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Understanding Simple & Compound Interest in Engineering Economics

School
Batangas State University**We aren't endorsed by this school
Course
CABEIHM ECON101
Subject
Economics
Date
Dec 12, 2024
Pages
49
Uploaded by CoachRabbit1849
Engineering Economics (Simple & Compound Interest) 549 ENGINEERING ECONOMY [SIMPLE & COMPOUND INTEREST) BASIC TERMS Economics is a science which deals with the attainment of the maximum fulfilment of society's unlimited demands for goods and service Engineering Economy is the branch of economics which deals with the application of economics laws and theories involving engineering and technical projects or equipments. Consumer goods and services refer to the products or services that are directly used by people to satisfy their wants. Examples are food, clothing, shelter or home, eic, Producer goods and services are those that are used to produce the consumer goods and services. Examples are buildings, machines, factories, etc. Utility refers to the satisfaction or pleasure derived from the consumer goods and services. This also means the power to salisfy human wants and needs. Luxury products are those products that have an income-elasticity of demand greater than one. This implies that as income increases, more income will be spent on these products. Examples are appliances, entertainment, vacations, etc Supply the amount of goods or products that are available for sale by the suppliers. Demand the want or desire or need for a product using money to purchase it Law of supply and demand. “"When free competition exists, the price of the product will be that value where supply is equal to the demand.” Competition is a form of market structure where the number of suppliers is used to determine the type of the market Perfect competition a market situation wherein a given product is supplied by a very large number of vendors and there is no restriction of any additional vendor from entering the market. Market is the place where the vendors or the sellers and vendees or the buyers come together.
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550 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas The following are the different market situations: Market situation Sellers Buye Perfect competition many | many | Monopoly one many Monopsony many one Dilateral monopoly one one Duopoly two many Duopsony many two Oligopoly few many Oligopsony many few Bilateral oligopoly few few SIMPLE INTEREST Interest is the amount of money or payment for the use of a borrowed money or capial. Simple interest () is defined as the interest on a loan or principal that is based only on the orginal amount of the loan or principal. This means that the interest charges grow in a linear function over a penod of time, it can be caiculated using the formula 1 = Pin where: P = principal | = interest per period n = number of interest penod Ordinary simple interest is based on one banker's year. One banker's year is equivalent to 12 months = - 8 I=Pi of 30 days each. Also, 1 banker's year = 360 days. o N Exact simple interest is based on the exact number of days in a given year. An ordinary year has 365 days while a leap year (which occurs once every 4 years) has 366 days. =385 For ordinary or normal year e For leap year COMPOUND INTEREST Compound interest is defined as the interest of loan or principal which is based not only on the original amount of the loan or principal but the amount of the loan or
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Engineering Economics (Simple & Compound Interest) 551 principal plus the previous accumulated interest. This means that the interest charges grow exponentially over a period of time. Compound interest is used frequently in commercial practices than simple interest. A. Total amount, F Fw=P141)" I where: P = principal [ 4} N L, 2 i = |nt8f85l per per'(x’ P ........................................................>.. F n = number of periods Cash flow of P B. Present worth, P 0 | 2 3 n T AT s « = ¢ r’ - 1 \1 i) ‘ P b s=sensssrosasasanicraniesensasoncssnsassonassntoransase f NOMINAL AND EFFECTIVE RATES OF INTEREST Rate of interest is the cost of borrowing money. It also refers to the amount earned by a unit principal per unit time. Nominal rate of interest is defined as the basic annual rate of interest while effective rate of interest is defined as the actual or the exact rate of interest eamed on the principal during 1 year penod. For example: 5% compounded quarterly In this example, the nominal rate is 5% while the effective rate is greater than 5% because of the compounding that occurs four times during a year The effective rate of interest may be calculated using the following formula. ER = (141" -1 where: m = number of interest penods per year Discount refers to the difference between the future worth of a negotiable paper and its present worth. It also refers to the sale of stock or share at reduced price. Discount may refer to the deduction from the published price of services or goods. discount = future worth - preséent worth
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552 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas (&= Tip: Determination of leap year: To determine a year whether a leap year or not, just divide the year by 4. If exactly divisible by 4, then it is a leap year, However years ending with two zeros or century years (1.e. 1800, 1800, etc.) must be divided by 400 not 4. If exactly divisible by 400 it is a leap yea otherwise its not Did pou know that... The Gregorean Calendar we are usSing NOW was named after 3 former teacher of law at the University of Bologna, Ugo Buoncompagn! who became Pope Gregory Xl in 1572! in February 24, 1582, he issued a Papal edict {irecting the former Jullan Calendar be allowed to catch up with the Lord’s Time and that aside from leap year évery four years leap year be once In every four centennial years, |.e. every 400 years! Proceed to the next page for your 22™ test. GOODLUCK | =
