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Engineering Economics Prof.Dr. Cengiz Kahraman ITU Industrial Engineering Department.

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Presentation on theme: "Engineering Economics Prof.Dr. Cengiz Kahraman ITU Industrial Engineering Department."— Presentation transcript:

1 Engineering Economics Prof.Dr. Cengiz Kahraman ITU Industrial Engineering Department

2 Engineering Economics  Interest Tables  In order to simplify the routine engineering economy calculations involving the factors, tables of factor values are prepared for some certain interest rates:  1-Discrete Tables  2-Continuous Tables

3 Engineering Economics  Compound Interest Rates 1- Nominal Interest Rate  Nominal interest is the annual interest rate without considering the effect of any compounding. where  interest rate / interest period  m = number of compoundings per year

4 Engineering Economics 2- Effective Interest Rate (EIR)  Effective interest is the annual interest rate, taking into account the effect of compounding during the year. where  i m = interest rate / interest period  m = number of compoundings per year

5 Engineering Economics  Example:  Consider the situation if a person deposited $100 in a bank that pays 5% interest, compounded semi-annually. How much would be in the savings account at the end of one year?  Solution:  5% interest, compounded semi-annually, means that the bank pays 2.5% every six months.  The total money at the end of one year is  What annual interest rate yields the same amount of money, $105.06?

6 Engineering Economics takes into account the effect of compounding during the year. So, it is EIR. 5% interest, compounded semi-annually does not take into account the effect of compounding during the year. So, it is NIR.

7 Engineering Economics  Using the formulas of EIR and NIR,  Example:  If a savings bank pays 1.5% interest every three months, what are the nominal and the effective interest rates? (NIR=6%, EIR=6.1%)  Example:  An engineer deposits $1,000 in a savings account at the end of each year. If the bank pays interest at the rate of 6% per year, compounded quarterly, how much money will have accumulated in the account after 5 years? ($5,652.40)

8 Engineering Economics  Interpolation  Sometimes it is necessary to locate a factor value for an interest rate i or number of periods n that is not in the interest tables. When this occurs, the desired factor value can be obtained in one of two ways: 1- by using the formulas, or 2- by interpolating between the tabulated values.  The value obtained through interpolation is not exactly the correct value, since we are linearly interpreting nonlinear equations.  Nevertheless, interpolation is acceptable and is considered sufficient in most cases as long as the values of i or n are not too distant from each other.

9 Engineering Economics  Example:  Determine the value of the A/P factor for an interest rate of 7.3% and n of 10 years, that is,.  Solution:  We have the following situation:  The correct factor value is % %X 8%

10 Engineering Economics  Example:  A university credit union advertises that its interest rate on loans is 1% per month. Calculate the effective annual interest rate and use the interest factor tables to find the corresponding P/F factor for n=8 years. 12% %P/F 14%  The interpolated value is while the correct factor value is

11 Engineering Economics  Periods  Adjusting Periods  To use interest tables, the payment and compounding periods must agree.  Single Payment Factor Requirements  i and n must be in agreement  If not, either i or n must be adjusted

12 Engineering Economics  Adjusting Uniform Series  If payment period (PP) and compounding period (CP) agree, solve as usual.  If CP is more frequent than PP  Find the effective i for the PP  Count the number of PP's and use as n  Use values in Standard Factor Notation  If PP's are more frequent than CP's, no interest is paid.

13 Engineering Economics  Continuous Compounding  Continuous compounding can be thought of as a limiting case of the multiple- compounding situation. If we define r as the nominal annual interest rate and m as the number of interest periods per year, then the interest rate per interest period and the number of interest periods in n years is mn. The single payment compound amount formula may be rewritten as

14 Engineering Economics:Continuous Compounding  If we increase m, the number of interest periods Per year, without limit, m becomes very large and approaches infinity and r/m becomes very small and approaches zero.This is the condition of continuous compounding, that is, where the duration of the interest period decreases from some finite duration to an infinitely small duration dt and the number of interest periods Per year becomes infinite. In this situation of continuous compounding:  and

15 Engineering Economics:Continuous Compounding  This is the continuous compounding single payment compound amount formula. The continuous compounding single payment present worth formula is.  Example: A bank offers to sell savings certificates that will pay the purchaser $5,000 at the end of ten years, but will pay nothing to the purchaser in the meantime. If interest is computed at 6% compounded continuously, at what price is the bank selling the certificates?

