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Lecture # 4 gradients factors and nominal and effective interest rates
1.
Lecture # 4 Gradient
Factors/Shifted gradients 1-1 Dr. A. Alim
2.
Gradients Gradients are
special cases where a series of cash flows consists of regular, unequal amounts that increase or decrease following a specific pattern Gradient factors are the factors used to calculate equivalent P, A and F for such series of cash flows. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-2 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
3.
ARITHMETIC GRADIENT An arithmetic
gradient is a cash flow series that either increases or decreases by a constant amount. The cash flow, whether income or disbursement, changes by the same constant amount each period. The amount of the increase or decrease is the gradient “G”. For example, if a manufacturing engineer predicts that the cost of maintaining a machine will increase by $500 per year until the machine is retired, a gradient series is involved and the amount of the gradient is $500. GEOMETRIC GRADIENT It is also common for cash flow series, such as operating costs, construction costs, and revenues, to increase or decrease from period to period by a constant percentage, for example, 5% per year. This uniform rate of change defines a geometric gradient series of cash flows. In addition to the symbols i and n used thus far, we now need the term g. g = constant rate of change, in decimal form, by which amounts increase or decrease from one period to the next Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-3 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
4.
Arithmetic Gradient Factors (P/G)
and (A/G) Cash flow profile 0 1 2 3 n-1 n A1+G A1+2G A1+(n-2)G A1+(n-1)G Find P, given gradient cash flow G CFn = A1 ± (n-1)G Base amount = A1 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-4 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
5.
Gradient Structure As
we know, arithmetic gradients are comprised of two components 1. Gradient component 2. Base amount When working with a cash flow containing a gradient, the (P/G) factor is only for the gradient component Apply the (P/A) factor to work on the base amount component P = PW(gradient) + PW(base amount) Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-5 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
6.
Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-6 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
7.
Use of the
(A/G) Factor 0 1 2 3 n-1 n G 2G (n-2)G (n-1)G Find A, given gradient cash flow G CFn = (n-1)G Equivalent A of gradient series A A A . . . A A A = G(A/G,i,n) Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-7 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
8.
Use of the
(A/G) Factor A = G(A/G,i,n) The general equation for PT and AT : PT = PA ± PG AT = AA ± AG Important: First A is end of year 1, while first G is end of year 2. Same n is used for both equations ! Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-8 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
9.
The P/G and
A/G factors Remember, there are only two ways to determine these factors: * From tables * From formulas No EXCEL functions for these factors ! Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-9 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
10.
1-10Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved The P/G and A/G factors
11.
1-11Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved The P/G and A/G factors’ formulas
12.
Geometric Gradient Series
Factor Geometric Gradient Cash flow series that starts with a base amount A1 Increases or decreases from period to period by a constant percentage amount This uniform rate of change defines… A GEOMETRIC GRADIENT Notation: g = the constant rate of change, in decimal form, by which future amounts increase or decrease from one time period to the next Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-12 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
13.
Typical Geometric Gradient A1 A1(1+g) A1(1+g)2 .
. . . 0 1 2 3 n-2 n-1 n A1(1+g)n-1 Required: Find a factor (P/A,g %,i %, n) that will convert future cash flows to a single present worth value at time t = 0 Given A1, i%, and g% Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-13 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
14.
Two Forms to
Consider… 1 (1 ) g nA P i 1 1 1 1 g i n g g i P A i g Case: g = i Case: g = i A1 is the starting cash flow There is NO base amount associated with a geometric gradient The remaining cash flows are generated from the A1 starting value No Excel function or tables available to determine this factor…too many combinations of i% and g% to support tables The easiest way to get equivalent A or F is to use A/P and F/P factors with the P shown above. To use the (P/A,g%,i%,n) factor Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-14 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
15.
Shifted payment series/gradients
Shifted uniform series. Shifted arithmetic gradients. Shifted geometric gradients. By definition, a shifted series/gradient means its present value is NOT at time zero. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-15 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
16.
1 - Shifted
Uniform Series 0 1 2 3 4 5 6 7 8 A = $-500/year Consider: Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-16
17.
1 - Shifted
Uniform Series 0 1 2 3 4 5 6 7 8 A = $-500/year Consider: P of this series is at t = 2 (P2) P2 = - 500 (P/A,i%,4) P0 = P2 (P/F,i%,2) P2 P0 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-17
18.
Example of Shifted
Series P and F A = $-500/year • F for this series is at t = 6; F6 = A(F/A,i%,4) • P0 for this series at t = 0 is also P0 = - F6(P/F,i%,6) 0 1 2 3 4 5 6 7 8 P2P0 F6 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-18
19.
2 - Shifted
Arithmetic Gradients The Present Worth of an arithmetic gradient (linear gradient) is always located: One period to the left of the first cash flow in the series ( “0” gradient cash flow) or, Two periods to the left of the “1G” cash flow Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-19 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
20.
