hp-19c & 29c solutions finance 1977 b&w

7/29/2019 HP-19C & 29C Solutions Finance 1977 B&W

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INTRODUCTION

This HP-19C/HP-29C Solutions book was written to help you get the most from your calculator.The programs were chosen to provide useful calculations for many of the common problemsencountered.They will provide you with immediate capabilities in your everyday calculations and you willfind them useful as guides to programming techniques for writing your own customized software.The comments on each program listing describe the approach used to reach the solution andhelp you follow the programmer's logic as you become an expert on your HP calculator.You will find general information on how to key in and run programs under "A Word about ProgramUsage" in the Applications book you received with your calculator.We hope that this Solutions book will be a valuable tool in y ~ u r work and would appreciate yourcomments about it.

The program material contained herein is supplied without representation or warranty of any kind.Hewlett-Packard Company therefore assumes no responsibility and shall have no liability,consequential or otherwise, of any kind arising from the use of this program material or any partthereof.

HEWLETT PACKARD COMPA NY 1977


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TABLE OF CONTENTS

I. DIRECT REDUCTION LOAN AMORTIZATION SCHEDULE 1This program generates the values for a loan amortization schedule2.

given the interest rate, amount of payment, loan amount and period.INTERNAL RATE OF RETURN -- UP TO 26 CASH FLOWS This program calculates the periodic internal rate of return giventhe initial investment and up to 26 uneven cash flows.

. 5

3. STRAIGHT LINE DEPRECIATION SCHEDULE . . 8This program generates the values for a depreciation schedule using4.

the straight line method given the starting book values, salvagevalue, and useful life expectancy.SUM-OF-THE-YEARS'-DIGITS DEPRECIATION SCHEDULE I I This program generates the values for a depreciation schedule usingthe sum-of-the-years'-digits method, given the starting book value,salvage value, and useful life expectancy.

.12

5. VARIABLE RATE DECLINING BALANCE DEPRECIATION SCHEDULE 16This program generates the values for a depreciation schedule usingthe declining balance method, given the starting book value, salvagevalue, useful life expectancy, and declining rate factor.6. CROSSOVER POINT--DECLINING BALANCE TO STRAIGHT LINE . 20This program calculates the optimum point in the useful life where aswitch from the declining balance method to the straight line methodshould be made given the starting book value, salvage value, usefullife expectancy, and declining balance factor.7. NOMINAL TO EFFECTIVE/EFFECTIVE TO NOMINAL RATE CONVERSION 24Given either the nominal or effective rate (and the number of compounding periods in the case of finite compounding) the other rate iscalculated for either finite or continuous compounding.8.

9.

LEASE VERSUS PURCHASE . . I I I .27This program calculates the present value of the cost of purchasing,the present value of the cost of leasing, and the net differencebetween the two.REAL ESTATE RENTAL INVESTMENT ANALYSIS I I This program calculates monthly rent or cash flow, gross return oninvestment, and taxable income.

30

10. BREAK-EVEN ANALYS IS . . . . . . .' 33This program calculates the break-even point, the margin of safetyratio and profit, given total fixed costs, variable cost per unit,sales price per unit, and expected sales.


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DIRECT REDUCTION LOANAMORTIZATION SCHEDULELoan Amt: $45,000 APR: 9.25%Monthly Payment: $385Payment Paid to Paid toNumber Interest Principal

13 343.20 41.8014 342.87 42.1315 342.55 42.4518 341.56 43.44

13 - 18 2054.29 255.71

RemainingBalance44480.8044438.6744396.22

.- -----44266.8944266.89

Given the periodic interest rate (i),periodic payment amount (PMT), loanamount (PV) and beginning and endingperiods (PI' P2 ) , this program willgenerate the values for a loan amortization schedule as pictured above,starting with the payment PI andending with P2 . After P2 has beenreached the program calculates theaccumulated interest and principle forthe payments made between PI -P 2inclusively. The schedule may bestarted at any point in the mortgages1ife.The data generated is valid for loansthat have a balloon payment as well asthose that are arranged to be fullyamortized.For loans with a balloon payment, theremaining balance of the last paymentperiod is the balloon payment due inaddition to the last periodic payment.For loans scheduled to be fullyamortized, the remaining balance afterthe last payment period may be slightlymore or less than zero. This isbecause the program assumes that all

payments are equal to the valueentered for PMT. In fact fo r mostloans the last payment is slightlymore or less than the rest.Equations:

n = - 1n (_ i PV +PMT 1)1 n (1 + i)

BAL pI _l PMT l -1 + i) P1-l-jINTPn= RND (BAL pn ixlOO )


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2Example:Duplicate the entries in the precedingamortization schedule.Solution:45@[email protected] ST02

9.25 ENTt12.eeSTG3

385.81, STC!413.88 STOeis.ee STOi

GSe8

381.14 H*44522.68 fH13.8& **:t.343.2& u*41.88 **t44488.88 *u14.88 ***342.87 **,:42.13 **t44438.67 *u15.8e ***J42.55 H*42.45 fU

44396.22 ***16.80 u*342.22 *u42.78 u-"44353.44 tt*17. Be * ' ~ "341. 89 tH43. j 1 u*44318.33 H t18 U l341.56 t: H43.44 U""44266.8' Ut:

mortgage amountannual percentage ratepayments per yearperiodic interest rateperiodic payment amountP1P2actual 1 fe of mortgageremaining bal. at P1-l*payment numberinterest portion of pmt.principle portion of pmt.remaining after 13th pmt.

2 B ~ 4 . 2 9 ft," acc. interest periods 13-18~ 5 . 7 1 tH acc. principle periods 13-1844266.89 ft . remaining balance

*When using the HP-29C all subsequentvalues are produced by pressing R/S.


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3User InstruetionsSTEP INSTRUCTIONS

1. Key in program2. Enter following data in any order

INPUTDATA/UNITS

r- -- ------- -_ ..---- Original amount borrowed PV--- Periodic interest rate i----f------- -- - - - - - - - - ------ - - - - - - - - - ----- Periodic payment amount PMT- - - - - - - - - - t - - - --

3. Key in- _ ~ ~ r t i ng pe)",_'_o_d_____I- Pl------------ -- - -+- - - - - - - - - j

1---- - - ~ I 1 ~ j ng_pE!ri 09. ______ P2.- - ----11------'------------1I- __ _ ' _ _ _ Ca 1 te lUe of mortgage ------------ --- t ---------- j

I - _ ~ _ ' _ _ _ Calculate remaining balance at Pl- Period number

-- - -----+-----------1

------------ -- - - - - - - - - - - - -I - -- - - _ ~ l " i n c : i P i i l f)ayme __t____________ _

- I n ! ~ ! ' ~ ~ L p C ! . Y m ~ l l t______________ - - - - - -1 r - - - - - - - - - - - - - - j___ - Remail}ing....balance __________ _ - - - - - - - - .- ----

- - - - - - - - - t - - - - - -- - -7. _ ~ f t e r l a s L r u ~ r i Q d iE'2.L_______ - ------ 1--- ----

Ca lculate _ accumulated.....:total s; -- - --Accumulated_interest ___ __________ -------- Accumulated principal

---------

- Remaining balance- -- -----

8. For a new payment period in same mortgagerepeat steps 3-7.

* If the program is run on the HP-19C, theR/S is not used.

