Engineering Economy Solution Manual PDF

THE BASICS OF ENGINEERING ECONOMY Engineering is the profession in which a knowledge of the mathematical and natural science gained by study, experience and practice is applied with judgement to develop ways to utilize, economically the materials and forces of nature for the benefit of manki...

THE BASICS OF ENGINEERING ECONOMY Engineering is the profession in which a knowledge of the mathematical and natural science gained by study, experience and practice is applied with judgement to develop ways to utilize, economically the materials and forces of nature for the benefit of mankind. In this definition, the economic aspects of engineering are emphasized, as well as the physical aspects. Clearly, it is essential that the economic part of engineering be accomplished well. In manufacturing, engineering is involved in every detail of a product’s production, from the conceptual design to the shipping. In fact, engineering decisions account for the majority of product costs. Engineers must consider the effective use of capital assets such as building and machinery. One of the engineer’s primary tasks is to plan for the acquisition of equipment (capital expenditure) that will enable the firm to design and produce products economically. Engineering economy is the discipline concerned with the economic aspect of engineering. It involves the systematic evaluation with the economic merits of proposed solutions to the engineering problems. To be economically acceptable (i.e., affordable), solutions to engineering problem must demonstrate a positive balance of long term benefits over long term cost. Engineering economics is the application of economic techniques to the evaluation of design and engineering alternatives. The role of engineering economics is to assess the appropriateness of a given project, estimate its value, and justify it from an engineering standpoint. The General Economic Environment There are numerous general economic concepts that must be taken into account in engineering studies. Engineering Economy Page 1   Consumer goods and services are those products or services that are directly used by people to satisfy their wants. Examples are foods, clothing, homes, cars, haircuts and medical services.  Producer goods and services are used to produce consumer goods and services and other producer goods. Examples are machine tools, factory buildings, buses and farm machinery.  Price of goods and services is defined to be the present amount of money or its equivalent which is given in exchange for it.  Demand is a quantity of certain commodity that is bought at a certain price at a given place and time.  Supply is a quantity of a certain commodity that is offered for sale at a certain price at a given place and time.  Perfect competition occurs in a situation in which any given product is supplied by a large number of vendors and there is no restriction in additional suppliers entering the market.  Perfect monopoly exists when a unique product or service is available from a single supplier and that vendor can prevent the entry of all others into the market.  Oligopoly occurs when there are few suppliers and any action taken by anyone of them will definitely after the course of action of the others.  Total Revenue is the product of the selling price per unit and the number of units sold.  Total Cost is the sum of the fixed costs and the variable costs.  Profit/ Loss is the difference between total revenue and the total costs. Cost Terminology Cost considerations and comparisons are fundamental aspects of engineering practice. Before the study of various engineering economic decisions problems, the concept of various costs must be understand. At the level of plant operations, engineers must make decisions involving materials, plant facilities and the in-house capabilities of company level. Engineering Economy Page 2   Fixed costs are those unaffected by changes in activity level over a feasible range of operations for the capacity or capability available. Examples are insurance and taxes on facilities, general management and administrative salaries, license fees, and interest costs on borrowed capital.  Variable costs are those associated with an operation that vary in total with the quantity of output or other measures of activity level. Examples are the costs of material and labor used in a product or service.  Incremental cost is the additional cost (or revenue) that results from increasing the output of the system by one or more units.  Recurring costs are those that are repetitive and occur when an organization produces similar goods or services on a continuing basis.  Nonrecurring costs are those which are not repetitive even though the total expenditure may become cumulative over a relatively short period of time.  Direct costs are costs that can be reasonably measured and allocated to a specific output or work activity. Examples are labor and material costs.  Indirect costs are those that are difficult to attribute or allocate to a specific output or work activity. Examples are the costs of common tools, general supplies, and equipment maintenance.  Overhead cost consists of plant operating costs that are not direct labor or direct material costs. Examples are electricity, general repairs, property taxes and supervision.  Standard costs are representative costs per unit of output that are established in advance of actual production or service delivery.  Cash costs are that involves payment of cash.  Noncash costs (book costs) are costs that does not involve a cash payment, but rather represent the recovery of past expenditures over a fixed period of time. Example is the depreciation charged. Engineering Economy Page 3   Sunk cost is one that has occurred in the past and has no relevance to estimates of future costs and revenues related to an alternative course of action.  Opportunity cost is incurred because of the use of limited resources such that the opportunity to use those resources to monetary advantage in an alternative use is foregone.  Life-cycle cost refers to a summation of all the costs, both recurring and nonrecurring, related to product, structure system, or services during its life span.  Investment cost is the capital required for most of the activities in the acquisition phase.  Working capital refers to the funds required for current assets that are needed for the startup and support of operational activities.  Operational and Maintenance cost includes many of the recurring annual expense items associated with the operation phase of the life cycle.  Disposal cost includes those nonrecurring costs of shutting down the operation and the retirement and disposal of assets at the end of the life cycle. These costs will be offset in some instances by receipts from the sale of assets with remaining value.  Economic life coincides with the period of time extending from the date of acquisition to the date of abandonment, demotion in use, or replacement from the primary intended service.  Ownership life is the period between the date of acquisition and the date of disposal by a specific owner.  Physical life is the period between original acquisition and final disposal of an asset over the succession of owner.  Useful life is the time period that an asset is kept in productive service (either primary or backup). It is an estimate of how long an asset is expected to be used in a trade or business to produce income. Engineering Economy Page 4  Chapter 1 The Economic Environment Engineering economy is the analysis and the evaluation of the factors that will affect the economic success of engineering projects to the end that a recommendation can be made which will insure the best use of capital. CONSUMER AND PRODUCER GOODS AND SERVICES Consumer goods and services are those products or services that are directly used by people to satisfy their wants Producer goods and services are used to produce consumer goods and services or other producer goods. NECESSITIES AND LUXURIES Necessities are those products or services that are required to support human life and activities that will be purchased in somewhat the same quantity even though the price varies considerably. Luxuries are those products or services that are desired by humans will be purchased if money is available after the required to support human life and activities that will be purchased in somewhat the same quantity even though the price varies considerably. DEMAND Demand is the quantity of a certain commodity that is bought at a certain price