EMIS 2360 Fall 2009
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Comparing Apples and Oranges with Ratios and Rates
Your bags of money: B1 and B2
You can only put one bag at risk in an investment at this time.
Bag B1 of your money could go to banker Apple.
Bag B2 of your money could go to banker Orange.
After a set period of time the chosen banker returns the bag to you with more money in it.
How does a (return) ratio operate? Call the ratio R.
How does a (growth) rate operate? Call the rate r.
R and r are only pure numbers and must operate on something (investment money in this example)
to make sense for comparisons.
Rudimentary Cash Flow Diagrams
(1) labeled time line
(2) directional arrow representation of economic exchange(s) from a single viewpoint
Notation: Ft = net cash flow at time t.
What is a project?
We take a project to be characterized only by its cash flows and
thus it is given completely by its cash flow diagram.
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Assumption: somehow we can specify a rate which reflects what money is worth to us over time.
Interest Rate Over a Single Period
Simple vs. Compound Interest
over Multiple Periods
Availiability diagram and its relationship to the associated simple cash flow diagram.
Doubling your money
at i = 100%: doubles in one period at simple or compound interest
To double in 2 years:
under simple interest solve 2P = P(1+2i) => 1 = 2i => i = 50%
under compound interest solve 2P = P(1+i)2 => 21/2 = 1+i => i = 41.4%
at i = 10% simple:
solve 2P = P(1+n(.1)) => 1 = n(.1) => n = 10
at i = 10% compound:
solve 2P = P(1+(.1))n => 2 = (1.1)n;
take ln = loge or log = log10
of both sides
using ln => n = ln(2)/ln(1.1) = 7.27254...;
in courses and textbooks like ours it is always assumed
that under compound interest,
interest only accrues at the end of a complete interest period (said to be not
prorated); so the answer is 8 periods.
Also, in courses and textbooks like ours it is always assumed
that under simple interest,
interest will accrue at points between complete periods
(said to be prorated); so in simple interest a fractional calculated answer for number of years is
not rounded up, but is correct.
Rule of 72 and Rule of 100:
n(100i) = 72 approximately for compound interest
n(100i) = 100 for simple interest
Or, if i is given in %,
n = 72/i approximately for compound interest
n = 100/i for simple interest
Availiability diagram and its relationship to the associated simple cash flow diagram.
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Homework 1
Example problems 1.11 and 1.16
The columns in the Compound Interest Factor Tables
with heading "Compound Amount F/P" are values of (1+i)n.
Doubling and the compound interest factor tables:
( 1 + i )n
Remember:
simple interest acts in an "additive" manner
and
compound interest acts in a "multiplicative" manner
The second sentence after [1.9] on p. 35 (it's wrong)
What's with these rates?
compound interest at i = 0%?
compound interest at i = -100%?
compound interest at -100% < i < 0%?
compound interest at i > 0% ?
compound interest at i < -100% ?
We never use any rates less than -100%!
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Homework 2
Example Problems 1.41 + modified(simple), 1.42 + modified(simple)
What is the end-of-period convention?
End-of-Period Convention (pp. 31-32)
All cash flows are assumed to occur at the end of an interest period.
When several receipts and disbursements occur within a given interest period,
the net cash flow is assumed to occur at the end of that interest period.
An alternative to the end-of-period convention:
the accountant's convention
Accountant's Viewpoint Convention
Disbursements move left and receipts move right (to the nearest accrual time point).
In this course, unless otherwise stated, we use the End-of-Period Convention
Interest-bearing account details -- not the end-of-period convention.
Equivalence
Equivalence vs Realization
Example Problem 1.16 (revisited, now that we know what equivalence is)
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[Projected]
Began repayment schedules
Repayment Schedules
Practice Repayment Schedule
Terminology
The Mnemonic Worth Factors
and their Algebra
F/P and P/F factors
(F/P,i,n) = F/P = (1+i)n
(P/F,i,n) = P/F = (1+i)-n
F/P = 1/(P/F)
P/F = 1/(F/P)
Example Problem 2.9
The factor F/A = (F/A,i,n)
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Homework 3
Derivation of the factor F/A = (F/A,i,n) = [(1+i)n - 1]/i
F = A(1+i)n-1 +
A(1+i)n-2 +
... +
A(1+i)2 +
A(1+i) + A
(1+i)F = A(1+i)n +
A(1+i)n-1 +
... +
A(1+i)3 +
A(1+i)2 +
A(1+i)
iF = (1+i)F - (1)F =
A(1+i)n - A
Why?