KEYS

CJCJL_JCJ

OUTPUTDATA/UNITS

~ [ u 1----lirQJ [ ~ ..[STQJ [4 UjC J [ ~I STO 11- -0-]l STO I --1-I

~ ~ J G : : JLBLi;]CJLRLs*] CJlJUs* ICJ[R/S* ]CJ

[ R L S J [ ~C ~ C J[ ~ [ : ~ ~[ _ ~ l ~ ~ ~

[ ~ [ - - - _J[ ~ ! s * J ~ _ J[R/s*l L__

[ ~ ~ ~ L__ J'--l ~ - - ~ JI II -lI 11 1

I " 1I J ' j1 II I

----- - - - - -

1--------- - - -

Rem. Bal.Pn N ------

I-Pn PRINPn1- - - INTPn BAl

1----- .-- --1-- - - - - - - -

--- -----TNT

I PRINRem. Bal.


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4 Program Listingsf11 tLBL8 51 PRTX ** Pn82 8 52 ISZ83 ST05 53 RCL984 ST06 54 RCLJ85 RCL3 55 486 EEX 56 EEX87 2 .', 788 58 +89 ST07 59 EEX18 RC!.2 f8 711 x 61 -........12 RCL4 .-- 62 PRTX ** Pn INT13 + ......... 63 ST+5,.....14 CHS >Ir- + 64 +...:::E: .--15 1 .,..... 0... .......... 65 RCL416 + .J.... t: LSTX-- 6617 LN t : 67-- -18 RCL7 II 68 PRTX ** Pn PRIN19 1 t : 69 ST+628 +

78 LSTX21 LN 71 +22 72 -23 CHS ** ** Pn Rem. Bal.73 PRTX24 SPC ** 74 ST09PRrx II Newll N **,. ..' 75 SPC26 RCL8 76 bT01'"' 1 ?7 tLBLJ28 - 78 RCL529 :cr 79 PRTX ** L INT.J8 - 88 r;:CL631 RCL? I I 81 PRTX **7" 1 :z L PRINI 82 RCL933 + .--I 83 PRTX ** Rem . BAL P234 X : ~ ' .--0... 84 R/$35 fX ..........,.....36 1 + .,......--37 x:r ..........I ** When using the HP-29C R/S must8 - .-- be used in the place of PRTX9 ReL7 , I48 and SPC shoulc be deleted.41 RCL4 r-:IE:42 x 0...43 ST0944 PRTX ** SAL PI 145 SPC **46 tLBLl47 RCLl48 RCL849 X.>Y?58 GSB3

REGISTERS0 PI 1 P2 2 PV 3 Periodic i 4 PMT 5 L INT6 L: PRIN 7 i/100 8 9 Rem. BALn .0 .1.2 .3 .4 .5 n 16 1718 19 20 21 22 2324 25 26 27 28 29


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INTERNAL RATE OF RETURNUP TO 26 CASH FLOWS

The interest rate that equates thepresent value of all future cash flowswith the original investment is knownas the Internal Rate of Return (alsocalled discounted rate of return oryield). Given the initial investmentand up to 26 cash flows, this programcalculates the periodic IRR.

Six End ofMonth Period

23456

5

Cash Flow$-50,000

011 ,00011 ,00013,000

280,000When using this program, cash receivedis positive and cash outlays are negative. Zero should be keyed in forperiods in which there are no cashflows.

(includes net proceedsThe answer produced is the periodicrate of return. If the cash flowperiod is other than annual, theanswer should be multiplied by thenumber of periods per year to determinethe annual internal rate of return.The program solves the followingequation iteratively for IRR:

nV ' " CF J.IN = L .,

j=l (1 + IRR)jwhere: n= number of cash flows

CF.= j th cash flowJNote:Problems involving a large number ofcash flows can often result in longexecution times.Example:A shopping center requires a $200,000investment, and will be sold at theend of 3 years. If this investmentresults in the semi-annual net cashflows shown, what is the internalrate of return?

Solution:GSP1-58888.88 R.S

8.0e R. .S11888.88 R..-"S

P/S13888.88 R.. S

28888e. @@ R. .S

from sale)

-288888.f@ GSB2 Cash paid out4.23 U.;: Semi-annual IRR (%)2. Bt! ):"8.4 H:1/: Annual IRR


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6 User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

1.1--- Key in the program DO2. Clear registers I f I[BJ3. Ini t ia l ize ~ U J 4I--4. Beginning with the f i rs t period, IRIS II I CF.-

enter all cash flows in sequence: D [ ~inflows posit ive, outflows negative DOand 0 for no cash flows. I ILJ

5. Enter ini t ia l investment using the INV IGSB II 2 I IRR(%)cash flow sign convention. DO

6. For a new case, go to step 2 I II IL ~ D0 1 IDO1-- - - - - - ~ , - . - - - [ ~ I II 101 - - - - ------ ----- D [ ~

- - ~ - - - - - - DO ----- -- -- - [ ~ [ ~

- - -------- ----- - -- -- - - - - ~ - DO-- --_.- - - _._._. -- . -- _ . - - ---- LJI I - - ~ - - - - - -------- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ - - - - - - - ------ ----------- [ - ~ [ ~

---- -------- -------- [--II I -- -- - -- - -- - - -- --- - - -- - -------- [ = ~ L ~ I - - ~ - - - - - - - - -

-- -- - ---- - - -- --- - - - - - - --- --- ------- c------ C ~ D 1--------- --- -- --- - -- -- - -- - - -- - - - -- ------- C ~ [ ~ ~ 1-------1 - - - - - -- --- - - - - -- - - -- - ---_.---- - -- ---- L_ [= _=J 1---------1 -Jr----J


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ProgralD Listings 7131 tLBL1 48 6 10- 682 A 49 UY?'t83 STaB 58 ~ T D 584 tLBLI3 store cash flows 51 RJ8S' P/S 52 RJ 1 + IRR136 STOi 53 187 ISZ 54 -88 HOO 55 EEXe9 ,LBL2 5 2113 liBS 57 x11 DSZ 58 R.'S ***IRR(%)12 STOt INV 59 tLBL513 RCLe n + 3 68 RJ.14 ST03 61 RJ. 1 + I R R ~15 1 initial value for 62 13.5TD216 EUTt 1 + IRR17 e 64 ~ T 0 6 I18 ST02 clear f (i) 65 tLBL7 CFi h 1 + IRR19 tLBL6 66

