Engineering Economy Module 1 PDF

An Engineering Science ENGINEERING ECONOMY University of Eastern Philippines College of Engineering Agricultural & Biosystems Engineering Department Engr. Elaido B. Jao Jr. Instructor 2020  E...

An Engineering Science ENGINEERING ECONOMY University of Eastern Philippines College of Engineering Agricultural & Biosystems Engineering Department Engr. Elaido B. Jao Jr. Instructor 2020  ES 414a: ENGINEERING ECONOMY MODULE 1: ENGINEERING ECONOMY ITS TIME VALUE OF MONEY AND IMPORTANCE OVERVIEW Engineering Economy is one of the most essential subjects in the engineering disciplines. This module is to give the students a sound understanding of the basic concepts of the course and some insights into approaches that can be used for making sound economic decisions. They shall also be able to familiarize and identify the different possible alternatives of economic studies in the different fields of specialization as presented mostly by Hipolito B. Sta. Maria. This module tackles on engineering science course particularly the essentials of Engineering Economy and their functions This 3-unit course is offered to all officially enrolled Engineering students with allotted lecture of three-hours per week. TOPICS: This module presents the following topics: 1. Economic Environment a. Consumer and Producer Goods and services; b. Necessities and Luxuries; c. The Law of Supply and Demand; d. Competition, Monopoly and Oligopoly; e. Law of Diminishing Returns 2. Interest and Money – Time Relationship a. Simple Interest; b. Compound Interest; c. Annuities LEARNING OUTCOMES At the end of this course, the student must be able to 1. Explain the time value of money and discounting at normal and inflationary conditions; a. Determine the future value of an investment and the present worth of a targeted future amount; b. Prepare using manual computations and computer the detailed investments, operating costs; financial an economic benefits of any engineering projects;  PRETEST Direction: Choose the letter of the correct answer. Do not write your answer on this manual. 1. It is the analysis and 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. a. Engineering Economy b. Economic Analysis c. Economics d. All of the above 2. These are those products that are directly used by people to satisfy their wants. a. Consumers goods b. Producers goods c. Dry goods d. Wet goods 3. These are goods used to produce consumer goods or other goods. a. Dry goods b. Wet goods c. Consumers goods d. Producers goods 4. These are those products or services that are required to support human life activities. a. Food b. Clothing c. Shelter d. Necessities 5. These are those products or services that are desired by humans and will be purchased if money is available after the required necessities have been obtained. a. Wants b. Luxuries c. Needs d. Excess 6. It is a quantity of a certain commodity that is bought at a certain price at a given time. a. Supply b. Demand c. Bulk purchased d. Panic Buying 7. In the cash-flow diagram shown in the given figure a. Equal deposits of Rs 3,000 per year (A) are made, starting now b. The rate of interest is 10% per year account c. The amount accumulated after the 7th deposit is to be computed d. All of the above 8. The interest calculated on the basis of 365 days a year, is known as: a. Interest b. Ordinary simple interest c. Exact simple interest d. None of these 9. If ‘P’ is principal amount, ‘I” is the rates of interest per annum and ‘n’ is the number of periods in years, the compound amount factor (CAF) is: a. (1 – i)n b. (1 – i)n/12 c. (1 – i/12)n d. (1 – ni) 10. Pick up the correct statement from the following: a. The change in the amount of money over a given time is called “time value” of money, a most important concept in engineering economy b. The manifestation of the time value of money is termed as current asset. c. Interest on borrowing = future amount owed-original loan. d. none of the above. Key Terms Time Value of Money. It is a fact that money makes money. This concept explains the change in the amount of money over time for both owned and borrowed funds. Economic Equivalence A combination of time value of money and interest rate that makes different sums of money at different times have equal economic value Cash Flow The flow of money into and out of a company, project, or activity. Revenues are cash inflows and carry a positive (+) sign; expenses are outfl ows and carry a  negative (−) sign. If only costs are involved, the − sign may be omitted, e.g., benefi t/cost (B/C) analysis. Loan An amount of borrowed capital or money from any entity such as banks or persons. End-of-Period Convention To simplify calculations, cash fl ows (revenues and costs) are assumed to occur at the end of a time period. An interest period or fi scal period is commonly 1 year. A half-year convention is often used in depreciation calculations. Cost of Capital The interest rate incurred to