Engineering economic analysis / Donald G. Newnan, Ted G. Eschenbach, Jerome . How to Use Price Indexes in Engineering Economic Analysis Library of Congress Cataloging-in-Publication Data Newnan, Donald G. Engineering economic analysis / Donald G. Newnan, Ted G. Eschenbach, Jerome P. The choice of an engine has important money consequences so would be (a) Yes. suitable for engineering economic analysis. Important economic- and.

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Engineering Economics. Cash Flow. Cash flow is the sum Engineering Economics. Discount Factors and . Comparison of Alternatives. Cost-Benefit Analysis. ENGINEERING ECONOMIC ANALYSIS r T ~'. ELEVENTH EDITION. Donald G. Newnan. San Jose State University. Ted G. Eschenbach. University of Alaska. The twelfth edition of the market-leading Engineering Economic Analysis offers comprehensive coverage of financial and economic decision making for.

Skip to main content. Log In Sign Up. Roger PhoneMaung. Uploaded by: Ebooks Chemical Engineering https: Important economic- and social- consequences. Some might argue the b Yes.

Chapter 6: Solving this series for A gives us the A for the infinite series. In this situation the annual capital recovery cost equals interest on the investment. If anyone doubts this, they should compute: This problem is much harder than it looks! The equipment download did not turn out to be desirable.

The problem must be segmented to use the 1. The bailer probably should be installed. Amount to deposit September 1: The fact that the tax payments are for the fiscal year, July 1 Through June 30, does not affect the computations. Ruarterly interest payments to the savings account could have an impact on the solution, but they do not in this problem. The solution may be verified by computing the amount in the savings account on Dec.

Now solve for the unknown n. Method 1: The payments would repay: The difference is due to four place accuracy in the compound interest tables.

Check solution using NPW: The analysis period is seven years, hence one cannot compare three years of A vs. Seventy or seventy- five years might be the range of reasonable estimates. Here we will use 71 years. Two possible solutions are provided below. Assuming the MS degree is obtained by attending graduate school at night while continuing with a full-time job: Assuming the MS degree is obtained by one of year of full-time study Cost: Whether working or at school there are living expenses.

Available interest tables obviously are useless. One may write: There are two possibilities: Convenience, improved quality of life, increased value of the dwellings, etc. Thus, the pipeline appears justified. By linear interpolation: Wait, leave your fund in the system until retirement.

Therefore, the increment is desirable. Select X. Therefore it is not a desirable increment of investment. Choose A. Select A. Select B. download Kicko. There is external investment until the end of the tenth year.

The internal interest rate is sensitive to the selected external interest rate: To search for positive rates of return compute the NPW for the cash flow at several interest rates. This is done on the next page by using single payment present worth factors to compute the PW for each item in the cash flow. Then, their algebraic sum represents NPW at the stated interest rate.

The revised cash flow becomes: Although there are three sign changes in the cash flow, the accumulated cash flow sign test, described in Chapter 18 indicates there is only a single positive rate of return for the untransformed cash flow.

Even though there is only one rate of return, there still exists the required external investment in Ruarter 1 for Ruarter 2.

On this basis the Part b solution appears to have more realistic assumptions than Part a. Before proceeding, we will check for multiple rates of return. This, of course, is not necessary here. Looking for the other possible rate of return: For further computations, see the solution to Problem This is only slightly different from the Tables could be produced, of course, for negative values.

Reject D and retain A. Reject A and accept B. Select Plan B. The rate of return for each Plan is computed. Two incremental analyses are performed. Reject Plan A. Retain Plan B.

Reject Plan C. Since at the same cost B produces a greater annual benefit, it will always be preferred over C. C may, therefore, be immediately discarded.

Retain A. Select Alternative A. MARR Test: Alternative Rate Test: C- A increment satisfactory Choose C. C- A increment satisfactory. Choose C. C- A increment unsatisfactory. Reject D and retain C. The B- C increment is undesirable. Reject B and retain C.