ST+228 RCLB 67 j(21 3 68 +...... X=YI) 69 l .+-, 1 + IRR....: ..j ~ T 0 8 78 S T ~ 224 RJ 71 '\,1 + IRR 1 + IRR7 , RTN 1 + IRR25 ;3.. t - i... ,.a:. { RCL; CFi28 GSB729 OSZ38 ~ T 0 631 *LBL832 RJ.JJ RJ.34 RCLJ35 STDB36 RJJ7 RCL2 f (i )38 PCL!39 -413 ....v, ;+ - ,4'l42 x I:,4J + 1 + IRR44 LSTX *** "Print x" rna' be inserted.45 liBS 1 1:,146 EEX47 CHS

REGISTERS0 i 1 INV 2 f (i ) 3 n + 3 4 CF 1 5 CF 26 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29 CF 26


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8

Year(End of)

12

STRAIG/{f LINE DEPRECIATHJJ SCHEDULE

Schedule - Straight-Line MethodStarting Book Value: $375,000 Salvage Value $30,000

Estimated Useful Life: 40.25 Years

Depreciation Remaining Remaining DepreciationAmount Depreciable Book Value To Date(DEP) Value (RDV) (RBV) (Reserve)8571.43 336428.57 366428.57 8571.438571.43 327857.14 357857.14 17142.86

"--........-.... .......... .....-..,.-....,-.....,--.,-1;",,.-......-.....-.....-.....-.....-.....--....-... ..... . - - . . . . . - - . . . . - . . . . . - . . . . . - . . . . . - . . . . . - . . . ~ . . - . . . . . - . . . . . - . . . . . - . . . . . - . . . . . - . . . . . . . . - . . . L . . . . . . . - . . . . . - . . . . . - . . . . . - . . . . . - . . . . . - - . ~

!

r - - ; , - - - 1 - - ; , 4 ~ . ~ ; - - T - - - - ; ; - l - ~ 0 ~ 0 ~ . ~ 0 - - r ~ 4 5 0 0 o . o ; - - , - ' 1

The annual depreciation allowance using this method is determined bydividing the cost or other basis ofvaluation (starting book value) lessits estimated salvage value by itsuseful life expectancy. This programdevelops the data shown in the exampleschedule, given the starting book value (SBV), salvage value (SAL), lifeexpectancy (LIFE), and first year ofthe schedule (YR). (The schedule maybe started at any point in the useful1 fe . )Fractional years lives must be enter-ed as an integer plus a fraction.Thus a life of 12 years 3 months wouldbe keyed in as 12.25 for LIFE.Values for the last year of an assetwith fractional years life (i.e., the21st year's values for an asset with20.5 years life) are calculated c o ~ rectly. However, al l other valuesrepresent a full year's depreciation.

For this reason only integer values(whole number, 1.0, 2.0, 17.0 etc.)may be entered for YR. The programmakes no checks on this value andgenerates invalid results if otherthan whole numbers are entered.EQUATIONS:

DEP = SBV - SALk LIFESBV - SALDEPk(last year)= ( LIFE ). F

RES = (K) (SBV - SAL)LIFERDV = (LIFE-K) (SBV - SAL)k LIFERBV k = RDVk + SAL


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whereRES = ReserveF Decimal portion of LIFEK = Value for YR

EXAMPLE: Complete the schedule shownfor the first two years.Then jump to the 41st yearand generate the data forthat year.

SOLUTION:38888. ee E ! - ; T - ~

37'58813.813 ENTt1.8B .851

p""SR---S

336428.57 :+11-' RDV I1;'''''S

366428.57 I'. RBV 1R,'8

8'571.43 .a:+ RES 18'571.43 DEP 2

~ - - - - s327857. 14 H::;: RDV 2

17142.86 .ff RES 241.ee sroe YR2142.86 .f. DEP 41

p""SR-5

!88B8. ee Ji:.:1I:p---S

345e8e.ee -f:Ji:'+ RESC;1

9


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10 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DO2. Enter data DO2a. Life of asset in years LIFE I ENT t lD2b. Salvage value SAL lENT t II I2c. Starting book value (beginning basis) SBV [Ei!iiJC l2d. Year for which depreciation is to be DO

calculated YR IGSB lei]3. RUN: I I [ ~ I3a. Depreciation M JD DEP3b. Remaining depreciable value IRjS II I RDV3c. Remaining book value M JD RBV3d. Total depreciation to date (Reserve) MJI I RES4. For next year's values, go to step 3a DO5. For another year's values, enter year and L ~ I I

go to step 3a YR GIQJGJ0 1 IDOI I L ~-- DO0 1 I

- L ~ D- 0 1 I -

-- ----------- L ~ [ ~- - --------------- [ __ D---- -- _0._0_- - - - - - - - - - [ _ ~ D---- I- - ----------- [ ~ C - - ] I-1---11 1


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&1 .:LBL182 STOB83 RJB4 ~ T l ' > f... ; Lf.i,B5 P'J.B6 ST02f!? RJ88 5T03e9 R.18 tLBL5f !12 .;.17 g; ' ')i.,-'14 fT0915 Ff..'C16 ST0417 ReL?18 ReLf}192821 \." \ ' f" it, ... : :'22 RCL42'324 ::.:25 ReL?26.,.,. peL1:" i"'''.:..,.: RCL22938 ST05317"... l.. R. .S77...1,_, ReL?34 PCLf35 ~ . ! ' \ v..,;",/: .36 Gi083?38 peL??948 PCL541 ::{42 ST0643 p. .. s44 .:LBL745 peLf46 f..'CL247 +

0 YR6 RDV 7.2 .318 1924 25

PrograUl Listings

Beyond useful life

Last year?

1 or F

*** DEP

Last year?

*** RDV

48 p. ....s49 ReLS58 ReL65152 TC.,. ..53 ps54 PCLf}


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12

SUM OF THE YEARS' DIGITS

Depreciation Schedule - Sum of the Years Digits MethodStarting Book Value: $375,000 Salvage Value: $30,000

Expected Useful Life: 40.25 Years

Depreciation Remaining Remaining DepreciationYear Amount Depreciable Book Value To Date(End of) (DEP) Value (RDV) (RBV) (Reserve)1 16,725.38 328,274.62 358,274.62 16,725.382 16,309.85 311 ,964.77 ) 341,964.77 33,035.23

''''"-''''''''--'''''''''"-'''..........'-' ' ' ......... ''''"-'''' . . . . . . . . . . . ''''"-'''' ..............................................................................'-''' ...............................'-''' ....................'-'' ' ' . . . . . . . . . .'-''' ........ I~ ................................................................ ..................................................................................... ........... '-''' ..................................................... ................................................................. ...........................................................................I 41 I 103.88 I 0.00 I 30,000.001 345 ,000.00 IThe sum-of-the-years' digits method isan accelerated form of depreciation,allowing more depreciation in the earlyyears of an asset's life than allowedunder the straight line method. Thisprogram generates the data shown inthe example schedule, given the starting book value (SBV), the salvagevalue (SAL), expected useful life inyears (LIFE), and beginning year (YR)for the schedule. (The schedule maybe started at any point in the useful1 fe. )Fractional years asset life must beentered as an integer plus a fraction.Thus a life of 12 years 3 months wouldbe keyed in as 12.25 for LIFE.Values for the last year of an assetwith fractional years life (i.e., the21st year's values for an asset with20.5 years life) are calculated correctly. However, all other valuesrepresent a full year's depreciation.