obtain capital investment funds. COC is usually a weighted average that involves the cost of debt capital (loans, bonds, and mortgages) and eq-uity capital (stocks and retained earnings). Minimum Attractive Rate of Return (MARR) A reasonable rate of return established for the evaluation of an economic alternative. Also called the hurdle rate, MARR is based on cost of capital, market trend, risk, etc. The inequality ROR ≥ MARR > COC is correct for an eco-nomically viable project. Opportunity Cost A forgone opportunity caused by the inability to pursue a project. Nu- merically, it is the largest rate of return of all the projects not funded due to the lack of capital funds. Stated differently, it is the ROR of the fi rst project rejected because of unavailability of funds. Nominal or Effective Interest Rate ( r or i ) A nominal interest rate does not include any compounding; for example, 1% per month is the same as nominal 12% per year. Effective interest rate is the actual rate over a period of time because compounding is imputed; for example, 1% per month, compounded monthly, is an effective 12.683% per year. Infl ation or defl ation is not con-sidered. Placement of Present Worth ( P ; PW) In applying the ( P/A , i %, n ) factor, P or PW is always located one interest period (year) prior to the fi rst A amount. The A or AW is a series of equal, end-of-period cash fl ows for n consecutive periods, expressed as money per time (say, $/year; /year). Placement of Future Worth ( F ; FW) In applying the ( F/A , i %, n ) factor, F or FW is always located at the end of the last interest period (year) of the A series. LEARNING PLAN READING RESOURCES AND INSTRUCTIONAL ACTIVITIES What to know Do you want to be a successful engineer-economist? Let’s find out how? Let's get started: Introduction The need for engineering economy is primarily motivated by the work that engineers do in performing analyses, synthesizing, and coming to a conclusion as they work on projects of all sizes. In other words, engineering economy is at the heart of making decisions. These decisions involve the fundamental elements of cash flows of money, time, and interest rates. This chapter introduces the basic concepts and terminology necessary for an engineer to combine these three essential elements in organized, mathematically correct ways to solve problems that will lead to better decisions. Economy Environment Engineering economy is the analysis an 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 direct use 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 desire by humans and will be purchased if money is available after the required necessities have been obtained. Law of Supply and demand Supply is the quantity of a certain community that is offered for sale at a certain price at a given place and time. Demand is he quantity of a certain commodity that is bought at a certain price at a given place an 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 the 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. The law of supply and demand is a theory that explains the interaction between the sellers of a resource and the buyers for that resource. The theory defines what effect the relationship between the availability of a particular product and the desire (or demand) for that product has on its price. Generally, low supply and high demand increase price and vice versa. Perfect examples of supply and demand in action include PayPal. The law of demand states that, if all other factors remain equal, the higher the price of a good, the less people will demand that good. In other words, the higher the price, the lower the quantity demanded. The amount of a good that buyers purchase at a higher price is less because as the price of a good goes up, so does the opportunity cost of buying that good. As a result, people will naturally avoid buying a product that will force them to forgo the consumption of something else they value more. The chart below shows that the curve is a downward slope. Like the law of demand, the law of supply demonstrates the quantities that will be sold at a certain price. But unlike the law of demand, the supply relationship shows an upward slope. This means that the higher the price, the higher the quantity supplied. Producers supply more at a higher price because selling a higher quantity at a higher price increases revenue. Unlike the demand relationship, however, the supply relationship is a factor of time. Time is important to supply because suppliers must, but cannot always, react quickly to a change in demand or price. So it is important to try and determine whether a price change that is caused by demand will be temporary or permanent. Let's say there's a sudden increase in the demand and price for umbrellas in an unexpected rainy season; suppliers may simply accommodate demand by using their production equipment more intensively. If, however, there is a climate change, and the population will need umbrellas year-round, the change in demand and price will be expected to be long term; suppliers will have to change their equipment and production facilities in order to meet the long-term levels of demand. Key Takeaways 1. The law of demand says that at higher prices, buyers will demand less of an economic good. 2. The law of supply says that at higher prices, sellers will supply more of an economic good. 3. These two laws interact to determine the actual market prices and volume of goods that are traded on a market. 