at a given place and time. Elastic demand occurs when a decrease in selling price result in a greater than proportionate increase in sales. Inelastic demand occurs when a decrease in selling price produces a less than proportionate increase in sales. Unitary elasticity of demand occurs when the mathematical product of volume and price is constant. Engineering Economy Page 5 PRICE PRICE FIGURE 1-1.General PricedemandFIGURE 1-2. Price-demand relationship forrelationship luxuries and necessities Competition, Monopoly and Oligopoly Perfect competition occurs in a situation where a commodity or service is supplied by a number of vendors and there is nothing to prevent additional vendors entering the market. Monopoly is the opposite of perfect competition. A perfect monopoly exists when a unique product or service available from a single vendor and that vendor can prevent the entry of all others into the market. Oligopoly exists when there are so few suppliers of a product or service that action by one will almost inevitably result in similar action by the others. The Law of Supply and Demand Supply is the quantity of a certain commodity that is offered for sale at ascertain price at a given place and time The law of supply and demand may be stated as follows: “Under conditions of perfect competition the price at which a given product will be supplied and purchased is the price that will result in the supply and the demand being equal” Engineering Economy Page 6  Supply Demands Price P Price Figure 1-3. General price- Figure 1-4.Price-supply supply relationship demand relationship The Law of Diminishing Returns “When the use of the one of the factors of the production is limited, either in increasing cost or by absolute quantity, a point will be reached beyond which an increase cost or by absolute quantity, a point will be reached beyond which an increase in the variable factors will result in a less than proportionate increase in output.” 4 3 2 1 0 1 2 3 4 5 INPUT IN KILOWATTS Figure 1-5. Performance curve of an electric motor. The effect of the law of diminishing returns on the performance of an electric motor is illustrated in Fig. 1-5. For the early increase in input through input of 4.0kw, the actual increase in output is greater than proportional; beyond this point the output is greater than proportional. In case the fixed input factor is the electric motor. Engineering Economy Page 7  Chapter 2 Interest and Money Time Relationships Capital The term capital refers to wealth in the form of money or property that can be used to produce more wealth. The majority of engineering economy studies involve commitment of capital for extended periods of time, so effect of time must be considered. It is recognized that a unit of principal today is worth more than a unit of principal one or more years from now because of interest (or profits) it can earn. Therefore, money has a time value. The following reasons why Php100 today is “worth” more than Php100 one year from today: 1. Inflation 2. Risk 3. Cost of money (interest) Of these, the cost of money is the most predictable, and, hence, it is the essential component of economic analysis. Cost of money is represented by money paid for the use of borrowed money, or return of investment. Cost of money is determined by an interest rate. It is established and measured by an interest rate, a percentage that is periodically applied and added to an amount of money over a specified length of time. When money is borrowed, the interest paid is the charge to the borrower for the use of the lender’s property; when the money is loaned or invested, the interest earned is the lender’s gain from providing a good to another. Interest maybe defined as the cost of having money available for use. Engineering Economy Page 8 Types of Capital 1. Equity Capital is that owned by individuals who have invested their money or property in a business project or venture in the hope of receiving a profit. 2. Borrowed Capital is obtained from lenders for investment, with a promise to repay the principal and interest on a specific date, whether or not the operations of the business have been profitable or not. In return, the lenders receive interest from the borrowers. Cash Flow Diagram Cash flow is the stream of monetary values – costs and benefits – resulting from a project investment. It is difficult to solve a problem is you cannot see it. The easiest way to approach problems in economic analysis is to draw a picture. Cash flow diagram is a graphical representation of cash flows drawn in a time scale. It has three elements: 1. A horizontal line represents time with progression of time moving from left to right. The period labels can be applied to intervals of time rather than to point on the time scale. A time interval is divided into an appropriate number of equal periods. 2. Arrows represent cash flow and are place at the specified period. If distinctions are needed to be made, downward arrows represents cash outflows (expenditures, disbursements) and upward arrows represents cash inflows (income). Engineering Economy Page 9  3. Depends on the person’s viewpoint.Unless otherwise indicated, all such cash flows are considered to occur at the end of their respective periods. The following symbols nomenclatures will be used: P = Present sum of money F = Future sum of money N = Number of interest periods i= Interest rate per period Interests the amount of the money paid for the use of borrowed capital or the income produced by money which has been loaned. Interest from the viewpoint of the lender is the income produced by money which has been borrowed or invested. For the borrower, it is the amount of money paid for the use of borrowed capital. Simple Interest When the total interest earned is linearly proportional to the amount of the loan (principal), the number of the interest rate per interest periods for which the principal is committed, and the interest rate per interest period, the interest is said to be simple. Is the calculated using the principal only, ignoring any interest that had been accrued in preceding periods. In practice, simple interest is paid on short term loans in which the time of the loan is measured in days I =Pni F = P + I = P + Pni F = P (1+ni) Where: I = interest P = principal or present worth n = number of interest periods i = rate of interest per interest period F = accumulated amount or future Types of Simple Interest 1. Ordinary Simple Interest (OSI) Engineering Economy Page 10  Based on one banker’s year which is equivalent to 300 days or 30 days in one month. is computed on the basis of 12 months of 30 days a year. 1 interest period = 360 days OSI = Pni Example: Determine the ordinary simple interest on P8,000 for 8 months at 7% per year. Given: P = 8,000 i = 7% / yr n = 8 monthly Required: OSI Solution: OSI = Pni OSI = 8000 (0.07) OSI = P373.33 Problem 1 Determine the ordinary simple interest on P700 for 8 months and 25 days if the rate of interest is 15%. Solution Number of days= (8) (30) + 15= 255 days I = Pni = P700 x x 0.15 = P74.38 2. Exact Simple Interest (ESI) Based in the exact number of days which is 365 days in ordinary year or 366 days for leap year. is based on the exact number of days in year, 365 days for an ordinary year and 366 days for a leap year. 1 interest period= 365 or 366days ESI = Pni ESI = P (i) for ordinary year Engineering Economy Page 11  ESI = P (i) for leap year Example: Determine the exact simple interest on P500 for the period from January 10 to October 28, 1996 at 16% interest. Solution Jan 10-31 = 21 (excluding Jan 10) February = 29 March =31 April =30 May =31 June =30 July =31 August =31 September =30 October =28 (including Oct.28) 292 days Exact simple interest =P500 (0.16) = P63.83 Example Determine the ordinary exact interest on P15,000 loan at 8% per year made from January 12 to September 24, 2007. Given: P = P15, 000 i = 8% per year n = January 12 to September 24, 2007 Required: a. OSI, ordinary simple interest b. ESI, exact simple interest Solution: Solving for the number of days, from Jan 2 to September 24, 2007, there are 256 days. Engineering Economy Page 12 (*Note: for a specific dates, to count the number of days , exclude the first day, include the last day.) a. Solving for the ordinary simple interest: OSI = P( )i =P15 000 (0.08/year) ( )year =P853.33 b. Solving for the exact simple interest: ESI = P( )i =P15 000(0.08/year)( )year =P853.34 Example What will be the future worth of money after 14months, if a sum of P10,000 is invested today at a simple interest rate of 