For this reason only integer values(whole numbers, 1.0, 2.0, 17.0, etc.)may be entered for YR. The programmakes no checks on this value andgenerates invalid results if other thanwhole numbers are entered.EQUATIONS:

SOYD - (W + l)(W + 2F)- 2DEP =(LIFE + 1 - K \ . (SBV-SAL)k \ SOYD -)RES =rl - (W-K+l)x(W-K+2F)l. (SBV-SAL)k L 2x{SOYD) JRDV = [(W-K+l )X(W-K+2F)] (SBV-SAL)k 2x{SOYD)RBV k = RDVk + SAL


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WhereK= value for YR

RES k = Reserve at period k= Integer portion of LIFEF = Decimal portion of LI FE

EXAMPLE: Complete the schedule shownfor the first two years.Then jump to the 41st yearand generate the data forthat year.

SOLUTION:

413.25389813. 98

375998.98! ee.1725. 70'Io.:t..,

'328274. 62358274.6216?25.J8

1 6 3 ~ 9 . 8 5

'311964. 77'341964.773?835r2341. 813

113'3.88

e.BB

'3ee98. BeJJ588B.e13

13

ENHENT1ENHr;SBlP,S

*** DEP 1R... e; RDV 1***.c, ... --'H* RBV 1R....S*** RESIR. S,:** DEP 2R"'"8*H RDV2R/S RBV 2H:1#:R....S*** RES 2STOeR.. S DEP 41H*R. . f'.;:

: , : : + : ~ : RDV 4 1R./S,:H: RBV 4 1/;' .s1#:H RES 4 1


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14 User InstruetionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

1. Key in the program DD----2. Enter data DD2a. Life of asset in year LIFE I ENT t lD2b. Salvage value SAL lENT t i l I 1-------2c. Starting book value (beginning basis) SBV L@iJLJ----2d. Year for which depreciation is to be DD------ 1---

calculated YR ~ O L C J3. RUN: I II I3a. Depreciation ~ D DEP3b.1-------- Remaining depreciable value IRjS I [ ~ RDV3c. Remaining book value ~ , D RBV3d. Total depreciation to date (Reserve) ~ I I RES4. For next year's values, go to step 3a DD5 For another vpar'c; 'La1ues, enter )lear and L ~ I I

00 tel_step 3a .._----- YR [STO J LJ------ D L ~

r--- - -- ~ - - - [ ~ D-- - - -- -- ---- D L ~

t - - - - - ---- -- --- [ ~ D1---- - - - - - - - -- .._- -- -- --- -- -- r ~ D 1------- -- --------- ------------------- -- - - ---------- - - - -- DL=J

- - -- - ---- ---- - - - - - - - - - - - - - - --------- 1--------- [ ~ - = J [ ~ 1-- - ----- - --- -- - -- - - - -- - -- - - - - - - ----- - -- -- --------- ----------- [ ~ - = J L ~ ----- - -------

----- ---- - ---------- ---- - -------1---------- [- - - J [ ~ 1--- ------- - - ---- -- -- - - - --- - - - - - - - -- - - - -- - -- - - - ------- [ ~ = J L = ~ ---- --- - ----- - - - - - - - - - -- -- - -- - - - -------------

[---11---- J- - - -- - - - - - - - - - --------,--- 11 -l


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PrograUl Listings 1581 tLBL! 48 peL5 *** RBV82 smB 49 ISZ83 N 58 R.'S *** RES1::1 f T05 Next4 STa! ... year or new"." tLBL9 year85 I V ~ \ . I --'';''; . , ~ I86 - 53 ENTt87 ST08 54 FPC88 RJ 55 EHTt+89 ST03 ",,it: 2F57 'x''''''18 f,SB9 "+-.11 Sm7 SOYD 58 INT12 R.'S 59 +68 ' C'TV13 tLBL5 L1..' ; i": Wor W-k14 ReL8 61 115 RCU 62 +1 1 63 .x.017 + tot 2.1 6518 X ; ~ J I ?19 Gro8 Beyond useful 1 fe 66 PTN28 It .V 67 t.LBLB"+-, 68 821 - Error6922 peL? 78 tLBL723 71 824 ReLS 1 25 72 STD5x *** RDV 0? RS =26 R'S *** DEP 74 Gm627 ReL?28 RCL829 -38 ,'


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16

VARIABLE RATE DECLINING BALANCEDEPRECIATION SCHEDULEDepreciation Schedule - Declining Balance Method

Starting Book Value: $375,000 Salvage Value $30,000Expected Useful Life: 40 Years Rate: 1.5

Depreciation Remaining Remaining i DepreciationYear Amount Depreciable Book Value Date(End of) (DEP) Value (RDV) (RBV) (Reserve)1 14,062.50 330,937.50 360,937.50 14,062.502 13,535.16 317,402.34 347,402.34 27,597.66~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ , ""'--""'--"' .......... '-"""""-""""--"''-'" ""--"'''''--''' .......................................................................

[ ~ - ~ - - - J - - ; , ; ~ ~ - - r ~ ; , ~ ~ ~ ~ - 1 - 2 D ~ 6 9 ~ 1 - - ~ 3 , 6 J O ' 4 9 - ------- --______1_______ ------________ JThe variable rate declining balancemethod is another form of accelerateddepreciation; as such i t provides formore depreciation in earlier years anddecreasing depreciation in later years.This program generates the data shownin the example schedule given the start-ing book value (SBV), salvage value(SAL), useful life expectancy (LIFE),the declining rate factor (FACT), andthe first year of the desired schedule(YR). The schedule may be started atany point in the useful life.The "variable rate" is indicated aseither a factor or percent with equalfrequency in the business community.Thus, "1.5 declining balance factor"and "150% declining balance" have thesame meaning. The number to be keyedin for FACT in this program, shouldbe in factor form, that is 1.25, 1.5,2, and not 125, 150 or 200.This method of depreciation is uniquein that i t may generate depreciationgreater than the depreciable value

for some assets, while i t may not generate sufficient depreciation forothers.This program will not allow an asset tobe depreciated below its salvage value.That is when the generated depreciationfor a year, usually the last, is great-er than the remaining depreciable value,the latter is displayed as the depreciation amount. Program 6 is provided toassist in determining the best time toswitch to straight line depreciation(tax laws permitting) so that an assetmay be fully depreciated.Fractional years lives must be enteredas an integer plus a fraction however.Thus, a life of 12 years 3 months wouldbe keyed in as 12.25.Values for the last year of an assetwith fractional years life (i .e. , the21st year's values of an asset with20.5 years life) are calculated correctly. However, all other values represent a full year's depreciation. Forthis reason only integer values (whole


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numbers 1.0, 2.0, 17.0, etc.) may beentered for YR. The program makes nochecks on this value and will generateinvalid results if other than whole num-bers are entered.EQUATIONS: k-l