4. Several independent factors can affect the shape of market supply and demand, influencing both the prices and quantities that we observe in markets. Competition, Monopoly and Oligopoly Perfect competition occurs in a situation where a commodity or service is supplied by a number of vendors an there is nothing to prevent additional vendors entering the market.. Monopoly is the opposite of the prefect competition. A perfect monopoly exists when a unique product or services is available from a single vendor and that vendor can prevent the entry of all others into the market. Oligopoly exists when there are a few suppliers of a product or services that action by one will almost inevitably result in similar action by the others What to process At this point, let us examine ourselves if we really capable of becoming an Economist. Let us know our chances. Activity #1: This is a self-examination activity. After knowing the basic concepts of economic process, examine yourself if you really understands economy itself. Simply put check ( ) if that particular process will you really do or an x (X) if you will not do it. Comment on your own status by giving possible options that can help you acquire it. Will I Do Realizations this? (What are my options?) Economic Process Producer’s goods or YES NO Consumers goods; ( ) (X) Necessities or Luxuries 1 I always buy classical books. Books are consumers good, classical books are Luxuries since I should rather buy engineering books 2 I rather photocopy a books than buying them 3 I always buy bags every semester as my lucky charm. 4 I only buy new bag when my old bag is unusable. 5 I always add more supplies of ingredients whenever my parents ask me for an errand for allowance purposes. 6 I always check the exact amount of ingredients needed to avoid wastage. 7 I always ride my motorcycle no matter how near it is. 8 I always ride my motorcycle no matter how far it is. 9 I intend to make my assignments written carefully to avoid erasure to save my money for bond papers. 10 It doesn’t matter whether I will rewrite my assignment as long as I can submit my work neatly. The next activity will test your knowledge on the law of supply and demand as well as the business establishment scenario undertaken. Write the letter corresponding the business establishment that practice whether Perfect Competition (P), Monopoly (M), or Oligopoly (O). Write your insight on their supply and demand analysis Economics Supply and Demand Business Establishments Practice analysis 1 Northern Samar Electric Cooperative M Electric supply is monopolized (Norsamelco) since it is the only electric establishment in the province. The electric demand is high during Christmas Season 2 Cable Television Services 3 Sari-sari Stalls 4 Caltex 5 Sugar Industry 6 Meat Section Market 7 Airlines 8 Gaisano Grand Mall 9 Pharmaceuticals 10 Dry Goods Section 11 Computer & Software 12 Coconut Industry Interest and Money – Time Relationships Interest is the manifestation of the time value of money. It is the amount of money paid for the use of borrowed capital or the income produced by the money which has been loaned. Computationally, interest is the difference between an ending amount of money and the beginning amount. If the difference is zero or negative, there is no interest. There are always two perspectives to an amount of interest—interest paid and interest earned. These are illustrated in Fig.1.. Interest is paid when a person or organization borrowed money (obtained a loan) and repays a larger amount over time. Interest is earned when a person or organization saved, invested, or lent money and obtains a return of a larger amount over time. The numerical values and formulas used are the same for both perspectives, but the interpretations are different. Interest paid on borrowed funds (a loan) is determined using the original amount, also called the principal, DIMENSIONAL ANALYSIS: Interest = amount owed now − principal (Eq. 1) When interest paid over a specific time unit is expressed as a percentage of the principal, the result is called the interest rate. Interest rate (%) = × 100% (Eq. 2) The time unit of the rate is called the interest period. By far the most common interest period used to state an interest rate is 1 year. Shorter time periods can be used, such as 1% per month. Thus, the interest period of the interest rate should always be included. If only the rate is stated, for