K-1)) / ((K-1)/i - K*n/(1+i))
        return i
    end
end

Iterations were setup with Newton's method.
But, denominator is not quite f'(i), resulting in better correction.

lua> g = pv_i(12, 5000/500) -- guess_i = 0.030555555555555558
lua> g(), g(), g()
0.02922075497755343        0.02922854050171087       0.029228540769134212

lua> g = fv_i(12, 5000/400) -- guess_i = 0.00749063670411985
lua> g(), g(), g()
0.007390656157378769      0.007390622804155507      0.007390622804147247

[Albert Chan]

(12-03-2020 08:53 PM)Albert Chan Wrote:  Iterations were setup with Newton's method.
But, denominator is not quite f'(i), resulting in better correction.

More important reason for modified denominator is safety.
Note: below proofs required Bernoulli inequality, i>-1, n ≤ 0, or n ≥ 1: \(\quad(1+i)^n ≥ 1 + n\,i\)

Quote:Case FV=0: a*i/(1-k) = 1       → f(i) = a*i - (1-k) = 0       → f'(i) = a - k*n/(1+i)

If we use f'(i) for Newton's denominator, and f'(i) ≤ 0, it is likely it would converge to trivial solution, i=0

Replacing with modified slope, (1-k)/i - k*n/(1+i), we remove this problem.

Proof:
Let K = 1/k = (1+i)^n ≥ 1 + n*i       → (K-1)/i ≥ n

Modified slope = \(\large {1-k \over i} - {k\,n \over 1+i}
= k \left({K-1 \over i} - {n \over 1+i}\right)
≥ k \left(n - {n \over 1+i}\right) = {k\,n\,i \over i+1} \)

If i > 0, n ≥ 1, then modified slope will be positive.

Quote:Case PV=0: a*i/(K-1) = 1       → f(i) = a*i - (K-1) = 0       → f'(i) = a - K*n/(1+i)

Same issue as above. But, this time we wanted negative "slope".
Replacing with modified slope, (K-1)/i - K*n/(1+i), we remove this problem.

Proof:

Modified slope = \(\large{(K-1)(1+i)\;-\;K\,n\,i \over i\,(1+i)}
= {K\over i\,(1+i)} \normalsize (1+(1-n)\,i - (1+i)^{1-n} )\)

Applying Bernoulli inequality, last term is non-positive.

If i > 0, n ≥ 2, then modified slope will be negative.

[Albert Chan]

From thread: Fun math algorithms:

(10-19-2020 11:05 PM)Albert Chan Wrote:  We can use \(C_±\) to solve Time value of money problem, if I is small, and N not too big.

XCas> C(d) := 1 + N*I*(N*I+d*6)/12 + I/2;       // compounding factor, d is direction
XCas> eqn := C(1)*PV + C(-1)*FV + N*PMT;       // variables follow cash flow sign convention

XCas> factor(symb2poly(eqn,I)); // quadratic coefficients, for I

[N^2*(FV+PV)/12 , (-FV*N+N*PV+FV+PV)/2 , N*PMT+FV+PV]

Divide coefficiens by (FV+PV):

[N^2/12 , (1 - N*(FV-PV)/(FV+PV))/2, PMT*N/(FV+PV) + 1]

Solving the quadratic, for the "small" root of I, we have:

Code:

function guess_i(n, pv, pmt, fv)    -- cash-flow sign convention
    local c = n/(fv+pv)
    local b = 4/((fv-pv)*c-1)
    c = (pmt*c+1)*b                 -- I coef = [b*n*n/12, -2, c]
    return c / (1+sqrt(1-c*b*n*n/12))   
end

(12-03-2020 08:53 PM)Albert Chan Wrote:  lua> g = pv_i(12, 5000/500) -- guess_i = 0.030555555555555558
lua> g(), g(), g()
0.02922075497755343        0.02922854050171087       0.029228540769134212

lua> g = fv_i(12, 5000/400) -- guess_i = 0.00749063670411985
lua> g(), g(), g()
0.007390656157378769      0.007390622804155507      0.007390622804147247

For above examples, guess_i() gives about 3 good digits.

lua> guess_i(12, 5000, -500, 0)
0.02919560045657818
lua> guess_i(12, 0, -400, 5000)
0.007391943630478942

[Albert Chan]

A faster version of guess_i(), replaced sqrt() by its Pade approximation
This is also safer, avoided sqrt of negative number issue.