16 Engineering Economics:Continuous Compounding  Solution:  F=$5,000r=0.06 n=10 years Therefore, the bank is selling the $5,000 certificates for $2,744.

17 Engineering Economics:Continuous Compounding  Discrete Payments  If interest is compounded continuously but payments are made annually, we can use the formulas given below:

18 Engineering Economics:Continuous Compounding  Example: A savings account earns interest at the rate of 6% Per year, compounded continuously. How much money must initially be placed in the account to provide for twenty end-of-year withdrawals, if the first withdrawal is $1,000 and each subsequent withdrawal increases by $200?

19 Engineering Economics:Continuous Compounding  Solution:

20  Continuous Payments  The other version of continuous compounding occurs when the total payment for 1 year is received in continuous, small, equal payments during that year. Engineering Economics:Continuous Compounding

21  Example:  At what rate must funds be continuously added to a savings account in order to accumulate $10,000 in 15 years, if interest is paid at 5%Per year, compounded annually?  Solution  per year.  $ must flow uniformly into the account each year.

22 Engineering Economics:Continuous Compounding  Example:  What effective annual interest rate corresponds to a nominal interest rate of 10% per year, compounded continuously? (Answer:10.52%)  Example:  Suppose that $2,000 is deposited each year, on a continuous basis, into a savings account that pays 6% per year, compounded continuously. How much money will have accumulated after 12 years? (Answer: $35,147.77)

23 Measures of Merit  Present Worth  Annual Worth  Future Worth  Internal Rate of Return  External Rate of Return  Benefit / Cost Ratio  Capitalized Cost  Payback Period

24 Present Worth Analysis  Most Popular Measure.  Other Names  Discounted Cash Flow Method  Discount Rate  Present Value  Net Present Value  Present worth analysis is most frequently used to determine the present value of future money receipts and disbursements.  The present worth (PW) or present value (PV) of a given series of cash flows is the equivalent value of the cash flows at the end of year 0 (i.e., at the beginning of year 1).

25 Net Present Worth  Net present worth = present worth of benefits – present worth of cost  NPW = PW of benefits – PW of cost  If NPW 0, the requested rate of return is met or exceeded and the alternative is financially viable.

26 Net Present Worth Plot  Typical NPW Plot for an Investment  The NPW – i function of a cash flow representing an investment followed by benefits from the investment.

27 Net Present Worth Plot Typical NPW Plot for an Investment NPW i

28 Net Present Worth Plot  Typical NPW Plot for Borrowed Money  There is a receipt of borrowed money early in the time period with a later repayment of an equal sum plus payment of interest on the borrowed money.

29 Net Present Worth Plot  Typical NPW Plot for Borrowed Money NPW i

30 Net Present Worth  ECONOMIC CRITERIA  One of the easiest ways to compare mutually exclusive alternatives is to resolve their consequences to the present time. The three criteria for economic efficiency are presented in the following table:

31 Net Present Worth: Economic Criteria

32 Net Present Worth  Analysis Period Situations  In present worth analysis careful consideration must be given to the time period covered by the analysis. There are three different analysis period situations encountered in economic analysis problems: 1-The useful life of each alternative equals the analysis period. 2-The alternatives have useful lives different from the analysis period. 3-There is an infinite analysis period.

33 Net Present Worth 1- The useful life of each alternative equals the analysis period  Example:  Make a present –worth comparison of the equal-service machines for which the costs are shown below, if i=10% per year.