(F/A,i,n) = F/A =
[(1+i)n - 1]/i
A/F, P/A, and A/P factors
(A/F,i,n) = 1/(F/A) = 1/(F/A,i,n) =
i/[(1+i)n - 1]
(P/A,i,n) = P/A = (P/F)(F/A) = (F/A)/(1+i)n
=
[(1+i)n - 1]/[(1+i)ni]
(A/P,i,n) = A/P = 1/(P/A) =
[(1+i)ni]/[(1+i)n - 1]
Example Problem
2.16
Example Problem 2.25:
Interpolation "without formulas".
Interpolation by averaging (the case of needing values for an argument
1/2 way in between given table values) or proportional parts for the
more general case.
Arithmetic Gradient -- the G pattern
the G pattern is equivalent to an A pattern with G/i flow amount + an F pattern with a -nG/i flow amount
You can use the equivalent A plus F patterns to generate (P/G,i,n), (F/G,i,n), and (A/G,i,n) factor formulas.
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Adding two A+G patterns produces a resultant A+G pattern:
a combination of A1 and G1
added to a
a combination of A2 and G2
produces a combination with
A = A1+A2
and
G = G1+G2
Example Problem 2.38
Obtaining formulas for (A/G,i,n),(P/G,i,n), and (F/G,i/n).
Example Problem 2.54
Example Problem 2.55
from the pad
Homework 4
A hack for starting a ROR search:
Example 7.2
Geometric Gradient
flow amount A1 at t=1 with flow amount growth at rate g from t=2 to t=n
Moving each individual cash flow to time n we have that
F = A1(1+i)n-1
+ A1(1+g)(1+i)n-2
+ A1(1+g)2(1+i)n-3
+ ...
+ A1(1+g)n-3(1+i)2
+ A1(1+g)n-2(1+i)
+ A1(1+g)n-1
Case 1: when i=g,
F = A1(1+i)n-1
+ A1(1+i)n-1
+ A1(1+i)n-1
+ ...
+ A1(1+i)n-1
+ A1(1+i)n-1
+ A1(1+i)n-1
F = nA1(1+i)n-1
= nA1(1+g)n-1
P
= nA1/(1+i)
= nA1/(1+g)
F
= nA1 (1+i)n-1
= nA1 (1+g)n-1
Case 2: when i is not g,
P = A1[1 - (1+g)n/(1+i)n]/(i-g)
F = A1[(1+i)n - (1+g)n]/(i-g)
Textbook Formula Sheet(+)
Example Problem 2.42
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Homework 5
Another hack for starting a ROR search:
Factor Approximation:
(A/P,i,n) is approximately 1/n + i
Example Problem 2.51
Example 2.16 (p.80) solutions using ingenuity
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Book's Ch. 3 philosophy on cash flow pattern technique: "chunkify"
Book's Ch. 3 philosophy on cash flow pattern technique vs. ingenuity
Example Problems 3.4, 3.5
Homework 6
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Example Problem 3.49
Begin Ch. 4
What exactly is an "effective rate"?
On a labeled cash flow diagram, the effective rate
is the rate which specifies what happens to (a representative) $1
in moving from one labeled time point to the next one,
relative to the underlying compound interest process.
Typically the labeling of the cash flow diagram has been done
to suit the purposes of the analyst.
Most likely this is to match the timing of the cash flows themselves,
not the compounding period.
Ex.
F2 = F4 = ... =
F48 = F50 = 100;
iyr = 10%. What is the equivalent future worth (at t=50)?
Converting given patterns to familiar ones via relabelling:
requires an effective rate for the relabelled diagram.
Traditional framework: nominal rate (annual)
is divided by the number of implied subperiods per year
in the interest specification to obtain the actual compounding
rate applied in each subperiod.
Ex. $1000 is borrowed for 5 years at 12% compounded monthly.
What are the payments?
What is the total interest paid?
Payoff computations.