Code:

function guess_i(n, pv, pmt, fv)    -- cash-flow sign convention
    local a = (fv+pv)/n
    local b = (fv-pv)-a
    local c = (pmt+a)/b
    b = n*n*a*c/b                   -- I coef = [b/c/3, -2, 4c]
    return c * (b-3)/(b-1.5)        -- pade(4c/(1+sqrt(1-4b/3)),b,2,2)
end

lua> guess_i(12, 5000, -500, 0)
0.029199705714052143
lua> guess_i(12, 0, -400, 5000)
0.007391911604953607

Update: formula adjusted so fv+pv=0 will give i = pmt/fv, instead of NaN

[Albert Chan]

pv_i() and fv_i() are similar, we can combine them.

Just keep pv_i(), using previous post guess_i() for the guess.

Code:

function pv_i(n, a, i)              -- a = -pv/pmt
    i = i or guess_i(n, a, -1, 0)
    return function()
        local k = (1+i)^-n
        i = i - (a*i-(1-k)) / ((1-k)/i - k*n/(1+i))
        return i
    end
end

lua> g = pv_i(12, 5000/500)
lua> g(), g()
0.029228537101221044      0.029228540769133633

For fv_i() problems, where PV=0, add negative sign for the input.
In other words, fv_i(n, a) ≡ pv_i(-n, -a)

lua> g = pv_i(-12, -5000/400)
lua> g(), g()
0.007390622809688779      0.007390622804149185

This setup has another advantage, with invariant: sign(i) = sign(n-a)
For above examples, 12 > 10, -12 > -12.5, thus positive interest rate.

Update: numbers adjusted with updated guess_i()

[Albert Chan]

(12-06-2020 02:32 PM)Albert Chan Wrote:  For fv_i() problems, where PV=0, add negative sign for the input.
In other words, fv_i(n, a) ≡ pv_i(-n, -a)

Proof:

Let K = (1+I)^N. TVM equation (NFV = 0):

PV*K + PMT*(K-1)/I + FV = 0

Travel backward in time: (N, PV, PMT, FV) := (-N, FV, -PMT, PV)

FV/K - PMT*(1/K-1)/I + PV = 0

Multiply by K, we recover the original TVM equation.

⇒ fv_i(n, a) = fv_i(n, FV/-PMT) = pv_i(-n, PV/PMT) = pv_i(-n, -a)

Since guess_i() is based on TVM, it can travel backward in time too.
Example taken from Fun Math Algorithms, car lease APR estimate.

lua> n, pv, pmt, fv = 36, 30000, -550, -15000
lua> guess_i(n, pv, pmt, fv) * 1200
6.9657545218584485
lua> guess_i(-n, fv, -pmt, pv) * 1200
6.9657545218584485

Update: numbers adjusted with updated guess_i()

[Albert Chan]

For getting interest rate for the general case, npmt() is very nice formula.

TVM equation, for "net payment", NPMT = NFV * i/(K-1):

Code:

require 'mathx'
expm1, log1p = mathx.expm1, mathx.log1p

function npmt(n,i,pv,pmt,fv)        -- npmt = (pv*K+fv) * i/(K-1) + pmt
    return ((fv+pv) / expm1(log1p(i)*n) + pv)*i + pmt
end

lua> n, pv, pmt, fv = 36, 30000, -550, -15000
lua> i1 = guess_i(n, pv, pmt, fv)
lua> p1 = npmt(n, i1, pv, pmt, fv)
lua> i1, p1
0.005804795434882038      -0.006442911074373114
lua> i2 = 1.001*i1                    -- p1<0, i1 is a bit low
lua> p2 = npmt(n, i2, pv, pmt, fv)
lua> i2, p2
0.005810600230316919      0.12838859778003098
lua> i1 - p1/(p2-p1) * (i2-i1)     -- secant's method
0.005805072816490459

