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ENGINEERING ECONOMICSIntroductionENGINEERING ECONOMICS - a social science concerned with man’s problem of issuing scarce resources to satisfy unlimited wants. - Comes from theancient Greek words( oikos , "house")+( nomos , "custom" or "law"), hence "rules of the house(hold) - uses mathematical formu...

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ENGINEERING ECONOMICS

The time value of money is the most import ant concept in eng ineering economy

Introduction ENGINEERING ECONOMICS - a social science concerned with man’s problem of issuing scarce resources to satisfy unlimited wants. - Comes from theancient Greek words(oikos, "house")+(nomos, "custom" or "law"), hence "rules of the house(hold) - uses mathematical formulas to account for the time value of money and to balance current and future revenues and costs - involves formulating, estimating, and evaluating the expected economic outcomes of alternatives designed to accomplish a defined purpose - a subset of economics for application to engineering projects Two Types of Economics a. Macroeconomics – the study of the entire entire system of economics (from Greek prefix “makros” meaning “large”+”economics ”) b. Microeconomics – study how the system affect one business or parts of the economic system. WHY ENGINEERING ECONOMY IS IMPORTANT TO ENGINEERS 1. Engineers design and create 2. Designing involves economic decisions 3. Engineers must be able to incorporate economic analysis into their creative efforts 4. Often engineers must select and implement from multiple alternatives 5. Understanding and applying time value of money, economic equivalence, and cost estimation are vital for engineers 6. A proper economic analysis for selection and execution is a fundamental task of engineering TIME VALUE OF MONEY (TVM) - explains the change in the amount of money over time for funds owed by or owned by a corporation (or individual) -

Corporate investments are expected to earn a return Investment involves money Money has a ‘time value’

STEPS IN AN ENGINERING ECONOMIC STUDY

Some Important Terms And Concepts NECESSITIES VERSUS LUXURIES ▪

▪

Necessities - products or services that are required to support human life and activities that will be purchased in somewhat the same quantity even though the prices vary considerably. Luxuries - products or services that are desired by humans and will be purchased if money is available after the required necessities have been obtained.

GOODS AND SERVICES ▪ Goods - anything that anyone wants or needs ▪ Services - performance of any duties or work for another, helpful of professional activity DIFFERENT KINDS OF GOODS A. CONSUMER GOODS - such as food and clothing that satisfy human wants or needs B. PRODUCER GOODS - such as raw materials tools, used to make consumer goods C. CAPITAL GOODS - machinery used in the production of commodities or producer goods. Include educational, health, communication, transportations and social services. Law of Supply - States that at higher prices, producers are willing to offer more products for sale than at lower prices - States that the supply increases as prices increase and decreases as price decreases - States that those already in business will try to increase productions as a way of increasing profits Law of Demand - States that people will buy more of a product at a lower price than at a higher price, if nothing changes - States that at a lower price, more people can afford to buy more goods and more of an item more frequently, than they can at a higher price. - States that at a lower prices, people tend to buy some goods as a substitute for others more expensive.

TYPES OF DEMAND 1. Elastic Demand – exist when there is a greater change in quantity demanded as a response to a change in price. 2. Inelastic Demand – exist when there is a lesser change in quantity demanded in response to change in price 3. Unitary – exist when there is an equal change in price and in quantity demanded (increase or decrease) FACTORS THAT INFLUENCES DEMAND 1. 2. 3. 4. 5.

Income Population Taste and preference Price expectation Price of related goods

FACTORS THAT INFLUENCE SUPPLY 1. 2. 3. 4. 5. 6. 7.