The A- C increment is undesirable. Reject A and retain C. Select alternative C. Alternative B is preferred over Alternative A. The C- B incremental rate of return of 6. Reject C. Select Alternative B. Select Y. Do nothing. Assuming this is not recognized, one would first compute the rate of return on the increment B- A and then C- B.

The problem has been worked out to make the computations relatively easy. Rate of Return on B- A: Decision Reject A.

Keep B. Select C. Effective Reject 2. Reject 1 and select 3 continue as is. Select Pump 1. Proceed with incremental analysis. Examine increments of investment.

So Alternative 1 can be rejected. This leaves alternatives 2 and 4. Examine increment. Choose Alternative D. Reject Company A. The correct choice is the Regular model. Since the B-A increment is not acceptable, Alternative B should not be adopted. Compute the amount of annual income for each alternative situation.

To maximize annual income, choose C. An alternate solution may be obtained by examining each separable increment of investment. Select Alternative C. Instead we can substitute: He should deposit: Reject A. Reject B.

Reject 5 stories. The 10 stories rather than 2 stories is desirable. Choose the story alternative. Alternative A: Choose B.

An incremental Uniform Annual Benefit becomes a cost rather than a benefit. Reject D. Reject E. Similarly, D, with greater benefits and identical cost, is preferred over B. Hence B may be rejected. On this basis C may be rejected. Therefore, do A. Alternative A may be considered if the investor is very short of cash and the short payback period is of importance to him.

The Present Worth method requires common analysis period, which is virtually impossible for this problem. The problem is easy to solve by Annual Cash Flow Analysis. In future worth analysis there must be a common future time for all calculations. In this case 12 years hence is a practical future time.

C Alt. Increment B- C Year Alt. B Alt. The altered cash flow becomes: Solutions for part b: Choose Alternative C. Thus each generates uniform annual benefits in excess of the cost, during the life of the alternative. From this is must follow that the alternative with a 2-year life has a payback period less than 2 years.

The alternative with a 4-year life has a payback period less than 4 years, and the alternative with a 6-year life has a payback period less than 6 years. Thus we see that the shorter-lived asset automatically has an advantage over longer-lived alternatives in a situation like this.

While Alternative A takes the shortest amount of time to recover its investment, Alternative C is best for long- term economic efficiency. The problem may be solved by inspection. Alternative z dominates Alternative y.

Alternative z has a positive rate of return actually Choose Alternative z. Two different equations that might be used. To maximize NFW, select F. To minimize payback period, select F. Here, however, we will include it.

At breakeven: For continuous compounding: Chapter Uncertainty in Future Events a Some reasons why a pole might be removed from useful service: The pole has deteriorated and can no longer perform its function of safely supporting the telephone lines 2.

The telephone lines are removed from the pole and put underground. The poles, no longer being needed, are removed. Poles are destroyed by damage from fire, automobiles, etc. The street is widened and the pole no longer is in a suitable street location. The pole is where someone wants to construct a driveway. Typical values for Pacific Telephone Co.

Optimistic Life: The salvage value drops by 5x8,x. The salvage value increases by 5x8,x. So the total probability of higher rates for year 2 is. An inspection of the Regular Season situation reveals that the sum of the probabilities for the outcomes enumerated is 0.

Thus one outcome win less than three games , with probability 0. This is not a faulty problem statement. The student is expected to observe this difficulty. Die 1 Die 2 2 6 3 5 4 4 5 3 6 2 The five ways of throwing an 8 have equal probability of 0. The probability of winning is 0. Here it is ignored. Remember, only the differences between alternatives are relevant. Probability that spillway capacity equaled or exceeded in any year is 0.

Damage if spillway capacity exceed: One must be careful not to confuse the frequency of a flood and when it might be expected to occur.