For comparison, with same i1, i2, and secant's method, for other TVM formula:

0.005805072843740277           -- NPV formula
0.005805072788863714           -- NFV formula

Solve npmt()=0, using Newton's method, we have: (guess_i code here)

Code:

function loan_rate(n, pv, pmt, fv, i)
    i = i or guess_i(n, pv, pmt, fv)
    return function()
        local x = expm1(log1p(i)*n)
        local y = (fv+pv)/x
        local z = (1+1/x)*i/(1+i)*n - 1
        i = i - ((pv+y)*i+pmt) / (pv-y*z)   -- Newton's method
        return i
    end
end

lua> g = loan_rate(n, pv, pmt, fv)
lua> g(), g()
0.005805072819567153      0.005805072819420132

lua> g = loan_rate(-n, fv, -pmt, pv) -- time symmetric, as expected
lua> g(), g()
0.005805072819567159      0.005805072819420132

Update: numbers adjusted with updated guess_i()

[Albert Chan]

(12-07-2020 06:55 PM)Albert Chan Wrote:  lua> i2 = 1.001*i1                    -- p1<0, i1 is a bit low

We can avoid thinking about direction, and how much adjustment, for i2

If zero interest, npmt = (pv+fv)/n + pmt, we can use this to get i2

lua> n, pv, pmt, fv = 36, 30000, -550, -15000
lua> i1 = guess_i(n, pv, pmt, fv)
lua> p1 = npmt(n, i1, pv, pmt, fv)

lua> function secant(x1,y1,x2,y2) return x1 - y1/(y2-y1)*(x2-x1) end
lua> i2 = secant(i1, p1, 0, (pv+fv)/n + pmt)
lua> p2 = npmt(n, i2, pv, pmt, fv)
lua> i3 = secant(i2, p2, i1, p1)

lua> table.foreachi({i1,i2,i3}, print)
1      0.005804795434882038
2      0.005805075946792827
3      0.005805072819418476

[Albert Chan]

(12-05-2020 01:05 AM)Albert Chan Wrote:  From thread: Fun math algorithms:

(10-19-2020 11:05 PM)Albert Chan Wrote:  We can use \(C_±\) to solve Time value of money problem, if I is small, and N not too big.

XCas> C(d) := 1 + N*I*(N*I+d*6)/12 + I/2;       // compounding factor, d is direction
XCas> eqn := C(1)*PV + C(-1)*FV + N*PMT;       // variables follow cash flow sign convention

XCas> factor(symb2poly(eqn,I)); // quadratic coefficients, for I

[N^2*(FV+PV)/12 , (-FV*N+N*PV+FV+PV)/2 , N*PMT+FV+PV]

I noticed exact definition of C, if we take away N*I/2, it turned to even function (of N)

XCas> C := I*N/(1 - (1+I)^-N)
XCas> simplify(C - C(N=-N))      → N*I
XCas> series(C - N*I/2, I, 0, 3, polynom) // even powers of N, as expected

1 + I/2 + (-1+N^2)/12*I^2 + (1-N^2)/24*I^3

If we define Ce = C - I*N/2, we avoided thinking of "backward in time" with negative (N,PMT)
Quoted C(d) dropped the term I^2/12, we might as well put it back in.

