Discount Rate
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04-04-2022, 01:54 PM
Post: #1
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Discount Rate
Find attached a two-page extract from Fundamentals of Infrastructure Engineering, Appendix A (pgs. 445 & 447) with respect to Computing the Discount Rate utilizing a generic calculator adaptation of the Newton-Raphson algorithm. The article suggests this listing needs only minor alterations for RPN deployment. We all await the response from the readership to this challenge.
[attachment=10502] [attachment=10503] BEST! SlideRule |
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04-08-2022, 07:44 PM
(This post was last modified: 04-09-2022 08:00 AM by Thomas Klemm.)
Post: #2
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RE: Discount Rate
There is a bug in the listing.
When calculating \( - n (i + 1)^{n - 1} \) the multiplication by \( n \) is missing. In the third column we find: Code: rcl 9 2 Instead it should be: Code: rcl 9 2 Quote:The machine should have a capacity of 98 keystrokes to perform this routine. With this fix we are at 100 keystrokes. Instead of using this suggestion I would rather follow how this is done in the HP-80. Initial Value \[i_0 = \frac{2(n-P)}{n(n+1)}\] Newton Iteration \[ \begin{align} Z &= (1 + i)^{-n} \\ \\ i_\text{next} &= i \left[ 1 - \frac{iP - (1 - Z)}{1 - Z - \frac{niZ}{1 + i}} \right] \end{align} \] Register R0: n R1: P = PV / PMT R2: i R3: u = 1 + i R4: Z = u-n R5: v = Z - 1 Program This program is for the HP-25, but can easily be adapted for other models: Code: 01: 24 00 : RCL 0 ; n Quote:The procedure is terminated when two successive values differ by less than a pre-set amount. This test can easily be built-in to the program by a conditional if-then-else command sequence if desired and if sufficient capacity is available. A round-off command should be programmed just prior to the test, of course. Example: True Interest Rate of a Loan What is the true interest rate on a 36 month loan for $3000 having monthly payments of $100? FIX 6 36 STO 0 3000 ENTER 100 / STO 1 CLEAR PRGM R/S 0.010190 R/S 0.010207 R/S 0.010207 Thus, after only three iterations, we get the result. We can now calculate the annual percentage rate: FIX 2 1200 × 12.25 References |
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04-08-2022, 10:21 PM
Post: #3
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RE: Discount Rate
Although time flow only from the present to the future, formula does not care.
We can use the *same* code, getting interest rate from PMT, FV (now, with PV=0) Just have the code run with negative input, traveling back in time Example, lets try n=36, PMT=100, FV=5000, find i lua> n = -36 -- note the negative sign, for both n and p lua> p = -5000/100 lua> i = 2*(n-p)/(n*(n+1)) lua> i 0.022222222222222223 Guess i formula is not designed for FV, but lets use it anyway. With better guess_i(n,pv,pmt,fv), we get guess i = 0.0179687 lua> z = (1+i)^-n lua> i = i - i*(i*p-(1-z))/(1-z-n*i*z/(1+i)) lua> i 0.018164680329101488 lua> z = (1+i)^-n lua> i = i - i*(i*p-(1-z))/(1-z-n*i*z/(1+i)) lua> i 0.017958086653757012 lua> z = (1+i)^-n lua> i = i - i*(i*p-(1-z))/(1-z-n*i*z/(1+i)) lua> i 0.017957584809347397 |
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04-09-2022, 12:50 PM
(This post was last modified: 04-09-2022 01:24 PM by Albert Chan.)
