Week 6

Report
INFO 630
Evaluation of Information Systems
Prof. Glenn Booker
Week 6 – Chapters 4-6
INFO630 Week 6
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The Business Decision-Making
Process
Slides adapted from Steve Tockey – Return on Software
INFO630 Week 6
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Overview
• For any technical problem
– Usually many viable technical solutions
• Goal for technical person is to:
– Make the most of the organization limited
resources
– By choosing the solutions that maximizes the
return on the software investment
• Why do care about this?
– Possibly large difference is cost and income for
the different solutions
– How come?
INFO630 Week 6
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Business Decision-making Process
Outline
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
4
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Comments on the Process
• This same process applies at all levels of
business decision
– Smaller scale decisions can be done less
formally
• The process is more fluid than implied
– Steps can be overlapped or parallel
– Steps can be done in different orders
INFO630 Week 6
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The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
6
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Understand the Real Problem
• Obvious
– but often overlooked
• In software, this is usually the
“requirements”
– Issues in contemporary requirements
• Ambiguity
• Incompleteness
• Mistaking a solution for the problem
– Analyze separate decisions separately
INFO630 Week 6
7
www.ischool.drexel.edu
The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
8
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Define the Selection Criteria
• Selection criteria need to be
– Unique
– Sufficient
– Meaningful
– Discriminating
INFO630 Week 6
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Typical Selection Criteria
•
Financial
–
–
–
–
–
•
Technical
–
–
–
–
–
•
Initial investment
Present worth (Net present value)
Internal rate of return
Discounted payback period
…
Performance
Reliability
Maintainability
Compatibility
…
Non-technical
– Reputable provider
– Creature comfort
– …
INFO630 Week 6
10
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The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
11
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Identify Reasonable Technically-feasible
Solutions
• We’re usually pretty good at this…
– Creative/lateral thinking helps (see
[DeBono92] or [vonOech98])
INFO630 Week 6
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The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
13
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Evaluate Each Proposal Against the
Selection Criteria
Proposals
Financial
Risk
Morale
Extend
$66,021
0.40
1.00
Fix defects
$58,056
0.20
0.50
Client-server
$76,605
0.50
0.80
INFO630 Week 6
14
www.ischool.drexel.edu
The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
15
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Select the Preferred Proposal
• Comparing proposal from a financial
perspective main topic of course
• More detail to follow
– For-Profit: Ch 10-17
– Non-Profit: Ch 18
INFO630 Week 6
16
www.ischool.drexel.edu
The Business Decision Making
Process
Define the
selection criteria
Understand the
real problem
Identify all
reasonable
technically -feasible
solutions
Select the prefe rred
proposal
INFO630 Week 6
Evaluate each
proposal against the
selection criteria
Monitor the
performance of the
selected proposal
17
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Monitor the Performance of the Selected
Proposal
• Quality of decision based on
“estimation”
– Bad estimate -> Bad decisions
– Close the loop
• compare original to actual
• Improve estimation technique
INFO630 Week 6
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Monitor the Performance of the Selected
Proposal
• Refine your estimation technique
– Look at where you’ve been
• Have you been meeting expectations?
• Cash-flow stream matching actual cash flow?
• If out of “sync” with reality, switch?
– Look at where you are
• Earned value
– Ratio of estimated effort and schedule for WBS tasks already
completed to the actual effort and schedule for the same tasks
– Look at where you’re going
• Improve future estimates
• Common mistake
– Resources available 100% to project
INFO630 Week 6
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Key Points
• There is a systematic process for making
business decisions
• The process applies at many scales
• The process is more fluid than implied
here
INFO630 Week 6
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Interest:
The Time Value of Money
Slides adapted from Steve Tockey – Return on Software
INFO630 Week 6
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Interest: Time-Value of Money
Outline
•
•
•
•
•
•
Time is money
Time value is quantifiable: interest
Naming conventions in interest formulas
Simple interest
Compound interest
Compound interest formulas
– Using interest tables
• Selecting an interest formula
INFO630 Week 6
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Time is Money
• Fundamental concept in business