DEP k = SBV ( l - [ ~ ~ ~ ) ( [ ~ ~ ~ )RES k = SBV r, _1 _ FACT) kl~ LIFE JRDVk = (SBV - SAL) - RES k

Wherek = Value for YRRES - Reserve at year k-

EXAMPLE: Duplicate the schedule shown.Also calculate the remainingdepreciable value in the lastyear.NOTE: Note that in the last year of theasset's life there would still bea total of $51,294.43 of remaining depreciable value on thebooks if this schedule were usedthroughout the asset's life.(See program 6)

Solution:4B. ee ENT1-

38888.Be ENT1-375888. e8 PITt1. 58 t;SB1

1.ee GSB21". .S14862. * : . ~ : , :

R .S338937. 5e H:iI:

R .S3689J7. 58 tHp. ....S

14862. 58 , :Hp./S13535. 16 :;:H

RS317482.34 .tH:p. ....S347482. 34 H:,:

R. .S27597. 66 u,:

IS.88 srae: ~ / . s

823'5.18 t.;:."! DEP 15ps181369.51 H:.;: RDV1 5P/S

211369.51 H:+: RBV 15R.. S163638.49 HI!: RES 15

48.88 STae2167.32 .H1i DEP 40

p. ..s51294.4J .':.;:jf: RDV 40

17


21/40

18 User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

1. Key in the program C ]C ]2. Enter data C ]C ]2a. Life of asset in years LIFE [BITt Ic]f--.- 2b. Salvage value SAL [ E N ~ I I2c. Starting book value (beginning basis) SBV LiliiJ [ - ] - -2d. Declining rate factor (1< FAC:s2) FACT ~ [ l ]3. Enter year for which depreciation is to be YR I GSB I L L J ~ - - - ~ -

calculated I II I4. RUN: C]c]4a. Depreciation Lrul I DEP4b. Remaining depreciable value l l i iJC] RDV4c. Remaining book value lliiJL_J RBV4d. Total depreciation to date

1--(Reserve) l l i iJC] RES

5. For next years values, go to--- step 4a L ~ I I --6. For another year's values, enter year and YR ISTO ]ca=J

go to step 4a ---- L_ L ~ ~- - - - ~ - ----- ---- - ------- [ _ ~ c ] 1----------

- _.- L ~ [ - J ------- ------ -- ---- - - - 1---- -- [=_J [ ---] -- - --- - -

- - - - -- -- - - ---- ----------1-- - - - - - I = ~ [--=-] ---- -- - - -- -- --- ---- -- - - - - - ~ - - - - - - ~ - - - - --- --- - -- -- ---- [ ~ ~ ~ ~ l l - -___ 1 -- - --

-- ---- -- ----- - - - - - - - - - ----- - ------ ---- - -- -- -- - - -- -- -[-- -II 1

-------- -- - -- ---- - , - - -- - [ I I- -- - -- --- -- - - - ---- -- ----

[ II II I II II II 11 J


22/40

81 tLBLl82 STC583 N84 STOt85 N86 ST028? RJ88 ST!J389 R-'-SIe tLBL211 ST!J812 R.-'S13 tLBL514 RCL815 RCD16 11? +18 X ~ \ ' ?19 GTD928 1" f.. RCLS22 RCn2324 ST0625 -26 STar2? RCLe28 129 -313 )IX31 RCL6.,., x,:a:.,33 PCL!34 x35 ST0836 RCL637 RCL138 x39 1413 RCL?41 RCLO42 yx43 -44 RCL!45 x46 .STO?4? Ret!

0 YR6 FACT/LIFE.21824

PrograDl Listings 1948 RCL249 - Depreciable value

X ) ~ ' ?bro8S2 STO? Depreciable value53 -

54 RCL855 -56 CHS DEP + depreciable5? 8 value - RES58 g)y?59 bT0668 Xty61 R.'S *** DEP62 tLBL?

ReL164 RCL265 -

Beyond useful life 66 RCL?67 -68 R/S '*** RDV69 RCLl78 ReL??1 -.,.,,


23/40

20

CROSSOVER POINT-DECLINING BALANCETO STRAIGHT LINE

As indicated in the description andexample for program 5, the decliningbalance method of depreciation may notfully depreciate an asset in the asset's1 fetime. In these circumstances thereis an optimum point in the useful lifewhere a switch from the decliningbalance method to the straight linemethod should be made. This is the"crossover point", the first year inwhich the depreciation by the straightline method is greater than if depreciation were continued using decliningbalance method. (In accordance withInternal Revenue Service publication534, the straight line depreciation isdetermined by dividing the remainingdepreciable value by the remaininguseful life.)Given the starting book value (SBV),salvage value (SAL), useful lifeexpectancy (LIFE), and decliningbalance factor (FACT), this programcalculates the last year that thedeclining balance method should beused, and the remaining life andremaining book value after this "lastyear" so that a switch to straight 1 nedepreciation can be made. Should therebe no optimum crossover point, a zerois displayed. This implies that thedeclining balance method is "best" forthe entire depreciable life.Thus, this program, the decliningbalance depreciation program (5) andthe straight line depreciation program(3) may be used as follows:A. This program is used to determinethe "crossover point" and associatedvalues.B. Program 5 is entered and a decliningbalance depreciation schedule isgenerated for the early years up toand including the year indicated asbeing the "last year".

Note that since the depreciationprograms use the same storageregister conventions, only a valuefor YR need be keyed in for program5.C. Finally, program 3 is entered. Theremaining book value at the end ofthe last "declining balance year"is keyed in for starting book valueand the remaining life is keyed infor the asset's life. A straightline depreciation schedule may now

be generated for the remainingyears.Note that for this portion of thedepreciation schedule the value for"total depreciation to date" (reserve) will be in error by anamount equal to the amount depreciated during the declining balancecalculations.

As in program 5 the declining balancefactor (FACT) should be entered infactor form (1.25,1.5, 2.0), not as apercent (125, 150, 200).Eguations:

k-l FACT BV k_lSBV(l _ FACT) (LIFE LIFE + l-kIFEBV =k SBV - SAL - RE\RES k = S B V ~ - (1 - FACT)JIFEK = the value for YRThe largest integer value for K whichmaintains the above inequality is the"last year" to use the DecliningBalance depreciation method.


24/40

Example:An asset has a starting book value of$375,000, a 40 year life expectancy,and a projected salvage value of$30,000. Using a 1.5 declining balancefactor:1. Determine the crossover point andthe associated remaining life andremaining book value.2. Generate the depreciation data forthe declining balance "last year"with program 5 (Normally the userwould generate a full schedulebeginning with the 1st year).3. Switching to the straight linemethod (program 3), generate thedepreciation data for the yearfollowing the declining balance"last year".Solution:

1. 58 P!Tl'48. ee Et-iTt.JeB88.68 ENTr

375e88. ee ,;SB!lB. fie 1ast year22. ee rema i ni ng 1 fe

IB8471.81 *** RBVKey in program 5

18.88 bSB2R"S

7342".B:: U t DEP 18R. ('158471. 81 RDV 18

r;, ...

