ABE_Engineering Economy.pdf

Philippine Copyright 2024 by MANILA REVIEW INSTITUTE, INC. 3/F Consuelo Building, 929 Nicanor Reyes St. (formerly Morayta), Manila Tel. Nos. 735-0112 736-MRII (6744)...

Philippine Copyright 2024 by MANILA REVIEW INSTITUTE, INC. 3/F Consuelo Building, 929 Nicanor Reyes St. (formerly Morayta), Manila Tel. Nos. 735-0112 736-MRII (6744) www.manilareviewinstitute.com All rights reserved. These handouts/review materials or portions thereof may not be reproduced in any form whatsoever without written permission from MRII. AGRICULTURAL ENGINEERING ENGINEERING ECONOMY INTEREST 1. Simple interest F = P (1 + ni) a) ordinary simple interest : one interest period = 360 days b) exact simple interest : one interest period = 365 days = 366 days 2. Compound interest F  F = P  , i , n  = P (1 + i ) n P  P  P = F  , i , n  = F (1 + i ) −n F  F   , i , n  = (1 + i ) = single payment compound amount factor n P  P   , i , n  = (1 + i ) = single payment present worth factor −n F  where: P = principal (present worth) F = accumulated amount (future worth) i = rate of interest per interest period n = number of interest periods Rates of interest 1. Nominal rate of interest - specifies the rate of interest and the number of interest periods in a year 2. Effective rate of interest ieff = (1 + i)n – 1 where n = number of interest periods in a year if compounded continuously ieff = er – 1 Manila Review Institute, Inc. Page 2 of 10 Problems: Simple Interest 1. Find the interest on P6800.00 for 3 years at 11% simple interest. a. P1,875.00 b. P1,987.00 c. P2,144.00 d. P2,244.00 2. How many years will it take the principal amount to double if it is invested in an account that gives 10% ordinary simple interest. a. 5 b. 8 c. 10 d. 16 3. A man borrows Php 800.00 from Pikachu and promised to repay him Php 850.00 after 1.5 years. What is the interest rate assuming ordinary simple interest? a. 3.7% b. 4.2% c. 2.1% d. 5.6% 4. From August 22, 2011 to February 22, 2012, determine the exact simple interest of the principal Php 5000.00 for 8%. a. Php 201.49 b. Php 721.21 c. Php 124.22 d. Php 219.89 Compound Interest SITUATIONAL PROBLEM 5 – 6 Pakito borrowed Php 2000 from a bank and promised to pay after 1 year. The bank deducted the interest of Php 80.00 in advance. 5. What is the interest rate? a. 4% b. 2% c. 4.2% d. 7% 6. What is the rate of discount? a. 4% b. 2% c. 4.2% d. 7% SITUATIONAL PROBLEM 7 – 8 Pakito borrows Php 8000.00 from a bank that has an interest rate of 2% compounded quarterly. 7. How much is the money after 8 years? a. Php 9384.34 b. Php 9348.34 c. Php 8523.66 d. Php 8325.66 8. How much is the money after 8 quarters? a. Php 9384.34 b. Php 9348.34 c. Php 8523.66 d. Php 8325.66 SITUATIONAL PROBLEM 9 – 10 A businessman borrows money from a bank and promised to pay Php 5000.00 after 12 months. How much money did the he borrow if: 9. Money is worth 8% compounded monthly. a. Php 6414.81 b. Php 1985.57 c. Php 4616.81 d. Php 1895.57 10. Money is worth 8% monthly. a. Php 6414.81 b. Php 1985.57 c. Php 4616.81 d. Php 1895.57 11. How long (in years) will it take for an investment to fivefold its amount if money is worth 14% compounded semi-annually? a. 11 b. 12 c. 13 d. 14 SITUATIONAL PROBLEM 12 – 14 What is the nominal rate of 4% compounded quarterly if compounded: 12. Monthly? a. 3.987% b. 4.02% c. 4.06% d. 4% 13. Semi – annually? a. 3.987% b. 4.02% c. 4.06% d. 4% Manila Review Institute, Inc. Page 3 of 10 14. Annually? a. 3.987% b. 4.02% c. 4.06% d. 4% 15. What is the present worth of money if Php 8000.00 is paid after 3 years? The money is worth 8% compounded continuously. a. Php 6293.02 b. Php 2963.02 c. Php 2693.02 d. Php 3296.02 16. Determine the effective rate of 8% compounded continuously. a. 8.22% b. 7.95% c. 8.33% d. 7.23% ANNUITY 1. Ordinary Annuity P   (1 + i ) n − 1   1 − (1 + i ) − n  P = A  , i, n  = A  =A    i (1 + i )  n A   i  F   (1 + i ) n − 1  F = A  , i, n  = A   A   i  uniform series present worth factor:  1 − (1 + i ) −n P (1 + i ) n − 1  , i, n  = = i (1 + i ) n A  i A  i i (1 + i ) n capital recovery factor =  , i, n  = −n = P  1 − (1 + i ) (1 + i ) n − 1  (1 + i ) − 1 n F uniform series compound amount factor =  , i, n  = A  i A  i sinking fund factor =  , i, n  =  (1 + i ) − 1 n F 2. Deferred Annuity P  P  P = A  , i, n   , i, m  A  F   1 − (1 + i ) − n   (1 + i ) −m P = A  i  3. Perpetuity A P= i 4. Uniform Arithmetic