In any year period, for example, there are 4 chances in 10 that a year flood or greater will occur. Since we are dealing with conditions of risk, it is not possible to make an absolute statement concerning which alternative will result in the least cost to the community.

Using a probabilistic approach, however, Alternative I is most likely to result in the least equivalent uniform annual cost. The P loss is unchanged at.

For example, the first and second rows' PWs are unchanged. The probability of a negative PW is. This illustrates why standard deviation alone is not the best measure of risk.

The standard deviation is higher, but the P loss has dropped by half. Sum-of-Years Digits Schedule is: Which method is preferred? One can also see this by inspection of the depreciation schedules above. Cost of Proceeds Undep. This is not a correct analysis of the situation. This may be illustrated by computing the Straight Line depreciation for Year 3, if DDB depreciation had been used in the prior years.

One would naturally choose to continue with DDB depreciation. For subsequent years: The resulting depreciation schedule is: On all income While the depreciation charges in any year may be different for different methods, the sum of the depreciation charges will be the same. The difference is not the amount of the taxes, but their timing. XYZ, Inc. Further calculations show actual rate of return to be approximately 4.

It does change the timing of these items. Calculator solution is Therefore the project should not be undertaken. Calculator solution: Unfortunately, people who are relocated often feel harmed, no matter how much money, etc.

Thus planners consider criterion a unworkable and use criterion b instead. In some cases, these projects are also used for urban renewal of decayed residential or industrial areas, which introduces other benefits. The costs of these projects include the money spent on the project, the time lost by travelers due to construction caused congestion, and the lost residences and businesses of those displaced. In some cases, the loss may be intangible as a road separates a neighborhood into two pieces.

In other cases, the loss may be due to living next to a source of air, noise, and visual pollution. To stay in the residence the rest of the year Food: To stay in the residence the balance of the first semester; apartment for second semester Housing: Move into an apartment now Housing: Jay appears to prefer Alternative 2, and he has sufficient money to adopt it.

While there are alternatives available, one appears so obvious that that foreman discarded the rest and asks to proceed with the replacement. One could argue that the foreman, or the plant manager, or both are making decisions. One of my students observed that his father would not fall for such a simple deception, and surely would insist on the weird shirt as a subtle form of punishment.

Alternatives to their current university program are likely to focus on other fields of engineering and science, but answers are likely to be distributed over most fields offered by the university.

Outcomes include degree switches, courses taken, changing dates for expected graduation, and probable future job opportunities. At best criteria will focus on joy in the subject matter and a good match for the working environment that pleases that particular student. Often economic criteria will be mentioned, but these are more telling when comparing engineering with the liberal arts than when comparing engineering fields.

Other criteria may revolve around an inspirational teacher or an influential friend or family member. In some cases, simple availability is a driver. What degree programs are available at a campus or which programs will admit a student with a 2.

At best the process will follow the steps outlined in this chapter. Nevertheless, it is a large mistake to not change majors when a student now realizes the major is not for them. This problem seems ideal for listing student ideas on the board or overhead transparencies. It is also a good opportunity for the instructor to add more experienced comments. On the other hand, Example shipping department downloading printing is a situation where the sub-problem does not lead to a proper complex problem solution.

Or simply, maximize net profit. The data from the graphs may be tabulated as follows: Since one cannot achieve maximum output with minimum input, the statement makes no sense.

One might begin, for example, assuming five hours of study on each course. The combined total score would be Decreasing the study of mathematics one hour reduces the math grade by 8 points from 52 to This hour could be used to increase the physics grade by 9 points from 59 to The result would be: Math 4 hours 44 Physics 6 hours 68 Engr.

Math 4 hours 44 Physics 7 hours 77 Engr. Open at , close at Set the first derivative equal to zero and solve for t. Since either home is really an individual plan selected by the homeowner, each should be judged in terms of value to the homeowner vs. On this basis the stock plan house appears to be the preferred alternative. The student can visually verify this from the figure.