Price of goods Cost of production Availability of resources Number of producers and sellers Technological advancement Taxes Subsidies

Market – the place where the vendors and buyers meet to transact Market Structures

INTEREST

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Cash Inflows (CI)– Revenues (R), receipts, incomes, savings generated by projects and activities that flow in. Plus sign used

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Cash Outflows (CO) – Disbursements (D), costs, expenses, taxes caused by projects and activities that flow out. Minus sign used

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Net Cash Flow (NCF) for each time period:

TIME VALUE OF MONEY o The “time value” of money is the most important concept in engineering economy o All firms make use of investment of funds o Investments are expected to earn a return o Money possesses a “time value”

NCF = CI-C0 = R – D

EQUIVALENCE Different sums of money at different times may be equal in economic value

What a typical cash flow diagram might look like:

SYMBOLS AND NOTATIONS •

t

time, usually in periods such as years or months

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P

value or amount of money at a time t designated as present or time 0

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F

value or amount of money at some future time, such as at t = n periods in the future

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A

series of consecutive, equal, end-ofperiod amounts of money

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N

number of interest periods; years, months

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i

interest rate or rate of return per time period; percent per year or month

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I

amount earned from borrowed money

CASH FLOW DIAGRAMS Depict the timing and amount of expenses (negative, downward) and revenues ( positive, upward)for engineering projects.

INTEREST RATE AND RATE OF RETURN •

The manifestation of the time value of money

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the amount of money paid for the use of borrowed capital or the income produced by money, which has been loaned.

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Difference between an ending amount of money and a beginning amount of money

𝑰= 𝑭−𝑷 Interest earned over a period of time is expressed as a percentage of the original amount o Borrower’s Perspective – interest rate paid o Lender or Investor’s Perspective – rate of return earned

TYPES OF INTEREST Simple Interest - Calculated using the principal only, ignoring any interest that had been accrued in preceding period

𝑰 = 𝑷𝒊𝒏 𝑭 = 𝑷+𝑰 ▪

▪

Ordinary Simple Interest - Under ordinary simple interest, it is assumed that each month contains 30 days and consequently, each year has 360 days Exact Simple Interest - Under exact simple interest, the exact days per month is used. There are 365 days per year on ordinary year and 366 days every fourth year called leap year

Sample Problem 1. $100,000 lent for 3 years at simple i = 10% per year. What is repayment after 3 years? Given: P=$100,000 n=3 years i=10% CA Solution: 𝑰=𝑷𝒊𝒏 I= 100,000(3)(0.10) = $30,000 Total due = $100,000 + $30,000 Ans. $130,000 2. A loan of P50,000 is made for a period of 13 months from April 1 to April 30 of the following year, at a simple interest rate of 20%. What future amount is due at the end of the loan period? Ans. P60, 833.33 3. What is the principal amount if the amount of interest at the end of 2 ½ year is 450 for a simple interest rate of 6% per annum? Ans. P3, 000 4. What will be the future worth of money after 12 months, if the sum of P25,000 is invested today at simple interest rate of 1% per month? Ans: P28,000

5. Determine the exact simple interest of P25000 for the period from Dec 27, 2001 to March 23, 2003, if the rate of interest is 10%? Ans. P3, 095.8 6. Determine the exact simple interest of 25000 for the period from Dec 27, 2001 to March 23, 2004, if the rate of interest is 10%? Ans. Php 36,161.20 Compound Interest - The interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous period. -

“ the interest on top of interest ”. The quantity (1 + 𝑖)𝑛 is commonly called the single payment compound amount factor

𝑭 = 𝑷(𝟏 + 𝒊)𝒏 Sample Problem 1. $100,000 lent for 3 years at i = 10% per year compounded. What is repayment after 3 years? Given: P=$100,000 n=3 years i=10% CA Interest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: T1 = 100,000 + 10,000 = $110,000 Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: T2 = 110,000 + 11,000 = $121,000 Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: T3 = 121,000 + 12,100 = $133,100 Ans: $133,100 2. What rate of interest compounded annually must be received if an investment of 5400 made now will result in a receipt of 7200 in 5 years? Ans. 5.92% 3. What amount will be accumulated by P4100 in 10 years at 6% compounded annually? Ans. P7342.48 4. How long it will take for the money to triple itself if invested at 8% compounded annually? Ans. 14.27 years

RATES OF INTEREST

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Effective Rate Of Interest - the actual or exact rate of interest on the principal during one year

𝒊 𝒆𝒇𝒇 = 𝑭 − 𝟏 𝒊 𝒆𝒇𝒇 = ( 𝟏 + 𝒊) 𝒎 − 𝟏

•

-

Nominal Rate Of Interest - specifies the rate of interest and a number of interest period in one year. Note: m = 1 = compounded annually (CA) m = 2 = compounded semi-annually (CSA) m = 4= compounded quarterly (CQ) m = 12 = compounded monthly (CM)