Post: #4
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RE: Discount Rate
(10-16-2020 04:02 PM)Albert Chan Wrote: XCas> C := I*N / (1 - (1+I)^-N) // C = |N*PMT/PV|, "compounding factor" This may be a better rate estimate, by dropping compounding factor O(I^2) With previous defined P, solve for I, we have: \(\displaystyle I ≈ \frac{2\;(N-P)}{P\;(N+1)}\) For previous 2 examples, formula give slightly better guess rate. Newton steps (denoted by →) converge faster. I(N= 36, P= 3000/100) ≈ 0.010811 → 0.010203 → 0.010207 I(N=-36, P=-5000/100) ≈ 0.016000 → 0.018003 → 0.017958 |
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04-09-2022, 05:47 PM
Post: #5
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RE: Discount Rate
(02-13-2020 10:42 AM)Gamo Wrote: To find the guess the HP-55 programs book used these formulas. Using our symbols here, we have: \(\displaystyle I ≈ \frac{1}{P} - \frac{P}{N^2}\) This is even better, and work well with big N. With above guess rate formula, we have: I(N= 36, P= 3000/100) ≈ 0.010185 → 0.010207 I(N=-36, P=-5000/100) ≈ 0.018580 → 0.017962 → 0.017958 Does anyone know how this is derived ? This is as good as first pade approximation of reverted C, but less complicated. \(\displaystyle I ≈ \frac{6\;(N-P)} {N\;(N-1) + 2P\;(N+2)}\) With above guess rate formula, we have: I(N= 36, P= 3000/100) ≈ 0.010169 → 0.010207 I(N=-36, P=-5000/100) ≈ 0.017751 → 0.017958 |
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04-10-2022, 04:03 AM
Post: #6
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RE: Discount Rate
(04-09-2022 05:47 PM)Albert Chan Wrote: \(\displaystyle I ≈ \frac{1}{P} - \frac{P}{N^2}\) (10-16-2020 04:02 PM)Albert Chan Wrote: XCas> C := I*N / (1 - (1+I)^-N) // C = |N*PMT/PV|, "compounding factor" This is my guess on how formula is derived, using continuous compounding. With small I, (1+I)^-N = exp(ln(1+I)*-N) ≈ exp(-I*N) Let x = I*N, and solve for x C = N/P ≈ x / (1 - e^-x) x = I*N = C + W(-C*exp(-C)) // W has 2 branches, but it does not matter here Let y = -C*exp(-C), solve for I I ≈ 1/P + W(y)/N If we can show -W(y) ≈ 1/C, we have I ≈ 1/P - 1/(CN) = 1/P - P/N^2, and prove is done. There are 2 LambertW branches, but we only want estimate when C close to 1 Both branches behave about the same, because y is very close to -1/e C = 1 + ε → y = 1/e * (-1 + ε^2/2 - ε^3/3 + ε^4/8 - ...) I borrowed lyuka e^W estimation formula, replace 0.3 by 1/3, for Puiseux series. e^W(y) ≈ 1/e + sqrt ((2/e)*(y+1/e)) + (y+1/e)/3 -W(y) = 1 - ln(e * e^W(y)) = 1 - ln(1 + sqrt(2*(e*y+1)) + (e*y+1)/3) = 1 - ln(1 + ε - 1/6*ε^2 + ...) = 1 - 2*atanh(ε/2 - 1/3*ε^2 + ...) ≈ 1 - ε + 2/3*ε^2 ≈ 1/(1+ε) = 1/C QED To confirm math is correct, at least numerically. lua> z = 1e-4 lua> c = 1 + z lua> y = -c*exp(-c) lua> -W(y), 1 - z + 2/3*z^2, 1/c 0.9999000066665401 0.9999000066666667 0.9999000099990001 |
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04-10-2022, 12:19 PM
(This post was last modified: 04-12-2022 11:16 PM by Thomas Klemm.)
Post: #7
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RE: Discount Rate
After a closer inspection of the assembler code I noticed that in fact the following formula is used:
Newton Iteration \[ \begin{align} Z &= (1 + i)^{-n} \\ \\ i_\text{next} &= i \left[ 1 + \frac{P + \frac{Z - 1}{i}}{\frac{Z - 1}{i} + \frac{nZ}{1 + i}} \right] \end{align} \] This saves two multiplications by \( i \) at the cost of one division. Register R0: n R1: P = PV / PMT R2: i R3: u = 1 + i R4: Z = u-n R5: v = (Z - 1) / i Program It allows to reduce the program by 2 steps: Code: 01: 24 00 : RCL 0 |
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04-10-2022, 11:29 PM
(This post was last modified: 04-10-2022 11:29 PM by Eddie W. Shore.)
