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Unformatted text preview: ENGINEERING ECONOMICS ECONOMICS
Prof. Dr. Cengiz Kahraman Prof. ITU Industrial Engineering Department Department INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION DEFINITION Engineering Economy is a collection of Engineering mathematical techniques that simplify economic comparisons when making moderate decisions. comparisons Engineering Economy is the art and the science of selecting the best course of action at any point in time. Every moment is a decision point, an opportunity for staying with the status quo or choosing a different course of action. Status quo alternative=Asis alternative=Donothing alternative: the current approach is nothing maintained. maintained. INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION
A Short History of Engineering Economy The Economic Theory of the Location of Railways, written by Arthur M. written Wellington in 1887, pioneered an engineering interest in economic evaluations. Wellington, a civil engineer, reasoned that the capitalized cost method of analysis should utilized in selecting the preferred lenghts of rail lines or curvatures of the lines. lines In the 1920s, C.L. Fish and O.B. Goldman looked at investments in In engineered structures from the perspective of actuarial mathematics. Fish formulated an investment model related to the bond market. In Financial Engineering, Goldman proposed a compoundinterest procedure for Engineering Goldman determining comparative values. determining The confines of classical engineering economy were staked out in 1930 by The Eugene L. Grant in Principles of Engineering Economy. Grant discussed Principles Grant the importance of judgment factors and shortterm investment evaluation as well as conventional comparisons of longrun investments in capital goods based on compoundinterest calculations. His many contributions resulted in the recognation that “Eugene L. Grant can truthfully be called the father of engineering economy.” engineering INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Classifying Investment Alternatives Dependent and Independent Investments: Assume there are two investment alternatives: A and B. The first investment proposal will be said to be The economically independent of the second if the cash flows economically expected from the first investment would be the same regardless of whether the second investment were accepted or rejected. accepted If the cash flows associated with the first investment are If affected by the decision to accept or reject the second investment , the first investment is said to be economically dependent on the second. economically INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Classifying Investment Alternatives Dependent and Independent Investments: If a decision to undertake the second investment will If increase the benefits expected from the first (or decrease the costs of undertaking the first without changing benefits), the second investment is said to be a complement of the first. complement If the decision to undertake the second investment If will decrease the benefits expected from the first (or increase the costs of undertaking the first without changing the benefits), the second is said to be a substitute for the first. substitute for INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Classifying Investment Alternatives Dependent and Independent Investments: In the extreme case where the potential benefits to be derived In from the first investment will completely disappear if the second investment is accepted, or where it is technically impossible to undertake the first when the second has been accepted, the two investments are said to be mutually exclusive. exclusive It is also possible to define an extreme case for investments It that are complements. Suppose that the second investment is impossible (technologically) or would result in no benefits whatsoever if the first investment were not accepted. Then the first investment can be said to be a prerequisite of the second. prerequisite INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Classifying Investment Alternatives Dependent and Independent Investments:
Graphic Representation for The Possible Relationships Between Graphic Investments Investments Prerequ. Inde. Inv.s Weak Comp. Weak .Subs Mu. Exc. Inv.s Strong Subs. Strong Comp. . INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Examples: Give your decision about the investment category for the following alternatives: category To build a toll bridge and operate a ferry across adjacent To points on a river. points The building of a new factory as one investment and the The purchase of an airconditioning unit for the factory as the second investment. second The XYZ Company is considering installing a solar The heating system in its manufacturing plant for space and hot water heating. There are three possible solar system alternatives in addition to the donothing alternative. alternatives INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION CATEGORY ECONOMIC CRITERION Fixed