XCas> Ce := 1 + I/2 + (N^2-1)/12*I^2 // truncated O(I^3) terms
XCas> NPMT := (Ce+N*I/2)*PV + (Ce-N*I/2)*FV + N*PMT
XCas> factor(e2r(NPMT,I)); // quadratic coefficients, for I

[(N^2-1)*(FV+PV)/12, (-FV*N+N*PV+FV+PV)/2 , N*PMT+FV+PV]

Only difference with previous version is N^2 replaced by (N^2-1)

Code:

function guess_i(n, pv, pmt, fv)    -- cash-flow sign convention
    local a = (fv+pv)/n
    local b = (fv-pv)-a
    local c = (a+pmt)/b
    b = (n*n-1)*a*c/b               -- I coef = [b/c/3, -2, 4c]
    return c * (b-3)/(b-1.5)        -- pade(4c/(1+sqrt(1-4b/3)),b,2,2)
end

Examples shown in this thread, estimated rate slightly improved.
With updated guess_i(n, pv, pmt, fv):

lua> guess_i(12, 5000, -500, 0) -- OP example
0.029209500177242102
lua> guess_i(12, 0, -400, 5000) -- OP example
0.007391057003072031
lua> guess_i(36, 30000, -550, -15000) -- car lease example
0.0058048457798059476

[Albert Chan]

We can estimate rate using simple formula, by "setting" PV=0, or FV=0

NPMT = (Ce+N*I/2)*PV + (Ce−N*I/2)*FV + N*PMT
= (Ce+N*I/2)*(PV+FV) + N*(PMT−FV*I)            // (PV,FV,PMT) → (PV+FV, 0, PMT−FV*I)
= (Ce−N*I/2)*(PV+FV) + N*(PMT+PV*I)            // (PV,FV,PMT) → (0, PV+FV, PMT+PV*I)

(05-10-2022 09:35 PM)Albert Chan Wrote:  lua> guess_i(36, 30000, -550, -15000) -- car lease example
0.005804845779805948

I = guess_i(36, 30000, -550, -15000) = guess_i(36, 15000, -550+15000*I, 0)      // "set" FV=0

N = 36
P = 15000 / (550-15000*I) = 1 / (11/300 - I)      → I = 11/300 - 1/P

(04-09-2022 12:50 PM)Albert Chan Wrote:  

(10-16-2020 04:02 PM)Albert Chan Wrote:  XCas> C := I*N / (1 - (1+I)^-N)       // C = |N*PMT/PV|, "compounding factor"
XCas> series(C,I,polynom)

\(1
+\frac{I(N+1)}{2}
+\frac{I^2 (N^2-1)}{12}
+\frac{I^3 (-N^2+1)}{24}
+\frac{I^4 (-N^4+20N^2-19)}{720}
+\frac{I^5 (N^4-10N^2+9)}{480}\)

This may be a better rate estimate, by dropping compounding factor O(I^2)
With previous defined P, solve for I, we have:

\(\displaystyle I ≈ \frac{2\;(N-P)}{P\;(N+1)}\)

We can solve I = 2*(N-P)/(P*(N+1)) for I, but there is no need.
For this formula, we had already done the work, when builiding guess_i()
guess_i() I coefs:

[(N^2-1)*(FV+PV)/12, (-FV*N+N*PV+FV+PV)/2 , N*PMT+FV+PV]

Drop I^2 terms, solve for I, we have Dieter's formula

I ≈ 2*(PV+FV + N*PMT) / (FV*(N-1) - PV*(N+1))
  = 2*(30000-15000 + 36*-550) / (-15000*35 - 30000*37)
  ≈ 0.005872

(04-09-2022 05:47 PM)Albert Chan Wrote:  \(\displaystyle I ≈ \frac{1}{P} - \frac{P}{N^2}\)

This is even better, and work well with big N.

I ≈ (11/300 - I) - 1/(11/300 - I) / 36^2
(2*I - 11/300)*36^2 * (I - 11/300) ≈ 1

2592*I^2 - 142.56*I + 0.7424 ≈ 0

Solve quadratics, for the small root, we have I ≈ 0.005824

Or, we can solve symbolically:

XCas> ICOEFS := [n^2*pv*fv, n^2*pmt*(-pv+fv), -(n*pmt-pv-fv)*(n*pmt+pv+fv)]
XCas> proot(ICOEFS(n=36, pv=30000, pmt=-550, fv=-15000))

[0.00582443202789,0.0491755679721] // keep the small root.