– A given amount of money at one time doesn’t
have the same value as the same amount of
money at a different time
– In other words
• Its value changes over time
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An Experiment
• Give one person $10 now
• Promise to give another person $10 later
• Questions:
– Who is better off, and why?
– How much better off are they?
– Would the $10 later person be willing to give
up some in order to get it now?
– Would the $10 now person be willing to wait if
we promised to give them more later?
INFO630 Week 6
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Time Value is Quantifiable
• Interest
– Money someone pays to use someone else’s
money
– Literally, a rental fee for money
• Evidence as early as 2000 BC
• Interest rate
– Specifies the rental fee as a percent of the
amount loaned, e.g., 6.825%
• Assumed to be annual unless noted
INFO630 Week 6
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Interest Rate
• What makes up an interest rate?
INFO630 Week 6
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The Lender’s Perspective
• Probability the borrower won’t repay
– $3 default per $100 loaned  3%
• Cost of setting up and administering
– $2 per $100 loaned  2%
• Compensation for loss of use of their own money
– 4.5%
• Probability prevailing interest rate will change
– 0.5%
• This lender should ask for 10% interest rate
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The Borrower’s Perspective
• Personal use
– E.g. vacation, new car, house, …
– How much is borrower willing to pay for
satisfaction now rather than later?
• Business use
– E.g., expand office space, fund new product
development, buy new equipment, …
– Expected return should be higher than loan
interest rate
INFO630 Week 6
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Interest Rate
• In general,
– Interest is thought of as a return that can be gained from
productive investment of money
• Will investigate different formulas of “interest”
– There are a set of standard formulas that allow you to convert
the value of money at one point in time to the value of a different
amount of money at some other time.
• First
– Terms used
INFO630 Week 6
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Naming Conventions in Interest Formulas
These are critical concepts!!!
•
P
– “Principal Amount”—how much is the money worth right now?
– Also known as “present value” or “present worth”
•
F
– “Final Amount”—how much will the money be worth at a later time?
– Also known as the “future value” or “future worth”
•
i
– Interest rate per period
– Assumed to be an annual rate unless stated otherwise
•
n
– Number of interest periods between the two points in time
•
A
– “Annuity”—a stream of recurring, equal payments that would be due at the end of
each interest period
INFO630 Week 6
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Simple Interest
• Interest is directly proportional to P, n, i
I  Pn i
unknown
0
known
1 2
3
n-1
n
F  P  I  P1  ni
• Borrowing $15k at 8% for 5 years
– I = $15k x 5 x 0.08 = $6k
– F = P + I = $15k + 6k = $21k
INFO630 Week 6
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Compound Interest
• Simple interest favors the borrower
– Lender wants to consider unpaid, accrued interest as part of the
loan (compound interest)
• Literally, lender wants to get interest on interest
– Hence simple interest is not used often
• Borrowing $15k at 8% for 5 years
–
–
–
–
–
–
Borrower owes $15.0k + $1.2k = $16.2k after 1st year
Borrower owes $16.2k + $1.3k = $17.5k after 2nd year
Borrower owes $17.5k + $1.4k = $18.9k after 3rd year
Borrower owes $18.9k + $1.5k = $20.4k after 4th year
Borrower owes $20.4k + $1.6k = $22.0k after 5th year
Compare this to simple interest case
INFO630 Week 6
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Compound Interest (cont)
I k+2
Ik+1
Ik
Pk
Tk
Pk+1
Tk+1
INFO630 Week 6
Pk+2
Tk+2
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Compound Interest Formulas
• Six different compound interest formulas
– Single-payment compound-amount (F/P)
– Single-payment present-worth (P/F)
– Equal-payment-series compound-amount
(F/A)
– Equal-payment-series sinking-fund (A/F)
– Equal-payment-series capital-recovery (A/P)
– Equal-payment-series present-worth (P/A)
INFO630 Week 6
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Single-Payment Compound-Amount (F/P)
• Most straight forward of the six compound interest
formulas
– Single payment at the end of a loan
– Includes all of the compounded interest
– Calculates unknown future value of some know present value
(F given P)
• Generic cash-flow diagram – no payments during
the loan!
unknown
0
• Example
known
1 2
3
n-1
n
– How much will be owed if you borrow $15k at 8% for 5 years?
INFO630 Week 6
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Single-Payment Compound-Amount (F/P)
• Deriving the formula
Year Owed at start
Interest accrued Owed at end of year
P
Pi
2
P(1+i)
P(1+i)i
P(1+i) + P(1+i)i = P(1+i)
3
P(1+i)2
P(1+i) i
P(1+i)2 + P(1+i) 2 i = P(1+i) 3
n
P(1+i)
n-1
P + Pi = P(1+i)
1
1
n-1
P(1+i)
2
2
i
P(1+i) n-1+ P(1+i)n-1i = P(1+i)n
INFO630 Week 6
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Single-Payment Compound-Amount (F/P)
• Formula
n