34 Net Present Worth The useful life of each alternative equals the analysis period Solution:  Situation : Fixed output  Criterion : Minimize the inputs  The cash-flow diagram is left to you. The PW of each machine is calculated as follows:  Type A is selected, since the PW of costs for A are less. Note the plus sign on the salvage value, since it is a receipt.

35 Net Present Worth  The alternatives have useful lives different from the analysis period  The alternatives must be compared over the same number of years. The equal-service requirement can be satisfied by either of two approaches: 1- Compare the alternatives over a period of time equal to the least common multiple (LCM) for their lives.  This procedure requires some assumptions be made about the alternatives in their subsequent life cycles:

36 Net Present Worth  The alternatives have useful lives different from the analysis period  The alternatives under consideration will be needed for the least common multiple of years or more.  The respective costs of the alternatives will be the same in all subsequent life cycles as they were in the first one.

37 Net Present Worth  The alternatives have useful lives different from the analysis period 2- Compare the alternatives using a study period of length n years, which does not necessarily take into consideration the lives of the alternatives. This is also called the planning horizon approach.  Select a time horizon.  Ignore any cash occured beyond the stated horizon.  Estimate a realistic salvage value at the end of the study period for both alternatives.

38 The alternatives have useful lives different from the analysis period  Example:  A plant superintendent is trying to decide between two excavating machines with the estimates presented below.  Salvage value of A =$1000 and of B =$2000.  Determine which one should be selected on the basis of a present-worth comparison using an interest rate of 15% per year? (,)  If a study period of 5 years is specified and the salvage values are not expected to change, which alternative should be selected? ( ; )  Which machine should be selected over a 6-year horizon if the salvage value of machine B is estimated to be $6,000 after 6 years? ( ;)

39 Present Worth Analysis  There is an infinite analysis period.  Some Definitions for Capitalized Cost  Capitalized cost (CC) refers to the present worth of a project that is assumed to last forever.  The sum of the first cost and the present worth of disbursements assumed to last forever is called a capitalized cost.  Capitalized cost is the present worth of a perpetual cash- flow sequence.  Some public works projects such as dams, irrigation systems, railways, tunnels, pipelines, and railroads fall into this category.  Equation for capitalized cost is derived from the factor when. As n approaches, becomes

40 Present Worth Analysis: Infinite analysis period.  Example:  Calculate the capitalized cost of a project that has an initial cost of $150,000 and an additional investment cost of $50,000 after 10 years. The annual operating cost will be $5,000 for the first 4 years and $8,000 thereafter. In addition, there is expected to be a recurring major rework cost of $15,000 every 13 years.Assume that i=15% per year. (Answer: $210,043)  Example:  A $500,000 gift was bequeathed to a city for the construction and continued upkeep of a music shell. Annual maintenance for a shell is estimated at $15,000. In addition, $25,000 will be needed every 10 years for painting and major repairs. How much will be left for the initial construction costs, after funds are allocated for perpetual upkeep? Deposited funds can earn 6 percent annual interest, and these returns are not subject to taxes. (Answer: $218,387)

41 Engieering Economics:Annual Cash Flow Analysis  Annual Cash Flow Analysis  The second of the three major analysis techniques.  In present worth analysis we resolved an alternative into an equivalent cash sum. This might have been an equivalent present worth of cost, an equivalent present worth of benefit, or an equivalent net present worth. But instead of of computing equivalent present sums we could compare alternatives based on their equivalent annual cash flows. Depending on the particular situation we may wish to compute the equivalent uniform annual cost (EUAC), the equivalent uniform annual benefit (EUAB), or their difference (EUAB-EUAC).  In annual cash flow analysis, the goal will be to convert money to an equivalent uniform annual cost or benefit.  Example: A woman bought $1,000 wort of furniture for her home. If she expects it to last ten years, what will be her equivalent uniform annual cost if interest is 7%?