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Engineering Economics (Simple & Compound Interest) 553 L4 4 0 A WEL Time element: 3.0 hours I — —— — = Problem 911: ME Board April 19958 P 4,000 is borrowed for 75 days at 16 % per annum simple interest. How much will be due at the end of 75 days”? i ¥ 413333 B P43333 C. P416866 D 415000 Problem 912: CE Board May 1997 A deposit of P 110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P 880 36 Find the rate of retum annuaily A 11 .98 % H 12.75% ™ 11,75 % "J 2 UK S - L T Problem 913: ME Board April 1993 Agnes Abanilla was granted a loan of P 20,000 by her employer CPM Industnal Fabricator and Construction Corporation with an inerest of 6 % for 180 days on the principal collectad in advance The corporabon would accept a8 promissory note for P 20,000 non-nterest for 180 days. If discountad at once, find the proceeds of the note A P 18600 8 P18.800 C. P 18000 D. P185.200 Problem 914: ECE loard November 1998 YWhat will be the future worth of m -.r..a, afler 12 months, if the sum of P 25,000 is nvesied today at szmpae interest rate of 1% per month? A P 30,000 8 P 29,000 " ":)} s ‘.‘ - QL f ':n;‘ - s " L‘ r £ (. ——
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Engineering Economics (Simpie & Compound Interest) 55§ B. P415.00 C. P551.00 D. P 450.00 Probiem 920! EE Board October 1997 A man borrowed P 100,000 al the interest rate of 12% per annum, compounded Quanerty. What is the effective rate” A 3 % B. 132% C. 12% r‘ . . ‘ff". Problem 921: ECE Board April 1999 Ywhat i1s the wnespomju‘-; effective rate of 18% compounded semi-quaneny” 1920 % 1940 % 18.46 % D 1895 % o > L) - Problem 922: ME Board October 1995, EE Board October 1997 Mandarin Bank advertises 9.5 % account that yields 9.84 % annually, Find how often the inleres! is compounded ’f)&:?y -‘41(“~1T‘:y C. Bi-monthly L. Quaneny )0 > Problem 923: EE Board October 1993 A bank pays one percent nterest on savings accounts four times a year. The effective annual interest rate 1s A 4 08 % 8 1.00 % . «04°% ), 3.36% —_— = Problem 924: ECE Board November 1998 The effective rate of 14% compounded semi-annually i1s A 1449 % B 1236% C 14.94 % D. 1488 %
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556 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas Problem 925: ME Board October 1996 An Interes! rale 15 quoled as being 7.5% compounded quarterly. What 15 the effective annual imerest rate”? A 771 % B 722% C. 15.70% D. 2181 % Problem 9261 ECE Board April 1998 The amount of P 12,800 in 4 years at 5 % compounded quarierly is A P 1478534 B P1561458 C P16311.26 D. P 15847 .33 Froblem 927: ECE Board April 1999 Find the present worth of a future payment of P 100,000 to be made In 10 years with an interest of 12% compounded quarterly A, P 30444 44 B P 3300000 C. P 3065568 D P3S40 Problem 928! EE Board June 1990 On his 6 birthday a boy s left an inherntance. The inhearitance will be paid n a lump sum of P 10,000 on his 21" buthday. What is the presemt valug of the inhertance as of the boy's 8 birthday, If the interest is compounded annually? Assume i = 4% A P86500 8 PB600 C P5500 D P7500 Problem 929: ECE Board April 1999 The amount of P 50.000 was deposided In the bank eaming at interast of 7 5% per annum. Determine the total amount at the end of 5 years, If the principal and interest were not withdrawn during the perod? A P71.781.47 B P7247523 C. P70374.90 D. P785365.34
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Engmneering Economics (Simple & Compound Interest) 557 Problem 930: ME Board April 1993 Alexander Michael owes P 25.000.00 due n 1 year and P 75000 due In 4 years. He agrees 10 pay P 50,000.00 today and the balance in 2 years. How much must he pay at the end of two years if money is worth 5% compounded semi annually? P 38 025 28 P 35,021.25 P 30,500.55 P 3902128 oOQ P Problem 931: ECE Board November 1998 Al an interest rate of 10% compounded annually, how much will a3 deposit of P 1.500 be in 15 years? A P610000 B P623406 C. P6266.87 D. P643790 Problem 932: CE Board May 1995 How long (In years) will 1 1ake money 10 quadrupde If it eams 7 % compounded semi-annually”? A. 20.15 B. 2630 C. 33.15 D 4030 Problem 933: ECE Board April 1999 In how many years s required for P 2,000 to increase by P 3,000 If interest at 12% compounded semi-annually? OO0 >» - - 1D O - Praoblem 934: ME Board April 1996 Consider a deposit of P 600.00 to be pald back In one vear by P 700 .00. What are the condions on the rate of nterest, 1% per year compounded annually such that the net present worth of the invesiment Is positive” Assume | 2 0 A 051<143% B 0s5i<16.7 % C. 125% 1< 143 % D. 16.7 % 100"
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$58 100/ Solved Problems in Engineering Mathematics by Tiong & Rojas Problem 935: ME Board October 1995 A company invests P 10,000 today to be repaid in 5 years in one lump sum at 12 % compounded annually. How much profit in present day pesos is realized? A PT78632 8 P 723 C. P 7326 D. P73682 Problem 936: ME Board April 1996 A fum borrows P 2,000 for 6 years at 8 %. At the end of 6 years. It renews the ioan for the amount due plus P 2,000 more for 2 years at 8 %. What is the lump sum oue '/ P 5,355.00 P 5.882.00 P 6,035.00 P 6,135.00 CC‘;CD)) Problem 937: ME Board October 1996 A deposit of P 1,000 is made in a bank accoun that pays 8 % interest compounded annually. Approximately how much money will be in the account after 10 yoars? A P 1.92500 B P 1.860.00 C. P234500 D. P2.160.00 Problem 938: CE Board May 1996 # 200,000 was deposited on January 1, 1888 at an interest rate of 24 % compounded semi-annually. How much would the sum be on January 1, 19937 A. P401.170 B P421170 C. P521.170 D. P621.170 Problem 939: CE Board November 1996 i ¥ 500,000 is deposited at a rate of 11.25 % compounded monthly, determine the compounded interest after 7 yaars and 9 months A. P 660,580 8. P670,650 C. P680,750 D P680850
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Engineering Economics (Simple & Compound Interest) 559 Problem 940: ME Board October 1996 Fifteen years 2ago P 1.000.00 was deposied In 8 bank account, and 1oday 1 s worth P 2.370.00. The bank pays interest sems-annually, What was the interest rate paid in this account? A. 38% B. 49% C. 50% D. S58% ; Problem 941: ME Board April 1998 P 500000 shall accumusate for 10 years at 8 % compounded gquaneny, ind the compounded interest at the end of 10 years P 6,005.30 P 6.,000.00 P 6.040.20 P 6.01020 wlolk P Problem 942: ME Board April 1998 A sum of P 1,000 is invested now and left for eight years, at which time the principal is withdrawn. The interest has accrued s left for another eight years, If the effective annual interest rate is 5 %, what will be the withdrawal amount at the end of the 16" year? A. P 70600 8 PS5S0000 C. P77400 N P79 00 Problem 943: ME Board April 1998 F 1.500.00 was deposited in a bank account, 20 years ago foday I is worth P 3.000.00. Interest 1s paid semi-annually. Determine the interes! rate paxd on this accoum ) 3 .“ g L el - . 3-_ » o > ~N O o) o0 - W [~ Problem 944: ME Board April 1998 A merchant puts in his P 2,000.00 to a small business for a penod of six years with a given interest rate on the nvestimeani of 15 % pel year, compounded annuaiy how much will he coliect af the end of the sixth year/ P 4,400.00 P 4,380.15 P 4,200.00 P 4,626.00 o0 m>»