Post: #8
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RE: Discount Rate
HP 32S/HP 32SII code:
N = n (R0) P = PV/PMT (R1) I = i (R2) U = 1+i (R3) Z = u^(-n) (R4) V = Z - 1 (R5) 1. Press XEQ I, enter N, enter PV N = n; P = PV/PMT (or) N = -n; P = -FV/PMT 2. Press XEQ J 3. Keep pressing R/S until desired accuracy is reached Code:
Size: I: 18.0 bytes, Checksum 32SII: 82BD J: 39.0 bytes, Checksum 32SII: 3EBB 57 bytes total |
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04-11-2022, 01:18 AM
Post: #9
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RE: Discount Rate
(04-10-2022 12:19 PM)Thomas Klemm Wrote: Newton Iteration When interest rate converging to true rate, numerator goes 0, or (Z-1)/i = -P I was reading https://brownmath.com/bsci/loan.htm#Newton. It's rate newton formula actually replaced denominator (Z-1)/i by -P The formula quoted above is good when guess rate is under-estimated. This explained why it has i0 = 2*(n-p)/(n*(n+1)) instead of 2*(n-p)/(p*(n+1)) Too high a guess can make Newton's step diverge. Example, with N=36, P=30, guess rate of 0.08, it barely able to converge. 0.08 → -0.071887 → -0.041499 → -0.014129 → 0.003911 → 0.009742 → 0.010205 → 0.010207 "-P" version had the opposite problem, and preferred over-estimated guess. Again, with N=36, P=30, guess rate of 0.004, it totally diverged 0.004 → -0.009120 → -0.003414 → -0.000780 → -0.000059 ... A better option is to solve NPMT=0 instead (now used for rate search in Plus42, see tvm_eq()) From NPMT=0 based loan_rate(n, pv, pmt, fv, i), simplified with FV=0, we get this: \( \displaystyle \large i_\text{next} = i \left[ 1 - \frac{(\frac{Z - 1}{i})/P + 1} {(\frac{nZ}{1 + i})/(\frac{Z - 1}{i}) +1} \right] \) This version, convergence is slightly faster, and *very* stable for guess i Again, with N=36, P=30, but ridiculous rate guess: 10+10 → 0.033333 → 0.012232 → 0.010228 → 0.010207 10−10 → 0.000231 → 0.010782 → 0.010209 → 0.010207 |
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04-11-2022, 02:37 AM
Post: #10
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RE: Discount Rate
(04-10-2022 11:29 PM)Eddie W. Shore Wrote: There's a 1 missing in line J22. Using my 2nd program could possibly save a byte or two. You could also consider using the solver. This made me wonder if there's a programmable HP calculator with recall arithmetic but without a solver (generic or specific for the TVM problem). |
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04-11-2022, 03:11 PM
(This post was last modified: 04-12-2022 03:53 PM by Albert Chan.)
Post: #11
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RE: Discount Rate
(04-11-2022 01:18 AM)Albert Chan Wrote: The formula [OP] is good when guess rate is under-estimated. To be more precise, OP formula preferred rate magnitude under-estimated. Plotting errors of both rate estimate, error(n version) ≈ -2 * error(p version) In other words, this is much better rate estimate. \(\displaystyle I_0 = \frac{2\;(N-P)}{N+1} \left( \frac{{1 \over N} + {2 \over P}}{3} \right) \) Examples: I0(N= 36, P= 30) = .010210 // true rate = .010207 I0(N=-36, P=-50) = .018074 // true rate = .017958 Nice. But, for OP formula, we wanted guess rate under-estimated. Perhaps give up some accuracy for safety in convergence, and make ratio 1:1 ? While we are at it, why not assume N+1 ≈ N, to simplify further ? \(\displaystyle I_0 = \frac{2\;(N-P)}{N} \left( \frac{{1 \over N} + {1 \over P}}{2} \right) = \frac{1}{P} - \frac{P}{N^2} \) This may be how I0 = 1/P - P/N² comes from Update: Perhaps formula designed also to match asymptote, when C is big. When compounding factor C is big, literally all payments goes to paying interest. C = N/P = I*N/(1-(1+I)^-N) ≈ N*I → I = 1/P Rate formula matched this behavior: I0 = 1/P - P/N^2 = (1 - 1/C^2)/P ≈ 1/P For better estimate, we can use continuous compouding for I, derived earlier. (04-10-2022 04:03 AM)Albert Chan Wrote: Let y = -C*exp(-C), solve for I When C = N/P is big, y is small, we have W(y) ≈ y I ≈ C/N + (-C*exp(-C))/N = (1-exp(-N/P)) / P |