42 Engieering Economics:Annual Cash Flow Analysis  Equivalent uniform annual cost =  Treatment of Salvage Value  In a situation where there is a salvage value, or future value at the end of the useful life of an asset, the result is to decrease the equivalent uniform annual cost.  In this case, the equivalent uniform annual cost may be solved by any of three different calculations:

43 Engieering Economics:Annual Cash Flow Analysis  EUAC =  The relationship above may be modified by using the equation:  Substituting this into the first equation, we obtain:

44 Engieering Economics:Annual Cash Flow Analysis  Or  3.  Example: Bill owned a car for five years. One day he wondered what his uniform annual cost for maintenance and repairs had been. He assembled the following data.

45 Engieering Economics:Annual Cash Flow Analysis  There is a direct relationship between the present worth of cost and the equivalent uniform annual cost. It is

46 Engieering Economics:Annual Cash Flow Analysis  ECONOMIC CRITERIA  One of the easiest ways to compare mutually exclusive alternatives is to resolve their consequences to the equivalent uniform annual worth. The three criteria for economic efficiency are presented in the following table:

47 Engieering Economics:Annual Cash Flow Analysis  ECONOMIC CRITERIA

48 Engieering Economics:Annual Cash Flow Analysis  Example: A firm is considering which of two devices to install to reduce costs in a particular situation. Both devices cost $1,000 and have useful lives of five years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year, but will decline $50 annually, making the second year savings $350, the third year savings $300, and so forth. With interest at 7% which device should the firm purchase?

49 Engieering Economics:Annual Cash Flow Analysis  Solution: Device A:EUAB =$300  Device B :EUAB =$  Select B.  Example: Three alternatives are being considered for improving an operation on the assembly line. The cost of the equipment varies as do their annual benefits compared to the present situation. Each of the alternatives has a ten-year life and a scrap value equal to 10% of its original cost.

50 Engieering Economics:Annual Cash Flow Analysis If interest is 8%, which alternative, if any, should be adopted? Answer: Select A.

51 Engieering Economics:Annual Cash Flow Analysis  Analysis Period  Analysis Period Equal to Alternative Lives  When the analysis period for an economy study coincides with the useful lives for each alternative we have an ideal situation which causes no difficulties. The economy study is based on this analysis period.  Analysis Period a Common Multiple of Alternative Lives  Under the assumption of identical replacement it is appropriate to compare the annual cash flows computed for alternatives based on their own service lives. There is no need to compute a CMAL.

52 Engieering Economics:Annual Cash Flow Analysis  Example: Two pums are being considered for purchase. If interest is 7%, which pump should be bought?  Solution:  The annual cost for 12 years of pumpA:  EUAC =

53 Engieering Economics:Annual Cash Flow Analysis  The annual cost for six years of pump B:  EUAC=  For a common analysis period of twelve years we need to replace pump B at the end of its six year useful life. If we assume that another pump can be obtained, with the same $5,000 initial cost, $1,000 salvage value and six year life, the cash flow will be as follows:  For the 12 year analysis period the annual cost for pump B: EUAC = = $909 For B the annual cost for the six-year analysis period is the same as the annual cost for the twelve-year analysis period.

54 Engieering Economics:Annual Cash Flow Analysis  Infinite Analysis Period  As given before, the formula in present worth analysis for infinite analysis period is  In equivalent uniform annual cash flow analysis, the formula above is converted to use as

55 Engieering Economics:Annual Cash Flow Analysis  Example: In the construction of an aqueduct to expand the water supply of a city there are two alternatives for a particular portion of the aqueduct. Either a tunnel may be constructed through a mountain or a pipeline laid to go around the mountain. If there is a permanent need for the aqueduct, should the tunnel or the pipeline be selected for this particular portion of the aqueduct ? Assume 6% interest rate.  Solution: EUAC for pipeline = $317,000  EUAC for tunnel = $330,000  Select pipeline.