𝒓 𝒊= 𝒎 Where: i= rate of interest / interest period r = nominal interest period m= number of compounding periods

Where: 𝑖 𝑒𝑓𝑓 = effective rate F = future worth rate i = rate of interest/ interest period m = number of compounding periods Sample Problem: 1. If 1000 becomes 5734 after 15 years, when invested at an unknown rate of interest compounded semiannually, determine the unknown nominal rate and corresponding effective rate. Ans. i=12.36% ; r=12% 2. What is the equivalent nominal rate compounded monthly of 15% nominal rate compounded semiannually?

CONTINOUS COMPOUNDING INTEREST The interest continuously accumulates

𝑭 = 𝑷𝒆𝒊𝒏 POINTS TO REMEMBER •

Interest rates are stated different ways must know how to get effective rates

•

For single amounts, make sure units on i and n are the same

Sample Problem 1. What effective annual interest rate corresponds to the following situations: a. Nominal interest rate of 10% compounded semiannually b. Nominal interest rate of 6% compounded monthly c. Nominal interest rate of 8% compounded quarterly Ans. 10.25%; 6.17%; 8.24%

𝒊𝒆𝒇𝒇 =

𝒓 𝒎 𝒆

−𝟏

Where: F = future worth i = rate of interest/ interest period n = number of compounding periods Sample Problem: 1. Philip invested 100 on bank. The bank offers 5% interest compounded continuously in a savings account. Determine (a) how long will it require for him to earn $5 (b) the equivalent simple interest rate for 1 year bank? Ans. 0.9758 year; 5.127%

EQUATION OF VALUE - obtained by setting the sum of values on a certain comparison or local date of one set of obligations equal to the sum of the values on the same date of another set of obligations Sample Problem: 1. Jay wishes his son, Jayson to receive 1000000 twenty years from now. What amount should he invest now, if it will earn interest of 12% compounded annually during the first 5 years and 10% compounded monthly for the remaining years. Ans. P5 = P224, 521.34; P0 = P127, 399.44

2. A machine worth P50,000 is expected to last for 3 years. During its operation, a maintenance cost of P1,000 is needed after the 1 st year of operation and P2,000 at the end of the 2nd year. What present amount is required to operate the machine, if money is worth 16% compounded quarterly? Ans. P52, 316.18 ANNUITY – a series of equal payments occurring at equal periods of time

2. Find the present worth of a future payment of P300,000 to be made in 10 years with an interest rate of 10% compounded annually. What will be the amount if it will be paid on the 15th year? Ans. P115, 662.99; P483,153.01 DISCRETE PAYMENTS - The solution in discrete payments or number of transactions occurring at different periods is taking each transaction to the base year and equating each value.

Annuities are established for the following purposes. 1. To extinguish a present debt by a series of equal payments made at equal time intervals, this is also known as amortization 2. To accumulate a required amount in the future by depositing equal amounts at equal intervals in time, such deposits are referred to as sinking funds 3. To replace a future lump-sum payment with equal periodic payments TYPES OF ANNUITIES • • • •

Ordinary Annuity Deferred Annuity Annuity Due Perpetuity

1. ORDINARY ANNUITY - Payments are made at the end of each period Sample Problem: 1. Acosta holdings borrowed P9000 from smith corporation on January 1, 1998 and P12000 on January 1, 2000. Acosta holdings made a partial payment of P7000 on January 1, 2001. It was agreed that the balance of the loan would be amortized by two payments. One on January 1, 2002 and one January 1, 2003. The second being 50% larger than the first. If the interest rate is 12%, what is the amount of each payment? Ans. P9137.18; P13 705.77

2. DEFERRED ANNUITY - The first payment is made several periods after the beginning of the annuity

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Cash flow occurs in consecutive interest periods Cash flow amount is same in each interest period P is one period ahead of first A value Last cash flow occurs in same period as F