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Ex.
F2 = F4 = ... =
F48 = F50 = 100;
the interest rate is 12% compounded monthly.
What is the equivalent future worth (at t=50)?
Example Problem 4.4 + effective rates
Terminology
Example Problem 4.3 + computing yearly effective rate iyr
Example Problem 4.28
Homework 7
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Example Problem 4.42
Subannual Compounding
Continuous Compounding
Approaches to continuous compounding problems
I. Find iyr = er - 1 and use regular formulas.
II. Substitute i = er-1 and 1+i = er into the
regular formulas to create your own continuous compounding formulas.
Then use those new formulas with the given nominal rate r.
Effective Continuous Rates:
for a fractional part f of a year, rf = (f)(ryr)
Example problem using [F/P,r,1.5]
Example Problem
4.51
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Strategy for dealing with interest rates that change with time:
accumulate worth at the boundaries where rates change.
Example Problems
4.53, 4.55
Start Ch. 5
Properties of groups or bundles of projects:
Independence, Mutual Exclusivity
One of our major themes: project selection
Three main questions: Q1, Q2, Q3
Q1: is a given project worth doing?
Q2: given two projects, which would we prefer to do?
Q3: given a group of possible projects, which combination of projects, if any, should we do?
Homework 8
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Example Problem 5.4
Feasible patterns for mutually exclusive projects.
Contingency:
(1) a project A is contingent upon another project B
if in order for project A to be selected project B must also be selected
(2) a project A is contingent upon condition C
if in order for project A to be selected condition C must be satisfied
Q3: given a group of possible projects, which combination of projects, if any, should we do?
Example Problem 12.5
Homework 9
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Example Problem 12.5 continued
Q2: given two projects, which would we prefer to do?
Q2 methodology
1. Ranking using one of the worths PW(MARR),FW(MARR), or AE(MARR)
-- byproduct: Q1 is answered for both projects
2. answering Q1 for the incremental project -- byproduct: Q1 is unknown for both projects
Incremental Analysis Example: C vs. E from problem 12.5
How do we handle differing time spans (lives)?
Projects must be compared over equal time spans (lives)
Notation: A[k] is the project A repeated k times with k-1 extra copies (A[1] = A)
Repeatability assumption
If two projects A and B having differing lives
n1 and n2
are repeated for comparison,
they should be repeated to a common life.
Repeated project A[n2] and
repeated project B[n1]
have a common life of
n1n2,
but this is in general too large a horizon
for efficienct computations.
Best common life = LCM(n1,n2)
= the least common multiple of the lives.
Note:
LCM(n1,n2)
can be obtained by producing the prime factorizations of
n1 and n2: the least common multiple
will be the product of the maximum number of the distinct primes
occuring in both factorizations.
Ex. 12 = 2*2*3 , 18 = 2*3*3 , LCM(12,18) = (2*2)*(3*3)
Example Problem 5.13
from the pad
from the pad -- quiz 6
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Computing Annual Equivalent for
projects over differing time spans (lives): Annual equivalent is the same for repeated
projects as for the original projects and the capital recovery formula makes computation
more efficient.
Capital Recovery Formula: CR(i) = (P-F)(A/P,i,n)+Fi (presented in CH. 6)
Applies to the skeleton of a project: initial cost P (labels a down arrow)
and salvage value F (labels an up arrow)
CR(i) is a cost; it will label a down arrow
More effective strateges for repeated projects:
(1) compute AW of original projects; then the AW of the repeated project is the same;
if you want PW or FW use a P/A or F/A factor over the time horizon of the repeated project
(2) use PW or FW of original projects; find effective rate for the repeated period;
use appropriate P/A or F/A factors to generate PW or FW of repeated projects
Infinite Horizon Formulas
Infinite A pattern factors:
(P/A,i,infinity) = 1/i or (A/P,i,infinity) = i
when i > 0
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quiz 7
More Infinite Horizon Formulas
Infinite G pattern factor:
(P/G,i,infinity) = 1/(i2) when i > 0
Infinite geometric gradient (A1,g) pattern factors:
(P/A1,g,i,infinity) = 1/(i-g) when i > g
Capitalized Equivalent CE(i) or Capitalized Cost CC(i)
Example Problem 5.30
What about salvage value for an infinite project?