input: Maximize the benefits or The amount of money or other outputs other input resources are fixed. Fixed output Minimize the costs or other inputs Neither input nor output fixed Maximize (benefits costs), or maximize profit INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Examples: make your decision about the category and the economic criterion of each situation in the following: criterion a) A project engineer has a budget of $350,000 to overhaul a portion of a a) petrolium refinery. petrolium b) A civil engineering firm has been given the job to survey a tract of land b) and prepare a “Record of Survey” map. and c) You have $100 to buy clothes for the start of school. d) An outomobile battery is needed. They are available at different prices, d) and although each will provide the energy to start the vehicle, their useful lives are different. lives e) You wish to purchase a new car with no optional equipment. f) A book publisher is about to set the list price (retail price) on a textbook. If f) they choose a low list price, they plan on less advertising than if they select a higher list price. The amount of advertising will affect the number of copies sold. g) At an auction of antiques, a bidder for a particular statue would be trying g) to......................... to......................... INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Nature of Problems Simple Problems These are defined as decisions that have all important information These readily available and it is easy to understand the information. Intermediate Problems This level of decision is defined as not having all important This information readily available and the most important issue is probably money. Most of the information is findable or could be calculated as well as finite. Complex Problems They represent a mixture of economic, political, and humanistic They elements.The third category of problems is defined as not having much of the important information or the important information is intangible or hard to determine. Usually there are many variables that are very complex and have many and variable potential impacts. INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Examples: Decide the categories of the following problems: a) The selection between a manual or a semiautomatic machine. b) The preparation of the annual budget of Turkey c) Would it be better to buy an otomobile with a diesel engine or a c) gasoline engine? gasoline d) One of the people you might marry has a job that pays very little d) money, while another one has a professional job with an excellent salary. Which one should you marry? salary. e) Decide on the components of a breakfast. f) Ahmet backed his car into a tree, damaging the fender. He has f) automobile insurancethat will pay for the fender repair. But if he files a claim for payment, they may change his “good driver” rating downward , and charge him more for car insurance in the future. downward g) How many people smoke cigarettes although they realize that g) smoking may be harmful to health. smoking INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Basic Terms Time Value of Money Time Equivalence Equivalence Interest Simple Interest Compound Interest Cash Flow Cash Cash Flow Diagram Economic Life Economic Before Tax Cash Flow Before Before Tax Rate of Return Before Purchase Cost Purchase Salvage Value Salvage Disposal Costs Disposal INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS The Time Value of Money The change in the value of money over time as a The result of its earning power is called the time value of money. value Since money has the ability to earn interest, its Since value increases with time. Since money increases in value as we move Since from the present to the future, it must decrease in value as we move from the future to the present. present. INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Equivalence Different sums of money at different times are Different equal in economic value. equal When we are indifferent whether we have a When quantity of money now or the assurance of some other sum of money in the future , or series of future sums of money, we say that the present sum of money is equivalent to the future sum or equivalent series of future sums. series INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Interest A fee that is charged for the use of fee someone else’s money. The size of the fee will depend upon the total amount of money borrowed and the length of the time over which it is borrowed. time Interest is the cost of using capital. Interest is the rent paid by a borrower to a Interest lender. lender. INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Interest Example: An engineer wishes to borrow Example: $20,000 in order to start his own business. A bank will lend him the money provided he agrees to pay $920 per month for two years. How much interest is he being charged? How The total amount of money that will be paid to The the bank is 24x$920=$22,080. Since the original loan is only $20,000, the amount of interest is $22,080  $20,000 = $ 2,080. $22,080 INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Interest Whenever money is borrowed or invested, one party Whenever acts as the lender and another party as the borrower. Lender is the owner of the money, and the borrower pays interest to the lender for the use of the lender’s money. money. Example: When money is deposited in a savings When account, the depositor is the lender and the bank is the borrower. The bank therefore pays interest for the use of the depositor’s money. the The bank will then play the role of the lender, by loaning The this money to another borrower , at a higher interest rate. this INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Interest Rate If a given amount of money is borrowed for a specified period of If time, a certain percentage of the money is charged as interest. This percentage is called the interest rate. interest Percentage of the original amount per unit of time. Percentage Example: Example: (a) A student deposits $1,000 in a savings account that pays (a) interest at the rate of 6% per year. How much money will the student have after one year? (Answer: $1,060) have (b) An investor makes a loan of $5,000, to be repaid in one lump (b) sum at the end of one year.. What annual interest rate corresponds to a lumpsum payment of $5,425? (Answer: 8.5%) to INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Interest Period Time unit used to express rate. Time Simple Interest Simple interest is defined as a fixed percentage of the principal (the Simple amount of money borrowed), multiplied by the life of the loan. I = Pni Pni where I = total amount of simple interest n = life of the loan life i = interest rate (expressed as a decimal) P=principal It is understood that n and i refer to the same unit of time. It Interest = Principal * Number of Periods * Interest Rate At the end of the loan period , the total amount (the principal + the At accumulated interest) to be repaid can be expressed as F = P + I = P (1 + ni ) INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Simple Interest Example: A student borrows $3,000 from his uncle in order to finish school. His uncle agrees to charge him simple interest at the rate of 5.5% Per year. Suppose the student waits two years and then repays th e entire loan. How much will he have to repay? (Answer: $3,330) loan. Example: What is the annual rate of simple interest if $265 is earned in four months on an investment of $15,000?(Answer:5.3%) $15,000?(Answer:5.3%) Example: Determine the principal that would have to be invested to provide $200 of simple interest income at the end of two years if the annual interest rate is 9%? end (Answer:$1111.119) INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS INTRODUCTION Compound Interest When interest is compounded, the total time When period is subdivided into several interest periods (e.g., one year, three months, one month). Interest is credited at the end of each interest period. and is allowed to accumulate from one interest period to the next. interest For the first interest period, the interest is For determined as determined and the total amount accumulated is I = iP 1 F = P + I = P + iP = P (1 + i ) 1 1 INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS For For the second interest period, the interest is determined as is I = iF = i (1 + i ) P 2 1 and the total amount accumulated is F = P + I + I = P + iP + i (1 + i ) P = P (1 + i ) 2 2 12 INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS For the third interest period, 2P I = i (1 + i ) 3
F = P (1 + i ) 3 3 and so on. In general, if there are n interest periods, we and have n Fn = P (1 + i ) Fn: the total amount of money accumulated, increases the exponentially with n, the time measured in interest the periods. periods. INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Compound interest Example: A student deposits $1,000 in a savings account that pays interest at the rate of 6% per year, compounded annually. If all of the money is allowed to accumulate, how much will the student have after 12 years? Compare this with the amount that would have accumulated if simple interest had been paid. (Answer: $2,012.20 and $1,720.00) $2,012.20 Example: Compare the interest earned from an investment of $1,000 for 15 years at 10% per annum simple interest, with the amount of interest that could be arned if these funds were invested for 15 years at 10% per year, compounded annually. (Answer: $1,500 and $3,177.25) $3,177.25) INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS compounded annually. What is the minimum amount of money that would have to be invested for a twoyear period in order to earn $300 in interest? (Answer: $1,428.57) order Example: How long would it take for an investor to double his money at 10% interest per year, compounded annually? (Answer: 7.27 years8 years) (Answer: Example : Suppose that a man lends $1,000 for four years at 12% per year simple interest. At the end of the four years, he invests the entire amount which he then has for 10 years at 8% interest per year, compounded annually. How much money will he have at the end of 