186471.81 tU RBV 18t'.E.86528.99 RES 18

(Note agreementwith RBV above)

Key in program 322.88 EHTf38888.88 ENTt (the first year of

21

188471.81 EHTt straight 1 ne depre-1.88 bSBl YR ciation)R."S7283.23 *** DEP 19R....SlS126??e. *** RDV 19R'"181267.78 ttt. RBV 19

etc.


25/40

22 U ser Instruet ionsSTEP INSTRUCTIONS INPUTDATA/UNITS KEYS

OUTPUTDATA/UNITS

1 Key in the program L ~ J l -J- ---------------- - - -- -2 Enter data: l - - ~ ~ J ____J------ -- - ----- - - - - - - - -_. - - - --2a Declining rate factor (1


26/40

81 *LPU82 STO!82 RJ84 ('Tn':>"_, ; L'i...B5 N-86 CTn- ',-,j87 pI.....88 ST05891e STOe11 ReLS12 RCLJ1?14 STC7lr :.....16 ST0817 *LBLe18 ISZ19 ReLe28 ReL?2122 +2? g ~ y ?24 r;ra925 ReL8

' : > ~.. o ReL8.,., \IX.,,,LC29J8 RCL!7 ' )::,-'.i.32 RCLlJ? +24 PCL23536 ~ T O (?7 RCL?7C> ReLe.. \.'J94841 ReL842 ReL84? \IX44 RCL?45 X4f RCL147 y

0 K -6 BV21824

ProgralD Listings

SBV

48 v'.\.',",r,o- i ;49 r;rae iterate513 ReLe51 R.'S ***last year52 ReL?5? ReLe5455 p ....S ***rem. 1 fe56 ReLf

start with K = 3 57 RCL258 +59 R/8 ***RBV68 tLBL9

K - 1

no crossover point

BV

right side ofi nequal ity

6162

left side of inequalityREGISTERS2 SAL 3 LIFE

8P/5 display "0"

***"Print x" may be inserted torepl ace "R/S".

4 5 FACT7FACT/LIFE 81_ FACT/LIFE 9 .0 .1.3 .4 .5 16 1719 20 21 22 2325 26 27 28 29

23


27/40

24

NOMINAL TO EFFECTIVE/EFFECTIVE TONOMINAL RATE CONVERSION

An annual effective interest rate dem-onstrates the effect of compounding fora full year of compounding periods ata particular periodic interest rate.The periodic interest rate to be usedis determined by dividing the number ofcompounding periods in a year into thestated annual nominal interest rate.The effect is such that if the nominalrate is held constant, as the numberof compounding periods per year is in-creased ihe annual effective interestrate will increase. The ultimate orupper limit in this process is tohave an infinite number of compoundingperi ods ina year, common1y ca 11 ed continuous compounding.The first part of the program addressesfinite compounding, that is quarterlycompounding, monthly compounding, etc.Given the number of compounding periodsin a year and one of the rates (nominalor effective) the other rate can becalculated. If for example, you requirethe periodic interest rate for a cal-culation, given the effective rate,use this program to determine the annual nominal rate first. Dividing thenominal rate by the number of compound-ing periods in a year will give the re -quired periodic interest rate.The latter part of the program is forcontininuous compounding. Given eitherrate, the other can be calculated.The most common and straightforwarddefinition of effective interest ratehas been implemented. Occasionallyother definitions will be used and theresults will not compare exactly withthose calculated by these programs.For example, since the maximum annualnominal rate that savings institutionscan offer is regulated by law, theymay modify the process (also regulated)

so that the effective rate is even higher (e.g., for daily compounding, theperiodic rate may be divided by 360 andthen compounding accomplished for 365periods). I t is important then, whenattempting to match results, to understand the process employed.EQUATIONS:

finite compoundingEFF = (1 + NOM) c _C

continuous compoundingEFF = (eNOM - 1)

EXAMPLES:1. An investment with monthly cashflows (implying monthly compounding)is said to have an annual effectiveyield (interest rate) of 21%. Whatannual (nominal) yield and periodicyield does this represent?2. A bank offers a savings plan witha 5% annual nominal interest rate.What is the annual effective rate ifcompounding is continuous?SOLUTIONS:

1138.813 STOfl12.88 STOt21. 88 ST03

&SB!19.21 ifH:

5.1313 ST04f;SB45.13 ,-Jl::t

Nom(;j) ,annual


28/40

25User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

l . Key in the program DD2. Enter constant and go to either step 3a 100 ~ ~for finite compounding or step 3b for DD --

continuous compounding L ~ I I3a. Enter [ ~ L J - -

- The number of compounding periods/year c/yr ~ Q - - ~and one of the following: LJL J

- Annual nominal rate Nom % ISTO II 2 Ior DD ..- -- Annual effective rate Eff% [STU [ iJ - - ~ - -

4a. Calculate the remaining rate C ~ D- Annual nominal rate ~ I 1 I ~ %or DD- Annual effective rate L@[]i 2 I Eff %-- - ~ -

3b. Enter one of the following: CD- Annual nominal rate Nom % [STU [_iJ1------ ~ ~ - - - ~ -or D [ = ~- - - - - ~ - - ~ . - -- - - - - - - - - - - ~ - - -- Annual effective rate for continuous Effcont% ~ T O ~ [ - ( ]._- --- -- - - - - ~ ~ - - - - - - - -compounding [ ~ ~ [ ~- - - - - - - - - - - - - - - - - - - - ----

[ ~ - ~ J ~ - Jb. Calculate the remaining rate- - - - - - ~ ~ - -- - 1---- - - - - ~ - ------ - - ----- - - - - - --

- Annual nominal rate [ ~ ? B l l - ~ ~ ] Nom %- - - ~ .- - - - - ~ ~ ~ -- - ------------ ---- - ----or [- ~ _ : JI 1-- -- 1------- - ------ - - - ~ ~ -------- ---- - ------- Annual effective rate for continuous IGSB J I 4 I _Effcont%---- - ~ - - - - -- - --- - - -------- - - - - - - -- - - - -

compounding I II I- ------ -- ------ --- - - - - --- - -- - -- - [ II I

[ II I[ II I


29/40

26

01 ftBL 182 RCL]83 RCLBIN8586 +87 RCL!88 1- .,v,",89 IIlX181112 RCL!lJ14 R C U ~1'5 .x:.l t, R. S17 ~ : L B L 218 PGL219 PCL!28 RCL821 .X.'222324 +2'5 RCL!26272829 RCL/]38 ::{71 R."S. ''-32 ,:LBLJ33 J : . I f " ~, . ' _ L ~ '34 ReL/]"" , ' ...