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§60 1001 Solved Probiems in Engineering Mathematics by liong & Kojas Problem 948: ECE Board November 1998 now considering interest at 8% compounded quarterly? A P13859.12 B P1365833 C P138752% — St P 1326583 Problem 946: CE Board November 1994 P S00000 was deposiled 2015 years ago &t an Interest rate of 7% compounded semi-annually. How much is the sum now? A P 2000000 8 PF2000150 C. P 2000300 0. P 2000500 Problem 947: ME Board October 1998 in year zor0, you invest P 10.000.00 in a 15% secunty for 5 years. Dunng that ime, the average annual inflation s § %. How much, in terms of year zero pesos will be in the account at matunty? P 12.020 1303 P 14 040 15030 Om> v O 0 Problem 948 ECFE Roard April 1998 By the condibon of a wil, the sum of P 20,000 s left to a gl to be heid in trust fund by her guardian until & amounts to P 50,000. When will the girl receive the money If the fund is nvested 3t 8 % compounded quanerty” A. 7088 years B. 10.34 years C. 1157 years D 1045 years Problem 9491 ME Board October 1996 You bormow P 2 500.00 for one year from & fnend at an interest rate of 1.5 % per month msicad of aking a oan rom a r L bank al a rate of 18% per year. Compare how monegy you will save or iose on the bransaction > You will pay P 155.00 more if you borrowed from the bank You will save P 55 00 by borrowing from your fnend ) 0 You will pay P 85.00 more il you borrowed from the bank D You will pay P 55.00 less If you borrowed from the bank
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Engineering Economics (Simple & Compound Interest) 561 Problem 950: ME Board April 1996 What is the present worth of two P 100 payments at the end of the third year and fourth year? The annual inlerest 'atv 15 8% A P153 B. P180 C. P162 0. P127 . — - TN ERT ANSWER KEY ’ RATING 91T A 921 B 831.C 941.C — 9012 C 922.D 932 A 942 A LN %R0 Tapnatctan 8138 H23 A H33. A 843 C | 814 C 924 A 534 B 044 D | ' ck~-33 Passer 15 C 8256 A 835 A 845 D wr 816 A 9268 938.C 948 B ‘ c0~25 Conditional 9ir. B8 927.C 937.D 847 D 918.0 9286.C 938, D 948 C L] 0-19 Failed 919.A 929 A 9835.D S840 D | .U 8300 840.D 950 A | If FAILED, repeat the test
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562 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas SOLUTIONS TO TEST 22 m F=P{1+ n) = 4000 |1+ 0.18{ — | 360 ) Y 4 F=413333 m 860.36 = 0.8 | = 1112595 | = Pin SuUbsiitute 31 ) 1112.95 = 110,000 (| | \ 3680 ) I=11.75 % interest = 0.06 (20.000) = 1,200 Froceeds on the note = 20,000 - 1200 - 18 800 F=P(1+in) » 25,000 11+ 0.01012)) F=28000 FaP1+in) . o\ 50.000=P|1 + 012 - 2) P =48 728 97 Note: From the choices, the nearest answer is 46 730 m F=P(1+in) 1250 = p[a + 0,08 20 | \ 380 P=P1233.55 917 [ 11,200 = 68 .80001)(1) 1 =16,28% B o codiinn - i = MPmn 3200 = (20,000 - 3200)(IK 1) } =195 m P(1 +in) 300{1 + 0.12(3)) F - F =408
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8§ B B B & Engincering Economics (Simple & Compound Interest) Kl ", 4 : | N ER= 14 - 1 =114 ‘.' 7 “ & 4 CR- C . —- -1t w 9 - : l r @5 ’n 00084 = |1+ 29991 _, \ n J 1.0984 = [ 14 2099 N Sytrialand emor. n= 4 Thus, the mode of interest is quanearly ER =14 — 1={1+0.01" -1 4 ER = 4 .068% I Y )14 Y ff{ = 14 i 1 = “C ] ) A ?:|‘ :I ER = 14 48% ' ! 4 ' . 0075 ER= 14+~ 1="--~U; -1 “ 4 ER=7T.71% =P ”" where | = 000/ = 00125 n=4(4)=16 12,800(1 + 0.0125)" 15614 59 - - F=P(1+ |)" where: | =0.12/4 =0.03 ne=410) = 40 100,000 = P(1 + 0.03)* P = 30,655.68 563
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S64 100! Solved Problems in Engin i 10.000 = P(1 + 0.04)" P = 4388 336 I;' - — - Fe2 P14 i Fi = 4388.336(1 + 0.04)" = 5552 645 Note: From the choices. the nearest answer I8 5.500 F=P+) = 50.000(1 » 0.075) F=7178147 20,000 P =50 AESR 25.000 7S (KN) Solving for the effective rate per yoar ' | Y D05 : 1= : 25000 7500 50,000 + ) 4 S . 3 . (Y4 1) {1+ 1) (1+ 1) P 25000 75000 90,000 + —— z ’ | NENRE L RPP— : {1+ 0.050625) (1+0050625)' (1+0 Fes P14+ = 3500(1 + 0.1 F = 6,26587 1= Q.07/2=0035 FuP(1+)" 4P = P(1 + 0.035)° 4 = (1.030) Take log on both sides 0g 4 = 10g(1.039) og 4 =2n log 1.035 1= 20,15 years \- n UOA 0625)
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Engineering Economics (Simple & Compowund Interest) 565 m i =0.12/2 = 0,08 Fe=pP(14+i) 2000 + 3000 = 2000 (1 » 0.08)° L' S; — i:' L‘_’:‘ " Take log on both sides og 2.5 = log(1.08)" og 2.5=2nlog 1.06 n = 7.66 years, approximately 8 years F=P(1+) 700 =600(7 +1) =186 FaepP+i)" = 10,000(1 + 0.127 F=17062342 Profit=F - P = 17623.42 - 10,000 FProfl = 7 623 .42 ! ™ . = P(1+)"+P(1 +0) 2000(1 « 0.08)" + 2000(1 + 0.08)° F=6034 665 100041 + 0.08)'° F=210892 Note: From the choices. the nearest answer is 2 1680 r n=2(1983 - 1988) = 10 I = 0242 =012 F=P(1+1) = 200.000(1 + 0.12)" £ =821 170 ] [ ’ 2( | 1 ) y " - N+9=983 125/12 = 0.008375 ! 25 n e A " P(1+i) 500,000 (1 + 0.008375)" 1,190 848,73 = interest = F — P = 1,190,848.73 - 500,000 Interest = 600 B48 73
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566 [00] Solved Probiems in Engineering Mathematicy by Tiong & Rojas Note: From the choices, the nearest answer is 880, 850 n=15) = 30 = P(1+ 0" : 1,000(1 + ¥2)™ 1 +05)" 8% N FJL\- W~ -y O N [l ua~ ne=10(4) = 40 = 008BM =0 02 FeP(l+) ' 5.000¢1 + 0.02)" F=11.04020 Inerest =F - P = 110402 - 5.000 iIMerest = 5,040 .20 FePe flh = 1,000(1 + 0.05)" F=1477.455 Money left after the princpal is withdrawn = 1, 477 455 - 1000 = 477 455 Let: Fyg = total amount after the end of 168" year Fiu =P 4 F W - T00.4 Note: From the choices, the nearest answer 18 706 n=20(2) = 40 F=P(1+)) 3,000 = 1,500(1 + i2)* 2=(1+05" i = 3.5% F et " « 2.000(1 + 0.15)° F = 4626 n=§4) =32 i= 0.08/4 = 0,02 F=P(1+i)" 25,000 = P(1 + 0.02)™ P = 1326583