It is never realized!
Infinite Horizon Example:
i = 7%;
F0 = 0, F1 = F2 = 1400,
F3 = F4 = ... = 2100.
Solution 1: CE(7%) = PW(7%) = (2100/.07)-700(P/A,7%,2)
Solution 2: CE(7%) = PW(7%) = (1400/.07)+(700/.07)(P/F,7%,2)
Solution 3: CE(7%) = PW(7%) = 1400(P/A,7%,2)+(2100/.07)(P/F,7%,2)
Example Problem 5.32
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Homework 11
Begin Payback
Payback occurs at time point t if the accumulated worth there is nonnegative
and the accumulated worth at any time point s < t is negative.
The interest rate i used must be specified.
Example Problem 5.39
Any worth computation can be used: AE(i) or FW(i) or PW(i) computed for cash flows
only from time point 0 to time point t, keeping in mind
that this is computed for a "truncated" project: one whose cash flows stop
after time point t.
Practically speaking, payback at time t means that the initial investment has been made up
(in worth) by the cash flows up to the that point or the project has "broken even" at that point.
If the interest rate used is greater than 0% the payback is
payback with interest
or
discounted payback
.
If the interest rate used is 0% the payback is
payback without interest
or
undiscounted payback
.
Begin Bonds
Bond particulars: par or face value, maturity date or period,
interest particulars, selling or purchase price
Interest particulars typically include a (nominal) rate and an indication of the number of
subperiods per year where bond interest payments are made
I like to organize bond information in tabular form:
Bond Format Example
Example Problem 5.47 (set up)
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Homework 12
Bond cash flows fall naturally into a skeleton with P = purchase price and F = par value
and a uniform A = periodic interest payment.
Some Bond quantitites of interest: current or coupon rate, yield to maturity
current rate = (total interest payments per year)/(selling price)
yield to maturity - an instance of Rate of Return
Example Problem 5.49
Begin Ch. 6 (not much new for us)
Capital Recovery (Sect. 6.2 pp. 220-223)
Capital Recovery Formula: CR(i) = (P-F)(A/P,i,n)+Fi
Applies to the skeleton of a project: initial cost P (labels a down arrow)
and salvage value F (labels an up arrow)
CR(i) is a cost; it will label a down arrow
Derivation moving the F portion of P = (P-F)+F at time 0 forward in time, dropping off the Fi part
at each subsequent time point.
(A/F,i,n) = i/[(1+i)n - 1]
(A/P,i,n) = i(1+i)n/[(1+i)n - 1]
(A/P,i,n) - i
= i(1+i)n/[(1+i)n - 1] - i
= i(1+i)n/[(1+i)n - 1] - i[(1+i)n - 1]/[(1+in - 1]
= [i(1+i)n - i(1+i)n + i]/[(1+in - 1]
= i/[(1+in - 1]
= (A/F,i,n)
(A/F,i,n) = (A/P,i,n) - i => P(A/P,i,n) - F(A/F,i,n) = (P-F)(A/P,i,n) + Fi
The problem with p. 221
Alternative to the "repeatability assumption" approach to differing lives:
the study period approach.
In a service environment the study period approach often requires that estimates be obtained.
Problems 6.1 and 6.2
are related to
the study period approach.
Example Problem 6.1
Set up Problem 6.2
Example Infinite Horizon Problems from Ch. 6:
6.18, 6.19
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Homework 13
Begin Ch. 7
The definition of ROR
Your book's definition:
ROR is the rate earned on the unrecovered balance of an
investment so that the final receipt brings the balance
to exactly zero with interest considered.
Another (better) definition:
A rate of return is any solution obtained by equating to zero any worth function for a project.
PW(i) = 0
FW(i) = 0
AE(i) = 0
All give the same solution(s)
Future worth function:
FW(i) =
F0(1+i)n +
F1(1+i)n-1 +
... +
Fn-2(1+i)2 +
Fn-1(1+i) +
Fn
Using the transformation x = 1+i (reverse transformation i = x-1)
we can form the future worth polynomial from FW(i).
P(x) =
F0xn +
F1xn-1 +
... +
Fn-2x2 +
Fn-1x +
Fn .