example, 8.5%, a 1-year interest period is assumed. Fig. 1 (a) Interest paid over time to lender. (b) Interest earned over time by investor. EXAMPLE 1 An employee at Gaisano Grand Mall borrows ₱10,000 on May 1 and must repay a total of ₱10,700 exactly 1 year later. Determine the interest amount and the interest rate paid. Given: borrowed amount = ₱ 10,000 Repayment = ₱ 10,700 Required Interest Solution The perspective here is that of the borrower since ₱10,700 repays a loan. Apply Equation Interest = amount owed now − principal To determine the interest paid. Interest paid = ₱10,700 − ₱10,000 = ₱700 Then determine the interest rate paid for 1 year. Interest rate (%) = × 100% Percent Interest rate (%) = × 100% = 7% per year EXAMPLE 2 Stereophonics, Inc. plans to borrow ₱20,000 from a bank for 1 year at 9% interest for new recording equipment. (a) Compute the interest and the total amount due after 1 year. (b) Construct a column graph that shows the original loan amount and total amount due after 1 year used to compute the loan interest rate of 9% per year. Solution (a) Compute the total interest accrued by solving Equation for interest accrued. Interest = ₱20,000(0.09) = ₱1800 The total amount due is the sum of principal and interest. Total due = ₱20,000 + 1800 = ₱21,800 (b) Figure 2 shows the values used in Equation : $1800 interest, $20,000 original loan principal, 1-year interest period.  Fig. 2. Values used to compute an interest rate of 9% per year. ENGINEERING ANALYSIS: Simple interest is 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 (Eq 3) F = P + I = P + Pni F = P (1 +ni) (Eq. 4) 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 worth (a) Ordinary simple interest is computed on the basis of 12 months of 30 days each or 360 days a year. 1 interest period = 360 days (b) Exact simple interest is based on the exact number of days in a year, 365 days for an ordinary year and 366 days for leap year. 1 interest period = 365 or 366 days EXAMPLE 3 Determine the ordinary simple interest on ₱ 700 for 8 months and 15 days if the rate of interest is 15% Given: P = ₱ 700, n = 8 mos & 15 days i = 15% Solution: Number of days = (8) (30) + 15 = 255 days From Eq.1: I = Pni I = ₱ 700 x x 15% = ₱ 74.38 EXAMPLE 4 Determine the exact simple interest on ₱ 500 for the period from January 10 to October 28, 2016 at 16% interest Given: P = ₱ 500, n = Jan 10 – Oct 28 i = 16% Required Exact interest Solution Jan 10-31 = 21 (excluding 10) February = 29 March = 31 April = 30 May = 31 June = 30 July = 31 August = 31 September = 30 October = 28 ( including 10) 292 days Exact simple Interest = ₱ 500 x x 16% = ₱ 63.83 EXAMPLE 5 What will be the future worth of the money after 14 months, if a sum of ₱ 10,000 is invested today at a simple interest rate of 12% per year? Given: P = ₱ 10,000, n = 14 mos i = 12% Required Future worth (F) Solution: F = P (1 + ni) = ₱ 10,000 1 + (0.12) = ₱ 11,400.00 Cash Flow Diagram 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. This represents a receipt (positive cash flow or cash inflow) This represents a disbursement (negative cash flow of cash outflow) Illustration 1: A loan of ₱ 100 at simple interest of 10% will become ₱ 150 after 5 years ₱ 150 l l I i 0 1 2 3 4 5 ₱ 100 Cash flow diagram on the viewpoint of the lender ₱ 100 l l l I 0 1 2 3 4 5 ₱ 150 Cash flow diagram on the viewpoint of the borrower Compound Interest In calculations of compound interest, the interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous periods Thus compound interest means 'interest on top of interest.' P l l l- - - - n-1 n 0 1 2 3 F Table 1. Compound Interest (Borrower's Viewpoint) 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 P (1 + i)2 i P(1 + i)2 + P(1+ i)2i = P(1 + i)3 --- --- --- --- n P (1 + i)n-1 n-1 P (1 + i) i P (1 + i) n F = P(1+ i)n (Eq. 5)  The quantity (1 + i)n is commonly called the ֞ single payment compound amount factor ֞ and is designated by the functional symbol F/P, i%, n. Thus, F = P (F/P, i%, n) (Eq. 6) The symbol F/P, i%, n is read as ֞F given P at i percent in n interest periods.֞ From Eq. 6 P = F(1 + i)-n (Eq. 7) The quantity (1 + i)-n is 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) (Eq. 8) The symbol P/F, i%, n is read as ֞P given F at i percent in n interest periods.