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F’L‘.‘-’f.’('t'r"“f,\: hL",'n(n"ll: 5 (.\.‘l."c"l';lit' & { .u."-[‘uh'.‘u/ 4’".'("".\,: 50.. ne=2015(2" =403 iw DO7/2=0.035 F=P(1+)) = 500 000(1 + 0.035)¢ F = 200166 Note. From 1he chocas, tThe nearest answer 1s 200 150 Let F = value of the account after 5 years considenng thare was NO NBanon ¥ = yalue of the account in today's peso due 10 inflation FePit+))" = 10.000(1+ 0.15)° F=2011357 F=P(1+i) 20.113.57 = P'(1 + 0.00)" P = 15030 i = 0,08/4 = 0.02 e P e c:'“ 50.000 = 20.000(1 + 0.02)" 25=(1.0)" Take log on both sides p— 2 K = b '“;4 0g £.0 = ogl1.ud) iog 2.9 =4njog 1.02 n=11.57 years 1 Compute for the amount due afler one year a. Borrow money from a fnend =P 4 i F = 3,500(1 + 0.015)"" = 4,185 b. Borrow money from a bank F=P(1+)) = 3,500(1 + 0.018) = 4,130 NUS, YOU Wil pay MO0 1ess Dy DOMMowing the monay rom the bank P= p‘. * P‘- g F. f-; (1+U3 (1-0)‘ 100 100 P=_ =4 = 163 {1+ 0.08)° (1+0.08)
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§68 100! Solved Problems in Engineering Mathematics by Tiong & Rojas | DAY 23 ENGINEERING EGONOMY UITY, DEPRECIATION, BONDS, - JREAKEVEN ANALYSIS, ETC.) Y » . .‘u - ~ame - q ‘-&’ ANNUITY Annulity s defined as a senes of equal payments occumng at equal interval of ime When an annuity has a fixed time span, it is known as annuity certain, The folioweng are annuity certam 1. Ordinary annuity is a type of annuity where the payments are made the end of each penod beginning on the first penod 2) Sum of ordinary annuity i { ! 2 3 4 n A+l -1 b : A A A A A T : > 5 D) Fresem wornn of omlnaf\_/ annuiry s T o Vg where. P = principal interest per panod number of parods uniform payment ')( : 5 - i u n - 2. Arnmuity due s the type of annuty where the payments are made at the beginning of each period starting from the first penod cash flow of annulty due
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Engineering Lconomics (Annuity, [)\"!'h’(.’u'l.'.’ufl, Donds, Breakeven, etc) 569 3. Deferred annuity is the one where the first payment does not begin until some later date in the cash flow U ) 2 3 4 5 D N : . . l I _I I l \ A A A A cash flow of deferred annuin When an annuity does not have a fixed time span but continues indefinitely, then it is referred 10 as a perpetulty The sum of a perpetuity i1s an infinite value Prasent worth of a perpetuity TN . m’_l 2 -3 X ‘ EREEE . A A A A A where: A = uniform payment : | = interest per penod P e CJD!(R'iICd costof a property refers 10 the sum of its first COos! and cost of perpetua maintenance. Thus Capitalized cost = first cost + cost of perpetlual mainenance BONDS Bond is a long-term note or a financial security Issued Dy businessas or corporation and guaranteed on certain assets of the corparation or its subsidianes ponds are repayable on maturity and bear a fixad nominal rate of interest Bond rate refers 1o the rate of interest that is quoted in the bond Bond value is the present worth of the future payments that will be received P, c 9 | : 3 T Sy nT 4 - W5 e Y | Fr Fi Fr Fr Fr Fr Fr l' o R av—— "
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S70 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas Where | Vi, = value of the bond n periads prior 1o redemption C = redempbion or amount at maturty (usually equal to F) F = par value of the bongd N = number of penods pnor o redamption | = bond yweid nterest per panod I = bond rate per interest penod DEPRECIATION Depreciation is the reduction or fall in the vaiue of an asset or physical property gduring the course of its working iife and due to passage of time Value is the money worth of an asset or product. it also refers to the present worth of all future profits that are 10 be received through ownership of a particular property Market value is the amount a willing buyer will pay 10 8 wing seier for a property where each has equal advantage and neither one of them is under the COMpulison 10 buy or sall Book value is the worth of the property as reflected in the book of records of the company Use value is the amount of the property which the owner belleved 1o be its worth as an operating unit Fair value is the worth of the property determined by a disinterested person In order 10 estabiish an amount which is fair 1o both the buyer and the seller Salvage value the amount obtained from the sale of the property. This is also known as resale value. Salvage value implies that the property will still be use for e purpose 1 1s infendeq. Differant mathods of compuding depraciafinn nf a proparty 1. STRAIGHT LINE METHOD 8) Annual depreciation charge, d Ca~C, =874 = n -n = first cost = value after n years (salvage vaiue or scrap value) n = life of the property r\ i where.
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W Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 571 b) Book value after *'m" years. Cm where. D, = lolal depreciation after “m” years Cay =6 -0y Us =dim) SINKING FUND METHOD 2) Annual depreciation charge D) Book value at the end of “m” years D = 0o ~-C, where: Dm = total depreciation after *m” years o - d+if 1) m ! DECLINING BALANCE METHOD. Also known as Diminishing Balance Method or Constant Percentage Method Use the Matheson Formula Note: This method is not applicable if the salvage or scrap value is zero SUM-OF-YEARS' DIGITS METHOD. Commonly known as SYD Method 8) Sum of the years' dight, I e D) Respeciive depreciation charges First year: JlIk S Second year dy = Gy - C, )
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572 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas Therd year gy = (C ook And s0 on BREAK EVEN ANALYSIS Break-even refers to the situation where the sales generated (iIncome) i just enough 1o cover the fixed and variable cost (expenses). The level of production where the fotal ncome Is equal lo the 1olal expenses is known as break-even point Break-even chart is a diagram which shows relationship between volume and fixed costs, vanable costs, and income. The following is an example of a break-even chan moome hreak-even "-‘(v'i'.’! ! ’ ['r"l"o' ) revenue “ 5 vanable costs r--q |~e ‘ : — \ ‘I AR B AR AR M AT BN R AR TN, AT AL LT, _'£ — f "",".4".'1« fion LEGAL FORMS OF BUSINESS ORGANIZATIONS I'he legal forms of business organizations are the folloming: ] Sole proprietorship - considered as the simples! type of businass organization wherein the firm is owned and controdled by a singie person <2 Partnership - is a firm owned and controlled by two or more persons who are bind to a partnership agreement