This indicates that production below units per year is most undesirable, as it costs more to produce units than to produce units. Check the sign of the second derivative: Since we do not know the Selling Price, we cannot know Marginal Revenue, and hence we cannot compute the optimum level of output.

From the graph we see that there are three break-even points: Mathematically the break-even points are: Sneaking a peak at the figure below we see that if: There will be two solutions: Break-even at 14 and 77 units.

Maximum profit at 45 units. Below is a list of possible recurring and non-recurring costs. Students may develop others. Book costs are not represented as before- tax cash flows. Engineering economic analyses can involve both cash and book costs. Cash costs are important in such cases. For the engineering economist the primary book cost that is of concern is equipment depreciation, which is accounted for in after-tax analyses.

By life-cycle costs the authors are referring to any cost associated with a product, good, or service from the time it is conceived, designed, constructed, implemented, delivered, supported and retired.

Firms should be aware of and account for all activities and liabilities associated with a product through its entire life-cycle. These costs and liabilities represent real cash flows for the firm either at the time or some time in the future. The key point being that most costs are committed early in the life cycle, although they are not realized until later in the project.

The implication of this effect is that if the firm wants to maximize value-per-dollar spent, the time to make important design decisions and to account for all life cycle effects is early in the life cycle.

The point of this comparison is that the early stages of the design cycle are the easiest and least costly periods to make changes. Both figures represent important effects for firms. In summary, firms benefit from spending time, money and effort early in the life cycle. Effects resulting from early decisions impact the overall life cycle cost and quality of the product, good, or service.

An integrated, cross-functional, enterprise-wide approach to product design serves the modern firm well. Each of these factors could influence the estimate, or the estimating process, in different scenarios in different firms. One-of-a-kind estimating is a particularly challenging aspect for firms with little corporate-knowledge or suitable experience in an industry. Estimates, bids and budgets could potentially vary greatly in such circumstances. This is perhaps the most difficult of the factors to overcome.

Time and effort can be influenced, as can estimator expertise. One-of-a-kind estimates pose perhaps the greatest challenge. As an example: Chapter 3: Understanding interest and its impact is important in many life circumstances. Examples could include some of the following: Selecting the best loans for homes, boats, jewellery, cars, etc. Many aspects involved with businesses ownership payroll, taxes, etc.

Using the best strategies for paying off personal loans, credit cards, debt! Making investments for life goals downloads, retirement, college, weddings, etc. We find the interest rate at which the two cash flows are equivalent by: Thus, as a decision maker she would be indifferent. On way would be: Since the compound interest factor is non-linear, linear interpolation will not produce an exact solution.

Therefore, it takes 18 months to repay the loan. Below are two of them: Alternative 1: Column 1 shows the number of interest periods. Column 2 shows the equal annual amount as computed in part a above. To compute the interest portion for year one, we must first multiply the interest rate in decimal by the remaining balance: This completes the year 1 row.

The other row quantities are computed in the same fashion. The interest portion for row two, year 2 is: Interest is computed on the remaining balance at the end of the preceding year and not on the original principal of the loan amount. The rest of the calculations proceed as before. Also, note that in year 7, the remaining balance as shown on Table is approximately equal to the value calculated in a using a formula except for round off error.

Owed Int. EOP This pmt Pmt.

The simplest solution is to draw a diagram of the situation and then proceed to solve the problem presented by the diagram. We need the Present Worth at April 1, We can use either interest rate, the quarterly or the semiannual.

There was no simple formula, or even a complicated formula, to arrive at the solution. While the actual calculations were not difficult, there were several steps required to arrive at the correct solution. The real rate of return is closer to 6. This is the cumulative PW in the last column below. The period with monthly figures is 34 months rather than the 35 months indicated below.

Chapter 5: In this case P is calculated as: Do not download equipment. Parts b and c assume earlier payments, hence their PW of Cost is greater. First compute equivalent A. Solving one portion of the perpetual series for A: Thus after 25 years all costs are identical.