F  P 1 i
• Solving the sample problem
– How much will be owed if you borrow $15k at 8% for 5 years?
F  $15k1  0.08  $22k
5
• Shorthand notation
F/P, i, n
F=P(
)
INFO630 Week 6
F/P, 8%, 5
F = $15k (
)
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How much will be owed if you borrow
$15k at 8% for 5 years?
Year
Owed at start
Interest accrued
Owed at end of year
1
$15K
$15K * .08 = 1.2K
$16.2K
2
$16.2K
$16.2K * .08 = 1.3K
$17.5 K
3
$17.5 K
$17.5 K * .08 = 1.4K
$18.9 K
4
$18.9 K
$18.9 K * .08 = 1.5K
$20.4 K
5
$20.4 K
$20.4 K * .08 = 1.6K
$22 K
www.ischool.drexel.edu
Interest Tables –
Conversion Factor (Page 498)
Table B-12
8% Interest Factors for Discrete Compounding
Single-Payment
Compound- PresentAmount
Worth
Find F Given P
Find P Given F
Equal-Payment-Series
CompoundAmount
SinkingFund
PresentWorth
CapitalRecovery
Find F Given A
Find A Given F
Find P Given A
Find A Given P
n
(F/P,i,n)
(P/F,i,n)
(F/A,i,n)
(A/F,i,n)
(P/A,i,n)
(A/P,i,n)
1
2
3
4
5
1.0800
1.1664
1.2597
1.3605
1.4693
0.9259
0.8573
0.7938
0.7350
0.6806
1.0000
2.0800
3.2464
4.5061
5.8666
1.0000
0.4808
0.3080
0.2219
0.1705
0.9259
1.7833
2.5771
3.3121
3.9927
1.0800
0.5608
0.3880
0.3019
0.2505
6
7
8
9
10
1.5869
1.7138
1.8509
1.9990
2.1589
0.6302
0.5835
0.5403
0.5002
0.4632
7.3359
8.9228
10.6366
12.4876
14.4866
F/P, 8%, 5
0.1363
0.1121
0.0940
0.0801
0.0690
4.6229
5.2064
5.7466
6.2469
6.7101
0.2163
0.1921
0.1740
0.1601
0.1490
So
F = $15k (
)
= $15K (1.4693) = $22.0395
INFO630 Week 6
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Single-Payment Present-Worth (P/F)
• P given F (P/F)
– Calculates the unknown present value needed to return a future
value
• Generic cash-flow diagram
known
0
unknown
1 2
3
n-1
n
• Example
– How much would need to be deposited at 8% to end up with
$15k after 5 years?
INFO630 Week 6
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Single-Payment Present-Worth (P/F)
• Deriving the formula
– Rearrange the Single-payment CompoundAmount formula…
F  P1  i 
n
– … to solve for P
 1

P  F
n 


1

i


INFO630 Week 6
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Single-Payment Present-Worth (P/F)
• Formula
 1

