1 / 35

Engineering Economics

Engineering Economics. November 3, 2004. Engineering Economy. It deals with the concepts and techniques of analysis useful in evaluating the worth of systems, products, and services in relation to their costs. Engineering Economy. It is used to answer many different questions

Pat_Xavi
Download Presentation

Engineering Economics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Engineering Economics November 3, 2004

  2. Engineering Economy • It deals with the concepts and techniques of analysis useful in evaluating the worth of systems, products, and services in relation to their costs

  3. Engineering Economy • It is used to answer many different questions • Which engineering projects are worthwhile? • Has the mining or petroleum engineer shown that the mineral or oil deposits is worth developing? • Which engineering projects should have a higher priority? • Has the industrial engineer shown which factory improvement projects should be funded with the available dollars? • How should the engineering project be designed? • Has civil or mechanical engineer chosen the best thickness for insulation?

  4. Basic Concepts • Cash flow • Interest Rate and Time value of money • Equivalence technique

  5. Cash Flow • Engineering projects generally have economic consequences that occur over an extended period of time • For example, if an expensive piece of machinery is installed in a plant were brought on credit, the simple process of paying for it may take several years • The resulting favorable consequences may last as long as the equipment performs its useful function • Each project is described as cash receipts or disbursements (expenses) at different points in time

  6. Categories of Cash Flows • The expenses and receipts due to engineering projects usually fall into one of the following categories: • First cost: expense to build or to buy and install • Operations and maintenance (O&M): annual expense, such as electricity, labor, and minor repairs • Salvage value: receipt at project termination for sale or transfer of the equipment (can be a salvage cost) • Revenues: annual receipts due to sale of products or services • Overhaul: major capital expenditure that occurs during the asset’s life

  7. Cash Flow diagrams • The costs and benefits of engineering projects over time are summarized on a cash flow diagram (CFD). Specifically, CFD illustrates the size, sign, and timing of individual cash flows, and forms the basis for engineering economic analysis • A CFD is created by first drawing a segmented time-based horizontal line, divided into appropriate time unit. Each time when there is a cash flow, a vertical arrow is added  pointing down for costs and up for revenues or benefits. The cost flows are drawn to relative scale

  8. Drawing a Cash Flow Diagram • In a cash flow diagram (CFD) the end of period t is the same as the beginning of period (t+1) • Beginning of period cash flows are: rent, lease, and insurance payments • End-of-period cash flows are: O&M, salvages, revenues, overhauls • The choice of time 0 is arbitrary. It can be when a project is analyzed, when funding is approved, or when construction begins • One person’s cash outflow (represented as a negative value) is another person’s inflow (represented as a positive value) • It is better to show two or more cash flows occurring in the same year individually so that there is a clear connection from the problem statement to each cash flow in the diagram

  9. An Example of Cash Flow Diagram • A man borrowed $1,000 from a bank at 8% interest. Two end-of-year payments: at the end of the first year, he will repay half of the $1000 principal plus the interest that is due. At the end of the second year, he will repay the remaining half plus the interest for the second year. • Cash flow for this problem is: End of year Cash flow 0 +$1000 1 -$580 (-$500 - $80) 2 -$540 (-$500 - $40)

  10. Cash Flow Diagram $1,000 1 2 0 $540 $580

  11. Time Value of Money • Money has value • Money can be leased or rented • Thepayment is called interest • If you put $100 in a bank at 9% interest for one time period you will receive back your original $100 plus $9 Original amount to be returned = $100 Interest to be returned = $100 x .09 = $9

  12. Compound Interest • Interest that is computed on the original unpaid debt and the unpaid interest • Compound interest is most commonly used in practice • Total interest earned = In = P (1+i)n - P • Where, • P – present sum of money • i – interest rate • n – number of periods (years) I2 = $100 x (1+.09)2 - $100 = $18.81

  13. Future Value of a Loan With Compound Interest • Amount of money due at the end of a loan • F = P(1+i)1(1+i)2…..(1+i)n or F = P (1 + i)n • Where, • F = future value and P = present value • Referring to slide #10, i = 9%, P = $100 and say n= 2. Determine the value of F. F = $100 (1 + .09)2 = $118.81

  14. Notation forCalculating a Future Value • Formula: F=P(1+i)n is the single payment compound amount factor. • Functional notation: F=P(F/P,i,n) F=5000(F/P,6%,10) • F =P(F/P) which is dimensionally correct.

  15. Notation forCalculating a Present Value • P=F(1/(1+i))n=F(1+i)-n is the single payment present worth factor. • Functional notation: P=F(P/F,i,n) P=5000(P/F,6%,10) Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period

  16. Spreadsheet Function P = PV(i,N,A,F,Type) F = FV(i,N,A,P,Type) i = RATE(N,A,P,F,Type,guess) Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means end-of-period cash payments, Type = 1 means beginning-of-period payments, guess is a guess value of the interest rate

  17. Equivalence • Relative attractiveness of different alternatives can be judged by using the technique of equivalence • We use comparable equivalent values of alternatives to judge the relative attractiveness of the given alternatives • Equivalence is dependent on interest rate • Compound Interest formulas can be used to facilitate equivalence computations

  18. Technique of Equivalence • Determine a single equivalent value at a point in time for plan 1. • Determine a single equivalent value at a point in time for plan 2. Both at the same interest rate and at the same time point. • Judge the relative attractiveness of the two alternatives from the comparable equivalent values.

