Depreciation - different results on HP30b and on paper

[FabioBrasil]

Hello.
I come again to kindly ask for help from you who know a lot about HP. I cannot make the result of the calculation of decreasing depreciation (DecBal) provide the same result on the HP30b calculator and on paper.

The problem consists of the following:
- Cost (Czero)= $ 25000
- Salvage = $ 5000
- Life = 15 years
- Factor =10.174%

On paper D (n) = Czero * (1- Factor) ^ n
in year 1 I have 22,456.50
in year 3 I have 18,119.49
in year 6 I have 13,132.64
in year 12 I have 6,898.65
(n = year use as Factor 10.174%)

On the HP30b calculator (use as Factor 10.174%):
in year 1 I have 24,830.70 (R.Book)
in year 3 I have 24,494.94
in year 6 I have 24,000.08
in year 12 I have 23,040.16
Where am I going wrong? I couldn't get the answer on the Learning Modules.

Thanks!

[SlideRule]

Methods of Depreciation: Formulas, Problems, and Solutions

Straight Line Method

Declining Balance Method
Declining Balance Method is sometimes called the Constant-Percentage Method or the Matheson formula. The assumption in this depreciation method is that the annual cost of depreciation is the fixed percentage (1 - K) of the Book Value (BV) at the beginning of the year.
Annual Rate of Depreciation(K): SV = FC (1 - K)ⁿ

source:Depreciation Methods in Engineering Economics

Any SME's with the formula/equation deployed in the HP-30B ?

BEST!
SlideRule

[Werner]

(25000 - 5000)*(1 - 10.174/1200) + 5000 = 24830.43
So it would seem it works per month i.o. per year.
Werner

[toml_12953]

(03-12-2021 12:05 AM)FabioBrasil Wrote:  Hello.
I come again to kindly ask for help from you who know a lot about HP. I cannot make the result of the calculation of decreasing depreciation (DecBal) provide the same result on the HP30b calculator and on paper.

The problem consists of the following:
- Cost (Czero)= $ 25000
- Salvage = $ 5000
- Life = 15 years
- Factor =10.174%

On paper D (n) = Czero * (1- Factor) ^ n
in year 1 I have 22,456.50
in year 3 I have 18,119.49
in year 6 I have 13,132.64
in year 12 I have 6,898.65
(n = year use as Factor 10.174%)

On the HP30b calculator (use as Factor 10.174%):
in year 1 I have 24,830.70 (R.Book)
in year 3 I have 24,494.94
in year 6 I have 24,000.08
in year 12 I have 23,040.16
Where am I going wrong? I couldn't get the answer on the Learning Modules.

Thanks!

It looks like your paper version is correct.

Code:


DECLARE FUNCTION D

INPUT PROMPT "Cost? ":Czero
INPUT PROMPT "Salvage? ":Salvage
INPUT PROMPT "Factor? ":Factor

LET n = 1

DO UNTIL D(n) <= Salvage
   PRINT n,USING$("$##,###.##",D(n))
   LET n = n + 1
LOOP

FUNCTION D(n)
   LET D=Czero*(1-Factor)^n
END FUNCTION

END

Cost? 25000
Salvage? 5000
Factor? .10174
 1                      $22,456.50
 2                      $20,171.78
 3                      $18,119.50
 4                      $16,276.02
 5                      $14,620.10
 6                      $13,132.65
 7                      $11,796.53
 8                      $10,596.35
 9                      $ 9,518.28
 10                     $ 8,549.89
 11                     $ 7,680.03
 12                     $ 6,898.66
 13                     $ 6,196.79
 14                     $ 5,566.33
 15                     $ 5,000.01

[SlideRule]

FabioBrasil

Did we help you answer your question?

BEST!
SlideRule

[FabioBrasil]

Thank you for your help! It is good to have good people.

[paul0207]

Fabio,
This is how I get the same answer on the HP12c (I don't have an hp30b to check it out, sorry).
25000 -> PV
15 -> n
152.61 -> i (that is 10.174 * 15)
Both PMT and FV = 0.
Then 1 DB yields
2543.50
x<-->y
22456.50 (which is your first year answer on paper).

Summing up, try i=152.61 (i*n) on the hp30b for the Declining Balance method.

Hope this helps,

Paul

[Albert Chan]

(03-13-2021 04:25 AM)paul0207 Wrote:  Summing up, try i=152.61 (i*n) on the hp30b for the Declining Balance method.

Yes! This is from HP-12C Appendix D, formula used for DecBal:

DPN_j = RBV_j-1 * FACT / (100L), for j = 1, 2, ..., L

We can also consider DecBal as TVM problem, with negative interest rate:

10.174 [CHS] [I]
25000 [CHS] [PV]
0 [PMT]

Book values, for different years:

1 [N] [FV]       → 22456.50
3 [N] [FV]       → 18119.50
6 [N] [FV]       → 13132.65
12 [N] [FV]      →  6898.66
15 [N] [FV]      →  5000.01

We can derive book value formula directly, by fitting the power law:

book_value / cost = (salvage / cost) ^ (years / life)

when exponent = 0, book_value = cost
when exponent = 1, book_value = salvage

For above example, we have:

book_value = 25000 * 0.2 ^ (years/15)

[FabioBrasil]

Paul0207, Thank you very much!
You solved the problem with Factor = rate * lifetime. In the Declining Balance mode (DecBal) this made me arrive at the same values of the exercise. Many, many thanks to all the colleagues who taught me a lot!

Paul0207, most things are the same on the HP12C and the HP30b. So much the same, that I use programs written for the HP12c on the HP30b with few modifications. And you don't need to apologize for anything, I appreciate your help!