P  F
n 
 1  i  
• Solving the sample problem
– How much would need to be deposited at 8% to end up with
$15k after 5 years?
P  $15k
1
 $10.2k
1  0.085
• Shorthand notation
P/F, i, n
P=F(
)
INFO630 Week 6
P/F, 8%, 5
P = $15k (
)
42
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Interest Tables –
Conversion Factor (Page 498)
Table B-12
8% Interest Factors for Discrete Compounding
Single-Payment
Equal-Payment-Series
Compound- PresentAmount
Worth
Find F Given P
CompoundAmount
Find P Given F
SinkingFund
PresentWorth
CapitalRecovery
Find F Given A
Find A Given F
Find P Given A
Find A Given P
n
(F/P,i,n)
(P/F,i,n)
(F/A,i,n)
(A/F,i,n)
(P/A,i,n)
(A/P,i,n)
1
2
3
4
5
1.0800
1.1664
1.2597
1.3605
1.4693
0.9259
0.8573
0.7938
0.7350
0.6806
1.0000
2.0800
3.2464
4.5061
5.8666
1.0000
0.4808
0.3080
0.2219
0.1705
0.9259
1.7833
2.5771
3.3121
3.9927
1.0800
0.5608
0.3880
0.3019
0.2505
6
7
8
9
10
1.5869
1.7138
1.8509
1.9990
2.1589
0.6302
0.5835
0.5403
0.5002
0.4632
7.3359
8.9228
10.6366
12.4876
14.4866
0.1363
0.1121
0.0940
0.0801
0.0690
4.6229
5.2064
5.7466
6.2469
6.7101
0.2163
0.1921
0.1740
0.1601
0.1490
P/F, 8%, 5
So
P = $15k (
)
= $15K (.6806) = $10.209K
INFO630 Week 6
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Equal-Payment-Series Compound-Amount (F/A)
• F given A (F/A)
– How much end up with based on series of equal payments made
over time
• Retirement account
• Generic cash-flow diagram
unknown
0
1
2
3
n-1
n
known
• Example
– How much would you end up with if you invested $1k at 8% at
the end of each of the next 5 years?
INFO630 Week 6
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Equal-Payment-Series Compound-Amount (F/A)
• Deriving the formula
F = A(1) + A(1+i) + … + A(1+i)
n-2
+ A(1+i)
n-1
Multiply by (1+i)
F(1+i) = A(1+i) + A(1+i) + … + A(1+i)
n-1
+ A(1+i)
n
Subtract the first equation from the second
F(1+i)
n-1
=
A(1+i) + A(1+i) 2 + … + A(1+i) + A(1+i)
2
n-1
-F = -A - A(1+i) - A(1+i) - … - A(1+i)
n
A(1+i) n
F(1+i) - F = -A +
Solve for F
 1  i n  1 
F  A

i


Note: A1 = last payment – has no interest
A(1+i) is the 2nd to last payment, has interest
INFO630 Week 6
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Equal-Payment-Series Compound-Amount (F/A)
• Formula
 1  i n  1 
F  A

i


• Solving the sample problem
– How much would you end up with if you invested $1k at 8% at
the end of each of the next 5 years?
5

1  0.08  1
F  $1k
 $5.9k
0.08
• Shorthand notation
F/A, i, n
F=A(
)
INFO630 Week 6
F/A, 8%, 5
F = $1k (
)
46
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Interest Tables –
Conversion Factor (Page 498)
Table B-12
8% Interest Factors for Discrete Compounding
Single-Payment
Equal-Payment-Series
Compound- PresentAmount
Worth
Find F Given P
CompoundAmount
Find P Given F
SinkingFund
PresentWorth
CapitalRecovery
Find F Given A
Find A Given F
Find P Given A
Find A Given P
n
(F/P,i,n)
(P/F,i,n)
(F/A,i,n)
(A/F,i,n)
(P/A,i,n)
(A/P,i,n)
1
2
3
4
5
1.0800
1.1664
1.2597
1.3605
1.4693
0.9259
0.8573
0.7938
0.7350
0.6806
1.0000
2.0800
3.2464
4.5061
5.8666
1.0000
0.4808
0.3080
0.2219
0.1705
0.9259
1.7833
2.5771
3.3121
3.9927
1.0800
0.5608
0.3880
0.3019
0.2505
6
7
8
9
10
1.5869
1.7138
1.8509
1.9990
2.1589
0.6302
0.5835
0.5403
0.5002
0.4632
0.1363
0.1121
0.0940
0.0801
0.0690
4.6229
5.2064
5.7466
6.2469
6.7101
0.2163
0.1921
0.1740
0.1601
0.1490
So
F = $1k (
7.3359
8.9228
10.6366
12.4876
14.4866
F/A, 8%, 5
)
= $1K (5.8666) = $5.8666K
INFO630 Week 6
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Equal-Payment-Series Sinking-Fund (A/F)
• A given F (A/F)
– How much you want to end with, and you are trying to figure out
how much to deposit each time
• Generic cash-flow diagram
known
0 1
2
3
n-1
n
unknown
• Example
– How much would need to be invested at 8% at the end of each
of the next 5 years to finish with $5k?
INFO630 Week 6
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Equal-Payment-Series Sinking-Fund (A/F)
• Deriving the formula
– Rearrange the Equal-payment-series
Compound-Amount formula…
 1  i n  1 
F  A

i


– … to solve for A


i
A  F

n


1

i

1


INFO630 Week 6
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Equal-Payment-Series Sinking-Fund (A/F)
• Formula


i
A  F

n


1

i

1


• Solving the sample problem
– How much would need to be invested at 8% at the end of each
of the next 5 years to finish with $5k?