The cost for option 2 will now be higher. Replace with untreated ties.

Replace with treated ties. Therefore, maximize PW of benefits. By inspection, one can see that C, with its greater benefits, is preferred over A and B.

Similarly, E is preferred over D. The problem is reduced to choosing between C and E. But the student should recognize that this is a faulty criterion.

It would download 83 shares of Spartan Products, but only 42 shares of Western House. The criterion, therefore, is to maximize NPW for the amount invested.

download Spartan Products. Thus we use 12 years and assume repeatability of the cash flows.

Problem has the same effective interest rate as , but the rate on is lower. B 1 Payment 2 N 3 Interest rate 0. B 1 Payment 2 N 30 3 Interest rate 6. This problem can also be solved by using NPV function: Chapter 6: Solving this series for A gives us the A for the infinite series. In this situation the annual capital recovery cost equals interest on the investment.

If anyone doubts this, they should compute: This problem is much harder than it looks! The equipment download did not turn out to be desirable. The problem must be segmented to use the 1. The bailer probably should be installed. Amount to deposit September 1: The fact that the tax payments are for the fiscal year, July 1 Through June 30, does not affect the computations. Ruarterly interest payments to the savings account could have an impact on the solution, but they do not in this problem.

The solution may be verified by computing the amount in the savings account on Dec. Now solve for the unknown n. Method 1: The payments would repay: The difference is due to four place accuracy in the compound interest tables. Check solution using NPW: The analysis period is seven years, hence one cannot compare three years of A vs. Seventy or seventy- five years might be the range of reasonable estimates.

Here we will use 71 years. Two possible solutions are provided below. Assuming the MS degree is obtained by attending graduate school at night while continuing with a full-time job: Assuming the MS degree is obtained by one of year of full-time study Cost: Whether working or at school there are living expenses.

Available interest tables obviously are useless. One may write: There are two possibilities: Convenience, improved quality of life, increased value of the dwellings, etc. Thus, the pipeline appears justified. By linear interpolation: Wait, leave your fund in the system until retirement. Therefore, the increment is desirable. Select X. Therefore it is not a desirable increment of investment. Choose A. Select A. Select B. download Kicko. There is external investment until the end of the tenth year.

The internal interest rate is sensitive to the selected external interest rate: To search for positive rates of return compute the NPW for the cash flow at several interest rates. This is done on the next page by using single payment present worth factors to compute the PW for each item in the cash flow. Then, their algebraic sum represents NPW at the stated interest rate. The revised cash flow becomes: Although there are three sign changes in the cash flow, the accumulated cash flow sign test, described in Chapter 18 indicates there is only a single positive rate of return for the untransformed cash flow.

Even though there is only one rate of return, there still exists the required external investment in Ruarter 1 for Ruarter 2. On this basis the Part b solution appears to have more realistic assumptions than Part a. Before proceeding, we will check for multiple rates of return. This, of course, is not necessary here. Looking for the other possible rate of return: For further computations, see the solution to Problem This is only slightly different from the Tables could be produced, of course, for negative values.

Reject D and retain A. Reject A and accept B. Select Plan B. The rate of return for each Plan is computed. Two incremental analyses are performed.

Reject Plan A. Retain Plan B. Reject Plan C. Since at the same cost B produces a greater annual benefit, it will always be preferred over C. C may, therefore, be immediately discarded. Retain A.

Select Alternative A. MARR Test: Alternative Rate Test: C- A increment satisfactory Choose C. C- A increment satisfactory. Like this presentation? Why not share! An annual anal Embed Size px. Start on. Show related SlideShares at end. WordPress Shortcode. Jassy Follow. Published in: Full Name Comment goes here. Are you sure you want to Yes No. Be the first to like this. No Downloads. Views Total views. Actions Shares.