Gradient  ( l + i ) n − 1 − in  P =G   i ( l + i ) 2 n  G  (l + i ) n − 1  F=  − n i  i  Manila Review Institute, Inc. Page 4 of 10 5. Geometric Gradient  1 − (1 + i ) − n (1 + g ) n  P = A1   ,i  g  i−g  nA1 P= ,i = g 1+ i  (1 + i ) n − (1 + g ) n  F = A1  , i  g  i−g  F = nA1 (1 + i ) n −1 , i = g 6. Annuity Due P = A + A 1-(1+i)-n i Problems: Cash Flow Diagrams 17. Tsupul borrows P10,000 at an interest rate of 8% compounded annually and wishes to repay the loan over a 4 year period with annual payments such that the second is P500 greater than the first payment, the third payment is P1000 greater than the second payment and the fourth payment is P2000 greater than the third payment. The amount of first payments is: a. P1854 b. P1574 c. P1474 d. P1754 18. A sum of P1000 is invested now and left for eight years at which time the principal is withdrawn. The interest that has accrued is 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 th year? a. P706 b. P500 c. P774 d. P799 Annuity 19. Determine the present worth of an annual payment of Php 2500.00 at the end of each year for 12 years at 8% compounded annually. a. P 32, 225.52 b. P 18, 840.20 c. P 47, 442.82 d. P 20, 288.21 20. A contractor bought a concrete mixer at Php 120,000 if paid in cash. The mixer may also be purchased by installment to be paid within 5 years. If money is worth 8%, the amount of each annual payment, if all payments are made at the beginning of the year is: a. P 27, 828.50 b. P 29, 568.00 c. P 31, 005.00 d. P 32, 555.00 21. A man wishes to establish a fund for his son’s education. He did this so that his son will receive Php 5000.00 every year, for 5 years, starting his son’s 18th birthday. If this fund is established by series of annual payments for 5 years, determine the annual payment if money is worth 8%. a. P 1351.34 b. P 3151.34 c. P3251.34 d. P 3521.34 22. Find the annual equivalent worth for the following infinite cash flow series at an interest rate of 10%. Net n Cash Flow 0 0 1 - 10 Php 400 11 - ∞ Php 500 a. P 461.20 b. P 438.60 c. P 445.10 d. P 985.40 Gradient 23. An arithmetic gradient with series of end of period payments of P 10,000 on the 1st year, P 8,000 on the 2nd year, P 6,000 on the 3rd year, P 4,000 on the 4th year and P 2.000 on the 5th year. If interest rate is 12%, what is its present worth? a. P 12,125 b. P 23,007 c. P 22,565 d. P 23,254 24. A newly acquired machine requires an annual maintenance cost of 10,000 pesos. If the annual maintenance cost increases by 20% each year for 10 years, what is the estimated present worth of the maintenance costs if money is worth 15%? a. P 106,100 b. P 126,100 c. P 116,100 d. P 136,100 Manila Review Institute, Inc. Page 5 of 10 BONDS Methods of Bond Retirement 1. Using sinking fund A+I=F i +r (1 + i)n – 1 where: (A + I) = total periodic expense A = periodic deposit to the sinking fund I = interest on bonds r = bond rate i = sinking fund interest rate 2. Issuing callable bonds (serial bonds) r=i A+I= F i +i (1 + i)n – 1 A+I= F i = F i(1+i)n 1 – ( 1+ i) –n (1 + i)n – 1 Bond Value P=I 1 – (1+i) –n + C(1+i)-n i P = F.r 1 – (1+i) –n + C(1+i)-n i where: P = value of bond n-periods before redemption C = redemption price (disposal price = often equal to F) F = face or par value i = investment rate or yield per period Problems: Bond Value and Capitalized Cost 25. A Php 100,000, 6% bond pays dividends semi-annually and will be redeemed at 110% after 3 years. Find the bond value if money is worth 4% compounded semi-annually. a. Php 114,481.14 b. Php 412,481.14 c. Php 141,814.14 d. Php 142,481.14 26. A machine costs P 300,000 new, and must be replaced at the end of each 15 years. If the annual maintenance required is P 5,000, find the capitalized cost, if money is worth 5% and the final salvage value is P 50,000. a. P 631,711.44 b. P 316,171.44 c. P 613,171.44 d. P 613,711.44 DEPRECIATION Depreciation methods I. Uniform methods 1. Straight line V − Vs d= n a Da = ad = (V − Vs ) n Va = V − Da Manila Review Institute, Inc. Page 6 of 10 2. Sinking fund (V − Vs ) i d = (1 + i ) n − 1  (1 + i ) a − 1  Da = (V − Vs )    (1 + i ) − 1  n Va = V − Da II. Non Uniform methods 1. declining balance da = V (1 − f ) a −1. f Va = V (1 − f ) a Vs f = 1− n V 2. double declining balance 2 da = V (1 − n2 ) a −1. n Va = V (1 − ) a 2 n 3. Sum of the Years digit  n − a + 1 da =  n  (V − Vs )  2 ( n + 1)  Va = V − Da = V − ( d1 + d 2 +... + da ) 4. Output method Qa da = (V − Vs ) Va = V – Da = V – (d1 + d2 + d3 + ……+dn) QT 5. Working Hours method Ha da = (V − Vs ) Va = V – Da = V – (d1 + d2 + d3 + ……+dn) HT where: d = annual depreciation V = original cost Vs = salvage value N = economic life A = number of years in actual use Da = accumulated depreciation Va = asset (book) value f = fixed percentage Qa = number of units a machine can produce for period a QT = number of units a machine can produce during its economic life Ha = number of hours a machine is utilized for period a HT = number of hours a machine is utilized during its economic life Manila Review Institute, Inc. Page 7 of 10 Problems: Depreciation 27. A machine cost P 8,850 has a life of 8 years. Determine the depreciation at the end of the third year using double declining balance method. a. P 1,070 b. P 1,244 c. P 1,370 d. P 1,422 SITUATION PROBLEM 28-31 A machine cost P 4,850 has a life of 10 years and a salvage value of P 500. Using straight line method: 28. Determine the book value at the end of the fourth year. a. P 3,110 b. P 435 c. P 1,305 d. P 2,321 29. Determine the total depreciation at third year. a. P 3,110 b. P 435 c. P 1,305 d. P 2,321 30. Find the depreciation at the 5th year. a. P 3,110 b. P 435 c. P 1,305 d. P 2,321 31. What is the rate of depreciation in percent? a. 8% b. 9% c. 10% d. 11% SITUTATION PROBLEM 32-34 A machine costing 720,000 pesos is estimated to have a life of 10 years. If the annual rate of depreciation is 25%, using declining balance method, determine the: 32. Total depreciation at the end of its economic life. a. P 679,454 b. P 695,125 c. P 687,584 d. P 702,135 33. Depreciation at the end of the fifth year. a. P 56,953 b. P 21,452 c. P 52,213 d. P 40,546 34. Salvage value. a. P 56,953 b. P 21,452 c. P 52,213 d. P 40,546 SITUTATION PROBLEM 35-37 A machine is bought for Php 20,000.00 with a useful life of 10 years. Other initial charges amounted to Php 2000.00. The book value after the useful life is Php 5000.00. Using Sum-of-the-Year’s Digit method, determine: 35. Book value at the second year. a. Php 16127.27 b. Php 10509.09 c. Php 2163.64 d. Php 4212.21 36. Depreciation at the fourth year. a. Php 16127.27 b. Php 10509.09 c. Php 2163.64 d. Php 4212.21 37. Total depreciation at the end of the fourth year. a. Php 16127.27 b. Php 10509.09 c. Php 2163.64 d. Php 4212.21 SITUATION PROBLEM 38-40 A machine, with a useful life of 15 years, is bought for Php 20,000.00. The money is worth 8% and the scrap value is 25% of the original price of the machine. Using sinking fund method, determine: 38. The annual depreciation. a. Php 281.29 b. Php 2489.37 c. Php 552.44 d. Php 11,997.04 39. Accumulated depreciation for the 4th year. a. Php 281.29 b. Php 2489.37 c. Php 552.44 d. Php 11,997.04 Manila Review Institute, Inc. Page 8 of 10 40. Book value at the tenth year. a. Php 281.29 b. Php 2489.37 c. Php 552.44 d. Php 11,997.04 41. A telecommunication company purchased an equipment for P 53000 and paid P 1500 for freight and delivery charges to the job site. The equipment has a normal life of 10 years with a trade-in value of P5000 against the purchase of a new equipment at the end of the life. Determine the annual depreciation cost by the sinking fund method. Assume interest at 6.5% compounded annually. a. Php 6326.35 b. Php 1356.42 c. Php 3668.18 d. Php 6323.63 BREAK-EVEN Break Even Analysis - involves investment of capital wherein at a certain level of production the total income of the company would just be equal to total expenses, thus resulting in no loss nor profit Output at break-even = FC where: SP - VC FC = fixed cost U U SP/U = selling price per unit VC/U = variable cost per unit Problems: Break-even Analysis 42. Compute for the number of blocks that an ice plant must be able to sell per month to break even based on the following data: Cost of electricity per block……… P 20.00 Tax to be paid per block………… P 2.00 Real Estate Tax…………………. P 3500.00/mo Salaries and Wages……………… P 25,000/mo Others……………………………… P 12,000/mo Selling Price of ice…………………P 55.00/block a. 1228 b. 2921 c. 1412 d. 1129 OBJECTIVE TYPE QUESTIONS 1. The decrease in the value of property due to the passage of time. a. salvage value b. depreciation c. depletion d. devaluation 2. It occurs when there are few suppliers and any action taken by anyone of them will definitely affect the course of action of others a. rate of return b. oligopoly