56 Engieering Economics: Rate of Return (ROR) Analysis  Rate of Return (ROR) Analysis  Other equivalent terms  Internal Rate of Return  Return On Investment  Breakeven Rate Of Return  Some important terms in ROR Analysis:  Rate of Return  is the rate of interest paid on the remaining balance of borrowed money.  or  The rate of interest earned on the unrecovered balance of an investment (loan) so that the final payment or receipt brings the balance to zero with interest considered.

57 Engieering Economics: Rate of Return (ROR) Analysis  Example: At i=6% per year, a $5,000 investment is expected to produce a net cash flow of $1187 for each of 5 years:  This represents a 6%-per-year rate of return on the unrecovered balance.

58 Engieering Economics: Rate of Return (ROR) Analysis Table: Unrecovered Balances Using a Rate of Return of 6%

59 Engieering Economics: Rate of Return (ROR) Analysis  The minimum Attractive Rate of Return (MARR) (or the minimum acceptable rate of return):  is a lower limit for investment acceptability set by organizations or individuals. It is generally accepted that the lower bound for a minimum required rate of return should be the cost of capital. MARR is also referred to as the hurdle rate for projects.

60 Engieering Economics: Rate of Return (ROR) Analysis Opportunity Cost: is rate of return on the best rejected project For example, if MARR =12% and proposal 1 with an expected ROR = 13% cannot be funded due to to a lack of capital funds, while proposal 2 has an estimated ROR=14.5% and can be funded from available capital, only proposal 2 is undertaken. Since proposal 1 is not persued due to the lack of capital, its estimated ROR of 13% is referred to as the opportunity cost. The internal rate of return(IRR): is the rate on the unrecovered balance of the investment in a situation where the the terminal balance is zero. The external rate of return (ERR): is the rate of return that is possible to obtain for an investment under current economic conditions. In engineering economy studies, the external interest rate most often will be set to the MARR.

61 Engieering Economics: Rate of Return (ROR) Analysis Calculation of Rate of Return To calculate a rate of return on an investment we must convert the various consequences of the investment into a cash flow. Then we will solve the cash flow for the unknown value of i. This value of i is the rate of return. Five forms of the cash flow equation are PW of benefits – PW of costs = 0 Net Present Worth = 0 EUAB – EUAC = 0 PW of Costs = PW of Benefits

62 Engieering Economics: Rate of Return (ROR) Analysis  Example: A $8,200 investment returned $2,000 per year over a five year useful life. What was the rate of return on the investment?  Solution:  From the interest tables we find  No interpolation is needed as the rate of return for this investment is exactly 7%.

63 Engieering Economics: Rate of Return (ROR) Analysis Example: If 5,000 is invested now in common stock that is expected to yield $100 per year for 10 years and $7,000 at the end of 10 years, what is the rate of return? Solution: PW of Costs = PW of Benefits 5,000 = 100(P/A, i*, 10)+7,000(P/F, i*, 10) By using trial-and-error method and interpolation i*=5.16%. Or AW of Costs = AW of Benefits 5,000(A/P,i*,10)= ,000(A/F, i*, 10) By using trial-and-error method and interpolation i*=5.16%.

64 Engieering Economics: Rate of Return (ROR) Analysis  Problem: An investor has purchased 5 building lots at the beginning of 1983 for $7,000 each, and another 5 lots at the beginning of 1990 for $10,000 each. She plans to sell one lot at the end of 1994 for $20,000; and three lots at the end of 1999 for $30,000 each. She plans to sell the remaining lots at the beginning of 2004 (assuming she is still alive) for $57,000 each. What rate of return on her investment is she expecting?

65 Engieering Economics: Rate of Return (ROR) Analysis  Problem: A small oil company constructs oil drilling works at the Hibernia oil fields with a capital cost of $15,000,000. The useful life of the venture is 20 years. The cost of operation and maintenance (O&M) at the end of the first year is estimated to be $200,000 and is expected to go up by $100,000 per year. If the price of oil is fixed at $20.00 per barrel for the 20 years of operation, what minimum constant production (number of barrels to be produced) per year from the oil well is necessary to earn the company a rate of return of 15%?