3637 +70 LN.. \, ;39 RCLe48 :.:41 p . S42 tLBL44! PCL444 pet84'546 eX.1"1

0 1006.21824

7.31925

ProgralD Listings

*** Nom

*** Eff

*** Nom

C/vr 2 Nom8.420

26

4849 ReLe'58 .x:'.! l

REGISTERS3 Eff9.52127

4 Nom.0162228

*** Effcont

5 Effcont.1172329


30/40

27

LEASE VERSUS PURCHASE

This program calculates the presentvalue of the cost of purchasing, CP,the present value of the cost ofleasing, CL, and the net differenceusing the following equations:CP = P + M (1 + i)n -

i(l + i)nCL L (1 + i) n

;(1 + i)nNet Difference = CP - CLwhere

P purchase price

SV

M= maintenance costs, per periodthe opportunity interest rate,per periodn = the number of periods (useful1 fe)

SV salvage valueL lease paymentsIt also calculates the cost of purchasing after leasing for n periods.OP = P - Credits (L - M)(l + i)n -- - - - -+(1 + i)n i(l + i)nwhere Credits = rental credits appliedtoward purchaseThis program is adapted from HP-65Users' Library program #01093A byRobert Dudugjian.

Example:Suppose a purchase price of $14,972,maintenance of $15/month, a salvagevalue at the end of 84 months of$1,000 and lease payments of $325/mo.Suppose further an opportunity rate ofinterest of .00757543 per month.Suppose further that the equipment isleased for 12 months and then purchasedwith $900 rental credits. What is thecost of doing this above the cost ofoutright purchasing? Suppose i t isleased for 24 months with $2000 rentalcredits.Solution:

14972.88 STU!325.88 ST04

8.88757543 ST!J515.88 ST06

1888.88 STC?84.88 (;S8115371.15 *u CPR'S-4771.33 *u

988.88 EHTt12.88 (;S821424.73 ***2888.88 ENTt24.88 (;582

Net (since the answer isless than 0, i t impliesi t is cheaper to purchase rather than tolease by $4771.33)OPI

2638.41 H * OP2 Cost of leasing for24 months beforepurchasing


31/40

28 User InstruetionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

1 Key in the program DO----- - - - - -- -2 Store data: I IC l - ~ - -2a Purchase price P ~ L l ]---- ---- ---------2b Lease payments L I S T ~ ~ ~

---- . _ . _ - " " -1 , per2c Opportunity rate of interest period [i@J1 5 I - -2d Maintenance cost m, per ~ L Oeriod2e Salvage value SV CST-OJ [__J - - -3 Enter useful 1ife n, period5 I II I4 Calculate net cost of purchasing ~ [ T I CP5 Calculate net difference [R/S J ~ CP - CLr 1--- - - - - -6 For a new case, go to step 2 DO -- --7 (Optional) Enter: Dr I f--------7a Rental credits applied to purchase price Credits I ENT t lD7b Periods of interim lease m, peri od5 C ~ C J f----------8 Calculate cost of leasing fo r m j:>eriods, W i J L 1 ~ OP

c - - - - - - r--before j:>urchasing [__ [ ~ - . _ ..9 For another c a s e ~ go to stgp 7 --- D r ~ = - _ - ] r - - - - - - - - -- - -

----- -----" - - - - - - - - - - --- [---] [-]---- -- -- - - - - - - ------ ----- ------ -_. , - -------------- - -- -- - -- --- L ~ = = =J [ = = ~ J c- -- - -- --

- - -- ---- ---- [ ~ _ - l l - - ] -- - ----------------- - - ---- -------- - -- [ __ 11 I

I- --

II I-- -- f- - - - - -- -- - - ---- - - - - - - - - -- - -- -- -- ------- - [ I I

- -[ II II I I1 II II II I


32/40

131~ 13283

13485861378889Ie1112f 7.1 ...141516f".i.;181921321222324... t:27':'0~ t . :293&7 f. .i.32'J'J347e :. ..j t :77_'i70",'t..'39484142434445464?

06 M.21824

tLBLlGSPBpeL!PCL?PCL2xpeL6peL?

..

+p's

PCL4peL3x

p'stLBL2(;8813peL.:!peLf

)(.... "!

l i . f - !PCL!PCL2

~ . : .

PCL!r; ",c, ......

~ : L B L f !peL5+

1/,1;'8T02LSTXENH

peL5

7.31925

PrograUl Listings 2948 8TOJ49 PTN

*** CP

*** CP - CL

Credits

*** OP

n

*** "Print x" may be inserted.

REGISTERSP 2 1/ (1 + i)" 3 (1 + i)"-1 4 L 5sv 8 i(l + i)n .0

.4 .5 16 1720 21 22 2326 27 28 29


33/40

30

REAL ESTATERENTAL INVESTMENT ANALYSIS

Using the equations below, this programsolves for monthly rent or cash flow,gross return on investment, and taxableincome.Cash flow %= Rent/month - Cost/monthInvestment/l2Gross growth return = Cash flow %+

(P x Inflation rate) + Equity build-upInvestmentDepreciation/yr(book value) P - Value of landdepreciable life

.-../. .7 P=2( jTaxable gain = Actual cash flowdepreciationTaxable income (shelter if negative)= Taxable gain x tax bracket

This program is adapted from HP-65Users' Library program #012l6A byJohn Feemster

Example:A house is for sale for $30,000 with anassumable 6 3/4% FHA Loan paid down to$23,500. Payments of principle,interest, taxes, and insurance are$239.17 per month. The place will rentfor $275.00/month. The investor is inthe 30% tax bracket. The inflationrate is 7%. Determine the cash flow %,gross growth return, and taxableincome from the investment.Solution:

Jae!)e.ee STet23'5el).l3e6':88. ee U:.;:

ST022J9.17 STOJ27'5.13e ST04

GSBlGSB3.61 ~ : H : Cash flow (%)

2J588. 88 fNTt6.7'5 fHTt7.88 I;SB458.67 u* Return (%)38.88 R,"S199.13 *** Taxable income ($)GTN


34/40

User Instructions 31INPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

1 Key in the program CJCJ --2 Store data: CJCJ I - - - -2a Purchase price P , $ ITi2J[O-- I - - - -2b Investment I , $ ~ Q j l 2 I I---2c Payment/month Pmt, $ [riQ][U-2d Either Rent/month Rent, $ ITi2J[Q

or Cash flow (% of Investment) Cash F, % [STQ][51 --3 Initialize I SB II 1 I - -4 Calculate either Rent lliiJ [L ] I -Rent lm9L t

1---- or Cash f l ow [ciJ[lJ rash flow,%5 Calculate gross return on investment CJCJ ---5a Loan amount Loan, $ I NTl] CJ5b Interest rate i , % lliUJCJ I - -Gross5c Inflation factor INF % ~ B J I 4 I rphJ rn o/n6 Ca 1cul ate rax Shelter ( if negative) C ~ C J - -Taxable income L = : J [ ~ ~-- - I - --6a Tax Bracket TB, % [RiO [-_:=J $ taxable/

1----- ------- - - - - - - ' i L - - ~ ~ -

- - [ = ~ = J [ ~ ~ J - ---- - --1------ -- - - - - - - - - [ = ~ [ - ~ ----- - - - -I---- -- 1----------- ------ ------- ------ ~ ~ - - - - - [ ~ = J [ ~ ~ = = l -- -