[Albert Chan]

(05-11-2022 01:07 PM)Albert Chan Wrote:  XCas> ICOEFS := [n^2*pv*fv, n^2*pmt*(-pv+fv), -(n*pmt-pv-fv)*(n*pmt+pv+fv)]
XCas> proot(ICOEFS(n=36, pv=30000, pmt=-550, fv=-15000))

[0.00582443202789,0.0491755679721] // keep the small root.

There is a weakness with I ≈ 1/P - P/N^2, extended with non-zero FV

For guess_i(), to avoid solving quadratics with "√", I use pade approximation.
(it is not just for speed, we simply don't want complex rate guess)

[b/3/c, -2, 4c]      → "small" root = 4c/(1+√(1-4b/3)) ≈ c*(b-3)/(b-1.5)

Updated guess_i() coefs (there was an eariler version with (N^2-1) replaced with N^2)

[(N^2-1)*(FV+PV)/12, (-FV*N+N*PV+FV+PV)/2 , N*PMT+FV+PV]

Above pade approx won't work if linear term goes 0, or FV = PV*(N+1)/(N-1)
With N≥1, this implied PV and FV have same sign, and FV bigger in size.
For rate with unique solution, this almost never happen      (*)

Now, compare I ≈ 1/P - P/N^2 derived rate coefs:

[PV*FV, -PMT*(PV-FV), ((PV+FV)/N+PMT)*((PV+FV)/N+PMT)]

If PMT=0 or PV=FV, linear term goes 0
Even if linear term is nonzero (but small, relative to others), pade approximation may fail.

Example: N=32, PV=-6000, PMT=1, FV=10000, True Rate = 0.016223

With guess_i() coefs:

[341000,-254000,4032] --> roots = [0.016228, 0.728640] --> paded = 0.016227

With "1/P-P/N^2" coefs:

[-60000000,16000,15876] --> roots = [-0.016134, 0.016400] --> paded = -0.488317 ?

"Small" root is the wrong root. How to pick ?
Apply pade on wrong root make it worse

c = -0.4796125
b = -11162.8125
c*(b-3)/(b-1.5) ≈ c*b/b = c

Normally, we expected c*(b-3)/(b-1.5) ≈ c*3/1.5 = 2c
Example, for guess_i(), with n,pv,pmt,fv = N,P,-1,0

c = (N-P)/(P*(N+1))
b = (1-N)*c

2c matched a good rate guess formula

(04-09-2022 12:50 PM)Albert Chan Wrote:  

(10-16-2020 04:02 PM)Albert Chan Wrote:  XCas> C := I*N / (1 - (1+I)^-N)       // C = |N*PMT/PV|, "compounding factor"
XCas> series(C,I,polynom)

\(1
+\frac{I(N+1)}{2}
+\frac{I^2 (N^2-1)}{12}
+\frac{I^3 (-N^2+1)}{24}
+\frac{I^4 (-N^4+20N^2-19)}{720}
+\frac{I^5 (N^4-10N^2+9)}{480}\)

This may be a better rate estimate, by dropping compounding factor O(I^2)
With previous defined P, solve for I, we have:

\(\displaystyle I ≈ \frac{2\;(N-P)}{P\;(N+1)}\)

(*) Rate is unique if 1 sign change: (PV, PMT+FV) with opposite sign.

(01-06-2020 01:57 PM)Albert Chan Wrote:  Some real roots for rate are meaningless, say with r ≤ -100%.
Let x = 1+r, and consider only positive x as valid, we get:

\( NFV = FV + PV x^n + PMT\left({x^n-1 \over x-1}\right)\)

\(\large {NFV \over PMT} = \left({PV \over PMT}\right) x^n + x^{n-1} + x^{n-2} + \;... +\;x + \left(1 + {FV \over PMT}\right)\)

If above has one sign change, we have exactly one positive solution for x.

For 2 sign changes, x has 0 or 2 positive roots (see Descartes' sign rules)