0.08
A  $5k 
  $853
5
 1  0.08   1
• Shorthand notation
A/F, i, n
A=F(
)
INFO630 Week 6
A/F, 8%, 5
A = $5k (
)
50
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Interest Tables –
Conversion Factor (Page 498)
Table B-12
8% Interest Factors for Discrete Compounding
Single-Payment
Equal-Payment-Series
Compound- PresentAmount
Worth
Find F Given P
CompoundAmount
Find P Given F
SinkingFund
PresentWorth
CapitalRecovery
Find F Given A
Find A Given F
Find P Given A
Find A Given P
n
(F/P,i,n)
(P/F,i,n)
(F/A,i,n)
(A/F,i,n)
(P/A,i,n)
(A/P,i,n)
1
2
3
4
5
1.0800
1.1664
1.2597
1.3605
1.4693
0.9259
0.8573
0.7938
0.7350
0.6806
1.0000
2.0800
3.2464
4.5061
5.8666
1.0000
0.4808
0.3080
0.2219
0.1705
0.9259
1.7833
2.5771
3.3121
3.9927
1.0800
0.5608
0.3880
0.3019
0.2505
6
7
8
9
10
1.5869
1.7138
1.8509
1.9990
2.1589
0.6302
0.5835
0.5403
0.5002
0.4632
7.3359
8.9228
10.6366
12.4876
14.4866
0.1363
0.1121
0.0940
0.0801
0.0690
4.6229
5.2064
5.7466
6.2469
6.7101
0.2163
0.1921
0.1740
0.1601
0.1490
A/F, 8%, 5
So
A = $5k (
)
= $5K (0.1705) = $852.50
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Equal-Payment-Series Capital-Recovery (A/P)
• A given P (A/P)
– Standard formula for loan payments
– Amount borrowed today. How much of an equal payment over
the next N periods, will reduce the amount to zero at last
payment.
• Generic cash-flow diagram
unknown
0
known
1 2
3
n-1
n
• Example
– If you borrowed $25k at 7% for 5 years, what would your annual
payments be?
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Equal-Payment-Series Capital-Recovery (A/P)
• Deriving the formula
Start with Equal-payment-series Sinking-fund and Single-payment Compound-amount


i
A  F

n


1

i

1


F  P1  i 
n
Substitute for F

i
n
A  P 1  i  

n
 1  i   1 
Simplify
 i 1  i n 
A  P

n


1

i

1


INFO630 Week 6
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Equal-Payment-Series Capital-Recovery (A/P)
• Formula
 i 1  i n 
A  P

n


1

i

1


• Solving the sample problem
– If you borrowed $25k at 7% for 5 years, what would your annual
payments be?
 0.071  0.075 
A  $25k
  $6098
5
 1  0.07  1 
• Shorthand notation
A/P, i, n
A=P(
)
INFO630 Week 6
A/P, 7%, 5
A = $25k (
)
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Equal-Payment-Series Present-Worth (P/A)
• P given A (P/A)
– How much money today would be equivalent to future series
of equal payments made over N periods at some interest
rate?
• Generic cash-flow diagram
known
0
unknown
1 2
3
n-1
n
• Example
– How much would need to be deposited today at 7% to give
10 end-of-year payments of $100 each, leaving $0 after the
last payment?
INFO630 Week 6
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Equal-Payment-Series Present-Worth (P/A)
• Deriving the formula
– Rearrange the Equal-payment-series Capitalrecovery formula…
 i 1  i n 
A  P

n


1

i

1


– … to solve for P
 1  i n  1 
P  A

n


i
1

i


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Equal-Payment-Series Present-Worth (P/A)
• Formula
 1  i n  1 
P  A

