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ISE 211 Engineering Economy Chapter Four Spreadsheets for Economic Analysis Chapter 5 >> Present Worth Analysis

Application of Spreadsheets in Engineering Economy  Constructing Tables of cash flows  Using annuity functions to calculate P, F, A, N, or i.  Using a block function to find present worth or internal rate of return for a table of cash flows.  Making graphs for analysis and convincing presentation.  Calculating “what if” for different assumed values of problem variables

Spreadsheets Annuity Functions  To find the equivalent P-PV(i,N,A,F,Type)  To find the equivalent A-PMT(i,N,P,F,Type)  To find the equivalent F-FV(i,N,A,P,Type)  To find NNPER(i,A,P,F,Type)  To find i RATE(N,A,P,F,Type, guess)

Spreadsheets Annuity Functions (cont’d) Guess is your guess for what the rate will be.  If you omit guess, it is assumed to be 10%. If RATE does not converge, try different values for guess. RATE usually converges if guess is between 0 and 1.

Example 1 A new engineer wants to save money for down payment on a house. The initial deposit is $685, and $375 is deposited at the end of each month. The savings account earns at an annual nominal rate of 6% with monthly compounding. How much is on deposit after 48 months?

Example 2 A new engineer buys a car with 0% down financing from the dealer. The cost with all taxes, registration, and license fee is $15,732. If each of the 48 monthly payments is $398, what is the monthly interest rate?

Spreadsheet Block Functions  Cash flows can be specified period-by-period as a block of values.  These cash flows are analyzed by block functions that identify the row or column entries for which a present worth or an internal rate of return should be calculated.  Note: The cash flows for 1 to N are assumed to be end-of- periods flows.

Example 1 Consider the following cash flow at 8% interest rate. 3 $2,500 6,000 8,000 10,000 12, a) Calculate the present worth value. b) Calculate the internal rate of return.

Example 2 Graph the loan payment as a function of the number of payments for a possible new car loan. Let the number of monthly payments vary between 36 to 60. The nominal annual interest rate is 12%, and the amount borrowed is $18,000.

Exercises (Chapter # 4)

Project Assignment #1  Problems & (using Excel).  Due via  Excel File Format: PA01_LastName.xls.  Problems should only be solved using spreadsheet functions as discussed at the end of Chapter 4 (or lecture notes).

Chapter 5 Present Worth Analysis

Economic Criteria  Therefore, we will examine ways to resolve engineering problems, so that criteria for economic efficiency can be applied.

Economic Criteria (cont’d)  Now we know how to convert a series of cash flows to another form that is desired.  What form do we want to compare mutually exclusive alternatives?  Chapter 5 -- Compare Present Worths  Chapter 6 – Compare equivalent Uniform Annual cash flow  Chapter 7 – Figure out what interest rate allows benefits to be bigger than costs

Applying Present Worth Techniques

Applying Present Worth Techniques (cont’d)  Present Worth Analysis is used to determine the present worth value of future money receipts and disbursements.  Choose the option with the “Best” Present Worth.  There are three different analysis-period situations that are encountered in economic analysis problems: 1) The “useful life” of each alternatives equals the analysis period 2) The alternatives have useful lives different from the analysis period 3) There is an infinite analysis period, n = .

1) Useful Lives Equal the Analysis Period Example 1: A firm is considering which of two mechanical devices to install to reduce costs in a particular situation. Both devices cost $1000 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?

Example 2 Wayne County will build an aqueduct to bring water in from the upper part of the state. It can be built at a reduced size now for $300 million and be enlarged 25 years hence for an additional $350 million. An alternative is to construct the full-sized aqueduct now for $400 million. Both alternatives would provide the needed capacity for the 50-year analysis period. Maintenance costs are small and may be ignored. At 6% interest, which alternative should be selected?

Example 3 A purchasing agent is considering the purchase of some new equipment for the mailroom. Two different manufacturers have provided quotations. An analysis of the quotations indicates the following: The equipment of both manufacturers is expected to perform at the desired level of (fixed) output. For a five-year analysis period, which manufacturer’s equipment should be selected? Assume 7% interest and equal maintenance costs.

