Engineering Economics 2 Lecture (Week 5 and 6 Annotation)

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School of Mechanical & Manufacturing Engineering MMAN4400 Week 5 Engineering Economics II
Question of the Day When Robby is 3 months old, his parents decide to make a regular deposit of $500 every 3 months, starting with first one when Robby is 3 months old in an account that earns interest of 8% p.a., the interest being paid every 3 months. No more payment are made into the account after Robby turns 15 and no withdraws are made. Then, Robby decides that he will withdraw a regular amount of money from this account each birthday, starting with his 16 th birthday. Robby wishes to keep withdrawing money out until he reaches 30 years old. How much money should he withdraw each year?
Week 5 Outline Engineering Economics 2 • Nominal vs Effective interest rate • Present worth (PW) analysis • Annual worth (AW) analysis
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Interest Rate Statements The terms ‘nominal’ and ‘effective’ enter into consideration when the interest period is less than one year. • Interest period (t) – period of time over which interest is expressed. For example, 1% per month. • Compounding period (CP) – Shortest time unit over which interest is charged or earned. For example,10% per year compounded monthly. • Compounding frequency (m) – Number of times compounding occurs within the interest period t. For example, at 𝑖 = 10% per year, compounded monthly, interest would be compounded 12 times during the one year interest period. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Nominal Interest Rate A nominal interest rate ( ? ) is obtained by multiplying an interest rate that is expressed over a short time period by the number of compounding periods in a longer time period: That is: ? = 𝑖??????? ???? ??? ???𝑖?? ? ?????? ?? ????????𝑖?? ???𝑖??? Example: If 𝑖 = 1% per month, nominal rate per year is: ? = (1)(12) = 12% ??? ???? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Effective Interest Rate Effective interest rates ( 𝑖 ) take compounding into account (effective rates can be obtained from nominal rates via a formula to be discussed later). IMPORTANT: Nominal interest rates are essentially simple interest rates. Therefore, they can never be used in interest formulas. Effective rates must always be used hereafter in all interest formulas. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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More About Interest Rate Terminology Sample Interest Rate Statements Comment • 𝑖 = 2% per month. • 𝑖 = 12% per year. When no compounding period is given, rate is effective. • 𝑖 = 10% per year, compounded semi-annually. • 𝑖 = 3% per quarter, compounded monthly. When compounding period is given and it is not the same as interest period, it is nominal. • 𝑖 = effective 9.4% per year, compounded semi-annually. • 𝑖 = effective 9.4% per quarter, compounded monthly. When compounding period is given and rate is specified as effective, rate is effective over stated period. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Effective Annual Interest Rates Nominal rates are converted into effective annual rates via the equation: 𝑖 𝑎 = 1 + 𝑖 ? − 1 Where: • 𝑖 𝑎 = effective annual interest rate. • 𝑖 = effective rate for one compounding period. • ? = number of times interest is compounded per year. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Effective Annual Interest Rates Example For a nominal interest rate of 12% per year, determine the nominal and effective rates per year for (a) quarterly, and (b) monthly compounding. (a) Nominal ? per year = 12% per year. Nominal ? per quarter = 12/4 = 3.0% per quarter. Effective 𝑖 per year = 1 + 0.03 4 − 1 = 12.55% per year. (b) Nominal ? per month = 12/12 = 1.0% per year. Effective 𝑖 per year = 1 + 0.01 12 − 1 = 12.68% per year. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Effective Interest Rates Nominal rates can be converted into effective rates for any time period via the following equation: 𝑖 = 1 + ? ? ? − 1 Where: • 𝑖 = Effective interest rate for any time period. • ? = Nominal rate for the same time period as 𝑖 . • ? = no. of times interest is compounded in period specified for 𝑖 . Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Effective Interest Rates Example For an interest rate of 1.2% per month, determine the nominal and effective rates (a) per quarter, and (b) per year. (a) Nominal ? per quarter = (1.2)(3) = 3.6% per quarter. Effective 𝑖 per quarter = 1 + 0.036 3 3 – 1 = 3.64% per quarter. (b) Nominal 𝑖 per year = (1.2)(12) = 14.4% per year. Effective 𝑖 per year = 1 + 0.144 12 12 – 1 = 15.39% per year. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Equivalence Relations: PP and CP Payment Period (PP) – Length of time between cash flows. 0 1 2 3 4 5 $1000 6 7 8 9 10 11 𝑀???ℎ? 12 𝑃𝑃 = 1 ????ℎ 𝐶𝑃 = 6 ????ℎ 𝐶𝑃 = 6 ????ℎ Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Single Amounts with PP > CP For problems involving single amounts, the payment period (PP) is usually longer than the compounding period (CP). For these problems, there are an infinite number of 𝑖 and ? combinations that can be used, with only two restrictions: 1. The 𝑖 must be an effective interest rate, and 2. The time units on n must be the same as those of 𝑖 . (i.e., if 𝑖 is a rate per quarter, then n is the number of quarters between 𝑃 and 𝐹 ) Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Single Amounts with PP > CP (cont’d) There are two equally correct ways to determine 𝒊 and 𝒏 Method 1: Determine effective interest rate over the compounding period CP, and set n equal to the number of compounding periods between P and F. Method 2: Determine the effective interest rate for any time period t, and set n equal to the total number of those same time periods. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Example: Single Amounts with PP ≥ CP How much money will be in an account in 5 years if $10,000 is deposited now at an interest rate of 1% per month (compounded monthly)? Use three different interest rates: (a) monthly, (b) quarterly , and (c) yearly. (a) For monthly rate, 1% is effective: ? = (5 ?????) × (12 𝐶𝑃 ??? ????) = 60 𝐹 = 10,000 𝐹/𝑃, 1%, 60 = $18,167 (b) For quarterly rate, effective 𝑖 per quarter, 1 + 0.03 3 3 − 1 = 3.03% 𝐹 = 10,000 𝐹/𝑃, 3.03%, 20 = $18,167 (c) For annual rate, effective 𝑖 per year, 1 + 0.12 12 12 − 1 = 12.683% 𝐹 = 10,000 𝐹/𝑃, 12.683%, 5 = $18,167 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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15 • ? = no. of times interest is compounded in period specified for 𝑖 .
16
Series with PP ≥ CP For series cash flows, first step is to determine relationship between PP and CP. Determine if PP ≥ CP, or if PP < CP. When PP ≥ CP, the only procedure (2 steps) that can be used is as follows: 1. Find effective 𝑖 per PP. Example: if PP is in quarters, must find effective 𝑖 /quarter. 2. Determine ? , the number of A values involved. Example: quarterly payments for 6 years yields ? = 4 × 6 = 24 . Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Example: Series with PP ≥ CP How much money will be accumulated in 10 years from a deposit of $500 every 6 months if the interest rate is 1% per month? First, find relationship between PP and CP PP = six months, CP = one month; Therefore, PP > CP. Step 1. Since PP > CP, find effective i per PP of six months. 𝑖 /6 ????ℎ? = 1 + 0.06 6 6 – 1 = 6.15% Step 2. Next, determine ? (number of 6-month periods). ? = 10(2) = 20 ?𝑖? ????ℎ ???𝑖??? Finally, set up equation and solve for F. 𝐹 = 500(𝐹/𝐴, 6.15%, 20) = $18,692 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
In-class Example Problem 3.37 Anderson-McKee Construction expects to invest $835,000 for heavy equipment replacement 2 years from now and another $1.1 million 4 years from now. How much must the company deposit into a sinking fund each 6 months to provide for the purchases, if the fund earns a rate of return of 12% per year compounded semi-annually? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
20 to invest $835,000 for heavy equipment replacement 2 years from now and another $1.1 million 4 years from now. How much must the company deposit into a sinking fund each 6 months to provide for the purchases
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Series with PP < CP Two policies: (1) Inter-period cash flows earn no interest (most common). (2) Inter-period cash flows earn compound interest. For policy (1), positive cash flows are moved to beginning of the interest period in which they occur and negative cash flows are moved to the end of the interest period. For policy (2), cash flows are not moved and equivalent P, F, and A values are determined using the effective interest rate per payment period. Note: The condition of PP < CP with no inter-period interest is the only situation in which the actual cash flow diagram is changed. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example Problem 3.53 Income from recycling the paper and cardboard generated in an office building has average $3000 per month for 3 years. What is the future worth of the income at an interest rate of 8% per year compounded quarterly? Annuity occurs every month Interest is compounded every 3 months We assume no inter-period interest Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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23