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& | XONI0005 158 _£Z oA o) aled xau ay) o) pesddud 4 » . '3 & [ | : ' i . ' ’ ' - ' - - - o At 1 - - S ( b i ) b ‘e é ; 4 sty ’ ’ l b 4 » ) - \ ' " ‘ - 5 ’ - ] ’ AN "v}. 3 > ) » ks , r . ‘i ' : 4 "‘0 g . A 2 4 ) - - "y - IV ] g PRI P 2 o d ] M | - . n - » " 4 4 ) . Y — ’ - A i & . » | . L‘n - : . ) 0.’ ” ) - . — - - - . . . - Besne s 144 s 4 F N ’ Yt ‘(' o 4% -t pYWi 4 wisd Y, ‘l',' GRLAI0DN L) S1E)qoud Faay) 348 35y | ol aouy nod Qg SARRISdo0D B 20 Auedwod ¥oois-uol S8 umouy SOLLNSWOS 81 SIY 1 'Op PINCO UOsuBd B8l B WO uogoesuss ssousng Aue v abebus U PUR §I SUMO OUM SIBNDIAPUI aul wou) seredas Aus 1elay RUSIP & SE DaUyep OS(E &1 1| “SaIeys JO BQuINU au] O] dn PaDIMD S L2IUMm JO |enaeD aUl pue siaployazeys Asuipio o anoub B AQ POUMD LY € 1 - uonesodion £L8 (210 ‘UIIYDALY "SPUOG UONDIDAIIACT "ANNuuy ) SONUCR00T SutipaSuy
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§74 W] Soived Problems in Engineering Mathematics by FNong & Rojas e R Ry SR IVOL &Y Time element: 4.0 hours Problem 951: EE Board October 1997 A man purchased on monthly instaliment 2 P 100,000 worth of land. The interest rate 15 12 % nominal and payable in 20 years. What is the monthly amorctization? A P1.101.08 B P1.121.0 C 1.152.15 D P1.12812 Problem 952: ECE Board April 1998 Money bormowed today s 10 be paid In 6 equal payments &t the end of & guarters. If the interest 1s 12 % compounded quarierly. How much was initially borrowed I quanerly payment s P 2000 007 A. P10834 38 8 P10.38280 C. P10586.89 D. P1020056 Problemm 9531 ME Board October 1996 You need P 4000 per year for four years 10 go 1o college. Your father invested H 5,000 In 7 % accoun for your aducation when you were bom. If you withdraw P 4,000 at the end of your 17", 18", 18" and 20" birthday, how much will be left in the account at the end of the 21" year? A P1.700 B. P2500 C. P340 D. P4000 Problemm 9541 ECE Board November 1998 wWhat 15 the accumulated amount of five year annuity paying P 6000 at the end of each year, with interest at 15 % compounded annually? P 40.454.29 P 4111429 P 41,454 29 P 40,544.29 SoOW >
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 575 Problem 955: ME Board April 1998 How much must be deposited at 6% sach year bagnning on January 1, year 1 in order o accumulate P 5,000 on the date of the last deposit, January 1, year 67 A. P 751.00 B P717.00 K P 715 00 D P72500 Problem 956t ECE Board November 1998 A debt of P 10.000 with 10 % interest compounded semi-annually is 10 be amortized by semi-annual payment over the next 5 years. The first due in © months Determine the seme-annual payment A P1.200.00 B P 120505 C. P1.153. %0 0D P 140045 Problem 957: EE Board October 1997 A young engineer borrowed P 10,000 at 12 % interest and paid P 2,000 per annum for the last 4 years. What does he have 10 pay at the end of the fifth year in order fo pay off his oan A, P8819528 B P 567400 C *5.074 00 D. P3206.00 Problem 968; EE Board April 1997 Mr. Cruz plans 1o deposit for the education of his 5 years old son, P 500 at the and of each month for 10 yvears at 12% annual interest compounded monthly. The amount that will be availlable In two years s A P 13.000 B P 14500 C. P13.500 D. P 14000 Problem 999 ME Board October 1994 if yvou obtain a loan of P 1M at the rate of 12% compounded annually in order o build a house. how much must you pay monthly 1o amortize the loan within a penod of ten years? A, P13.584.17 B. P 1285521 C. P1585545 D. P12950025
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576 [00] Solved Problems in Engineering Mathematicy by Tiong & Rojas Problem 960: ECE Board April 1998 How much must you nvest today in order to withdraw P 2 000 annually for 10 years if the interes! rate is 8%7 A. P12853.32 8 P1288137 C. P 1238532 L. P1283532 Problem 961: ECE Board April 1998 A person buys a plece of lot for P 100,000 downpayment and 10 deferred seny annual payments of P 8,000 each, starting three years from now. What is the present value of the investment if the rate of interest 15 12 % compounded semi-annually? M 134 666.80 M 143,989 08 P 154 606.80 P 164 9589 BD RO R Problem 962: CE Board May 1998 A man loans P 187 400 from a bank with mterest at 5% compounded annually He agrees to pay his obligations by paying 8 equal annual payments, the first being due at the end of 10 years. Find the annual payments A, P 4458204 B. P56143.03 C P6233462 D P 3823604 Problem 963: A housewrfe bought a brand new washing machine costing P 12,000 if paid in cash. However, she can purchase it on Installment basis 10 be paid within 5 vears, If money s worth 8% compounded annually, what is her yearly amortization if all payments are 1o be made at the beginning of each year? A, P 278285 H P28725 C. P2400.00 D. P282758 Problem 964: ME Board October 1996 Mr. Ayala borrows P 100,000 at 10% effective annual interest. He mus! pay back the loan over 30 years with uniform monthly payments due on the first day of each month. What does Mr. Ayala pay each month? o P 870.00 B. P 846.00 C. P878.00 D. P839.00
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 577 Problem 965: ME Board April 1998 A house and ot can be acquired by a downpayment of # 500,000 and a yearly payment of P 100,000 at the end of each year for a period of 10 years, starting at the end of 5 years from the date of purchase. If money is worth 14% compounded annually, what 1S the cash pnce of the property” A ” 810,100 P 808 811 C. P 801,800 0. P 805,502 Problem 966: ME Board April 1998 A plece of machinery can be bought for P 10,000 cash or for P 2 000 down and payments of P 750 per year for 15 years. What is the annual interest rate for the time payments’ A 461 % 8 381% C. 56711 % D. 110% Problem 967: CE Board November 1996 A man inherited a regular endownment of P 100,000 every end of 3 months for 10 years. However, he may choose 10 gel a single lump sum paymen! st the end of 4 years. How much i this lump sum if the cost of money s 14% compounded Quarteny’/ A P 3802862 B. P3702838 C. P 3502546 D. P3602431 Problem 968: ME Board April 19986 A parent on the day the child s borm wishes 10 determine what lump sum would have 10 be paid into an account beanng interest at 5 % -;‘.mr.p:)unded annuasy, n order 1o withdraw P 20,000 each on the child's 18", 19", 20" and 21" birthdays How much is the lump sum amount? A P35941.7 8 P 3364173 C FJ0 88173 D P2584173 Problem 969: ME Board April 1998 An instructor pdans to retire in exactly one year and want an account that wl pay ham P 25,000 a year for the next 15 years. Assuming a 6 % annual effective inforest rate, what is the amount he would need 10 deposit now? (The fund will be depleted after 15 years) A. P 249000