Example 4 A firm is trying to decide which of two alternate weighing scales it should install to check a package filling operation in the plant. The scale would allow better control of the filling operation and result in less overfilling. If both scales have lives equal to the six-year analysis period, which one should be considered? Assume 8% interest rate. Alternate Cost Uniform Annual Salvage Value Benefit $$$ Atlas Scale > $2000 $450$100 Tom Thumb> $3000 $600$700

2) Useful Lives Different from the Analysis Period Example 1: Consider the previous example 3. Which alternative is best? ManufacturerCost ($)Useful life (years) End-of-useful life salvage value ($) Speedy Allied

What if one item lasts 11 yrs and the other lasts 13 years?  Use the least common multiple. It seems unrealistic to use this as the analysis period, so use the situation to decide on an appropriate period.  Example, use __________, and assign some terminal value to each alternative at the end of ____________.  You will be exposed to this kind of problems in the homework.

3) Infinite Analysis Period – Capitalized Cost  Some projects need to be maintained permanently (forever): roads, dams, graves, pipelines.  This particular analysis is called Capitalized Cost.  Capitalized Cost is the present amount of money required to be invested at some interest rate to provide the service (or whatever) for the project forever (indefinitely).  In this situation, you use the interest each year to pay for the project, so you need to maintain the principal amount in the bank.  The cost each year is assumed to be the same – so each year we withdraw A.

Illustration End of Year PrincipalInterest Earned Withdraw

Example 1 You need $50/year for your grave maintenance after you die. If you earn 4%, how much do you need to set aside by one year before you die?

Example 2 A city plans a pipeline to transport water from a distant watershed area to the city. The pipeline will cost $8 million and have an expected life of seventy years. The city anticipates it will need to keep the water line in service indefinitely. Compute the capitalized cost assuming 7% interest.

Assumptions in Econ Analysis 1)End of year conventions  Remember tables and formulas assume cash flow at the end of the period  Spreadsheets can handle cash flows at beginning or end of period – but be careful! 2)Use the viewpoint of the whole company / large entity – like shipping department VS entire company. 3)Sunk costs – ignore them! 4)Borrowed Money  We assume that money spent in the problems is borrowed at the same rate and conditions given in the problem.

Assumptions in Econ Analysis (cont’d) 5) Effect of Inflation and Deflation  For now, assume no inflation or deflation – prices are stable 6) Income Taxes  For now, no income taxes effects/consequences

Multiple Alternatives  For more than two alternatives, just get all the present worth values and choose the best alternative for your situation. Example 1: A contractor has been awarded the contract to construct a six-miles-long tunnel in the mountains. During the five-year construction period, the contractor will need water from a nearby stream. He will construct a pipeline to convey the water to the main construction yard. An analysis of costs for various pipe sizes is as follows: The pipe and pump will have a salvage value at the end of five years equal to the cost to remove them. The pump will operate 2000 hours per year. The lowest interest rate at which the contractor is willing to invest money is 7%. (The minimum required interest rate for invested money is called the Minimum Attractive Rate of Return, MARR). Select the alternative with the least present worth of cost.

Example 2 An investor paid $8,000 to a consulting firm to analyze what he might do with a small parcel of land on the edge of town that can be bought for $30,000. In their report, the consultants suggested four alternatives. Assuming 10% is the minimum attractive rate of return, what should the investor do?

Example 3 A piece of land may be purchased for $610,000 to be strip-mined for the underlying coal. Annual net income will be $200,000 per year for ten years. At the end of the ten years, the surface of the land will be restored as required by a federal law on strip mining. The cost of reclamation will be $1,500,000 more than the resale value of the land after it is restored. Using a 10% interest rate, determine whether the project is desirable.

Example 4 Two pieces of construction equipment are being analyzed. Based on an 8% interest rate, which alternative should be selected?

Using Spreadsheets Example 1: NLE Construction is bidding on a project whose costs are divided into $30,000 for startup and 12 equal payments of $20,000 at the end of each month. If the annual interest rate is 12.5%, compounded monthly, what is the present worth?

Using Spreadsheets (cont’d) Example 2: Regina Industries has a new product whose sales are expected to be 1.2, 3.5, 7, 5, and 3 million units per year over the next 5 years. Production, distribution, and overhead costs are stable at $120 per unit. The price will be $200 per unit for the first 2 years, and then $180, $160, and $140 for the next 3 years. The R&D costs are $400 million. If the annual interest rate is 15%, what is the present worth of the new product?

Homework # 05 (Chapter # 5)