12% per year? Solution: F = P(1+ni) = P10,000 [ 1 + (0.12)] F = P 11,400 Problem: Determine the exact simple interest on P50 000 periods for the period June 25, 2005 to September 2, 2006, if the rate of interest is 12% per year? Solution: I = Pni = 50,000 (.12) I = P1131.15 Engineering Economy Page 13 Problem: A loan of P2000 is made for a period of 13 months, from January 1 to January 31 the following year, at simple interest rate of 20%. What future amount is due as the end of the loan period? Solution: F = P (1 + ni) = P2000( 1+( )(.20)) F = P2433.33 Problem: If you borrow money from your friend with simple interest of 12%, find the present worth of P20, 000, which is due at the end of nine months. Solution F = P (1 + ni) 20,000 = P (1 + (.12) 20,000 = P (1.09) P = 18,348.62 Cash-Flow Diagrams A cash-flow diagram is simply a graphical representation of cash flows drawn on a time scale. Cash-flow diagram for economic analysis problems is analogous to that of free body diagram for mechanics problems. Receipt (positive cash flow or cash inflow) Disbursement (negative cash flow or cash outflow) Engineering Economy Page 14  A loan of 100 at simple interest of 10% will become P150 after 5 years. Cash flow diagram on the viewpoint of the lender P100 0 1 2 3 4 5 P150 Cash flow diagram on the viewpoint of the borrower Compound Interest Whenever the interest charge for any interest period is based on the remaining principal amount plus any accumulated interest charges up to the beginning of that period the interest is said to be compounded. In calculations of compound interest, the interest for an interest period isn calculated on the principal plus total amount of interest accumulated in previous periods. Thus compound interest means “interest on top of interest.” Engineering Economy Page 15  P 0 1 2 3 n-1 n F Compound Interest (Borrower’s Viewpoint) Discrete Compounding The formulas are for discrete compounding, which means that the interest is compounded at the end of each finite length period ,such as a month or a year. Discrete Cash Flows The Formulas also assume discrete (i,e., lump sum) cash flows spaced at theend of equal time intervals on a cash flow diagram. Derivation of Formula Interest Principal at Interest Earned Amount at End Period Beginning of During Period of Period Period 1 P Pi P+Pi = P(1+ni) 2 P(1+i) P(1+i) i P(1+i) + P(1+i)i = P(1+I) ² 3 P(1+i)² i P(1+i) ² i P(1+i) ² + P(1+i) ²i =P(1)............ n P(1+i) n-1 P(1+i) n-1i P(1+i) ⁿ Engineering Economy Page 16  F = P (1+i)n The quantity (1+i)nis commonly called the “single payment compound amount factor” and is designed by the functional symbol F/P, i%, n. Thus, F = P (F/P, i%, n) The symbol F/P, i% n is read as “F given P at per cent in n interest period.” From Equation (2-3), P = F(1+i)-n The quantity (1-i)-nis called the “single payment present worth factor” and is designated by the functional symbol P/F, i% ,n. Thus, P = F (P/F i% n) The symbol P/F, i%, n is read as “P given F at i percent in n interest periods.” The Five Types of Cash Flows Whenever patterns are identified in cash flow transactions, these patterns can be used to develop concise expressions for computing either the present or future worth of the series. Cash flows can be classified into five categories: 1. Single Cash Flows: The simplest case involves the equivalence of a single cash present amount and its future worth. Thus, the single-cash flow formula deal with only two amounts: the single present amount P, and its future worth F. 2. Equal Uniform Series: Probably the most familiar category includes transactions arranged as a series of equal cash flows at regular intervals, known as an equal-payment series (or uniform series) 3. Linear Gradient Series consists of cash flows that increase or decrease by uniform amount each periods. 4. Geometric Gradient Series consist of cash flows that increase or decrease by a fixed percentage. 5. Irregular Series consists of cash flows that change with no pattern. Engineering Economy Page 17 Problem: How long will it take inventory to triple itself if invested of 12% per unit? Solution: F = P (1 + i)n P = 3P (1.12)n = (1.12)n 3 = (1.12)n ln3 = ln(1.12) (. ) n = n = 9.69 years Problem: You deposit $3,000 in a savings account that 9% simple interest per year. How many years will it take to double your balance? If instead you deposit the $3,000 in another savings account that earns 8% interest compounded yearly, how many years will it take to double your balance? Solution: F = Pni, when i= 9% 6000 = 3000 n (.09) n = 22.22 years F = Pni , when i=8% 6000 = 3000 n (.08) n = 9 years Interest Interest from the viewpoint of the lender is the income produced by money which has been borrowed or invested. For the borrower, it is the amount of money paid for the use of borrowed capital. Is the amount of the money paid for the use of borrowed capital or the income produced by money which has been loaned Rate of Interest is defined as the amount earned by one unit of principal during a unit of time Engineering Economy Page 18 (a) Nominal rate of interest The nominal rate of interest specifies the rate of interest and a number of interest periods in one year. Nominal rate of interest specifies the rate of interest and the number of compounding period in one year. It has because customary to wrote interest rates on an annual basics followed by the compounded period if different from one year in length. Even if a financial institution. i= where: i = rate of interest per interest period r = nominal interest rate m = number of compounding periods per year n = Tm where: n = no. of interest per periods T = no. of years m = number of compounding periods per year If the nominal rate o interest is 10%compounded quarterly, then i= 10%/4= 2.5%, the rate of interest per interest period. The following list the different compounding periods and the occurrence in one year annually once a year semi- annually twice a year Quarterly every 3 months Bi – monthly every 2 months Monthly every month Semi- monthly twice a month Weekly Daily Continuously Engineering Economy Page 19 Example: The sum of P10, 000 was deposited to a fund earning interest 10% per annum compounded quarterly, what was the principal in the fund at the end of a year? Given: P = 10,000 r= 10% per year compounded quarterly n= 3 years (4) = 12 quarters Required: The amount in the fund at the end of 3 years Solution: Solving for the rate of interest per period. I= = = 2.5 % per quarter (b) Effective rate of interest Effective rate of interest quotes the exact rate of interest for one year it should be noted that the effective interest rates are always expressed on an annual basis. Effective rate of interest is the actual or exact rate of interest on the principal during one year. If P1.00 is invested at a nominal rate of 15% compounded quarterly, after one year this will become.. 4 P1.00 1 + = P1.1586 The actual interest actual interest earned is 0.1586, therefore, the rate of interest after one year is 15.86%. Hence, Effective rate = F1 – 1= (1+i) m – 1 where: F1 = the amount P1.00 will be after one year Engineering Economy Page 20  F1 0 1 2 m 1 year P 1.00 Effective Rate of Interest Example Consider, one unit of principal for one unit a time invested in a nominal interest of 12% compounded monthly, P= 1.00 N= 12 months I = 12/12 = 1%/ month The accumulated amount of P for one year will be F = P(1+i)ⁿ =1(1.01)ⁿ = P1.1268 The implication is that, for one unit of principal. The interest earned will be I = 0.1268 and in terms of effective or actual annual interest (ERI) the interest can be rewritten as a percentage of the principal amount: i= I/ P = 0.1268 / 1 = 12.68% Therefore, to determine the effective rate of interest for a given nominal rate of interest, ER = (1+i)ᵐ -1 In other words, paying 1% interest per month for 12 months is equivalent to paying 12.68% interest just one time each year. Engineering Economy Page 21 Example A P2000 loan was originally made 8% simple interest for 4 years. At the end of this period th loan was extended for 3 years, without the interest being paid, but the new interest rate was made 10% compounded semi annually. How much should the borrower pay at the end of the 7 years? Solution F7 F4 0 1 2 3 4 5 6 7 Simple interest compound interest P2000 F4 = P(1 + ni ) = P2000 [1 + (4) (0.08) ]= P 2,640 F7 = F4 (1 + i )n = P2,640 (1 +0.05)6 = P3,537.86 Example Find the normal rate which if converted quarterly could be used instead of 12% compounded monthly. What is the corresponding effective rate? Solution: Let r= the unknown nominal rate For two or more nominal rates to be equivalent , their corresponding effective rates must be equal. Nominal rate Effective rate 4 r% compounded quarterly 1+ -1. 