---- ------ ----------- - - ~ - - - - - - - -- -- - -- 1--- - ------ [ ~ _ ~ - : l [--I - -1--- - - I - - - ----- ------- - - - ---- . - ------ - - - - - - -- - -- - -- - -- -

[ --- II- I1 - - --- ---- - - - - - - - - - - - ----- - - - - I II I

- -- - -- - - --- - - -- - -- I " I- -- - -- - I " II II II II I


35/40

32 PrograDl Listings81 tLPL1 initialize 48 RCL6e2 1 49 +8j 2 S8 RCL284 S T . ~ ( 3 annual payments Sl85 EEX S2 RCLS86 2 S3 +87 STOe S4 RCL888 ~ T - ~ SS x ***gross return- " , , : " :- . -M R. .S S6 R/S1e *LPL2 calculate rent S7 RCLB11 RCL5 S812 PCL2 S9 STaB13 :..:" 68 RCU bldg. value61 approx.14 PCL:! .15 + 62 716 1 63 x1? 2 64 2 depreciation, 2018 6S 8 year basis19 ST04 662e R ~ *** rent/month 67 CHS21 tLPL:! calculate cash flow 68 RCLS22 PCL4 69 RCL22J 1 78 x24 2 71 +25 72 RCL?

Ret! 73 +,;:'t- 74 RCL827 - 7S x *** taxable $28 RCL2 76 R/S29Je srG531 PCLt!:2 :.:"

R . 8 *** Cash flow34 .f.LPL4 calculate gross7


36/40

33

BREAK-EVEN ANALYSIS

This program solves the following equations for Break-Even point in units(BEPu), Break-even point in dollars(BEPD), Margin of Safety Ratio (M),and Profit or Loss:Computation Based on Units

1) BEP = Fu S-V2) M= u-BEPu- - -u3) Profit or Loss = u(S-V)-F

Computation Based on Dollars1) BEP D = F-R-2) M= D-BEPDD3) Profit or Loss = DR-F

whereF = Total fixed costsV = Variable cost per unitS = Sales price per unitu = Expected sales in unitsD = Expected sales in dollarsR = Marginal income ratio = (S-V)/S

NOTE: The margin of safety will generally have no meaning i f expected sales are less than salesat the break-even point.

This program is adapted from HP-65Users' Library program #01275A byLouis Martinez.

EXAMPLE:The Delux Publishing Company publishesa magazine with variable costs of$0.40 and a sales price of $0.50. Thecompany has annual fixed cost of$1,000,000.Compute the following:1) Break-even point in (a) units and(b) doll ars.2) (a) Profit or loss and (b) Marginof safety ratio for expected salesof 12,500,000 magazines.3) (a) Profit or loss and (b) Marginof safety ratio for expected salesof $20,000,000.4) Sales volume in (a) units and (b)dollars needed to generate a profitof $5,000,000.SOLUTION:

8.48 fNTt8.58 fHTt

1888088.88 bSB118888088.88 tUR/S5888888.80 Ut

12588888.88 bSB2258888.88 t uR.'S8.28 tU28888888.88 bSBJ3888888.88 tUR.S

8.75 t l t t5888888.88 ST+lbSB468888888.88 . tuP.S38888888.88 tU

ProfitM

ProfitM


37/40

34 User InstruetionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITSl . Key in the program DD- f----- - -2. Enter data: L ~ C J-- c - - - . ~ - ~ - -2a. Variable cost per unit V I ENi] [_ ] r----.---2b. Sales price per unit S l i i I I ~ L ~- 1-------..-2c. Fixed costs F D L ~- - f-- --3. Compute break-even point in units ~ J [ ~ I J BEP u -4. Compute break-even point in dollars [ RjS I I - - ~ - - ~-S. ENTER: I II I - - - - -Sa. Expected sales in units u DD - Profit . -Sb. C o m ~ u t e ~ r o f i t (or loss if negative) ~ i J [ = U or LossSc. ComQ.ute margin of safety ratio [RIiJD M1- .__ _---6a. Expected sales in dollars D DD~ [ 3 J Profil----6b. C o m ~ u t e ~ r o f i t (or loss if negative) or Lossf--.-----6c ComDutp marain of safety ratio [lUi] - - ~ ._ _M_. ._ _. -7. (Optional) To compute sales volume 0[=]---. - - f---- -_ . -C-] [---]ecessary to provide a desired profit: - - - " ----- 1- - - - .- - -- - ..-7a. Add the desired profit to fixed costs DES. [STO] [+1 ]rofitf--- .. -- --------7b. Compute break-even point in units L ~ = ~ ~ 1 ~ ~ ] BEP u- . - ~ - - - ~ - - - - - - - - - . .- - -- - .--. - ~ - -7c. Go to step 4 [- _ = _ ~ 1 -l

. - - ~ - - - - - - - - - - - - --- - - ----- --- ---- . ---- --I -- ----- I -- - -I. For a new case, go to step 2-- ----- _. -- - - - - -- ------ - - - - ------ ------- .. - -- . -I - I - I- - - -- ------ ---- ----- - - - - - - - - - - -- - .- - --- -- ---- -- - - - - -- I II I

. - - I II II II II II II II II II I


38/40

ProgralD Listings 3581 tLPLl82 STlJl *** M83 IN84 STfJ28S XtY86 ST0387 - s-v8S STD489 RCL21811 STOS R12 tLBL41? RCL114 PCLS1516 STO? BEPD17 PCL1fO.it..: PCL41928 STaG21 P..S *** BEP u22 \1"'\": , ~ :., 7 R.. S *** BEP D:.. ....24 tLBL22S STOS26 PCL427 x.,0 peL!: . . ~ :29 -38 P,.S *** Profit or Loss:1 RCLS32 peL 633 -34 PCLS3'536 R. .S *** M37 tLBLJ38 STDS39 peLS *** "Printx" may b inserted to repl aCI48 x "RIS".41 pelt42 -42 P'S *** Profit or Loss44 peLS45 peL?46 -47 ReLS

REGISTERS0 1 F 2 S 3 V 4 s-v 5 R6 BEP u 7 BEPD 8 U or D 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


39/40

In the Hewlett-Packard tradition of supporting HP programmable calculators with quality software, the followingtitles have been carefully selected to offer useful solutions to many of the most often encountered problems in yourfield of interest. These ready-made programs are provided with convenient instructions that will allow flexibility ofuse and efficient operation. We hope that these Solutions books will save your valuable time. They provide you with atool that will multiply the power of your HP-19C or HP-29C many times over in the months or years ahead.

Mathematics SolutionsStatistics SolutionsFinancial Solutions

Electrical Engineering SolutionsSurveying SolutionsGames

Navigational SolutionsCivil Engineering Solutions

Mechanical Engineering SolutionsStudent Engineering Solutions


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hp-19c & 29c solutions finance 1977 b&w

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