66 Engieering Economics: Rate of Return (ROR) Analysis Decision for a Single Alternative The calculated rate of return is compared with a preselected minimum attractive rate of return (MARR). MARR is the same value of i used for present worth and annual cash flow analysis. If ROR MARR, the alternative is economically desirable.Reject the alternative when the rate of return is less than the minimum attractive rate of return (ROR). Incremental Rate of Return Analysis When there are two or more alternatives, rate of return analysis is performed by computing the incremental rate of return on the difference between the alternatives. If the incremental rate of return is grater than or equal to the minimum attractive rate of return, choose the higher cost alternative. If the incremental rate of return is less than the minimum attractive rte of return, choose the lower cost alternative.

67 Engieering Economics: Rate of Return (ROR) Analysis The important steps in incremental rate of return analysis are: 1. Compute the rate of return for each alternative. Reject any alternatives where the rate of return is less than the minimum attractive rate of return. 2. Rank the remaining alternatives in their order of increasing PW of cost. If any higher cost alternative has a rate of return greater than the rate of a lower cost alternative, then the lower cost alternative may be immediately rejected.

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83 The negative net cash flow in year 4 is the result of an upgrade to the asset. Determine the number of i* roots and estimate their values graphically.

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86 BENEFIT / COST RATIO ANALYSIS

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91 PAYBACK PERIOD ANALYSIS

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94 REPLACEMENT ANALYSIS  Retirement and replacement decisions should always be based on economics rather than on whether or not the equipment has reached the end of its physical service life. A piece of equipment may have many years of service life remaining beyond the point at which it has become uneconomical to operate it.  Why do we want to replace?  Reduced Performance  Parts are wearing out.  Reliability: the asset does not work when we need it.  Productivity: the asset does not work as fast or as accurately as it used to.  Altered Requirements  Chanced specs so the old machine can’t meet them.  Obsolescence  New equipment on the market is faster, better, and cheaper.

95 REPLACEMENT ANALYSIS  Basic Concepts  Defender  The equipment we already own.  Challenger  One or more alternatives to replace our equipment.  Pretend you are an outside consultant.  Company must invest the current market value in anything we already own.  Throw away all of the estimates that were made when the equipment was bought and re-estimate costs, useful life.

96 REPLACEMENT ANALYSIS  Very Important Concepts  You can’t get back money you have already spent.  The past is the same for all alternatives.  Sunk costs = book value – market value  Book value  What you say something is worth.  Market value  What the market says it is worth (what some else is willing to pay for it).  Don’t include sunk costs in a replacement analysis.

97 REPLACEMENT ANALYSIS  Study Period (Planning Horizon)  The numbers of years used to compare alternatives.  2 Cases  1. Life left in the defender equals the life of the challenger.  In this case, use any of the capital budgeting techniques.  2. Life left in the defender is shorter than the challenger.  Need to pick a study period.  Common practice is to use the life of the challenger.  Management Scepticism  Leads to abbreviated study period.  The thing we bought last time didn’t last for the estimated life.  Why should we expect the thing we are buying now to last for its estimated life?

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107 INFLATION – DEFLATION – INTEREST RATE  Inflation is the effect that causes the price of a good or service to increase in monetary units over time. The opposite effect is deflation, i.e. negative inflation. Over long periods of time the historical record shows that inflation is a persistent but unsteady effect.  The inflation rate tends to be compounded because it is generally calculated as a percentage of the previous period. Forecasting inflationary effects is uncertain, and especially difficult for extended periods. Inflationary effects vary considerably from country to country at the same times. They are effected by national monetary policy and other factors such as trading in national currencies.  Inflation is an effect like income taxes that must be considered in every realistic time value analysis. The analysis is easier if you ignore inflationary effects. However the chance of the analysis leading to a serious error is increased if inflation or any other persistent effect is ignored.

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