12 12% compounded monthly 1+ -1 1+ − 1 = 1(0.01)12-1 1 + = (1.01)3 =1.0303 r=0.1212 or 12.12% compounded quarterly Engineering Economy Page 22 Example Find the amount at the end of two years and seven months if P100 is invested at 8% compounded quarterly using simple interest for anytime less than a year interest period. Solution: % For compound interest: i = = 2%, n =(20) (40) =8 For simple interest: i = 8%, n = F2 F1 0 1 2 2 yrs, 7months. compound simple interest P1.00 F1= P (1 + i)n = 1000(1+0.02)8= p1171.66 F2 = F1(1+ ni) = P1171.66 [ 1= 7/12(0.08)] = P 1,226.34 Example: A loan of P10,000 for 3 years at 10% simple interest per year. Determine the interest earned and the accumulated amount. Given: P = P10,000 n = 3 years i = 10% per year Required: a. Total Interest, I c. Accumulated amount, F Engineering Economy Page 23 Solution: a. The total interest earned: I = Pni I = P10,000 (3 years) (0.1/year) I = P3,000 The accumulated amount at the end of the interest period: F= P + 1 F = P10,000 + 3,000 F = P13,000 Amortization Schedule - A table showing the mode of payment. Period Principal Interest Payment 1 10, 000 1, 000 0 2 10, 000 1, 000 0 3 10, 000 1, 000 13, 000 Total 3, 000 Problem: Annie buys a television set from a merchant who asksvP12,500. At the end of 60 days (cash in 60 days). Annie wishes to pay immediately and the merchant offers to compute the cash price in the assumption that money is worth 8% simple interest. What is the cash price today? Solution: F = P (1 + ni) 12,500 = P (1 + (.08) 12,500 = P (1.013150685) P = 12,337.75 Problem: What will be the future worth of money after 14 months of a sum of P10, 000 is invested today at a simple interest rate of 12% per year? Engineering Economy Page 24 Solution: F = P (1 + ni) = 10,000 (1+ (.12)) F = P11,400 Problem: What is the amount of interest earned on $2,000 for five years at 10% simple interest per year? Solution: I = Pni = 2,000 (5)(.10) I = P100 Problem: Determine the effective rates of interest for the following nominal rates of interest a. 9% compounded annually b. 9% compounded semi annually c. 9% compounded quarterly d. 9% compounded monthly e. 9% compounded daily a). ER = (1 + r)m - 1 d). ER = (1 + r)m - 1. = ( 1 +.09)1 – 1 = ( 1+ )12 – 1 ER = 0.09 or 9% ER = 0.0938 or 9.38% b). ER = ( 1+ r )m – 1 e). ER = (1 + r)m - 1.. = ( 1+ )2 – 1 = ( 1+ )360 – 1 ER = 0.092025 or 9.20% ER = 0.0942 or 9.42% c).ER = ( 1+ r )m – 1. =( 1+ )4 – 1 ER = 0.0931 0r 9.31% Engineering Economy Page 25 Problem If you are investing which is better 15% compounded semiannually or 14.5 compounded monthly? Solution: I= = 15%/ 2= 7.5 % / semi- annual I= = 14.5 %/ 12 = 1.21/ month Problems: What nominal rate of interest compounded quarterly can be used instead of 8% compounded monthly. (r =8% compounded monthly but the required interest rate, i, is per quarter.) Solution: ERmonthly= ERquarterly (1 + r)m – 1 = (1 + r)m – 1 ( 1 + 0.08 )12 – 1 = ( 1 + r )4 – 1 √2.52 = (1 + )4 1.26 = 1 + r r = 26% Problem: Supposed that you have just borrowed P75,000 at 12% nominal interest compounded quarterly. What is the total lump-sum, compounded amount to be paid by you at the end of a 10 – year loan period? Solution: F = P (1 + I )n. = P75,000 ( 1 + )40 F = P244,652.83 Engineering Economy Page 26 Problem: A sum of P1,000 is invested now and left for 8 years at which time the principal is withdrawn. The interest has accrued is left for another 8 years. If the effective annual interest rate is 5%, what is the amount of the account at the end of the 16th year? Solution: F = P (1+i) n F = 1477.46 – 1000 F = 477.46 (1.05)8 F = 1000 (1.05) 8 F = 477.46 F = 705.43 F = 1477.455 P = 477.46 F = 1477.455 Problem: A businessman wants to have P500,000 4 years from now. What amount should he invest now if it will earn interest of 6% compounded quarterly for the first two years and 8% compounded semi – annually during the next two years? Solution: 500, 000 P 0 2 4. P = 500,000 1 + P = 427,402.096. P = 427,402.96 1 + P = 379,409.5945 Engineering Economy Page 27 Problem: How long will it take you to become a millionaire if you invest P10, 000 at 20% compounded quarterly? Solution: % 1,000,000 = 10,000 1 + , , % , = 1+ , , % ln = [(1 + ] , , , ( ) n= , % ( ) n = 94.39 n = tm. =T T = 24 years Problem: How many years will it take an investment to triple if the interest rate is 8% compounded (a). Quarterly? (b). Monthly? (c). Continuously? Solution: a.)P = 10 F = P(1 + i). F= 3P 30 = 10 1 +. F= 3(10) ln = n ln 1 + F= 30 n= 55.48165.34 n= tm t= T= 13.78 yrs. or 14 yrs. t= 13.87 or 14 days Engineering Economy Page 28  b.)F= P(1 + i) c.)F=.. ( ) 30 =10 1 + 30 = 10e. ln = 1+ ln = 0.08 n lne. ln = n ln 1 + n= 13.73 or 14 yrs.. 1.098612289 = nln 1 + n=165.34 Equation of Value If cash flows occur on different periods, comparison of such should be made on a same focal date. Equation of values is obtained by setting the sum of one set of obligation on a certain comparison or focal date equal to the sum of another set of obligation on the same date. An equation of value is obtained by setting the sum of the values on a certain comparison or focal date of one set of oblifations equal to the sum of the values on the same date of another set of obligations. Application of Compound Interest Deposits ------- withdrawals Loans ---------- repayments Investment ---- income Cash inflow --- cash outflow Example A man bought a lot worth P800,000 if paid in cash. On the installment basis, he paid a down payment of P20,000; P300,000 at the end of one year, P400,000 at the end of three years and a final payment at the end of five years. What was the final payment if interest was 20%. Engineering Economy Page 29 Solution P 800,000 0 1 2 3 4 P300,000 300,000(P/F,20%,1) P400,000 P400,000(P/F,20%,3) Q Q(P/F,20%,5) Using today as the focal date, the equation of value is P800, 000 = P300. 000 (P/F,20%,1) + P400,000 (P/F,20%,3) + Q (P/F, 20%,5) P800, 000 = P300,000 (1.20)-1 + P400,000 (1.20)-3 + Q (1.20)-5 P800, 000 = P300,000 (0.8333)+ P4000, 000 (0,5787) + Q (0.4019) Q= P792,560 Problems: A student plans to deposit P1, 500 in a bank now and another P3,000 for the next two years. If he plans to withdraw P5, 000 three years after his deposit for the purpose of buying shoes. What will be the amount of money left in the bank one year after his withdrawal? Effective rate is 10%. Solution: 1500 (1.10)6 + 3000 (1.10)4 + X = 5000(1.10)1 X = P1,549.64 Engineering Economy Page 30 Problem: A man purchases a lot worth 1M if paid in cash. On installment basis, he paid P200,000 down payment, P300,000 at the end of the year, P400,000 at the end of the fourth year and the last payment at the end of the 5th year. If money is worth 16%, what is the worth of the last payment? Solution: 1M 200, 000 300, 000 400, 000 X Continuous Compounding and Discrete Payments In discrete compounding the interest is compound at the end of each finite- length period, such as a moth, a quarter or a year. In continuous compounding, it is assumed that cash payments boccur once per year, but the compounding is continuous throughout the year. F 0 1 2 3 mn n years P Engineering Economy Page 31  Continous Compounding (Lender’s Vieswpoint) r = nominal rate of interest per year = rate of interest per period m = number of interest periods per year n = number of interest periods in n years mn F=P 1+ Let = k, then m = rk, as m increases so must k mn rkn rn 1+ = 1+ = 1+ The limit of 1 + as k approaches infinite is e rn 1+ = ern Thus, F= Pern P = Fern Example Compare the accumulated amounts after 5 years of P1,000 invested at the rate of 10% per year compounded (a) annually, (b) semiannually, (c) quarterly, (d) monthly, I daily, and (f) continuously. Solution Using the formula, F=P(1+i)n (a) F = P 1,000 (1+0.10)5 = P1,610.51. 10 (b) F = P 1,000 1 + = P1,628.89. 20 I F = P 1,000 1 + = P1,638.62. 60 (d) F = P 1,000 1 + = P1,645.31. 