14year period? (Answer: $3,195.21) $3,195.21) Compound interest Example: Suppose that the interest rate is 10% per year, INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Rule of 72 Estimated n = 72 divided by i Estimated Estimated Interest Rate = 72 divided by Estimated n The time required for an initial single The amount to double in size with compound interest is approximately equal to 72 divided by the rate of return value (in percent): percent): 72 Estimated n = i INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Rule of 72 The compound rate The i in percent required for money to double in a specified period of time n can be estimated by dividing 72 by the specified n value: 72 Estimated i = n INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Rule of 72 : to double the amount Rate of Return, %per year 1 2 5 10 20 40 Ruleof72 Actual Estimate Years 72 36 14.4 7.2 3.6 1.8 70 35.3 14.3 7.5 3.9 2.0 INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Symbols  Notation P = $ at time 0 F = $ at a future time A = $ at each period End of period, Uniform series amounts, Equal amounts of money G = $ at each period End of period A uniform gradient is a cashflow series which either increases or uniform decreases uniformly. The cash flow, whether income or disbursements, changes by the same arithmetic amount each interest period. The amount of the increase or decrease is the gradient. gradient. n = Number of interest periods i = Interest rate per period INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Example: Which symbols represent the numerical values given in the following problems? values If an automobile manufacturer predicts that the cost of If maintaining a robot will increase by $500 per year until the machine is retired. the A company expects income to decrease by $3,000 per company year for the next 5 years. year How much money should you be willing to pay now for How an investment that is guaranteed to yield $600 per year for 9 years starting next year, at an interest rate of 16% per year? per How much money must Mehmet deposit every year How starting 1 year from now at 5.5% per year in order to accumulate $6,000 seven years from now? accumulate INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Cash Flow A net cash flow is the difference between total cash receipts (inflows) and total cash disbursements (outflows) for a given period of time. period INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Cash Flow Diagrams The individual cash flows are represented as vertical The arrows along a horizontal time scale. Positive cash flows are represented by upward–pointing Positive arrows, and negative cash flows by downward –pointing arrows. arrows. The length of an arrow is proportional to the magnitude The of the corresponding cash flow. of Each cash flow is assumed to occur at the end of the Each respective time period. respective Arrow direction Up for inflows of cash Up Down for outflows of cash Down INTRODUCTION TO ENGINEERING ECONOMIC ANALYSIS ANALYSIS Example: A company plans to invest $500,000 to manufacture a new product. The sale of this product is expected to provide a net income of $70,000 a year for 10 years. Draw the cash flow diagram for this proposed project. project. Example: Suppose that you have a savings plan covering the next ten years, according to which you put aside $600 today, $500 at the end of every other year for the next five years, and $400 at the end of each for remaining five years. As part of this plan , you expect to withdraw $300 at the end of every year for the first 3 years, and $350 at the end of every other year thereafter. thereafter. Tabulate your cash flows. Draw your cash flow diagram. Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations 1. Single –Payment, CompoundAmount Factor Single payment formula to calculate the future worth of a present amount is The ratio is called the singlepayment, compoundamount factor, but it is usually referred to as the F P factor. In functional notation, the single payment, compound amount formula is n F = P (1 + i ) n F P = 1+ i F = P ( F P , i%, n) Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations Example: A student deposits $1,000 in a savings account that pays interest at the rate of 6% per year, compounded annually.If all of the money is allowed to accumulate, how much money will the student have after 12 years? money Solution: Example: A student deposits $1,000 in a savings account that pays interest at the rate of 6% per year, compounded annually.If all of the money is allowed to accumulate, how much money will the student have after 12 years? 12 Solution: F = P ( F P , i%, n) = $1,000( F P ,6%,12) = $1,000( 2,0122) = $2,012.20 Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations 2. 