c. inventory d. profitability 3. The amount of money paid for the use of borrowed capital or the income produced by money which has been loaned a. asset b. profit c. bond d. interest 4. It exists when a certain product is offered for sale by many vendors or suppliers and there is no restriction against other vendors from entirely the market a. valuation c. present worth b. perfect competition d. profitability 5. The process of determining the value of certain properties or equipment for certain reasons. a. asset b. appreciation c. amortization d. valuation 6. Those funds that are required to make the enterprise or product a going concern a. banking c. working capital b. accumulated amount d. principal or present worth Manila Review Institute, Inc. Page 9 of 10 7. The length of time during which the property may be operated at a profit a. length b. life c. physical life d. economic life 8. Worth of the property shown in the accounting records of an enterprise a. fair value b. use value c. book value d. market value 9. Cost of things that are neither labor nor materials a. construction cost c. labor cost b. expenses d. investment 10. The series of equal payments at equal intervals of time a. interest b. depreciation c. annuity d. amortization 11. An index of short term paying ability is called a. current ratio c. acid test ratio b. receivable turnover d. profit margin ratio 12. Used to produce consumer goods a. producer goods b. supply c. consumer goods d. cash flow 13. It is a series of equal payments occurring at equal interval of time where the first payment is made several periods after the beginning of the payment a. deferred annuity c. progressive annuity b. delayed annuity d. simple annuity 14. The amount of a property in which a willing buyer will pay to a willing seller for the property when neither one is under the compulsion to buy or sell a. fair value b. goodwill value c. book value d. market value 15. A type of annuity where the payments are made at the start of each period, beginning from the first period a. ordinary annuity b. annuity due c. deferred annuity d. perpetuity 16. The intangible item of value from the exclusive right of the company to provide a specific product or service in a stated region of the country a. market value b. book value c. goodwill value d. franchise value 17. The length of time during which it is capable of performing the function for which it was designed and manufactured. a. economic life b. business life c. insular life d. physical life 18. Products that are directly used by people to satisfy their wants. a. supply b. consumer goods c. producer goods d. cash flows 19. A change in cost for a small change in volume of production. a. fixed cost b. sunk cost c. first cost d. differential cost 20. An amount which has been spent and for some reasons cannot be recovered. a. sunk cost b. first cost c. increment d. fixed cost 21. The ratio between the average demand and the maximum amount demand. a. load factor c. plant factor b. demand factor d. rate of return (ROR) 22. The ratio of the output power in watts and the product volts and amperes. a. load factor b. diversity factor c. power factor d. demand factor 23. The decrease in the value of property due to the gradual extraction of its content. a. depletion b. depreciation c. amortization d. book value Manila Review Institute, Inc. Page 10 of 10 24. The sum of the first cost and the present and the worth of all cost of any property of replacement, operation and maintenance for long time or forever. a. fixed cost b. capitalized cost c. sunk cost d. increment cost 25. The difference between the book value and the actual lower resale value is a. salvage value b. resale value c. sunk cost d. fixed cost 26. It occurs when a unique product or service is available only from a single supplier and entry of all other possible suppliers presented. a. competition b. monopoly c. inventory d. profitability 27. A place where buyer and seller come together a. market c. department store b. shop d. parlor 28. Ratio of annual net profit and the capital invested a. proportion b. rate of return c. load factor d. use factor 29. The price of which a willing buyer will pay to a willing seller for a commodity a. resale value b. market value c. book value d. scrap value 30. The price of property when sold for a junk a. scrap value b. book value c. resale value d. market value