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§78 1007 Solved Problems in Engineering Mathematics by Tiong & Rojas B. P 242800 C. P 248500 0. P 250400 Problem 970: EE Board October 1997 An investment of P 350.000 s made 10 be followed by payments of P 200,000 each yoor for 3 yaare What 8 tha annual rate of rmiurm on nvestimant for the project? A 41V T7T% 8 K ¥ ) C o % D 15% Problem 971: EE Board April 1997 A small machine has an initial cost of P 20,000, a saivage value of ¥ 2 000 and a fife of 10 years. If your cost of operation per year is P 3, 500 and your revenues per year is P 9,000, what is the approximate rate of retum (ROR) on the investment? A 2! 0 )'u B 225% C 239 % D 248% Problem 972: CE Board November 1996 A man paid 10% down payment of F 200,000 for a house and ot and agreed 10 pay the balance on monthly instaliments for “x” years at an mieres! rate of 15% compounded monthiy. if the monthly instaliment was ¥ 42 821 87, ind the vaiue of x7 DO m>» -~ D - Probliem 973 ME Board April 1998 A manufactunng firm wishes 1o give each 50 employees a holiday bonus, How much 8 needed 10 Invest monthly for 8 year al 12 % nominal inlerest rate compounded monthly, so that each employee will receive a P 2,000 bonus? B F 12,008 8 P 12610 C. P12800 D P12.300 ~ Problem 974: CE Board November 199§ Find the present value in pesos, of a8 perpetuity of P15,000 payable sems annually if money is worth 8% compounded quanerly A P3ITL537
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 579 B P374977 C. P373,767 D P37 287 Problem 975: A fund donated by a wealhy person 0 HEE 0 provide annual scholarships 10 deserving EE students. The fund will gramnt ¥ 5,000 Tor each of the hwsl ive years, M 8,000 for the next 5 years and P 10,000 for each year thereafier. The scholarship will start one yeoar after the fund IS estabished. If the fund eams 8% Inlerest, whal IS the amount of the donation” ~ P 101 605.71 B. P 10150521 C. P100.506.21 U P 9860171 Problem 976: ME Board April 1998 A company issued 50 bonds of P 1,000.00 face value each, redeemabie al par at the end of 15 years o accumuiate the funds reguired for redemption. The firm established a sinking fund consisting of annual deposits, the inerest rate of the fund being 4 %. What was the pancipal in the fund at the end of the 12" year? P 35,883.00 P 38,378.00 P 4145300 D. P37518.00 OO0 > Problem 977: ME Boasrd April 1992 A unit of welding machine cost P 45 000 with an estimated life of 5 years. Its salvage value is P £2.500. Find #is depreciation rate Dy straight-iine method A 17.75% B, 19.88 % C. 18.89% D. 15.56% Problem 978: EE Board April 1997 A machine has an iniial cost of P 50,000 and a salvage value of P 10,000 after 10 years. Find the book value after 5 years using straight-ine depreciation A P12500 B. P 30,000 C. P16400 D. P22300
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Engineering Economics (Anmuity, Depreciation, Bonds, Breakeven, etc) 3581 Problem 984: CE Board November 1995 A machine costing F 720,000 is esiimated to have a book value of P 40 54573 when retred at the end of 10 years, Depréeciation cost is compuled using a constant percentage of the deciining book value. What is the annual rate of depreciation in %7 A 28 8 25 C. 16 D. 30 Problem 985: CE Board May 1996 A machine costing P 45 000 is estimatad 10 have a book value of P 4 350 when retirad at the end of 6 years Depreciation cost is computed using a constant percentage of the decining book value, What is the annual rate of depreciation in %7 A 3325% 8 32.25% C. 3525% V. 34245'% Problem 986! ECE Board November 1998 ABC Corporation makes it a policy that for any new equipment purchased, the annual depreciabon cost should not exceed 20% of the first cost al any time with no salvage value. Determine the length of service life necessary If the depreciabon used s the SYD method 8 years 10 years 14 years 19 years o000 » Problem 987: ME Board April 1998 A company purchases an asset for P 10.000 00 and plans to keep it for 20 years. If the salvage value is zero st the end of 20" year. what is the depreciation in the third year? Use SYD method A P 1,00000 8B P857.00 C. P837.00 D P 747 00 Problem 988: ECE Board April 1999 A Telephone company purchased a8 microwave radio equipment for P 6 million, freght and installation chargas amounted to 4% of the purchased price. If the equipment will be depreciated over a period of 10 years with a salvage value of 8% determine the depreciation cost during the 5 year using SYD A P 52626910 B PB62278607
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582 100! Solved Problems in Engineering Mathematics by Tiong & Rojas C. P6B38.27208 D P627.988 80 Problem 989: ME Board April 1998 An asset is purchased for P 8.000.00. its estimatad ife is 10 years after which it will be soid for P 1,000.00. Find the book value during the first year if sum-of-years’ digit (SYD) depreciation is used P8 00000 P 6.500.00 P 7.545.00 P 6.000.00 oO@m> Problem 990: EE Board April 1997 The maintenance cost for 2 sewing machine this year s expecied 1o be P 500 The cost will increase P 50 each year for the subseguent 8 years. The mterest is 8 % compounded annually. What 18 the spproximats present worth of maintenance for the machine over the full 10-year period? A. P4700 8. P 5300 C. P 4,300 Lt "‘ 5), l.)’ Problem 991: CE Board November 1996 Al 6%, find the capitalized cost of a bndge whose cost is P 250M and life is 20 years, Il the bridge must be partially rebullt at a cost of P 100M at the end of each 20 years A P275.3M B. P 2855M C. P2953M D. P2821M Problem 992: CE Board May 1997 A corporation uses a type of motor truck which costs P 5,000 with e of 2 years and final salvage value of P 800, How much could the corporation afford to pay for another type of truck of the same purpose whose life is 3 years with a final salvage value of P 1,000. Money is worth 4% A. PB4a50686 B P7.164 37 F 5. 398 24 D. P903456 )