Shifted Gradient A
Conventional Gradient has its present worth point at t = 0 A Shifted Gradient has its present value point removed from time t = 0 0 1 2 3 4 5 6 7 8 9 10 P 0 1 2 3 4 5 6 7 8 9 10 P3 P0 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-20 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
21.
Shifted Gradient: Numerical
Example Base Series = $500 G = $+100 0 1 2 3 4 ……….. ……….. 9 10 Cash flows start at t = 3 $500/year increasing by $100/year through year 10; i = 10%; Find P at t = 0 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-21 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
22.
Shifted Gradient: Numerical
Example PW of the base series Base Series = $500 0 1 2 3 4 ……….. ……….. 9 10 P2 = 500(P/A,10%,8) = 500(5.3349) = $2667.45 nseries = 8 time periods P0 = 2667.45(P/F,10%,2) = 2667.45(0.8264) = $2204.38 P2 P0 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-22 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
23.
Shifted Gradient: Numerical
Example PW for the gradient component 0 1 2 3 4 ……….. ……….. 9 10 G = +$100 P2 = $100(P/G,10%,8) = $100(16.0287) = $1,602.87 P0 = $1,602.87(P/F,10%,2) = $1,602.87(0.8264) = $1,324.61 P2 P0 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-23 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
24.
Example: Total Present
Worth Value For the base series P0 = $2204.38 For the arithmetic gradient P0 = $1,324.61 Total present worth P = $2204.38 + $1,324.61 = $3528.99 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-24 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
25.
To Find A
for a Shifted Gradient 1) Find the present worth at actual time 0 2) Then apply the (A/P,i,n) factor to convert the present worth to an equivalent annuity (series) 3) A = P(A/P, i, n) A = $3528.99 (A/P, 10%,10) = $547.36 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-25 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
26.
Using Spreadsheet Functions
(NPV) Excel is a powerful tool to calculate P0 for shifted series or gradients in one step. We use the NPV function. Net Present Value for a shifted series or gradient. Excel function is: =NPV(i%,second_cell : last_cell) + first_cell Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-26 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
27.
Spreadsheet Applications –
NPV Function The NPV function requires that all cells in the defined time range have an entry The entry can be $0…but not blank! A “0” value must be entered Incorrect results can be generated if one or more cells in the defined range is left blank Unlike the PV function, CF values do not need to be equal. In fact, they can be negative or zero values This makes the NPV function particularly suitable for shifted series and/or gradients. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-27 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
28.
Shifted Gradient: Numerical
Example Base Series = $500 G = $+100 0 1 2 3 4 ……….. ……….. 9 10 Cash flows start at t = 3 $500/year increasing by $100/year through year 10; i = 10%; Find P at t = 0 using the NPV function. Then find the equivalent A. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-28 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
29.
Using NPV Function Slide
Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-29 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved YEAR CF $ 0 0 1 0 2 0 3 500 present Worth = $3,529.20 4 600 5 700 Equivalent A = $574.36 6 800 7 900 8 1000 9 1100 10 1200
30.
3 - Shifted
Geometric Gradient Conventional Geometric Gradient 0 1 2 3 … … … n A1 Present worth point is at t = 0 for a conventional geometric gradient Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-30 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
31.
Shifted Geometric Gradient Shifted
Geometric Gradient 0 1 2 3 … … … n A1 Present worth point is at t = 2 for this example Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-31 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
32.
Example 3.7, page
83, Blank’s 7th edition Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-32 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
33.
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34.
35.
Using NPV Function YEAR
CF 0 $35,000 1 $7,000 2 $7,000 3 $7,000 Present Worth = $83,229.83 4 $7,000 5 $7,000 Equivalent A = $14,907.33 6 $7,840 7 $8,781 8 $9,834 9 $11,015 10 $12,336 11 $13,817 12 $15,475 13 $17,331 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-35 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
36.
Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-36 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved CHEE 5369/6369 Homework # 1 Thursday January 23, 2014 The following solved examples from Blank (seventh edition): Example 1.3 page 11 Example 1.4 page 11 Example 1.5 page 12 Example 1.8 page 14 Example 1.9 page 17 Example 1.11 page 18 Example 1.14 page 21 Example 1.15 page 22 Example 2.3 page 44 Example 2.5 page 47 Example 2.8 page 51 Example 2.12 page 60 Example 3.5 page 80 Example 3.8 page 84 Example 14.2 page 372
37.
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accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-37
38.
Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-38
39.
Nominal and Effective
Interest Rate Statements So far, we have learned: Simple interest and compound interest definitions Compound Interest – Interest computed on interest For a given interest period The time standard for interest computations – One Year Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-39 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
40.
Two Common Forms
of Quotation Two types of interest quotation: 1. Quotation using a Nominal Interest Rate 2. Quoting using an Effective Interest Rate Nominal and Effective interest rates are common in business, finance, and engineering economy Each type must be understood in order to solve problems where interest is stated in various ways Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-40 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
41.
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42.