֞ From Eq. 8 Rates of Interest Nominal and effective interest rates are similar to simple and compound interest rates, with a nominal rate being equivalent to a simple interest rate. All of the equations expressing time value of money are based on compound (i.e., effective) rates, so if the interest rate that is provided is a nominal interest rate, it must be converted into an effective rate before it can be used in any of the formulas. The first step in the process of insuring that only effective interest rates are used is to recognize whether an interest rate is nominal or effective. Table 1 shows the three ways interest rates may be stated. Table 2. Various interest statements and their interpretation Interest Rate Statement Interpretation Comment i = 12% per year i = effective 12% per year When no compounding compounded yearly period is given, interest rate i = 1% per month i = effective 1% per month is an effective rate, with compounded monthly compounding period i = 3-1/2% per quarter i = effective 3-1/2% per assumed to be equal to quarter compounded stated time period quarterly i = 8% per year, i = nominal 8% per year When compounding period compounded monthly compounded monthly is given without stating i = 4% per quarter i = nominal 4% per quarter whether the interest rate is compounded monthly compounded monthly nominal or effective, it is i = 14% per year i = nominal 14% per year assumed to be nominal. compounded compounded Compounding period is as semiannually semiannually stated. i = effective 10% per year i = effective 10% per year If interest rate is stated as compounded monthly compounded monthly an effective rate, then it is i = effective 6% per quarter i = effective 6% per quarter an effective rate. If compounded quarterly compounding period is not i = effective 1% per month i = effective 1% per month given, compounding period compounded daily compounded daily is assumed to coincide with stated time period. The three statements in the top part of the table show that an interest rate can be stated over some designated time period without specifying the compounding period. Such interest rates are assumed to be effective rates with the compounding period (CP) assumed to be the same as that of the stated interest rate. For the interest statements presented in the middle of Table 2, three conditions prevail: 1. the compounding period is identified, 2. this compounding period is shorter than the time period over which the interest is stated, and 3. the interest rate is not designated as either nominal or effective. In such cases, the interest rate is assumed to be nominal and the compounding period is equal to that which is stated. (We show how to get effective interest rates from these in the next section.) For the third group of interest-rate statements in Table 3, the word effective precedes or follows the specified interest rate and the compounding period is also stated. These interest rates are obviously effective rates over the respective time periods stated. Likewise, the compounding periods are equal to those stated. Similarly, if the word nominal had preceded any of the interest statements, the interest rate would be a nominal rate. Table 3 contains a listing of several interest statements (column 1) along with their interpretations (columns 2 and 3). Table 3. Specific examples of interest statements and interpretations (1) (2) (3) Interest Statement Nominal or Effective Compounding Interest Period 15% per year compounded monthly Nominal Monthly 15% per year Effective Yearly Effective 15% per year compounded Effective Monthly monthly 20% per year compounded quarterly Nominal Quarterly Nominal 2% per month compounded Nominal Weekly weekly 2% per month Effective Monthly 2% per month compounded monthly Effective Monthly Effective 6% per quarter Effective Quarterly Effective 2% per month compounded Effective Daily daily 1% per week compounded continuously Nominal Continuously 0.1% per day compounded continuously Nominal Continuously All of the formulas used in making time value calculations are based on effective interest rates. Therefore, whenever the interest rate that is provided is a nominal rate, it is necessary to convert it to an effective interest rate. As shown below, an effective interest rate, i, can be calculated for any time period longer than the compounding period. The most common way that nominal interest rates are stated is in the form 'x% per year compounded y' where x = interest rate and y = compounding period. An example is 18% per year compounded monthly. When interest rates are stated this way, the simplest effective rate to get is the one over the compounding period because all that is required is a simple division. For example, from the interest rate of 18% per year compounded monthly, a monthly interest rate of 1.5% is obtained (i.e., 18% per year/12 compounding periods per year) and this is an effective rate because it is the rate per compounding period. To get an effective rate for any period longer than the compounding period use the effective interest rate formula. i = (1+r/m)m - 1 (Eq. 9) where: i = effective interest rate per period r = nominal interest rate per period m = number of times interest is compounded per period This effective interest rate formula can be solved for r or r/m as needed to determine a nominal interest rate from an effective rate. For continuous compounding, the effective rate formula is the mathematical limit as m increases without bounds, and the formula reduces to i = er - 1. (Eq. 10) Example 6: For an interest rate of 12% per year compounded quarterly, what is the effective interest rate per year ? Given: r = 12% Required: effective rate Solution: An effective interest rate per year is sought. Therefore, r must be expressed per year and m is the number of times interest is compounded per year. Using Eq. 9. i = (1+r/m)m - 1 = (1 + 0.12 / 4)4 - 1 = 12.55 % Example 7: Find the nominal rate which if converted quarterly coul be used instead of 12% compounded monthly. What is the corresponding effective rates? Given: r (known) = 12% Required 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 r% compounded quarterly ( 1 + )4 – 1. 