Sample Problem: 1. A chemical engineer believes that by modifying the structure of a certain water treatment polymer, his company would earn an extra 5000 per year.A t an interest rate of 10% per year, how much could the company afford to spend now just to break even over a 5 year project period? Given: A=5000 i=10% CA n=5 years Req’d: P Solution:

Ans = $ 18, 953.93 2. What are the present worth and the accumulated amount of a 10 year annuity paying P100,000 at the end of each year with interest as 15% compounded semiannually? Ans. P=P491,300.79 F=P2, 086,972.6 3. A chemical engineer wishes to set –up a special fund by making a uniform semi-annual end- of period deposits for 20 years. The fund is to provide a $100,000 at the end of each of the last five years of a 20 year period. If interest is 8% compounded semi-annually, what is the required semi-annual deposit to be made? Ans. A = $6,193.99

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Cash flow occurs in consecutive interest periods Cash flow amount is same in each interest period The first A is deferred by k periods Last cash flow occurs in same period as F

Sample Problem: 1. What lumpsum amount is equal to a yearly deposit worth $8000 from years 3 through 10 at an interest rate of 10% per year? Given: A= $8000 n= 8 (year 3-10) i= 10% CA Reqd: P Solution:

Ans = $35, 272.24 2. If 10,000 is deposited each year for 9 years, how much annuity can a person get annually from the bank for 8 years starting 1 year after the 9th deposit is made. Cost of money is 14% Ans. A = P34, 675

3. A debt of P40,000 whose interest rate is 15% compounded semi-annually, is to be discharged by a series of 10 semiannual payment, the first payment is to be made 6 months after consummation of the loan. The first 6 payments will be P6,000 each while the remaining 4 payments will be equal and of such amount that the final payment will liquidate the debt. What is the amount of the last 4 payments? Ans: A = P5,454 3. ANNUITY DUE - Payments are made at the beginning of each period

2. A farmer bought a tractor costing P12,000 if paid in cash. The tractor may be purchased by installment to be paid within 5 years. Money is worth 8% compounded annually. Determine the amount of each annual payment if all payments are made: a. At the end of each year for 5 years b. At the beginning of each year for 5 years Ans. a. A = P3, 005.48; b. A = P2, 782.85 3. Yosef bought a lot worth P1, 000,000 and promised to pay in equal amount every month for 3 years at 5% compounded monthly rate. He missed 5 payments after paying his first 12 payments. What single amount must he pay by the time his 18th payment is due to extinguish all his debts? Ans. P699, 781.7 4. PERPETUITY - periodic payments continue indefinitely

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Cash flow occurs in consecutive interest periods Cash flow amount is same in each interest period P occurs on the same period of first A value F occurs one period after the last A value

Sample Problem: 1. What is the current and future value of a $50 payment to be made at the beginning of each year, for three years if the prevailing rate of interest is 7% CA? Given: A= $50 n= 3 i= 7% CA Reqd: P & F Solution:

Ans. P=$140.40; F =$172

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Cash flow occurs in consecutive interest periods Cash flow amount is same in each interest period P may occur one period ahead of first A value P may occur several periods ahead of first A value

Sample Problem: 1. ABC Corporation pays an annual dividend worth $3 estimates that they will pay the dividends indefinitely. How much are the investors willing to pay for the dividend with a required rate of return of 6%? Given: A= $3 i= 6% CA Reqd: P

Solution: The figure shows the cash flow diagram for arithmetic gradient increasing or decreasing by a constant amount G in every period. The G is the uniform gradient amount

Ans = $ 50 2. What amount of money invested today at 15% interest can provide the following scholarships; P30,000 at the end of each year for 6 years; P40,000 for the next 6 years and P50,000 thereafter? Ans. P241,277

Gradient Series as a Composite Series

GRADIENT •

Engineering economy problems sometimes involve a series of receipts or disbursements that increase or decrease in each succeeding period by varying amounts

• Gradient is a revenue or disbursement that grows or decrease by a fix amount (G) each period by a fix amount or rate •

Four possibilities for gradient o Change in every succeeding period is constant. And there is no disbursement on the first year o A>0 and G>0 means positive and increasing o A>0 and G...