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 583 Problem 993%: ME Board October 1995 A company must relocate one of its factones in three years. Equipment for the loading dock is being considered for purchase, The onginal cost i1s P 20.000. the salvage vaiue of the equipment after three years is P 8 000. The company's rate of retum on the money 18 10%. Determine the capital recovery rate per year A. P5115 B P4548 C. P5625 D. P4805 Problem 994: EE Board October 1998 The annual maintenance cost of @ machine shop I8 P 69994 If the cost of making & forging is P 56 per unit and its selling price is P 135 per forged unit, find the number of units to be forged 10 break-even A. 888 unis B 885 units C. 688 unis . 6868 units Problem 995: CE Board May 1998 A manufacturer produces certain flems at a abor cost of P 115 each. material cost of P 76 each and vanable cost of P2.32 each If the item has a unit price of P o000, how many number of units must be manufactured each month for the manutfacturer 1o break even If the monthly overhead is P 428 000 A 1053 8. 1138 C. 946 D. 1232 Problem 996: ME Board April 1996 Steel drum manufacturer incurs a yearly fixed operating cost of § 200,.000. Each drum manufactured cost § 180 10 produce and sells $§ 200 What 5 the manufacturer's break-even sales volume in drums per year? A 1250 B 2500 C. 5000 D. 1000 Problem 9971 JRT Indusines manufactures aulomatic voltage reguiators at a labor cost of P 80.00 per unit and madenal cost of P 350.00 per unit. The fixed charges on the business are M 15,000 per month and the variable costs are P 20.00 per unit. if the automatic voltage regulators are sold 10 retallers at P 580.00 each, how many units must be produced and sold per month 1o breakeven? A. 104
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584 100! Solved Problems in Engineering Mathematics by Tong & Rojas B 200 C. 120 D. 150 Problem 998: ME Board October 1990 Compute for the number of locks that an ice plant must be able to sell per month 1o break even basad on the following data Cost of elecincty per block Tax 1o be paxd per block Real Estate Tax Salanes and Wages Uthers Selling pnce of ice - P 2000 - P 2.00 P 3.500.00 per month - P 25,000 .00 per month P 12.000.00 per month P 55.00 per bloc A 1228 8. 1285 G 1973 D 1312 Problem 9991 EE Board October 1997 The annual maintenance cost of a machine 1 P 70,000, If the cost of making a forging is P 56 and its selling pnice s P 125 per forged unil, Find the number of unils {0 be forged to break even 1015 units 985 units 1100 units 1000 units o> Problem 1000: ME Board April 1998 XYZ Corporation manufaciures bookcases that selis for P 65.00 each. It costs XYZ Corporation P 35,000 per year 10 operate its piant. This sum includes remt depreciation charges on equipment, and salary payments. If the cost 10 produce one bookcase Is P 5000, how many cases mus! be sold each year for XYZ 10 avowd taking a loss” A 4 A ~hN o W W L W -
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 585 Problem 1001: ME Board April 1998 A company which manufactures electric motors has a production capacity of 200 motors a month. The vanable costs are P 150 .00 per motor. The average selling price of the motors is P 275.00. Fixed costs of the company amount to P 20,000 per month which incudes taxes The number of motors that must be sold each month 1o break even Is closest to 40 80 oOm>» 150 160 851 052 853 454 955 Ho0 :‘ f’ : 958 (_4{_,[_‘ o650 961 962 (_‘b' { > — e ST w2 OPRAO>D> P> ~ y > 1 ANSWER U4 D 977 p65. 8 978 966 A 979 967 8B 98( 968 C 981 908 8 982 970. 0 983 971. 8 584 gr2. C $985 873. D 886 974. 8 087 g9/5. 0 588 J9/6. 0 588 KE ™ 5 O>PO0>00>0>P000 w0 L » OO>PrO00> RATING _I W3-5) Topnotcher ‘ |as-3e Conditional :] 0-24 Failed I FAILED. repcat the test
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586 J00] Solved Problems in Engineering Mathematics by Tiong & Rojay SOLUTIONS TO TEST 23 BBl i-01212=001 n=12(20) = 240 0 / 2345 2 D= l\"oor 12 l I I II l AK1+0.01P* -4 P ieioiond 100,000 = _-b‘~_°.9_47_f_ {1+ 0.01"10.01) A=110108 1=0.12/4 =000 n=_6 o - Al 7-1] 20001 003f (+ifi (1+003/(0.03) P = 10834238 Fe=Py(1+i) p = 500(1 + 0.07)" : ’,"v Fy = 20,702.8¢ T . 0 16 |7 18 1920 2] F::A"lj_“l ’ ll 1 1 ) 400041+ 0,07)* 1' A A A A 0.07 B B Fz=17.758.772 LSy _’l_ Fa= F‘m +1) = 17,708,772 (1 + 0.07) 3= 19.002.95 Money left = F, « F; = 2070281 ~ 19.002.96 Money left = 1, 680 88, approximately 1,700 : x 4 0 Ty Ak B 954, [ Aft+ir-1] 60001+ 0.15F 1] i n 0.156 Y l 1 l 1 F = 40454 29 A4 A A A A
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 587 m {= 0102 =008 n=2(5 =1 i A1+ 1y-1] (1+if] e it ta ) 10,000 = At +0.05)°-1) (1+0.05)°{0.05) A= 120506 957. | Py + Py = 10,000 e { s J 42 Y ] .A}'." - 'j 1Ji ety -= 10,000 1 Yy l 1 (141 (1+9) i A A A 20001+ 0.12) -1 : : | | . h O‘ ) J b e = 10,000 P, - i (1+0.12)'(0.12) (1+0.12F Ps < j FE0 U/ 7L Note: From the choices, the nearest answer is 6.919.28 | = 0.12/12 = 0.01 n=12(2) = 24 F u A[‘." | :‘7' 1J & 5&)'3[‘7 + 0.0 1_"‘_4__1J | 0.01 F=13486.7 Note: From the choices. the nearest answer is 13.5 m Solving for the interest rate per month 0.12= (14| ~1 L 0.009488 12(10) = 120 - f'_\h\ +f 1J (1+iFi 12 n .a.[n + 0.000488)'* 1] (1+0.000488)"“* (0.000488) A=1399417 1,000,000 =
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588 [00] Solved Problems in Engineering Mathematics by Tiong & Rojas At + ir-1] _ 200041+ 0.00)"°1] (1411 (1+0.09)°(0.08) P=1283532 5. = A+ ir-1|_ s0oofi+ 0.08)° 1| (1+0.08)"(0.08) “ ' I’II P, = 5888069 Py= Pyt + i) ‘ 58 88069 =P, (1+0.06) P:= 43 900078 100,000 Total amount = 100,000 + P, 00,000 + 43 969 078 = 143 999 078 — Vs < | _ A[nur 1], Af1+0.05)" -1 (1+ifi (1+0.05 (0 .05) 6643 A e n pv Lo p‘-(“ ’ I}fl . 6643 A =187 400(1 + 0.5)° A =44 982 04 Cashprice=A+P _qf 12,000=A + A+ iy 1] {0 P HYMMEry ) /'\‘ c'(.' (1+ 11 A}:i +0.08F 1J (1+0.08)(0.0.08) A=2 400 12000 = A + Solving for the interest rate per month 14 _! f‘: ) 12 ) .01:i \
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, eic) 589 Borrowed monay = A + P Aff1+0.007 974P* - 1| 100,000 = A + R (1+0.007974)* *10.0.007974) A =838 S o . Al (1+ifi !0 payments oopoft o1 s (1+0.14)°(0.14) - 2 NN, - B e MY N L T s [ i1 pya Pi_o 5218115 00 SB L 7 i (1+if [(1+0.14) o =5 P;= 3088358 | T S &5 PISREVSR WRn | R Cash price = 500,000 + P; = 500,000 + 308 835 9 Cash price = 808, 835 8 Note: From the choices, the nearest answer is 808 811 Cash price = Down payment + Presént worth of the annusty 10,000 = 2000 + A1+ 1«11 75001+ 1) 10,000 = 2000 + L:h' .t ) 1+4)° 106667 = A1 =3 {1 +1) | By substitution, | = 4 §1% m | =0.14/4 = 0.035 16 payments 24 payments F’O B - il ¥ /} " 4 ‘l ‘, " (. )' 7 .' h‘ 3() J/} gy sy e meeg _ 10000041+ 0.035 1| i o A A i1 A A A (1+0.0351*(0.035) e ' A P = 1,605836.76 e i & ' ol - n—— — F-h“'l~’~~ p _ 100 CCQ}? +0.035)'% -1 0.035 el | = 2,087 102 97