1825 I F = P 1,000 1 + = P1,628.89 (f) F = Pern = P1,000I(0.10x5) = P1,648.7 Engineering Economy Page 32 Discount Discount is defined as interest deducted in advance. In negotiable paper, it is the difference between the present worth of the paper and its value sometime in the future. Discount on a negotiable paper is the difference between the present worth (the amount received for the paper in cash) and the worth of the paper at some time in the future (the face value of the paper or principal). Discount is interest paid in advance. Discount = Future Worth – Present Worth The rate of discount is the discount on one unit of principal for one unit of time. Rate of Discount Rate of discount is defined as the discount of one unit of principal for one unit of time. 1.00 (1+i)-1 In the cash flow diagram, P = (1+i)-1 And F = 1.00 Using the formula of discount, d=F–P d = 1 - (1+i)-1 d=1– Simplifying the equation will give us; i= and, Engineering Economy Page 33  d= where d = rate of discount for the given period i = rate of interest for the same period Example A man borrowed P5,000 from a bank and agreed to pay the loan at the end of 9 months. The bank discounted the loan and gave him P4,000 in cash. (a) What was the rate of discount? (b) What was the rate of interest? (c) What was the rate of interest for one year? Solution P4,000 P 0.80 P 5,000 P 1.00 , (a) d = = = 0.20 or 20% , Another solution, using equation (2-12) d = 1 – 0.80 = 0.20 or 20%. (b) i = =. = 0.25 or 25% Another solution, , i= =( , ) = 0.25 or 25% , (c) i = = = 0.3333 or 33.33% ( , ) Engineering Economy Page 34 Problem: Mr. Cadevida was grant a loan of P20,000.00 by his employer CPM Industrial Fabricator & Construction Corporation with an interest of 6% for 180 days on the principal collected in advance. The corporation would accept a promissory note for P20, 000 non interests for 180 days. If discounted at once, find the proceeds on the note. Solution: Given: Loan = 20,000 I = 6% for 180 days F = P(1+i)n Discount = 20,591.26 – 20,000 = 20,000 (20,000 +.06).5 = 591.26 = 20,591.26 Proceeds will be = 20,000 – 591.26 = 19,409.74 Problem: A man applied for a loan with 16% interest. The interest will be automatically deducted from the loan at the time the money is released, and after one year will have to pay the same amount as stated in the loan. Determine the interest actually charge to him. Solution: d = 1- (1 + I )-1 I=. = 1- (1 +.16/1 )-1 =. d = 0.1329 or 13.29 I = 15.99% Problem: How much interest is deducted in advance from a loan of P18,000.00 for one year and 6 months if the discount rate is 9%. How much is the proceeds after deducting the interest? Engineering Economy Page 35 Solution: i= d=.09 = , D = 1,620 Problem: Mr. de la Cruz borrowed money from a bank. He received from the bank P1, 342 and promised to repay P1, 500 at the end of 9 months. Determine the corresponding discount rate. Solution: Given: F = 1500 Discount = 1500- 1342 I= P = 1342 d = 158 I = 11.77% d=? d= i= ? d = 10.53% Problem: John is deciding between a loan charged at 5% bank discount, and one charged at 6% simple interest. Either would require John to repay P1000 on August 3, 2010. It is now August 3, 2010, which loan should John get? Solution: 6% 5% F = P (1 + ni) P = 1000 – (1000 x.05) 1000 = P (1 + (.06) P = 950 1000 = P (1.02) P = 980.39 Choose the 6% SI Engineering Economy Page 36 Problem: How much should the engineer borrow from a bank that charges 10% simple discount for 6 months if he received an amount of P14, 500? Solution: F = P (1 + ni) 14500 = P (1 + (.10)) 14500 = P (1.05) P = 13,809.52 Inflation Inflation is the increase in the prices for goods and services from one year to another, thus decreasing the purchasing power of money. FC = PC ( 1+ f)n where:PC = present cost of a commodity FC = future cost of the same commodity f = annual inflation n = number of years Example An item presently costs P1000. If Inflation is at the rate of 8% per year. What will be the cost of the item in two years? Solution FC = PC ( 1+ f)n = P1000 (1 + 0.08)2 = P116.40 In an inflationary economy, the buying power of money decrease as cost increase. Thus, F=( ) where F is the future worth, measured in today’s pesos, of a present amount P. Engineering Economy Page 37 Example An economy is experiencing inflation at the annual rate of 8%. If this continuous, what will P1000 be worth two years from now, in terms of today’s peso? Solution F= = = P857.34 ( ) (. ) If interest is being compounded at the same time that inflation is occurring. The future worth will be ( ) ( ) F= ( ) =P ( ) Example A man invested P10,000 at an interest rate of 10% compounded annually. What will be the final amount of his investment, in terms of today’s pesos, after five years, if inflation remains the same at the rate of 8% per year? Solution ( ) F = P( ) ( ) (. ) = P10,000 (. ) F = P10,960.86 ANNUITY An annuity is a series of equal payments made at equal intervals of time. Financial activities like installment payments, monthly rentals, life-insurance premium, monthly retirement benefits, are familiar examples of annuity. Annuity can be certain or uncertain. In annuity certain, the specific amount of payments are set to begin and end at a specific length of time. A good example of annuity certain is the monthly payments of a car loan where the amount and number of payments are known. In annuity uncertain, the annuitant may be paid according to certain event. Example of annuity uncertain is life and Engineering Economy Page 38 accident insurance. In this example, the start of payment is not known and the amount of payment is dependent to which event. Annuity certain can be classified into two, simple annuity and general annuity. In simple annuity, the payment period is the same as the interest period, which means that if the payment is made monthly the conversion of money also occurs monthly. In general annuity, the payment period is not the same as the interest period. There are many situations where the payment for example is made quarterly but the money compounds in another period, say monthly. To deal with general annuity, we can convert it to simple annuity by making the payment period the same as the compounding periods by the concept of effective rates. Types of Annuities In engineering economy, annuities are classified into four categories. These are (1) ordinary annuity, (2) annuity due, (3) deferred annuity, and (4) perpetuity. An annuity is a series of equal payments occurring at equal periods of time. Symbol and Their Meaning P = Value or sum of money at present F = Value or sum of money at some future time A = A series I periodic, equal amounts of money n = Number of interest periods I = Interest rate per interest period Ordinary Annuity An ordinary annuity is a series of uniform cash flows where the first amount of the series occurs at the end of the first period and every succeeding cash flow occurs at te end of each period. An ordinary annuity is one where the payments are made at the end of each period. Engineering Economy Page 39 Characteristics of ordinary annuity: a.) P (present equivalent value) - Occurs one interest period before the first A (uniform amount) b.) F (future equivalent value) - Occurs at the same time as the last A and n intervals after P c.) A (annual equivalent vale) - Occurs at the end of each period Finding P when A is given P 0 1 2 3 n-1 n A AAAA A(P/F,i%,1) A(P/F,i%,2) A(P/F,i%,3) A(P/F,i%,n-1) A(P/F,i%,n) Cash flow diagram to find P given A P = A (P/F, i%, 1) + A (P/F, i%, 2) + (P/F, i%, 3) +……. + A (P/F, i%, n-1) + A (P/F, 1%, n) P = A (1+i)-1 + A (1+i)-2 + …… + A (1+i)-(n-1) + A (1+i)-n Multiplying this equation by (1+i) result in P + Pi = A + A (1+i)-1 + A (1+i)-2 + …… + A (1+i)-n+2 + A (1+i)-n+1 Subtracting the first equation from the second gives P + Pi = A + A (1+i)-1 + A (1+i)-2 +…………… + A (1+i)-n+1 -P = - A (1+i)-1 - A (1+i)-2 +…………… + A (1+i)-n+1 – A (1+i)-n Pi =A – A (1+i)-n Engineering Economy Page 40 Solving For P gives ( ) ( ) P=A =A ( ) The quantity in brackets is called the “ uniform series present worth factor” and is designated by the functional symbol P/A, i%, n, read as “P given A at I percent in the interest periods. “ Hence Equation can be expressed as P = A (P/A, i%, n) Finding F when A is Given F 0 1 2 3 n-1 n A AAAA A(F/P,i%,1) A(F/P,i%,n-3) A(F/P,i%,n-2) A(F/P,i%,n-1) Cash flow diagram to find F given A F = A + A (F/A, i%, 1) + ……. + A (F/P, i%, n-3) + A (F/P, 1%, n-2) + A(F/P, i%, n-1) F = A + A (1+i) + ……… + A (1+i)-(n-3) + A (1+i)-n-2 + A (1+i)-n-1 Multiplying this equation by (1+i) results in F + Fi = A + A (1+i) + A (1+i)2 + …… + A (1+i)n-2 + A (1+i)n-1 + A (1+i)n Subtracting the First equation from the second gives F + Fi = A (1+i) + ………. + A (1+i) n-2 + A (1+i)n-1 + A (1+i)n - F = -A- A (1+i) + …… + A (1+i) n-2 + A (1+i)n-1 Fi = -A + A (1+i) n Engineering Economy Page 41 Solving for F gives ( ) F=A The quantity in brackets is called “Uniform series compound amount factor” and is designated by the functional symbol F/A, i%, n, read as “F given at I percent in n interest period. “ Equation can now be written