Resolving the book's ROR definition and ours
P(x) = FW(i) ; each root of P(x) corresponds to a rate of return.
If i* is a ROR then x* = 1+i* is a root of P(x) and vice-versa.
Example problem 7.7 (finding ROR by hand)
Descarte's Rule of Signs: if S = the number of sign changes
in the sequence of net cash flows
F0,
F1,
F2,
...,
Fn
(ignoring any zero net cash flows), then
the number of RORs > -1 is S or S-2 or ... (stop before S-2-...-2 is negative).
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[Projected]
quiz 8
Project Selection (Q1) using ROR as the figure of merit:
a project is worth doing if it has a unique rate of return (ROR) and
ROR >= MARR.
Note: if a project has multiple rates of return this figure of merit cannot be used.
Lorie-Savage Problem
Famous problem having to do with speeding up oil well flow.
(Reference: Lorie, J.H. and Savage, L.J., "Three Problems in Rationing Capital,"
Journal of Business, XXVIII(4), Oct. 1955).
F0 = -1,600 ,
F1 = 10,000 ,
F2 = -10,000
Descartes' rule of signs says there are either 2 or 0 positive roots of P(x).
P(x) =
-1,600 x2 +
10,000 x - 10,000
= -400 (4x2 -25x + 25)
= -400 (4x - 5)(x - 5)
Thus there are two roots:
x1 = 1.25 and
x2 = 5 .
The corresponding rates of return are
i1 = 1.25-1 = 25% and
i2 = 5-1 = 400% .
Since we do not have a unique ROR we cannot use rate of return as the
figure of merit for project selection in this case.
How to solve for ROR(s) using a computer:
Using goal-seeking, RATE, and IRR under EXCEL
Handout 1
Using a Rootfinder (Bairstow's Method)
Handout 2
How to solve for (a) ROR by hand:
(1) Obtain a guess for i* by some hack method
(2) Set up to solve one of the equations
PW(i) = 0
FW(i) = 0
AE(i) = 0
the intermediate worth at time t = IW(i)t = 0
You may wish to divide by some constant(s) to obtain "nice" numbers in the equations
(3) bracket i* between i values by a table search starting near the guess from (1)
(4) obtain a more accurate (one extra digit) value for i* by interpolation
Comments on the book's hack
for a guess for i* in Example
7.2 on pp. 245-246
F0 = -500k, F1=
...=F9 = 10k, F10=10k+700k
The 10k values were all moved to t=10 so that the hacked diagram became
F0 = -500k, F1=
...=F9 = 0k, F10=10(10k)+700k= 800k
500k(1+i)10 = 800k was solved for the guess for i* =
(800k/500k).1-1 = 4.8%
This hack is good when your flows can be viewed as small As and bigger P and F values
which are not close to equal.
Good hacks for bond ROR problems or anywhere the flows can be viewed as small
with bigger P and F values that are close to equal:
The Q&D approximation:
i = A/[(P+F)/2]
Helgason's approximation:
i = A/P - (1/n)(P-F)/P
or
i = [ A - (1/n)(P-F) ]/P
Hacks for 7.27 and 7.29
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Homework 14
Example problem 7.19
How to tell if a unique ROR is present
(1) Descarte's rule: if S = the number of sign changes
in the sequence of net cash flows
F0,
F1,
F2,
...,
Fn
(ignoring any zero net cash flows), then
the number of RORs > -1 is S or S-2 or ... (stop before S-2-...-2 is negative).
If S = 1, there is a unique ROR > -1 .
(2) Norstrom's criterion: if
PB(0%)0 < 0
and there is exactly one sign change in the sequence of project balances
PB(0%)0 ,
PB(0%)1 ,
... ,
PB(0%)n-1 ,
PB(0%)n
then there is a unique ROR > 0.
(3) Intermediate balances criterion:
if i* is known to be a ROR,
PB(i*)0 < 0
and
the intermediate project balances
PB(i*)1 ,
... ,
PB(i*)n-1
are all nonpositive,
then i* is the unique ROR.
Chapter 8
Examples showing that incremental analysis must be used when answering
question Q2 (which project would you prefer to do) using ROR as the
figure of merit (measure of goodness)
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