  19. Engineering Economic Analysis Calculation • Generally involves compound interest formulas (factors) • Compound interest formulas (factors) can be evaluated by using one of the three methods • Interest factor tables • Calculator • Spreadsheet

  20. Given the choice of these two plans which would you choose? To make a choice the cash flows must be altered so a comparison may be made.

  21. Resolving Cash Flows to Equivalent Present Values • P = $1,000(PA,10%,5) • P = $1,000(3.791) = $3,791 • P = $5,000 • Alternative 2 is better than alternative 1 since alternative 2 has a greater present value

  22. An Example of Future Value • Example: If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually? • Given P = 500, i = 6%, n = 3, use F = FV(6%,3,,500,0) = -595.91 • Note that the spreadsheet gives a negative number to find equivalent of P. If we find P using F = -$595.91, we get P = 500.

  23. An Example of Present Value • Example 3-5: If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account? • n = 4, F = $800, i = 5%, P = ? • P = PV(5%,4,,800,0) = -$658.16 • You should use P = $658.16

  24. Economic Analysis Methods • Three commonly used economic analysis methods are • Present Worth Analysis • Annual Worth Analysis • Rate of Return Analysis

  25. Present Worth Analysis • Steps to do present worth analysis for a single alternative (investment) • Select a desired value of the return on investment (i) • Using the compound interest formulas bring all benefits and costs to present worth • Select the alternative if its net present worth (Present worth of benefits – Present worth of costs) ≥ 0

  26. Present Worth Analysis • Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives • Step 1: Select a desired value of the return on investment (i) • Step 2: Using the compound interest formulas bring all benefits and costs to present worth for each alternative • Step 3: Select the alternative with the largest net present worth (Present worth of benefits – Present worth of costs)

  27. Present Worth Analysis • A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows: • First cost = $50,000, annual operating cost = $2000, annual income = $9,000, salvage value is $10,000, life = 10 years. Is this investment worth undertaking? • P = $50,000, A = annual net income = $9,000 - $2,000 = $7,000, S = 10,000, n = 10. • Evaluate net present worth = present worth of benefits – present worth of costs

  28. Present Worth Analysis • Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762 • Present worth of costs = $50,000 + $2,000(PA,9%,10) - $10,000(PF,9%,10)= $50,000 + $2,000(6..418) - $10,000(.4224) = $58,612 • Net present worth = $57,762 - $58,612 < 0  do not invest • What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?

  29. Present Worth Analysis • Present worth of benefits = A(PA,9%,10) = A(6.418) • Present worth of costs = $50,000 + $2,000(PA,9%,10) - $10,000(PF,9%,10)= $50,000 + $2,000(6..418) - $10,000(.4224) = $58,612 • A(6.418) = $58,612  A = $58,612/6.418 = $9,312.44

  30. Cost and Benefit Estimates • Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative

  31. Annual costs and Income for a Product • Annual product total cost is the sum of annual material, labor, and overhead (salaries, taxes, marketing expenses, office costs, and related costs), annual operating costs (power, maintenance, repairs, space costs, and related expenses), and annual first cost minus the annual salvage value. • Annual income generated through the sales of a product = number of units sold annuallyxunit price

  32. Rate of Return Analysis • Single alternative case • In this method all revenues and costs of the alternative are reduced to a single percentage number • This percentage number can be compared to other investment returns and interest rates inside and outside the organization

  33. Rate of Return Analysis • Steps to determine rate of return for a single stand-alone investment • Step 1: Take the dollar amounts to the same point in time using the compound interest formulas • Step 2: Equate the sum of the revenues to the sum of the costs at that point in time and solve for i

  34. Rate of Return Analysis • An initial investment of $500 is being considered. The revenues from this investment are $300 at the end of the first year, $300 at the end of the second, and $200 at the end of the third. If the desired return on investment is 15%, is the project acceptable? • In this example we will take benefits and costs to the present time and their present values are then equated

  35. Rate of Return Analysis • $500 = $300(PF, i, n=1) + 300(PF, i, n=2) + $200(PF, i, n=3) • Now solve for i using trial and error method • Try 10%: $500 = ? $272 + $247 + $156 = $669 (not equal) • Try 20%: $500 = ? $250 + $208 + $116 = $574 (not equal) • Try 30%: $500 = ? $231 + $178 + $91 = $500 (equal)  i = 30% • The desired return on investment is 15%, the project returns 30%, so it should be implemented

More Related