as F = A (F/A, i%, n Finding A when P is Given Taking Equation and solving for A, we have A= ( ) The quantity in brackets is called the “capital recovery factor.” It will be denoted by the functional symbol A/P, i%, n which is read as “ A given at I percent in an interest periods.” Hence A = P (A/P, i%, n) Finding A When F is Given A=F ( ) The quantity in brackets is called the “sinking fund factor”. It will be denoted by the functional symbol A/F, i%, n which is read as “ A given F at I per n interest periods.” Hence A = F (A/F, i%, n) Relation A/P, i%, n and A/F, i%, n ( ) ( ) +i= X ( ) ( ) ( ) ( ) +i= ( ) A/F, i%, n+1 = A/P, i%, n Thus, Sinking fund factor + i = capital recovery factor Example What are the present worth and the accumulated amount of a 10-year annuity paying P10,000 at the end of each year, with interest at 15% compounded annually? Engineering Economy Page 42 Solution A = P10,000 n = 10 i = 15% P F 0 1 2 3 9 10 P10,000P10,000P10,000P10,000P10,000 P10,000 (P/A,15%,10) P10,000 (F/A,15%,10) P = A(P/A, i%, n) = P10,000 (P/A, 15%, 10) (. ) = P10,000 [. ] = P50,188 F = A(F/A, i%, n) = P10,000 (F/P, 15%, 10) (. ) = P10,000 [. ] = P203,037 Example What is the present worth of P500 deposited at the end of every three months for 6years if the interest rate is 12% compounded semiannually? Solution Solving for the interest rate per quarter,. (1+i) 4 – 1 = 1+ –1.5 1 + i = (1.06) i = 0.0296 or 2.96% per quarter P = A (P/A, 2.96%, 24) (. ) = P500. = P500 (17.0087) = P 8,504 Engineering Economy Page 43 Example A businessman need P50,000 for his operation. One financial institution is willing to lend him the money for one year at 12.5% interest per annum (discounted). Another lender is charging 14% with the principal and interest payable at the end of one year. A third financier is willing to lend him P50,000 payable in 12 equal monthly installments of P4,600. Which offer best for him? Solution Compare the effective rate each offer and select the one with the lowest effective rate. First offer: Rate of discount, d = 12.5%. Rate of interest, i= =. = 14.29% per year Effective rate = 14.29% Another solution Amount received = P50,000 (0.875) = P43,750 P43,750 0 1 P50,000 , , Rate of interest = = 14.29% per year , Effective rate = 14.29% Engineering Economy Page 44 Second Offer: P50,000 P50,000(1.14) Effective Rate = 14% Third Offer: P50,000 0 1 2 11 12 P4,600P4,600P4,600P4,600 , P/A, i%, 12 = = , ( ) =10.6896 Try i = 1%, (. ) = 11.2551. Try i = 2% (. ). = 10.5753 1% 11.2551 X0.3855 i% 10.8696 0.6798 2% 10.5753 Engineering Economy Page 45. % =. x = 0.57% i = 1% + 0.57% = 1.57% per month Effective rate = (1+0.0157)12 – 1 = 20.26% The second is the best offer. Example A chemical engineer wishes to set up a special fund by making uniform semiannual end-of-period deposits for 20 years. The fund is to provide P100,000 at the end of each of the last five years of the 20-year period. If interest is 8% compounded semiannually, what is the required semiannual deposit to be made? Solution P100,000(F/A,8.16%,5) P100,000P100,000P100,000P100,000P100,000 15 16 17 18 19 20 years 0 1 2 30 31 32 33 34 35 36 37 38 39 40 Period A A AAAAAAAAAAA A(F/A, 4%, 40) For the deposits, i = = 4% For the withdrawals, i = (1+0.04) 2 = 0.0816 or 8.16% Using 20 years from today as the focal date, the equation of value is A (F/A, 4%, 40) = P100,000 (F/A, 8.16%, 5) (. ) (. ) A = P100,000.. A (95.0255) = P100,000 (5.8853) A = P6,193.39 Engineering Economy Page 46 Example Using a compound interest of 8%, find the equivalent uniform annual cost for a proposed machine that has a first cost of P100,000 an estimated salvage value of P20,000 an estimated economic life of 8 years. Annual maintenance will amount to P2,000 a year and periodic overhaul costing P6,000 each will occur at the end of the second and fourth year. Solution P20,000(P/F,8%,8) P20,000 0 1 2 3 4 8 0 1 2 3 4 8 P2,000P2,000P2,000P2,000 P2,000A AAAAA A (P/A, 8%, 8) P6,000 P600 (P/F,8%,2) P6,000 P600 (P/F, 8%, 4) P2,000(P/A, 8%, 8) P100,000 Let A = the equivalent uniform annual cost Using today as the focal date the equation of value is A (P/A, 8%, 8) = P100,000 + P2,000 (P/A, 8%, 8) + P6,000 (P/F, 8%, 2) + P6,000 (P/F, 8%, 4) – P20,000 (P/F, 8%, 8) A (5.7466) = P100,000 + P2,000 (5.7466) + P6,000 (0.8573) + P6,000 (0.7350) – P20,000 (0.5403) A = P19,183 Engineering Economy Page 47 Example A man purchased a house for P425,000. In the first month that he owned the house, he spent P75,000 on repairs and remodeling. Immediately after the house was remodeled, he was offered P545,000 to sell the house. After some consideration, he decided to keep the house and have it rented for P4,500 per month starting two months after the purchase. He collected rent for 15 months and then sold the house for P600,000. If the interest rate was 1.5% per month, how much extra money did he make or lose by not selling the house immediately after it was remodeled? Solution P600,000(P/F, 1.5%, 15) P600,000 P4,500(P/A, 1.5%, 15) P545,000P4,500 P4,500P4,500P4,500 1 0 0 1 2 3 4 16 P75,000 P425,000 P425,000(F/P, 1.5%, 1) Engineering Economy Page 48  The only focal date that can be used is one month after purchase. Profit if the house was sold immediately after remodeling (1 month after purchase) = P545,000 – P75,000 – P425,000 (F/P, 1.5%, 1) = P38,625 Profit if the house was sold later (16 months after purchase) = P600,000 (P/F, 1.5%, 15) + P4,500 (P/A, 1.5%, 15) – P75,000 – P425,000 (F/P, 1.5%,1) = P600,000 (0.7999) + P4,500 (13.3432) – P75,000 – P425,000 (1.015) = P33,610 Loss by not selling immediately after remodeling = P38,625 – P33,610 = P5,105 Another solution: Loss by not selling immediately after remodeling = P545,000 – P4,500 (P/A, 1.5%, 15) – P600,000 (P/F, 1.5%, 15) = P545,000 – P4,500 (13.3432) – P600,000 (0.7999) = P5,105 Example: Determine the present equivalent value of P5,000 paid every 3 months for period of seven years and the rate of interest is 12% compounded quarterly. Required: The present value, P ( ) Solution: P = A (. ) = P5, 000. = P93, 820.54 Engineering Economy Page 49 Example Today, you invest P100,000 into a fund that pays 25% interest compounded annually. Three years later, you borrow P50,000 from a bank at 20% annual interest and invest in the fund. Two years later, you withdraw enough money from the fund to repay the bank loan and all interest due on it. Three years from this withdrawal you start taking P20,000 per year out of the fund. How much was withdrawn? Solution P50,000(1.20)2 (F/P, 25%, 7) P50,000(1.20)2 P20,000(F/25%, 5) P20,000 each 0 1 2 3 45 6 7 8 9 10 11 12 P50,000 P50,000 (F/P, 25%, 9) P100,000 P100,000 (F/P, 25%, 12) Let Q = the amount withdrawn after 12 years Using 12 years from today as the focal date, the equation of value is Q + P20,000 (F/A, 25%, 5) + P50,000 (1.20)2(F/P, 25%, 7) = P100,000 (F/P, 25%, 12) + P50,000 (F/P, 25%, 9) (. ) Q + P20,000. + P50,000 (1.20)2 (1.25)7 = P100,000 (1.25)12 + P50,000 (1.25)9 Q + P20,000 (8.2070) + P50,000 (1.20)2(4.7684) = P100,000(14.5519)+P50,000 (7.4506) Q = P1,320,255 Engineering Economy Page 50 Problem: If P25, 000 is deposited now into a savings account that earns 6% per year, what uniform annual amount could be withdrawn at the end of each year for ten years so that nothing would be left in the account after the 10th withdrawal? Solution: A =? 