2. SinglePayment, PresentWorth Factor Factor Single payment formula to calculate the Single present worth of a future amount is present −n P = F (1 + i ) The The singlepayment, presentworth factor is the reciprocal of the singlepayment, compound amount factor: compound −1 = (1 + i ) − n P F = ( F P) Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations The functional notation for this quantity is And the single payment, present worth formula in And functional notation is functional ( P F , i%, n) Example: A certain sum of money will be deposited in a savings account that pays interest at the rate of 6% per year, compounded annually. If all of the money is allowed to accumulate, how much must be deposited initially so that $5,000 will have accumulated after 10 years? years? P = F P F ,i%,n Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations Solution: P= F P F,i%,n = $5,000 P F,%6,10 = $5,000(0.5 584) = $2,792.00 Formulas for Single Payment, Uniform Payments, and Uniform Gradient Payments and Their Functional Notations 3. UniformSeries, CompoundAmount 3. Factor Factor Let equal amounts of money, A, be Let deposited in a savings account at the end of each year over a period of n years. If the money earns interest at a rate i, compounded annually, how much money will have accumulated after n years? years? 3. UniformSeries, CompoundAmount 3. Factor Factor The first year’s deposit will have increased The in value to in n−1 F = A(1 + i ) 1 Similarly, the second year’s deposit will Similarly, have increased in value to have n− 2 F = A(1 + i ) 2 3. UniformSeries, CompoundAmount Factor Factor and and so on. The total amount accumulated will thus be the sum of geometric progression: progression: F =F + F +...+ Fn 1 2
F = A 1+ i n−1 + A 1+ i n− 2 + ...+ A n−1 + (1 + i ) n− 2 + ... + 1 = A ( 1 + i ) (1 + i ) n − 1 = A i 3. UniformSeries, CompoundAmount Factor 3. The ratio 1+ − i 1 F A= i n is called the uniformseries, compound amount factor. amount In functional notation, the uniformseries, In compoundamount factor is compoundamount F A = ( F A, i%, n) 3. UniformSeries, CompoundAmount Factor 3. and the uniformseries, compount amount and formula is formula F = A F A,i%,n Example: A student plans to deposit $600 each Example: year in a savings account, over a period of 10 years. If the bank pays 6% per year, compounded annually, how much money will have accumulated at the end of the 10year period? period? F = A( F A, i%, n) = $600( F A,6%,10) = $600(13.1808) = $7,908.48 4. UniformSeries, SinkingFund Factor 4. The uniformseries, sinkingfund factor is The the reciprocal of the uniformseries, compoundamount factor: compoundamount
−1 = A F = ( F A) (1 + i ) n − 1 i The functional notation of the uniform series, sinkingfund factor is ( A F , i%, n) . 4. UniformSeries, SinkingFund Factor 4. The uniformseries, sinking fund formula in functional The notation is notation A = F ( A F , i%, n) Example: Suppose that a fixed sum of money, A, will be Suppose will deposited in a savings account at the end of each year for 20 years. If the bank pays 6% Per year, compounded annually, find A such that a total of $50,000 will be accumulated at the end of the 20year period. accumulated Solution: A = F ( A F , i%, n) = $50,000( A F ,6%,20) = $50,000( 0.02718) = $1,359 5. UniformSeries, Capital5. Recovery Factor Suppose Suppose that a given sum of money , P, iis s deposited in a savings account where it earns interest at a rate i per year, compounded annually. At the end of each year a fixed amount, A, iis withdrawn. How s large should A be so that the bank account will just be depleted at the end of n years? years? 5. UniformSeries, Capital5. Recovery Factor
A 0 n P ………… A Previously defined factors to solve this Previously problem can be used, since problem A = P ( F P )( A F ) 5. UniformSeries, Capital5. Recovery Factor
n A = P (1 + i ) The ratio is i (1 + i ) n n − 1 = P (1 + i ) n − 1 (1 + i ) i (1 + i ) n i AP= n −1 (1 + i ) The uniformseries, capitalrecovery The formula in functional notation is formula A = P ( A P , i%, n) 5. UniformSeries, Capital5. Recovery Factor Example: Example: An engineer who is about to retire has accumulated $50,000 in a savings account that pays 6% Per year, compounded annually. Suppose that the engineer wishes to withdraw a fixed sum of money at the end of each year for 10 years. What is the maximum amount that can be withdrawn? can 5. UniformSeries, Capital5. Recovery Factor Solution:
A = P ( A P , i%, n) = $50,000( A P ,6%,10) = $50,000( 0,1359) = $6,795 6. UniformSeries, Present – 6. Worth Factor The uniformseries, presentworth factor is The the reciprocal of the uniformseries, capitalrecovery factor: capitalrecovery (1 + i ) n − 1 P A = ( A P ) −1 = (1 + i ) n i The functional notation for the uniformseries, presentworth factor is P A = ( P A, i%, n) 6. UniformSeries, Present – 6. Worth Factor The uniformseries, presentworth formula in functional The notation is notation Example: An engineer who is planning his retirement has decided An that he will have to withdraw $10,000 from his savings account at the end of each year. How much money must the engineer havein the bank at the start of his retirement, if his money earns 6% Per year, compounded annually, and he is planning a 12 year retirement (i.e. 12 annual withdrawals)? retirement Solution:
P = A( P A , i%, n) = $10,000( P A,6%,12) = $10,000( 8.3839) = $83,839 P = A( P A, i%, n) 7. Gradient Series Factor 7.