12% compounded monthly (1+ )12 – 1. Equating the Two Equation: ( 1 + )4 – 1 = ( 1 + )12 – 1  1 + = (1.01)3 = 1.0303 r = 0.1212 or 12.12% compounded quarterly Example 8: Find the amount at the end of two years and seven months if ₱ 1,000 is invested at 8% compounded quarterly using simple interest for anytime less than a year interest period. Given: P = ₱ 1,000, r = 8% compounded, n = 2yr & 7mos Required Future amount, F % Solution: For compound interest: i= = 2%, n = (2)(4) = 8 For simple interest: i = 8%, n = F2 F1 l 0 1 2 2 yrs 7 mos compound simple interest interest ₱ 1000 F1 = P (1 + i)n = ₱ 1000 ( 1 + 0.02)8 = ₱ 1,171.66 F2 = F1 (1 + ni) = ₱ 1,171.66 1 + (0.08) = ₱ 1,226.34 Example 9: A ₱ 2,000 loan was originally made at 8% simple interest for 4 years. At the end of this period the loan was extended for 3 years, without the interest being paid, but the new interest rate was made 10% compounded semiannually. How much should the borrower pay at the end of 7 years? Given: P = ₱ 2,000, i = 8% compounded, n1 = 4yrs n2 = 3yrs r = 10% Required F7 Solution: F7 F4 l l l l l l 0 1 2 3 4 5 6 7 simple compound interest interest ₱ 2,000  F4 = P (1 + ni) = ₱ 2,000⟦1 + (4)(0.08)⟧ = ₱ 2,640.00 F7 = F4 (1 + ni) = ₱ 2,640.00 (1 + 0.05)6 = ₱ 3,537.86 Equation of Value An equation of value is obtained by setting the sum of the values on a certain comparison or focal date of one set of obligations equal to the sum of the values on the same date of another set of obligations. Example 10: A man bought a lot worth ₱ 1,000,000 if paid in cash. On installment basis, he paid a down payment of ₱ 200,000, ₱ 300,000 at the end of one year; ₱ 400,000 at the end of three years and a final payment at the end of five years.. What was the final payment if the interest was 20%. Given: P = ₱ 1,000,000, CP1 = ₱ 200,000, CP2 = ₱ 300,000 @ 1yr, CP3 = ₱ 400,000 @ 3yrs i = 20% n = 5 yrs Required Present worth after 5yrs (P5) Solution: ₱ 800,000 0 1 2 3 4 5 l l ₱ 200,000 ₱ 300,000 P30,000(P/F,20%,1) ₱ 400,000 P40,000(P/F,20%,1) P5 P5 (P/F,20%,1) Using today as the focal date, the equation of value is ₱ 800.000 = ₱ 300,000 (P/F, 20%, 1) + ₱ 400,000 (P/F, 20%, 3) + P5 (P/F, 20%,, 5) ₱ 800,000 = ₱ 300,000 (1.20)-1 + ₱ 400,000 (1.20)-3 + P5(1.20)-5 ₱ 800,000 = ₱ 300,000 (0.8333) + ₱400,000 (0.5787) + P5(0.4019 P5 = ₱ 792,560 Continuous Compounding In discrete compounding the interest is compounded at the end of each finite – length period, such as a month, a quarter or a year. In continuous compounding, it is assumed that cash payments occur once per year Formula; F = Pem or P = Fe-m where m = number of interest periods Example 11: Compare the accumulated amounts after 5 years of ₱ 1,000 invested at the rate of 10% per year compounded (a) annually, (b) semiannually, (c) quarterly (d) monthly, (e) daily, and (f) continuously. Given: P = ₱ 1,000, r = 20% n = 5 yrs Required Future Worth (a) annually, (b) semiannually, (c) quarterly (d) monthly, (e) daily, and (f) continuously. Solution: Using the formula, F = P (1 + 1)n (a) F = ₱1,000 ( 1 + 0.10)5 = ₱ 1,610.51. (b) F = ₱1,000 ( 1 + )10 = ₱ 1,628.89. (c) F = ₱1,000 ( 1 + ) 20 = ₱ 1,638.62. (d) F = ₱1,000 ( 1 + )60 = ₱ 1,645.31. 1825 (e) F = ₱1,000 ( 1 + ) = ₱ 1,648.61 m 0.10x5 (f) F = Pe = ₱1,000 e = ₱ 1,648.72 Discount 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 time. Below shows rate of discount cash flow (1 + i)-1 0 1 ₱ 1.00 d = 1 – (1 + i)-1 (Eq. 11) i = (Eq. 12) where d = rate of discount i = rate of interest of the same period Example 12 A man borrowed ₱ 5,000 from a bank an agreed to pay the loan at the end of 9 months. The bank discounted the loan and gave him ₱ 4,000 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? Given: P = ₱ 5.000, discount = ₱5,000 - ₱ 4,000 = ₱ 1,000 Required (a) d ; (b) i ; (c) i @ 1 yr, Solution: ₱ 4,000 ₱ 0.08 0 9 mos 0 9 mos ₱ 5,000 ₱ 1.00 ₱ , (a) d = = = 0.20 or 20% ₱ , Another solution, using Eq. 11 d = 1 – (1 + i)-1 d = 1 – 0.80 = 0.20 or 20% ₱ , (b) i =. = = 0.25 or 25% ₱ ,. Another solution using Eq 12. i = = = 0.25 or 25%. ₱ , (c) i =.