25,000 A = P (A/P, i%, n). A = 25,000 (. ) A = P3396.70 Problem: You can buy a machine for P100, 000 that will produce a net income, after operating expenses, of P10, 000 per year. If you plan to keep the machine for four years, what must be the market resale value be at the end of four years to justify the investment? You must make a 15% annual return on your investment. Solution: ( %) 100,000 (1.15)4 – 10,000 % = P124,966.88 10,000 100,000 Engineering Economy Page 51 Problem: Sixty monthly deposits are made into an account paying 6% nominal interest compounded monthly. If the subject of these is deposits is to accumulate P100,000 by the end of the fifth year, what is the amount of each deposits? Solution: A = F (A/F, i%, n). A = 100,000 ( ). A = P1433.28 Problem: What is the principal remaining after 20 monthly payments have been made on a P20,000 five-year loan? The annual rate is 12% nominal compounded monthly. Solution: A = F (A/F, i%, n). A = 20,000 ( ). = P244.83 F = A (F/A, i%, n) (. ) = 244.83. F = 5392.21 P20,000 - P5392.21 = P14,607.75 Problem: How much should be deposited each year to 12 years if you wish to withdraw P309 each per years for five years beginning at the end of the 15 year? Let the interest rate be 8%per year. Engineering Economy Page 52 Solution: A2 =309 1 12 15 19 A1 = ? P = A (1 + P/A, i%, n-1) ( %) ( ) P2 = 309 % + 1 (1.08)-14 = P453.65. A = P453.65 (. ) = P60.20 Problem: An individual is borrowing P100, 000 at 8% interest compounded annually. The loan is to be repaid in equal payments over 30 years. However, just after eight payment is made, lender allows the borrower to triple the payment. The borrower agrees to this increase payment. If the lender is still charging 10% per year, compounded annually, on the unpaid balance of the loan, what is the balance still owed just after the 12 payment is made? Solution:. P = 100,000 (. ) P = P8,882.74 ( %) F8 = P8,882.74 % = P94482.43 F30 = 100,000 (1.08)8 = 185,093.021 Problem: A woman arranges to repay a P 102,000 bank loan into equal payment at a 10% effective annual interest rate. Immediately after her third payment, she borrows another P5, 000 also at 10% per year. When she borrows the P5, 000 she talks Engineering Economy Page 53 the banker into letting her to repay the remaining debt of the first loan and the entire amount of second loan in 12 equal payments. The first of these 12 payments would be made one year after she receives the P5, 000. Compute the amount of each of the 12 payments? Solution: 102,000 3 12 5,000. A = 102,000 (. ) = P4,769.86 -9 P3 = 102,000 (1.10) = 43,257.96 Problem: A loan of P10, 000 is to be repaid over a period of eight years. During the first four years, exactly half of the loan principal is to be rapid (along with the accumulated compound interest) by uniform series of payments of per year. The other half of the loan principal is to be repaid over four years with accumulated interest by a uniform series of payments of per year. If I = 9% per year, what are the value of and ? Solution: 10,000 1 2 3 4 8 A=?. A1 = 5000 ( ) = P1093.34. F = 5000(1.09)8 = 9962.81. A2 = 9962.81 ( ). A2 = P2178.56 Engineering Economy Page 54  PROBLEMS 1. What is the annual rate of interest if P265 is earned in four months on an investment of P15, 000? Solution: P = P15,000 I = P265 n = 4 months i= = ( , )( ) i = 0.053 or 5.3% 2. A loan of P2, 000 is made for a period of 13 months, from January 1 to January 31 the following year, at a simple interest of 20%. What is the future amount is due at the end of the loan period? Solution: i = 20% P = P2,000 n = 13 months F = P (1 + ni) = P2,000 1 + (20%) F = P2,433.33 3. If you borrow money from your friend with simple interest of 12%, find the present worth of P20, 000, which is due at the end of nine months. Given: Future worth: F = P20,000 Number of interest period: n = 9 months Simple interest: i = 12% Engineering Economy Page 55 Solution: F = P ( 1+ ni) 20,000 = P (1 + ( )(.12)) 20,000 = P (1.09) P = P18,348.62 4. Determine the exact simple interest on P5, 000 for the period from Jan.15 to Nov.28, 1992, if the rate of interest is 22%. Solution: January 15 = 16 (excluding Jan.15) February = 29 March = 31 April = 30 May = 31 June = 30 August = 31 September = 30 November = 28 318 days Solution: Exact simple interest = Pni =5,000× ×0.22 = P955.74 Engineering Economy Page 56 5. A man wishes his son to receive P200, 000 ten years from now. What amount should he invest if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years? Solution: Given: F= P200, 000; For compound interest: i= 10%; n=5 For compound interest % i= ; n= 5×4=20 P110,735.15 P20, 000 0 5 10 0 1 2 3 4 P1 P2 Solution: P2 = F (1+ i)-n P2 = P200, 000(1+0.03)-20 P2 = P110, 735. 15 P1 = P2 (1+i)-n = 110,735.15 (1+0.10)-5 P1 = P68, 757.82 Engineering Economy Page 57 6. By the condition of a will the sum of P25, 000 is left to be held in trust by her guardian until it amounts to P45, 000. When will the girl receive the money if the fund is invested at 8% compounded quarterly? Given: % P = P25, 000 i = = 2% F = P45, 000 m=4 Solution: F = P (1+i) n 45000= 25000 (1+0.02)4n 45000⁄25000= (1.02) 4n 1.8= (1.02)4n ln(1.8) = 4nln(1.02) 29.682 = 4n N = 7.42 7. At a certain interest rate compounded semiannually P5, 000 will amount to P20, 000 after 10 years. What is the amount at the end of 15 years? Given: P = P5, 000; n1 = 10 F1 = P20, 000; n2 = 15 P 20, 000 F2=? Solution: i= − 1 5 10 15 ( ) , i= − 1 , P 500 i = 7.177% or 7.18% F2 = P (1+ i)TM = 5000 (1+7.18%)15(2) F2 = P 40, 029.72 Engineering Economy Page 58 8. A man borrows money from a bank which uses a simple discount rate of 14%. He signs a promissory note promising to pay a P500 pre-month at the end of 4th, 6th and 7th months, respectively. Determine the amount he received from the bank? Solution: P500 P500P500 F = P (1+ni) + P (1+ni) + P (1+ni) = 500 1 + (14% ) + 500 1 + (14% ) + 500 1 + (14% ) F = P1,599.16 F = P (1+ni) P=( ) ,. = % P = P 1,478.43 9. An amount of P3, 000 is invested for 9 years at 6% compounded quarterly. Determine the amount of interest from the earned by the investment. Given: P = P3, 000 m = compound quarterly i = 6% F=? Solution: F = P (1+i) n % 9x4 F = 3,000 (1 + ) F = P5, 127.42 Engineering Economy Page 59 10. If P10, 000 is invested at 12%, interest compounded annually, determine how many years it will take to double? Given: P= P10,000 i = 12% m = compounded quarterly n=? Let x = P= P1,000 2x F = 2(P1,000) = P2,000 F= P(1+i)n , = (1+i) n ln % = n ln (1 +i) ln 1 + =n =n n = 6.12 years ( ) 11. Which is better for an investor, to invest at 5 ½% compounded semi-annually as 5% compounded monthly? Given: i1 = 5.5% i2 = 5% m1 = compounded semi annually m2 =compounded annually Solution: ER semi-annually = ER monthly (1 +i) m – 1 = (1 +i) m – 1 % 2 % 12 (1 + ) – 1 = (1 + ) –1. % % (1 + ) = (1 + ) –1 0.056 = 0.051 Answer: 5.5% compounded semi-annually, better Engineering Economy Page 60 12. Mr. W borrowed P2, 000 from Mr. Y on June 1, 1928 and P500 one June 1, 1930. Mr. W paid P500 on June 1, 1931, P400 one June 1932 and P700 on June 1, 1933. If money is worth 5% compounded annually, what additional payment should Mr. W pay on June 1, 1936 to discharge all remaining liability? Solution: Inflow = outflow P2,000(F/P, i%, n)+P500(F/P, i%, n) = P500(F/P, i%, n)+P400(F/P, i%, n) + P700(F/P, i%, n) + x P2,000(1 + 5%)8 + P500(1 + 5%)6 = P500(1+5%)5 + P400(1+5%)4 +P700(1+5%)3 + x X =P1, 690.28 13. How long will it take P500 to be four times in value if invested at the rate of 7% compounded semi-annually? What is the effective rate of interest? Solution: Given: P = P500 i = 7% F = P2, 000 Solution: F = P (1+i) n , = (1+i) n = 2t ln = n ln (1 + i) ln 1 + 3.5% =n t= 20.15 years ( ) Engineering Economy Page 61 14. The sum of P26, 000 was deposited in a fund earning interest at 8% interest per annum compounded quarterly. What was the amount in the fund at the end of 3 years? Solution: % P = P26, 000 i = = 2% F = P (1+i) n = P26,000 (1+2%) 4(3) = P 3,297.43 15. The Philippines Society of Mechanical Engineer is planning to construct its own building. Two proposals are being considered: a. The construction of the building now to cost P400, 000. Solution: a.) P = P400, 000 i = 12% F = P (1+i) n = P400,000 (1+0.20) 5 = P 995,328 b. The construction of a smaller building now to cost P300, 000 and at the end of 5 years, an extension to be added to cost P200, 000, which proposal is more economical, if interest is 20% depreciation is disregarded and by how much? Solution: b.) P = P300, 000 i = 20% F = F (1+i) n = P300,000 (1+20%) 5 F = P 946,496 Answer: Proposal B is more economical Engineering Economy Page 62  16 If the sum of P15, 000 is deposited in an account earning 4% per annum compounded quarterly, what will be the amount of the deposited at the end of 5 years? Solution: % P = P15, 000 i = = 1% F = P (1+i) n = P15,000 (1+1%) 5(4) F = P 18,302.85 17.Jones Corporation borrowed P9, 000 from Brown Corporation on Jan. 1, 1978 and P12, 000 on Jan. 1, 1980. Jones Corporation made a partial payment of P7, 000 on Jan. 1, 1981. It was agreed that the balance of the loan would be amortizes by two payments one of Jan. 1, 1982 and the other on Jan. 1, 1983, the second being 50%larger than the first. If the interest rate is 12%. What is the amount of each payment? Given: =12% P9, 000 12, 000 1978 1979 1981 1982 1983 P7, 000 X 50%X + X Solution: 9000(1+i)5 + 12,000(1+i)3 = 7,000(1+i)2 + (1+i)1 + 15,861.08 + 16,851.14 = 8,780.8 + 1.12X + 1.5X X1 = 9,137.18 Where X2 = X1 ( ) X2 = 9,137.18( ) X2 = 13,705.77 Engineering Economy Page 63 18.A woman borrowed P3,000 to be paid after 1 years with interest at 12% compounded semi-annually and P5,000 to be paid after 3 years at 12% compounded monthly. What single payment must she pay after 3 at an interest rate of 16% compounded