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590 1001 Soived Problems in Engineering Mathematicy by Tiong & Rojas Llumpsumprce =P + F = 16056836.76 + 2007102.97 Lump sum price = 3.702.8398.73 m Note: From the choices, the nearest answer is 12810 o = Alft+ir-1] 20001 0.08)y 1 (144 (1+0.05)°(0.08) 0 1718 1920 21 ...70919 ...............l l l 1 -'1"L,Y“;. A A AA _ 70919 . (1+0.05) P .; P =3094172 350000 B .. Ao ir-1). 2so00ft. 0081 Tk (1+i1F (1+0.08)°(0.08) P =242 808 v l i A 4 <70 IR 2 P Uk o (1+ifi - bospums & ,’_-q. ...................200,000§1+ if* -1) l1 0 !'J “' f'l. <‘ N S " By substituton, | = 32.7% S > } A T e t11 T /'z L2 3 e, & -5-!»“:-‘!——-]4--"— e o - 20 ()’f .4..]1 +1) 20 000 Note: A = revenues per year — oparating cost per year A= 000 -~ 3500 =5500 350,000 = Subatitudc 550001 + 1) 1] 2000 i‘ . — = 20,000 {1+ (141)"° By substitution, i = 24 8% Down payment = 10% of Cost of house and lot 200,000 = 0. 1(Cosl) Cosl -?DO(:OOO —— g
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Engineering Economics (Anmuity, Depreciation, Bonds, Breakeven, eic) Balance = Cost - Down paymemnt = 2 000.000 - 200.000 = 1.800 000 |=0.15/12 = 0.0125 A[‘1.‘,r. 1J “ ’ .‘ffi'l 42821 avkt + 0.0125)/%* -1 (1+0.0125)**(0.0125) 2543(1.0125)"™ 12 [ 1,800,000 = nm:m - " n 0.5 2.107 Take log on both sides 12x log 1.0125 = log 2. 1072 X =5 yoars BEEl - 01212001 n= 12 F = 80(2000) = 160,000 e "l‘.‘ *1]- fl ;Ju' +0.01) .ff S R acets RS- 0.01 A=12615.80 160,000 A ¥ ! ) | ER (1o1] =[1e] 4, 2) 4 2 P e I ) ba , —‘,1'2.l L = 0.404 F 5e A 15000 i 0 404 P =371,287.128 @4 . 2. 3% 5 oo A o1 Yy (1+1)' A A AAA _ 5000f1+0.08)° -1 prad o4 (1+0.08)(0.08) P, = 19.963.55 591
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592 1001 Solved Problems in Engineering Mathematics by Tiong & Rojas P;zAiLJ'JJ (1+if 0. 6 7 8 910 _ 5000f1+ 0.6 - RERR (1+0.08)(0.08) g o422 .44 P; = 31841 68 o | “< ................P, 3184168 o VLR IS i p! - e—— (14 1)5 (1 “().08}‘J 1, ar Pym 2173897 i tciimeent Ilf l l l Pe= = = 19990 4 126,000 AAAA s ) P, = ‘_‘p_l__:_ b, -JzS_OOCa P‘ .(- : X - (1+0)"° (1+0.08)"° TAR L a2 P:= 57809 186 Total = Py + Py + Pe =57899.186 « 21,738 97 + 18.063.55 Tolal = 99 601.71 A1+ ir 0 - JF. 0 14 15 e S TR 15 50(1,000) = Ak.’ . %00:‘_ .'J A A 4 1 4 A= 2.407 . VATV . ie J Let: Fy = value of the given annusty when n = 12 years Fi= A_h +if 1J ) k 2.497k1 +0.04)'7 - ,J T 004 Fi = 37,519 -C, _ 4500( 50C m dqu_ Cp _ 45000 25&):8‘500 n 5 d 8500 o Depreciation rate = . = —— x 100% Co 45000 Deprecation rate = 18 .89 %
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Engineering Economics (Annuity, Depreciation, Bonds, Breakeven, etc) 593 m dwS0-Cn , 5000010000 _ . ., n 10 Cm = Cp - d{m) = 50,000 — 4000(5) Cm = 30,000 m 4= Cg-(C, ~costto dismantie) _ n 10 d=P 76500 o = Cy - d(m) = 800,000 ~ 76,500(8) » = 341,000 ) r) 500,000 + 30,000 = 530,000 0.10{500,000) = 50,000 O O " Cg ~Cy _ 530,000 - n 5 50000 _ o0 0 Q Ll OO0 o = Co - d(m) = 530,000 ~ 96.000(4) 146,000 - - -~ o —C 00,000 0,000 m da Cp -G, _ 500,000 ‘l}‘flyf-{"PIG.OCO n :)‘:l Let: D = 10tal depreciation after *m” years | dm = 16,000(3) 43,000 Co~C 50000 - 10000 gn ——Aa = 4000 n 10 N - ) vm = LUo = dim) = 50,000 - 4000(5) —_ e e (Co~Cy )t _ (10000 - 500)0.04) (1+0.04)" -1 40,545.73 = 720,000(1 k) * (1-k)"=0.0683 k =0.25 or 25% m ":n C (% - ""— = 45000(1 - k) ' 0.09668 0.3225 or 32 25% 80,000 - (50,000 - 15,000)
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594 100] Solved Problems in Engineering Mathematics by Tiong & Rojas m Note: Using SYD, tha largest charge of depreciation is the first year s = (Co = C R ] Oy = 0™ /'u‘ ; "|'S years | |' i 0.2Cs=Co| = | \ dears. Y years = 5n -— Using formula for sumofan AP . nin+ 1) }_yea_q L L LED in = T( _} 2 IOn=2n+n"-n On=n' n=9years | - m dx:\'C;;—(:'f—". - i | ‘} years | Using formuda for sumof an AP nn+1) = 220+ 7 Y years = ~ - 210 ad 2 Substitute , [ 18 ! dy = (10000 - 0] - \ 210 s = 857 m g = (Co—Cn) | < \ - - n i T EA . ‘ iiyears, Using formula for sum of an A P nin+Y) T0(10+9) [ ) years = - 55 - » - .- Co = 6,000,000 + 0.04(6,000,000) = 6,240,000 Cn = 0.08(6,240,000 ) = 459,200 Subsiitute ds = (8.240,000 - 499,200) - ds = 626,269.10 — e ——— . &
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Engineering Economics (Annuity, Deprectation, Bonds, Breakeven, etc) B c.-c-o - a_— n d':{l\,[,—\_,.! I l S yoars nn+1 2 1X10+ 1) Y years = >— =85 - » | 1() ! d1= (8,000 1,000) [ = | = 1,454.54 L N Substitute Cet = 5,000 - 1 454 54 (;n| = 7 6454'3 m Using uniform gradeent formula Af1+i) __-.]’(.Jln.;r' - Pe , S 3R (1+1)' | FQ«i ) ‘4-»_*‘ + 0 0B o 1J 14 i .0 o =000 |2 008) =1 - (1+0.08)"(0.08) {0.08(1+008)" (0.08)1+008)" = ;) = 4 653 .88 Note: From the choices, the nearest answer s 4 700 : ™ (“." (,-‘v.. m Capitalized cost = Cy + 0 (140" -1 ) 100 250 + (14 0.08)*" _1 Capitalized cost = 295.3 miillion m Let: ACy = annual cost of the old motor truck AC: = annual cost of the new motor truck (Cas =G ACy = (Coy)l + =1~ (\ ) ’lil‘ - 3 = (5.00010.04) + (3000 - 800)(0.04) (14 0.04)° -1 A"J. - ".,258 82 o~ ':{:'C‘ ~ ;‘ AL " (Lol + — - ‘10 l'l Y . < 42 (Ca, ~ 1000)(0.04) ACs = (Ca){(0.04) + —¢ 4 st (1+0.04)° -1 b c——— §95
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$¥001Q 822 L Amjewxoudde ‘SO £ /221 = X Q0% OF = XEE OO0 ZL +000'GZ + 00S'E + XZ + X0Z = XG6 SESUSCXT = SLLOOU) LIUOW 180 PI0S 94 0} $H20IQ JO BqQuWINUy = X 387 m SHUN OZL = X 000Gl = XG2 D00 €l + XOF + XOCE « XCQ - ¥uo $ASUAdXT = SWOooUY ) - Jiuow Jad paonpoxd 8q 0f SpUN JO JAQWINU = X 197 m SIUN DDO'G = X 000 002 = XD X094 + 000 002 = X002 SOSUSOXT = MU0 ieak Jad 1IN0 Pros 8q 0} SHUN JO JBQUINU = X 187 m SPUN £G0° | Amjewmxosdde ‘'Zp 780’ = X 000 9Ty = X0 00% 000 8CF + XZE Z + XBL + X511 = X009 $9SUSdXT = SWOOU| Yiuow sed paimpejnuew aq o} SHun JO BqWnU = X 387 m SPUN ORE » X PO0 68 = X5 PE6 69 + X0G = XGE | SAasSUAdXT = SOV pabio} 99 O SHUN JO JequINU = X Ja7 m G29'G = sl Jad sl Aerooas |eyden b "Ol- D ' ‘." (0100008 - 00002) + 04000002 = by (Y~ 99) 1500 [enuuy = seah Jad ajes Alenooay epden SEPEL'L =) SEOZE —TOZE0 + POPO'0O = 28'852'2 b= P00+ e+ (PO'ON™D) = 2886 (+0'0M000L - “9D) SOV = YOV Wioy p Juoty Ag sonvwaopy JurpauSuy up swapQoid Pasos 100} 65
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r! v")/h.l',.l Lt x = INCome Bhx = income - - - £ Fdw » ng Aconomcy (Anmuih numbear of units 10 be forged = Expenses - ‘:l}x * T_; '-\AQJ » 70,000 = 10714 48 units Spproximaledy 1 015 units number of cases 10 be soid sach vear EXDENSEes 333 33 cases, approxamately 2 334 2585 iumber of motors 10 be soid each month = Expenses -~ A - ISR T £V &'u.\-. PO . 20 . UOUL 160 motors t » 1 ’ '» . L"Qt".l .-’l‘lt'.". .‘“ 'c»..'..-, Ar.'.rl L;-“ Vel ) | S L 597
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