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04-11-2022, 03:57 PM
Post: #12
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RE: Discount Rate
(04-11-2022 03:11 PM)Albert Chan Wrote: Plotting errors of both rate estimate, error(n version) ≈ -2 * error(p version) Instead of looking at plots, lets look at reverted series of I, from C = N/P More specifically, from u = (C-1)/(N+1) From A Series of Interest, by David W. Cantrell I = (2*u)*(1 - (N-1)/3*u * (1 - (2*N+1)/3*u * (1 - (11*N+7)/15*u * (1 - ... Assume 2 terms of I is good enough for estimating true I Assume N big enough, so that N+1 ≈ N-1 ≈ N Let C = 1 + ε I/(2u) ≈ 1 − (N/3) * (ε/N) = 1 − 1/3*ε I/(2u/C) ≈ (1+ε) * (1-ε/3) ≈ 1 + 2/3*ε Rate guess, p version is (2u), n version is (2u/C), rel error ratio ≈ -2 QED |
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04-11-2022, 08:48 PM
Post: #13
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RE: Discount Rate
Comparing the code in HP Journal to the US patent, I noticed something slightly interesting:
In the US patent DOWN ROTATE is used twice in several places, while in HP Journal a JSB DR2 is used. Here are the corresponding code snippets for comparison: US Patent Code: L3222: .....1...1 → L3004 JSB XTY HP Journal Code: L3224: .....1...1 → L3004 JSB XTY While I couldn't find the DR2 label in ROM 3 of the US patent, I was able to find similar code in ROM 5: Code: L5071: 11..1.1... DOWN3 : DOWN ROTATE Both DOWN2 and DOWN3 are used in this ROM. I wasn't aware of several ROM versions of the HP-80. Based on Bernhard's statement I assume that the code from the US patent is used in the HP-80: Quote:I compared my HP-80 ROM code and found, that is was correct. While some of the ROMs can be downloaded, the HP-80 unfortunately is missing. Therefore I can not verify that. If anyone can shed some light on the subject I would be very grateful. |
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04-12-2022, 11:11 PM
Post: #14
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RE: Discount Rate
(04-10-2022 12:19 PM)Thomas Klemm Wrote: After a closer inspection of the assembler code I noticed that in fact the following formula is used: For this I wrote a simulator for the CPU, similar to the Code Analyzer for HP Calculators: Code: @trace First we create some symbols: Code: from sympy import symbols, simplify, expand, latex Then we reset the registers: Code: A, B, C, D, E, F, M = [0] * 7 And now we just follow the assembler code. From time to time, we reassign variables to keep the expressions simple. PS: I had to split this post since it was too big. |
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04-12-2022, 11:13 PM
(This post was last modified: 04-13-2022 01:26 AM by Thomas Klemm.)
Post: #15
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RE: Discount Rate