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Continuous Compounding When the interest period is infinitely small, interest is compounded continuously. Therefore, PP > CP and m increases. Take limit as m → ∞ to find the effective interest rate equation: 𝑖 = ? 𝑟 – 1 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Formulating Alternatives Types of alternatives Mutually exclusive (ME) – only one viable project can be accepted. Do- nothing (DN) alternative is selected if none are justified economically. Independent – more than one project can be selected. DN is one of the projects. Do-nothing – maintain status quo/current approach. Types of cash flow estimates for an alternative Revenue – estimates include costs, revenues and (possibly) savings. Cost – only cost estimates included; revenues assumed equal for all alternatives. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Formulating Alternatives (cont’d) Much of the emphasis in professional engineering practice is on ME, cost alternatives. However, all tools in Eng. Econ. can be used to evaluate ME and independent alternatives that are revenue- or cost- based. Examples of both are included later. Notes: P value of cash flows is now called PW, or present worth P now represents first cost of an alternative Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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PW Selection Criterion For single project analysis: Calculate PW at stated MARR. Criterion: If PW ≥ 0, project is economically justified. For Equal-life ME Alternatives: Calculate PW of each alternative at MARR. Equal-service of alternatives is assumed. Selection criterion: Select alternative with most favourable PW value, that is, numerically largest PW value. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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When comparing alternatives: PW analysis FW analysis AW analysis IRR analysis 29
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In-class Example Problem 4.15 The bureau of India Affairs need to select the better of two medical X-ray system alternatives. At 5% per year, select the more economical system, using PW analysis A B First cost $ -250,000 -224,000 AOC (annual operation cost) $ -231,000 -235,000 Overhaul in year 3 $ - -26,000 Overhaul in year 4 $ -140,000 - Salvage value $ 50,000 10,000 Expected life, years 6 6 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Different-life Alternatives PW evaluation always requires equal-service between all alternatives. Two methods available: • Study period (same period for all alternatives). • Least common multiple (LCM) of lives for alternatives. Study period method is recommended. Evaluation approach: Determine each PW at stated MARR; select alternative with numerically largest PW. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Different- life Alternatives (cont’d) Study Period of length ? years (periods): • ? is same for each alternative. • If life > ? , use market value estimate in year n for salvage value. • If life < ? , estimate costs for remaining years. Estimates outside time frame of the study period are ignored. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example Problem 4.22 (modified) A company has 2 options to supply personal safety equipment to employees. The MARR is 10% p.a. Select from the two sales vendors over a study period of 3 years. You may assume that the annual upkeep cost for Vendor R doubles after exceeding the estimated life, and it is assumed that to extend the contract of Vendor R for 1 more year, you do not need to pay the initial cost. Use PW analysis. Vendor R Vendor T Initial cost, $ 75,000 125,000 Annual upkeep cost, $/year 27,000 12,000 Salvage value $ 0 30,000 Estimated life, years 2 3 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Different- life Alternatives (cont’d) LCM Method Assumptions (may be unrealistic at times) Same service needed for LCM years (e.g., LCM of 5 and 9 is 45 years!). Alternatives available for multiple life cycles. Estimates are correct over all life cycles (true only if cash flow estimate changes match inflation/deflation rate). Evaluation approach: obtain LCM, repeat purchase and life cycle for LCM years; calculate PW over LCM; select alternative with most favourable PW. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example Problem 4.22 (modified) A company has 2 options to supply personal safety equipment to employees. The MARR is 10% p.a. Select from the two sales vendors using PW analysis Vendor R Vendor T Initial cost, $ 75,000 125,000 Annual upkeep cost, $/year 27,000 12,000 Salvage value $ 0 30,000 Estimated life, years 2 3 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Future Worth Evaluation FW evaluation of alternatives is especially applicable for LARGE capital investment situations when maximising the future worth of a corporation is important. E.G., buildings, power generation, acquisitions. Evaluation approach: Determine FW value from cash flows or PW with an n value in F/P factor. • Equal to study period, or, • Equal to LCM of alternatives’ lives. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Life Cycle Costing (LCC) Another application of PW analysis. Useful when entire life cycle of a system is under evaluation. E.G., new car model or aircraft model; introducing new technology. PW evaluation must include cost estimates for all stages of the product or service: • Design (initial and detail) • Development • Production cost • Marketing cost • Operating costs • Warranty commitments • Phase-out costs • etc. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Capitalized Cost (CC) • PW of alternative that will last ‘forever’. • Especially applicable to public project evaluation (dams, bridges, irrigation, hospitals, police, etc.). • CC relation is derived using the limit as ? → ∞ for the P/A factor. 𝑃𝑊 = 𝐴 𝑃/𝐴, 𝑖%, ? = 𝐴 1 − 1 1 + 𝑖 ? 𝑖 𝐶𝐶 = 𝐴 1 𝑖 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Procedure to find CC 1. Draw diagram for 2 cycles of recurring cash flows and any nonrecurring amounts. 2. Calculate PW (CC) for all nonrecurring amounts. 3. Find AW for 1 cycle of recurring amounts; then add these to all A series applicable for all years 1 to ∞ (or long life). 4. Find CC for amount above using CC = AW/i. 5. Add all CC values (steps 2 and 4). Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example Problem 4.39 (modified) Two alternatives to incorporate improved techniques have been developed. Compare the alternatives below using PW analysis at 12% p.a. compounded quarterly. (3% per quarter) A B Initial cost, $ 2.0m 10m Net income, $ per quarter 0.3m 0.4m Salvage value, $ 50,000 - Estimated life, years 4 ∞ Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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CC Evaluation of Alternatives • For two long-life or infinite-life alternatives: • SELECT ALTERNATIVE WITH LOWER CC OF COSTS. • For one infinite life and one finite life: Determine CC for finite life alternative using AW of 1 life cycle and relation 𝐶𝐶 = 𝐴𝑊/𝑖 • SELECT ALTERNATIVE WITH LOWER CC OF COSTS. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Independent Projects • Situation: Select from several (m) projects. Revenue and costs are estimated for each. • Solution approach: Basically different from that for ME alternatives. • One-time projects; no equal-service evaluation necessary; LCM not necessary. • Two types of budget situations are possible -- no limit or stated limit. • No limit: select from none (DN alternative) to all m projects using criterion. SELECT ALL PROJECTS WITH PW ≥ 0 AT WACC Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Procedure for stated budget limited evaluation • No more than specified amount (b) can be invested and each project must demonstrate PW ≥ 0 at MARR. • Form ME bundles of projects which do not exceed limit. Include DN alternative. There are 2 m ME bundles Procedure: 1. Determine all bundles with total investment ≤ b. 2. Calculate PW of all projects included in bundles. (Note: any bundle with a PW < 0 project can be eliminated now). 3. Add project PW values to get total PW for each viable bundle. 4. Select bundle with largest PW value. These are the projects to accept. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example Problem 4.42 (modified) 3 independent vendor proposals are considered. The corporate MARR is 10% p.a. If a maximum of $4million can be spent today. Develop all possible bundles from the following table and then determined using PW analysis, the most economically justified alternatives. Vendor Initial investment $ Life, years Annual net revenue, $/year A 1.5m 8 360,000 B 3m 10 600,000 C 1.8m 5 620,000 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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