= = 0.3333 or 33.33% ₱ , 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 (Eq 13) where: PC = present cost of a commodity FC = Future cost of the same commodity f = annual inflation rate n = number of years In an inflationary economy, the buying power of money decreases as costs increase. Thus, F = ( ) (Eq. 14) Where F is the future worth, measured in today’s pesos, of a present amount P. If interest is being compounded at the same time that inflation is occurring, the future worth will be ( ) F = ( ) =𝑃 (Eq. 15) Example 13 An item presently costs ₱ 1,000. If inflation rate of 8% per year, what will be the cost of the item in two years? Given: PC = ₱ 1,000, f = 8%, n = 2yrs Required: F Solution: Using Eq. 13, we get FC = PC (1 + f)n = ₱ 1,000 (1 + 0.08)2 FC = ₱ 1,166.40 Example 14 An economy is experiencing, inflation at an annual rate of 8%. If this continues, what will ₱ 1,000 be worth two years from now, in terms of today’s pesos? Given: PC = ₱ 1,000, f = 8%, n = 2 yrs Required: FC Solution: Using Eq. 14, we get F = ( ) ₱ , = ( ). F = ₱ 857.34 Example 15 A man invested ₱ 10,000 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? Given: P = ₱ 10,000, i = 10%, f = 8% n = 5 yrs Required: F Solution: Using Eq. 15, we get F = 𝑃. = ₱ 10,000. F = ₱ 10,960.86 Annuities An annuity is series of equal payments occurring at equal period of time. Symbols and their meaning: P = value or sum of money at present F = value of money at some future time A = a series of periodic, equal amounts of money n = number ate per interest period Ordinary Annuity An ordinary annuity is one where the payments are made at the end of each period Finding P when A is Given Solving for P gives 1  1  1 n   1  i n  1 P =A   =A  n  (Eq. 16)  i   i 1  i   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 n interest periods', hence Equation 16 can be expressed as P = A (P/A, i%, n) When P is given and we need to find A gives  i  A= P  n  (Eq. 17) 1  1  i   The quantity in brackets is called the 'capital recovery factor' and is denoted by the functional symbol A/P, i%, n, read as 'A given P at i percent in n interest periods', hence Equation 17 can be expressed as A = P (A/P, i%, n) Finding F When A is Given Solving for F gives  1  i n  1 F =A   (Eq. 18)  i  The quantity in brackets is called the 'uniform series compounded amount factor' and is designated by the functional symbol F/A, i%, n, read as 'F given A at i percent in n interest periods', hence Equation 18 can be expressed as F = A (F/A, i%, n) When F is given and we need to find A gives  i  A= F   (Eq. 19)  1  i   1 n The quantity in brackets is called the 'sinking fund factor' and is denoted by the functional symbol A/F, i%, n, read as 'A given F at i percent in n interest periods', hence Equation 19 can be expressed as A = F (A/P, i%, n) Relation between A/P, i%, n and A/F, i%, n i  i 1  i   i 1  i  n n i  i =. 1  i n  1 i 1  i   1 n 1  i  n i i i = 1  i   1 n 1  1  i  n A/P, i%, n + 1 = A/F, i%, n Thus, sinking fund factor + 1 = capital recovery factor Example 16 What are the present worth and the accumulated amount of a 10-year annuity paying ₱10,000 at the end of each year, with interest at 15% compounded annually? Given: A = ₱ 10,000, i = 15%, n = 10 Required: (a) P  (b) F Solution: (a) Using Eq.16 P = A (P/A, i%, n) 1  1  i  n  P = A    i  1  1  0.1510  P = ₱10,000    0.15  P = ₱ 50,188 (b) Using Eq. 18 F = A (F/A, i%, n)  1  i n  1 F = A    i   1  0.15 10  1 F = ₱10,000    0.15  F = ₱ 203,037 Example 17 What is the present worth of ₱500 deposited at the end of every three months for 6 years if the interest rate is 12% compounded semiannually? Given: A = ₱ 500, r = 12%, n = 6yrs, m1 = 2 , m2= =4 Required: Present worth Solution: Solving for interest per quarter. (1 + i)4 – 1 = (1+ )2 – 1 1+i = (1.06)0.5 i = 0.0296 or 2.96% per quarter  P = A (P/A, 2.96%, 24) 1  1  0.0296 24  P = ₱ 500    0.0296  P = ₱ 500 (17.0087) P = ₱ 8,504.00 PROBLEMS Solve the following problems and show the cash flow and your step solution. 1. What is the annual rate of interest if ₱265.00 is earned in four months on an investment of ₱ 15,000.00 2. A loan of ₱ 2,000.00 is made for a period of 13 months, from January 1 to January 31 the following year, at a simple interest rate 20%. What future amount is due at the end of the loan period? 3. If you borrow money from your friend with simple interest of 12%, find the present worth of ₱ 20,000.00, which is due at the end of nine months. 4. Determine the exact simple interest on ₱ 5,000.00 for the period from Jan.15 to Nov.28, 1992, if the rate of interest is 22%. 5. A man wishes his son to receive ₱200,000.00 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? 6. By the conditions of a will, the sum of ₱ 25,000.00 is left to a girl to be held in trust by her guardian until it amounts to ₱ 45,000.00. When will the girl receive the money if the final is invested at 8% compounded quarterly? 7. At a certain interest rate compounded semiannually, ₱5,000.00 will amount to ₱20,000.00 after 10 years. What is the amount at the end of 15 years? 8. Jones Corporation barrowed ₱9,000.00 from Brown Corporation on Jan 1, 1980. Jones Corporation made a partial payment of ₱ 7,000.00 on Jan. 1, 1981. It was agreed that the balance of the loan would be amortized by two payments, one of Jan 1, 1982 and the other on Jan 1, 1983, the second being 50% large than the first. If the interest rate is 12%, what is the amount of each payment? 