Code: C, D, E = PV, PMT, n A: 0 B: 0 C: PV D: PV E: PMT F: n M: 0 : C_STACK A: PV B: 0 C: PV D: PMT E: n F: n M: 0 : STACK_A A: PV B: 0 C: PMT D: n E: n F: PV M: 0 : DOWN_ROTATE A: PV/PMT B: PMT C: PV/PMT D: n E: n F: PV M: 0 : DIV Code: print(latex(C)) \frac{PV}{PMT} \[ P = \frac{PV}{PMT} \] Code: A = C = P A: P B: PMT C: 0 D: n E: n F: PV M: P : C_EXCHANGE_M A: P B: PMT C: n D: n E: PV F: 0 M: P : DOWN_ROTATE A: P B: PMT C: n D: n E: n F: PV M: P : C_STACK A: n B: PMT C: 1 D: n E: n F: PV M: P : ONE A: n + 1 B: n C: n + 1 D: n E: n F: PV M: P : ADD A: n B: n + 1 C: n + 1 D: n E: n F: PV M: P : A_EXCHANGE_B A: n*(n + 1) B: n C: n*(n + 1) D: n E: n F: PV M: P : MPY A: n*(n + 1) B: n C: P D: n E: n F: PV M: n*(n + 1) : C_EXCHANGE_M A: n B: n C: P D: n E: PV F: PV M: n*(n + 1) : STACK_A A: n B: n C: P D: P E: n F: PV M: n*(n + 1) : C_STACK A: n B: n C: P D: P E: -n F: PV M: n*(n + 1) : CSN A: n B: n C: P D: -n E: PV F: P M: n*(n + 1) : DOWN_ROTATE A: -P + n B: n C: -P + n D: -n E: PV F: P M: n*(n + 1) : SUB A: -2*P + 2*n B: -P + n C: -2*P + 2*n D: -n E: PV F: P M: n*(n + 1) : ADD A: -2*P + 2*n B: -P + n C: n*(n + 1) D: -n E: PV F: P M: -2*P + 2*n : C_EXCHANGE_M A: (-2*P + 2*n)/(n*(n + 1)) B: n*(n + 1) C: (-2*P + 2*n)/(n*(n + 1)) D: -n E: PV F: P M: -2*P + 2*n : DIV Code: print(latex(simplify(C))) \frac{2 \left(- P + n\right)}{n \left(n + 1\right)} \[ i = \frac{2 \left(- P + n\right)}{n \left(n + 1\right)} \] Code: A = C = i A: i B: n*(n + 1) C: -2*P + 2*n D: -n E: PV F: P M: i : C_EXCHANGE_M A: i B: n*(n + 1) C: -n D: PV E: P F: -2*P + 2*n M: i : DOWN_ROTATE A: i B: n*(n + 1) C: i D: PV E: P F: -2*P + 2*n M: -n : C_EXCHANGE_M A: PV B: n*(n + 1) C: i D: P E: -2*P + 2*n F: -2*P + 2*n M: -n : STACK_A A: i B: n*(n + 1) C: 1 D: P E: -2*P + 2*n F: -2*P + 2*n M: -n : ONE A: i + 1 B: i C: i + 1 D: P E: -2*P + 2*n F: -2*P + 2*n M: -n : ADD A: i + 1 B: i C: i + 1 D: i + 1 E: P F: -2*P + 2*n M: -n : C_STACK A: i + 1 B: i C: i + 1 D: i + 1 E: i + 1 F: P M: -n : C_STACK A: i + 1 B: i C: i D: i + 1 E: i + 1 F: P M: -n : B_C A: i + 1 B: i C: -n D: i + 1 E: i + 1 F: P M: i : C_EXCHANGE_M A: (i + 1)**(-n) B: i + 1 C: (i + 1)**(-n) D: -n E: i + 1 F: P M: i : XTY Code: print(latex(C)) \left(i + 1\right)^{- n} \[ Z = \left(i + 1\right)^{- n} \] Code: A = C = Z A: Z B: i + 1 C: -n D: i + 1 E: P F: Z M: i : DOWN_ROTATE A: Z B: i + 1 C: -n D: -n E: i + 1 F: P M: i : C_STACK A: -Z*n B: Z C: -Z*n D: -n E: i + 1 F: P M: i : MPY A: -Z*n B: Z C: Z D: -n E: i + 1 F: P M: i : B_C A: -Z*n B: Z C: -n D: i + 1 E: P F: Z M: i : DOWN_ROTATE A: -Z*n B: Z C: i + 1 D: P E: Z F: -n M: i : DOWN_ROTATE A: -Z*n/(i + 1) B: i + 1 C: -Z*n/(i + 1) D: P E: Z F: -n M: i : DIV A: -Z*n/(i + 1) B: i + 1 C: P D: Z E: -n F: -Z*n/(i + 1) M: i : DOWN_ROTATE A: -Z*n/(i + 1) B: i + 1 C: Z D: -n E: -Z*n/(i + 1) F: P M: i : DOWN_ROTATE A: Z B: i + 1 C: 1 D: -n E: -Z*n/(i + 1) F: P M: i : ONE A: Z - 1 B: Z C: Z - 1 D: -n E: -Z*n/(i + 1) F: P M: i : SUB A: Z - 1 B: Z C: i D: -n E: -Z*n/(i + 1) F: P M: i : M_C A: (Z - 1)/i B: i C: (Z - 1)/i D: -n E: -Z*n/(i + 1) F: P M: i : DIV A: (Z - 1)/i B: i C: -n D: -Z*n/(i + 1) E: P F: (Z - 1)/i M: i : DOWN_ROTATE A: (Z - 1)/i B: i C: -Z*n/(i + 1) D: P E: (Z - 1)/i F: -n M: i : DOWN_ROTATE A: Z*n/(i + 1) + (Z - 1)/i B: (Z - 1)/i C: Z*n/(i + 1) + (Z - 1)/i D: P E: (Z - 1)/i F: -n M: i : SUB A: P B: (Z - 1)/i C: Z*n/(i + 1) + (Z - 1)/i D: (Z - 1)/i E: -n F: -n M: i : STACK_A A: P B: (Z - 1)/i C: Z*n/(i + 1) + (Z - 1)/i D: Z*n/(i + 1) + (Z - 1)/i E: (Z - 1)/i F: -n M: i : C_STACK A: P B: Z*n/(i + 1) + (Z - 1)/i C: (Z - 1)/i D: Z*n/(i + 1) + (Z - 1)/i E: (Z - 1)/i F: -n M: i : B_EXCHANGE_C A: P + (Z - 1)/i B: P C: P + (Z - 1)/i D: Z*n/(i + 1) + (Z - 1)/i E: (Z - 1)/i F: -n M: i : ADD A: P + (Z - 1)/i B: P C: Z*n/(i + 1) + (Z - 1)/i D: (Z - 1)/i E: -n F: P + (Z - 1)/i M: i : DOWN_ROTATE A: P + (Z - 1)/i B: Z*n/(i + 1) + (Z - 1)/i C: P D: (Z - 1)/i E: -n F: P + (Z - 1)/i M: i : B_EXCHANGE_C A: P + (Z - 1)/i B: Z*n/(i + 1) + (Z - 1)/i C: (Z - 1)/i D: -n E: P + (Z - 1)/i F: P M: i : DOWN_ROTATE A: P + (Z - 1)/i B: (Z - 1)/i C: Z*n/(i + 1) + (Z - 1)/i D: -n E: P + (Z - 1)/i F: P M: i : B_EXCHANGE_C A: (P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) B: Z*n/(i + 1) + (Z - 1)/i C: (P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) D: -n E: P + (Z - 1)/i F: P M: i : DIV A: (P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) B: Z*n/(i + 1) + (Z - 1)/i C: i D: -n E: P + (Z - 1)/i F: P M: i : M_C A: i*(P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) B: (P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) C: i*(P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) D: -n E: P + (Z - 1)/i F: P M: i : MPY A: i*(P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) B: (P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) C: i D: -n E: P + (Z - 1)/i F: P M: i*(P + (Z - 1)/i)/(Z*n/(i + 1) + (Z - 1)/i) : C_EXCHANGE_M Code: print(latex((M))) \frac{i \left(P + \frac{Z - 1}{i}\right)}{\frac{Z n}{i + 1} + \frac{Z - 1}{i}} \[ \Delta i = \frac{i \left(P + \frac{Z - 1}{i}\right)}{\frac{Z n}{i + 1} + \frac{Z - 1}{i}} \] Code: A = M = Δi A: i + Δi B: Δi C: i + Δi D: -n E: P + (Z - 1)/i F: P M: Δi : ADD A: i + Δi B: Δi C: Δi D: -n E: P + (Z - 1)/i F: P M: i + Δi : C_EXCHANGE_M Code: print(latex((M))) i + Δi \[ i_\text{next} = i + \Delta i \] |
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04-18-2022, 01:58 PM
Post: #16
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RE: Discount Rate
(04-11-2022 08:48 PM)Thomas Klemm Wrote: If anyone can shed some light on the subject I would be very grateful. Meanwhile I found the following comment in asm/80.asm of nonpareil: Quote:; The code from the patent may not match released code in actual HP-80Obviously I'm not the first to notice this discrepancy. |
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04-18-2022, 06:41 PM
Post: #17
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RE: Discount Rate
On (far too) many occasions, when I think I may be the first to discover something, upon checking, I have found that Eric had found it, documented it, shared it, and just as likely forgotten it, long before I got there.
Same with Joe Horn... <sigh>... --Bob Prosperi |
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04-18-2022, 07:03 PM
Post: #18
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RE: Discount Rate
Well, it's not the first time I've taken a close look at the assembly code from this patent.
Interestingly the bug is still there in asm/80.asm: Code: dd2: shift right c[w] |
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05-11-2022, 06:22 PM
Post: #19
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RE: Discount Rate
(04-11-2022 03:11 PM)Albert Chan Wrote: Plotting errors of both rate estimate, error(n version) ≈ -2 * error(p version) There are better ways to compensate the 1:2 error ratio. I recently updated guess_i(n, pv, pmt, fv) that keep C series terms upto I^2 Converting it with FV=0, I1 = guess_i(N, P, -1, 0): Rough estimate: \(\displaystyle I_0 = \frac{2\;(N-P)}{P(N+1)} \) With correction: \(\displaystyle I_1 = I_0 \left( \frac{6 + 1\;(n-1)\;I_0}{6 + 2\;(n-1)\;I_0} \right) = I_0 - \frac{I_0}{2 + \large \frac{6}{(N-1)\;I_0}} \) Redo examples: N= 36, P= 30 → I0 = .010811 → I1 = .010205 N=-36, P=-50 → I0 = .016000 → I1 = .017967 |
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