Examples of Nominal
Interest Rates 18% per year Same as: 18/12 = 1.5% per month Same as: 18/4 = 4.5% per quarter Same as: 18/0.5 = 36% per two years 1.5% per 6-month period Same as: (1.5%)(2 six-month periods) = 3% per year 1% per week Same as: (1%)(52 weeks) = 52% per year Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-42 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
43.
Effective Interest Rate Definition: The
effective interest rate is the actual rate that applies for a stated period of time. The compounding of interest during the time period of the corresponding nominal rate is accounted for by the effective interest rate. The effective rate is commonly expressed on an annual basis denoted as “ia” All interest formulas, factors, tabulated values, and spreadsheet relations must have the effective interest rate to properly account for the time value of money. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-43 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
44.
Time Based Units Time
Period – The period over which the interest is expressed (always stated). Ex: “8% per year”, or “1% per month” Compounding Period (CP) – The shortest time unit over which interest is charged or earned. Ex: “8% per year, compounded monthly” Compounding Frequency – The number of times m that compounding occurs within time period t. Ex: “10% per year, compounded monthly” has m = 12 Ex: “1% per month, compounded monthly” has m = 1 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-44 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
45.
Effective Interest Rates
for Any Time Period The following equation is the effective interest rate per time period as related to nominal int. rate and compounding frequency for the same time period. r i = (1+ ) 1 m m Effective where: r = nominal interest rate per time period m = number of compounding periods per time period r/m is the nominal; and also the effective interest rate per compounding period. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-45 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
46.
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47.
Annual Effective Interest
Rates ONLY in this case of ANNUAL rates, there are excel functions as follow: To calculate ia from r use EFFECT(r,m) To calculate r from ia use NOMINAL(ia,m) where: r = Annual nominal interest rate m = number of compounding periods per year r/m is the nominal; and also the effective interest rate per compounding period. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-47 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
48.
Effective Interest Rates
for Any Time Period Note: when the CP is equal to the time period: Then m = 1, and Effective i = r meaning nominal interest rate is also the effective interest rate. When we say interest rate of say 8% per year and do not refer to a compounding period, it usually means that the compounding is annual and the 8% is also the effective rate per year. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-48 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
49.
Example: Calculating the
Effective Rate Interest is 8% per year, compounded quarterly What is the effective int. rate per quarter? What is the effective annual interest rate? Use equation above with r = 0.08, m = 4 Effective quarterly int. rate = nominal quarterly int. rate = 0.08/4 = 0.02 Effective annual ia = (1 + 0.08/4)4 – 1 = (1.02)4 – 1 = 0.0824 or 8.24%/year OR: EFFECT(0.08,4) = 8.24% Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-49 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
50.
In the
previous example: What is the future value of $10,000 invested for 5 years? F = 10,000 (1.0824)5 = $ 14,859 OR F = 10,000 (1.02)20 = $ 14,859 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-50 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
51.
Common Compounding Frequencies
Interest may be computed (compounded): Annually – One time a year (at the end) Every 6 months – 2 times a year (semi-annual) Every quarter – 4 times a year (quarterly) Every Month – 12 times a year (monthly) Every Day – 365 times a year (daily) … Continuous – infinite number of compounding periods in a year. One Year is segmented into: 365 days, 52 weeks, 12 months One Half Year is segmented into: 182 days, 26 weeks, 6 months One Quarter is segmented into: 91 days, 13 weeks, 3 months Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-51 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
52.
Example: APR and
APY APR is Annual Percentage Rate ( Nominal) APY is Annual Percentage Yield ( Effective) Example: A credit card company charges 18 % APR. The Law requires that the APY must also be stated. What is the APY if interest is compounded daily? APY = (1 + 0.18/365)365 – 1 = 19.716 % Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-52 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
53.
Effective Interest Rate
for Continuous Compounding Recall that effective i = (1 + r/m)m – 1 What happens if the compounding frequency, m, approaches infinity? This means an infinite number of compounding periods within a time period, and The time between compounding approaches “0” A limiting value of i will be approached for a given value of r Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-53 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
54.
Continuous Compounding Effective
Rate The effective i for the same time period when interest is compounded continuously is then: Effective i = er – 1 To find the equivalent nominal rate given i when interest is compounded continuously, apply: ln(1 )r i Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-54 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
55.
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56.
Examples 4.11, p.150,
Blank’s 6th edition. Example 4.11 Given r = 18% per year, compounded contin., find: The effective annual rate The effective monthly rate r/month = 0.18/12 = 1.5%/month Effective monthly rate is e0.015 – 1 = 1.511% The effective annual interest rate is e0.18 – 1 = 19.72% per year. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-56 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
57.
Example 4.11 Note: The
following is critically important. You obtain the nominal monthly rate by dividing the nominal annual rate by 12. You CAN NOT divide the effective annual interest rate by 12 In order to obtain the monthly effective interest rate. You must follow this sequence: Nominal annual rate → Nominal monthly rate → Effective monthly rate. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-57 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
58.
Slide Sets to
accompany Blank & Tarquin, Engineering Economy, 7th Edition, 2012 1-58 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved Example 4.12, page 150, Blank’s 6th edition
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