9. A woman burrowed ₱3,000.00 to paid after 1 ½ years with interest at 12% compounded semiannually and ₱ 5,000.00 to be after 3 years at 12% compounded monthly. What single payment must she pay after 3 ½ years at an interest rate of 16% compounded quarterly to settle the two obligations? 10. Mr. J. dela Cruz borrowed money from a bank. He received from the blank ₱1,342.00 and promise to repay ₱1,500 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." 11. A man deposits ₱50,000.00 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? 12. What is the future worth of ₱600 deposited at the end every month for 4 years if the interest rate is 12% compounded quarterly? 13. 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 ₱2,000.00 in a fund at 15% annual interest. How old will she be when the fund has an accumulated value of ₱1,000,000.00? 14. Mr. Reyes borrows ₱600,000.00 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 payments? FEEDBACK If you solve the problems with difficulties, you need to back read the topics on " Interest and Money – Time Relationships." There is a need to understand fully the present and future worth. You may o some cross reference to fully understand those terminologies. Never memorize the ready-made solution since memorization is not the recommended way for you to learn. Instead, you must understand the underlying basic principles involved. You need to solve more problems since constant practice makes your ability to solve correctly. SUMMARY Engineering economy is the analysis and evaluation of the aspects that will mark the economic achievement of engineering ventures to the end that endorsement can be made which will indemnify the best use of investment. Simple interest is calculated using the principal only, ignoring any interest that had been accrued in preceding periods. I = Pni. In calculations of compound interest, the interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous periods. F = P(1+ i)n An annuity is series of equal payments occurring at equal period of time. 1  1  1 n   1  i n  1 P =A   or F = A    i   i  SUGGESTED READINGS Deferred Annuity of time value of Money Annuity Due of time value of Money. POSTTEST Direction: Choose the letter of the correct answer. Do not write your answer on this manual. 1. In a cash-flow diagram:. a. Time 0 is considered to be the present b. Time 1 is considered to be the end of time period 1 c. A vertical arrow pointing up indicates a positive cash flow d. All of the above 2. In the cash-flow diagram shown in the given figure a. Equal deposits of Rs 3,000 per year (A) are made, starting now b. The rate of interest is 10% per year account c. The amount accumulated after the 7th deposit is to be computed  3. The interest calculated on the basis of 365 days a year, is known as: ` a. Interest b. Ordinary simple interest c. Exact simple interest d. None 4. If ‘P’ is principal amount, ‘I” is the rates of interest per annum and ‘n’ is the number of periods in years, the compound amount factor (CAF) is: a. (1 – i)n b. (1 – i)n/12 c. (1 – i/12)n d. (1 – ni) 5. In the cash flow diagram shown in the given figure a. The disbursement occurs at the end of year 2 b. The second disbursement occurs at the end of year 4 c. The first receipt occurs at the end of year 1 d. All of the above. 6. The PICE student chapter wants to deposit a single amount of money today to purchase a set of transit costing P50,000 five years from now. If the money can be deposited into an account which earns an interest at 10% annually, how much amount of money must be deposited by the PICE student chapter?. a. ₱ 29,524.50 b. ₱ 80,525.50 c. ₱ 31,046.06 d. ₱ 84,675.44 7. What is the length of time required for a P10,000 to P67,275 in value at an interest rate of 10% per annum? a. 10 years b. 15 years c. 20 years d. 25 years 8. For an interest rate of 1.5% per month, what should be its effective semiannual rate? a. 9.34% b. 8.43% c. 11.33% d. 12.43% 9. One of the engineering student deposits P100 per month into an account which pays interest at a rate of 6% per year compounded monthly. What is the value of the account after five years? a. ₱ 6,977.00 b. ₱ 133.82 c. ₱ 3,298.76 d. ₱ 9,876.23 10. What the interest rate of effective 12% per year compound monthly? a. 11.39% b. 9.31% c. 12.19% d. 8.67%. REFERENCES: Besa Villa, V.I. Engineering Economy Revised Edition. VB Publisher. Copyright 1989. Hartman, J.C., Engineering Economy and Decision-Making Process. Prentice-Hall Pearson. 2007 Sta Maria, Hipolito B., Engineering Economy, Second Edition, National Bookstore, 1993 Sullivan, William G; Wicks, Elin M; Koeling , Patrick; Engineering Economy 16th Edition, Pearson Publishing, 2013 http://www.csun.edu/~ghe59995/docs/Glossary%20of%20Concepts%20&%20Terms%20i n%20Engineering%20Economy.pdf https://www.investopedia.com/terms/l/law-of-supply-demand.asp