quarterly to settle the two obligations? Given: P1 = 3,000 P2 = 5,000 t3 = 3.5 years N1 = 1.5 years n2 = 3 years i3 = 16% I1 = 12% i2 = 12% m3 = quarterly M1 = semi-annually m2 = monthly F1 = P1 (1+i1)n1 F2 = P2 (1+i2)n2 F3 = F1 (1+i2) F4 = F2 (1+i3) % % ( ) F1 = 3,000 1 + F2 = 5,000 1 + F3 = 3,573(1+i3)4(2) % (. ) F4 = 7,153.84 1 + F2 = 7,153.84 F3 = 4,889.96 F1 = 3,573.048 F4 = 7,739.59 F5 = F3 + F4 F5 = 4,889.96 + 7,739.59 F5 = 12,627.55 19. Mr. J. de la Cruz borrowed money from a bank. He received from the bank P1,342 and promised to repay P1,500 at the end of 9 months. Determine the simple interest rate and the corresponding discount rate or often referred to as the “Banker’s discount”. Given: F = P1,500 discount = P1,500 – P1,342 i = 158/1,342 = 11.77% P = P1,342 d = 158 d=? d = 158/1500 = 10.53% i=? Engineering Economy Page 64 20. A man deposits P50,000 in a bank account at 6% compounded monthly for 5 years. If the inflation rate of 6.5% per year continues for this period, will this effectively protect the purchasing power of the original principal? Given: P = 50,000 f = 6.5% i = 6% m = compounded monthly F=? % F = 50,000 (. %) F = 49,224.99; No 21. What is the future worth of P600 deposited at the end of every month for 4 years if the interest rate is 12% compounded quarterly? Given: A = 600 F=? ERmonthly= ERquarterly % (1+i)12 – 1 = 1 + −1 % (1+i)12 = 1 + % i= 1+ −1 i = 0.99% ( ) F=A (. %) ( ) F = 600. % F = 36,641.32 Engineering Economy Page 65 22. A young woman, 22 years old, has just graduated from college. She accepts a good job and desires to establish her own retirement fund. At the end of each year thereafter she plans to deposit P2,000 in a fund at 15% annual interest. How old will she be when the fund has an accumulated value of P1,000,000? Given: A = 2,000 i = 15% F = 1,000,000 T=? F = A (F/A, i%, n) ( %) 1,000,000 = 2,000 % , , ( %) = , % 15% (1,000,000) = 2,000 (1+15%)n – 2,000 150,000 + 2,000 = 2,000 (1+15%)n , ln = nln (1+15%) , , , = (1 + 15%) n = 30.98 ≈ 31 T = 22 + 31 = 53 years old Engineering Economy Page 66 23. Mr. Reyes borrows P600,000 at 12% compounded annually, agreeing to repay the loan in 15 equal annual payments. How much of the original principal is still unpaid after he has made the 8th payment? Given: P = 600,000 i =12% m = compounded annually n = 15 600,000 0 1 8 15 A = P (A/P, i%, n) % A = 600,000 ( %) A = 88,094.54 @ n =7 ( %) P = 88,094.54 % P = 402, 042.05 24. A man who won P300,000 in a lottery decided to place 50% of his winnings in a trust fund for the college education of his son. If the money will earn 14% a year compounded quarterly, how much will the man have at the end of 10 years, when his son will be starting his college education? Given: P = 300,000 50% x 300,000 = 150,000 i = 14% m=4 % F = 150,000 1 + = P593,888.96 Engineering Economy Page 67 25. A man borrowed P25,000 from a bank which he signed a promissory note to repay the loan at the end of one year. He only received the amount of P21,915 after the bank collected the advance interest and an additional amount of 85.00 for legal fees. What was the rate of interest that the bank charged? Given: F = 25,000 P = 21,915 i=? D = 25,000 – 21,915 = 3,085 D = 3,085 – 85(legal fees) D = 3,000 , i= = 0.1369 13.69% , 26. A man wishes to have P35,000 when he retires 15 years from now. If he can expect to receive 4% annual interest, how much must he set aside beginning at the end of each of the 15 years? Given: F = 35,000 i = 4% n = 15 years A = F (A/F, i%, n) % A = 35,000 ( = 1,747.94 %) 27. An investor deposits P1,000 per year in a bank which offers an interest of 18% per annum for time deposits of over 5 years. Compute how much the investor can collect at the end of 13 years, assuming that he never withdraws any amount before the 13th year. F 5 13 1,000 Engineering Economy Page 68  F = A (F/A, i%, n) ( %) F = 1,000 % F = 7,154.21 F = P (1+i)n F = 7,154.21 ( 1+18%)8 F = 26,891.67 28. A man wishes to prepare the future of his ten-year old son. Determine the monthly savings that he should make with interest at 5.41% per annum to amount to P120,000 at the time his son will be 18 years old. ERmonthly= ERper annum (1+i)m – 1 = (1+i)m – 1. % (1+i)12 -1 = 1 + −1 i = (1 + 5.41%) − 1 i = 0.44% A = F (A/F, i%, n). % A = 120,000 (. %) A = 1,007.24 29. A building and loan association requires that loans be repaid by uniform monthly payments which include monthly interest calculated on the basis of nominal interest of 5.4% per annum. If P5,000 is borrowed to be repaid in 10 years, what must the monthly payment be? ERmonthly= ERper annum (1+i)m – 1 = (1+i)m -1 (1+i)12 – 1 = (1+5.4%)1 -1 i = (1 + 5.4%) − 1 i = 0.44% A = P (A/P, i%, n). % A = 5,000 ( ( ). %) A = 53.72 Engineering Economy Page 69 30. You want to start saving for your ten-year old son’s college education. If you are guaranteed 6% interest compounded quarterly, how much would you have to save each month to amount to 12,000 by the time he is 18 years old? F = 12,000 i = 6% m=4 ERquarterly= ERmonthly A = F (A/F, i%, n) m 12. % (1+i) – 1 = (1+i) –1 A = 12,000 (. %) % 1+ − 1 = (1 + ) −1 A = 97.82 6% 1+ − 1= 4 i = 0.4975% 31. Find the annual payments to extinguish a debt of P10,000 payable in 5 years at 12% compounded annually? Given: P 10,000 i = 12% m=1 n=5 A = P (A/P, i%, n) % A = 10,000 ( %) A = 2,774.10 32. A man owes P10,000 with interest at 6% payable semi-annually. What equal payments at the beginning of each 6 months for 8 years will discharge his debt? Given: P = 10,000 A = P (A/P, i%, n) % i = 6% A = 10,000 ( %) n=8 A = 989.52 m=2 A= ? 33. The P100,000 cost of an equipment was made available through a loan which earns 12% per annum. If the loan will be paid in 10 equal annual payments, how much is the annual installment? Engineering Economy Page 70 Given: P = 100,000 A = P (A/P, i%, n) % i = 12% A = 100,000 ( %) A=? A = 17,698.42 34.An individual is borrowing P100, 000 at 8% interest compounded annually. The loan is to be repaid in equal payments over 30 years. However, just after eight payment is made, lender allows the borrower to triple the payment. The borrower agrees to this increase payment. If the lender is still charging 10% per year, compounded annually, on the unpaid balance of the loan, what is the balance still owed just after the 12 payment is made. ( %) P = 100,000 ( ) F8 = P8,882.74. % P = P8,882.7 F8 = P94482.43 F30 = 100,000 (1.08)8 =185,093.021 Deferred Annuity A deferred annuity is one where the first payment is made several periods after the beginning of the annuity. An ordinary annuity where the first cash flow of the series is not at the end of the lot period or it is deferred for some time m periods n periods 0 1 2 n-1 n 0 1 2 m A AAA A (P/A,i%,n) (P/F,i%,n) A (P/A,i%,n) Cash flow diagram given A to find P Engineering Economy Page 71  P = A (P/A, i%, n) (P/F, i%, m) ( ) P=A (1 + i) Where : m = deferred period n = number of annuities *Note that the counting of m is up to the period before the first A. Example On the day his grandson was born, a man deposited to a trust company a sufficient amount of money so that the boy could receive five annual payments of P80,000 each for his college tuition fees, starting with his 18th birthday. Interest at the rate 12% per annum was to be paid on all amounts on deposit. There was also a provision that the grandson could select to withdraw no annual payments and received a single lump amount on his 25th birthday. The grandson chose this option. (a)How much did the boy received as the single payments? (b)How much did the grandfather deposit? Solution Let P=the amount deposited X=the amount withdrawn X P8,000 P8,000P8,000P8,000P8,000 0 18 19 20 21 22 23 24 25 P Engineering Economy Page 72  The P80,000 supposed withdrawals are represented by broken lines, since they did not actually occur. Three separated cash flows diagrams can be drawn. (a) X P80,000 (F/A, 12%, 5) P80,000(F/A,12%,5)(F/P,12%,3) P80,000P80,000P80,000P80,000P80,000 17 18 19 20 21 22 23 24 25 X and the P80,000 supposed withdrawals are equivalent. Using 25 years of age as the focal date, the equation of value is X= P80,000 (F/ A, 12%, 5)(F/P, 12%, 3) (. ) = P80,000 (1+.12)3. = P714,023.4509 The other good focal dates are 17 and 22 years from today. (b) P 80,000(P/A,12%,5)(P/F,12%,17) P80,000(P/A,12%,5) P80,000 0 17 18 19 20 21 22 P Engineering Economy Page 73  P and P80, 000 supposed withdrawals are equivalent. Using today as the focal date, the equation of value is P = P80,000 (P/ A, 12%, 5) (P/F, 12%, 17) (. ) = P80,000. (1+.12)-17 = P42,001.22 Another solution: X(P/F, 12%, 25)