A+(n1)G A+2G A+G A Time P Cash flows of this form may be resolved Cash into two components: into 7. Gradient Series Factor 7.
(n1)G A A A A A G + 2G P=P +P 1 2 7. Gradient Series Factor 7. (1 + i ) n − 1 P = A n i = A( P A, i%, n) 1 (1 + i ) G (1 + i ) n − 1 1 P= − n n = G ( P G , i%, n) 2 i i (1 + i ) The gradient series factor is :
1 n AG= − n − 1 = ( A G , i%, n) i (1 + i ) 7. Gradient Series Factor 7. Uniform Gradient Increases or decreases by a uniform Increases amount each interest period. Increase or decrease is the gradient Increase amount. amount. Year one is the Uniform Series Amount. Not a part of the gradient. G = Gradient amount 7. Gradient Series Factor 7. Example: Example: An engineer is planning for a 15year retirement. An In order to supplement his pension and offset the anticipated effects of inflation, he intends to withdraw $5,000 at the end of the first year, and to increase the withdrawal by $1,000 at the end each successive year. How much money must the engineer have in his savings account at the start of his retirement, if money earns 6% per year, compounded annually? year, P = $106,116.59 7. Gradient Series Factor 7. Example: Example: How How much money must initially be deposited in a savings account paying 5% Per year, compounded annually, to provide for ten annual withdrawals that start at $6,000 and decrease by $500 each year? (Answer: $30,504.19) each 8. Geometric Series Factor 8. Oftentimes, Oftentimes, cash flows change by a constant percentage in consecutive paymnet periods. This type of cash flow, called a geometric series, is shown in general form in the following figure where g represents the geometric growth rate in decimal form. decimal 8. Geometric Series Factor 8.
C1(1+g)n1 C1 C1(1+g) 1 P 2 n F 1 − (1 + i ) − n (1 + g ) n P =C i 1 i−g (1 + i ) n − (1 + g ) n F =C i≠g 1 i−g ≠g nC nC 1= 1 P= 1+ g 1+ i i=g F = nC ( 1 + i ) n− 1 = nC ( 1 + g ) n − 1 1 1 i=g 8. Geometric Series Factor 8. Example: Example: Someone plans to deposit $2,000 in a money Someone market account starting 1 year from now and wants to increase annual deposits by 20% each year for the following 6 years. Assuming that deposits earn 9% annually, determine what equalpayment annuity would accumulate the same amount over the 7year period. (A=$3,458.53) (A=$3,458.53) Standard Notations Standard SPPWF = (P/F,i%,n) SPCAF = (F/P,i%,n) USPWF = (P/A,i%,n) CRF = (A/P,i%,n) SFF = (A/F,i%,n) USCAF = (F/A,i%,n) ...
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This note was uploaded on 05/12/2011 for the course INDUSTRIAL 312 taught by Professor Hjtuk during the Spring '11 term at Mitchell Technical Institute.
 Spring '11
 hjtuk
 Industrial Engineering

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