n


i
1

i


• Solving the sample problem
– How much would need to be deposited today at 7% to give 10 end-ofyear payments of $100 each, leaving $0 after the last payment?
 1  0.0710  1 
P  $100
 $702
10 
 0.071  0.07 
• Shorthand notation
P/A, i, n
P=A(
)
INFO630 Week 6
P/A, 7%, 10
P = $100 (
)
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Summarizing the Formulas
• Vary depending on whether you are
calculating
– Forward/Backward
• Forward from a known present into an unknown
future
• Backward from a known, desired future into an
unknown present situation
– Single-Payment/Equal Payment
• See next slide for help
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Selecting an Interest Formula
The unknown future is
also a single pa ym ent
The known present
is a single pay m ent
Find an unknown future
from a known present
?
?
Equal-pay ment-series
capital-recovery
The unknown future is a
(A/P,i,n)
series of equal paym ents
Equal-pay ment-series
com pound-am ount
(F/A,i,n)
The unknown present is
also a single pa ym ent
The known future
is a single pay m ent
Find an unknown present
from a known future
?
The known present is a
series of equal paym ents
?
Single-paym ent
com pound-am ount
(F/P,i,n)
Single-paym ent
present-worth
(P/F,i,n)
?
Equal-pay ment-series
sinking-fund
The unknown present is a
(A/F,i,n)
series of equal paym ents
The known future is a
series of equal paym ents
INFO630 Week 6
Equal-pay ment-series
present-worth
(P/A,i,n)
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Looks messy?
• The quantities ‘i’ and ‘n’ are usually given
• We’re solving for F, A, or P
• There are three pairs of formulas,
connected by being the inverse of each
other
A/P = 1 / (P/A)
F/P = 1 / (P/F)
F/A = 1 / (A/F)
Or ignore this slide if it’s confusing you. 
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Key Points
• “Amount of money” and “value” are different concepts
– The same amount of money has different values at different
times
• The time-value of money is quantified by interest
• Simple interest isn’t normally available
• Compound interest involves interest on interest
– Several compound interest formulas exist (six)
– Interest tables help solve compound interest calculations
INFO630 Week 6
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Other Interest Calculations
Chapter 6
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Toying with Chapter 5
• This chapter introduces few new
equations, but shows how you can work
with the key six equations from chapter 5
• Start by examining the interest rate, i
• We assumed it was known, and typically
given for the rate it was compounded
– 6% annual interest, compounded annually
– 1.2% monthly interest, compounded monthly
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Different interest period and
compounding frequencies
• What if that isn’t true?
• Convert the nominal interest rate (r) to the
effective annual interest rate (i)
i = (1 + r/m)^m -1
– Where ‘m’ is the number of compounding
periods per year (see Table 6.1, p. 75, and fix
m=365 for daily) and ^ is ‘to the power of’
INFO630 Week 6
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Different interest period and
compounding frequencies
• So 16% annual interest compounded
quarterly (m=4) is really
i = (1 + 0.16/4)^4 – 1 = 0.16985856 =
16.99% effective annual interest
• Or flip the equation around to solve for r
r = m( (1+i)^(1/m) – 1)
• So 6.5% effective interest converted to
monthly is r = 12( (1.065)^(1/12)-1)=6.31%
INFO630 Week 6
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Or even more arbitrary
• Or convert a nominal annual rate (r) based
on one compounding period to any other
compounding period (m) with
i = (1 + r/m)^c – 1
where ‘c’ is the number of compounding
periods in ‘i’ divided by the periods in ‘r’
– So 4% interest every 26 weeks compounded
weekly, converts to quarterly (13 weeks) as
i = (1+0.04/26)^13 – 1 = 2.02%
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APR
• APR (Annual Percentage Rate) shows up
in ads all the time – it’s the nominal
annual interest rate, but can be based on
any period or compounding rate
– Use the r formula on slide 65 to convert to the
nominal annual rate (APR=r)
– APR isn’t the effective annual percentage
rate, since compounding can be not annual
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Cash flow rate versus
compounding interval
• When the cash flow comes in faster or
slower than the compounding of interest,
or even at the same rate, use the formula
on slide 64 to get the actual interest rate
per compounding interval
• Then use the equations from chapter 5
with that value of ‘i’
INFO630 Week 6
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Continuous compounding
• There is such a thing as continuous
compounding, which means interest is
compounded every moment
• Formulas for F, A, and P involve
exponential terms (page 81)
• This is rarely used in reality
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Other special uses of Chapter 5
• Many of the other examples are just
working backward to solve for ‘i’
– Can interpolate on the tables in Appendix B,
or solve analytically
• For example, to pay off a loan early you
need to solve for the loan’s present worth
(P) when you’re ready to pay it off
INFO630 Week 6
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Loan interest
• Loan payments (At) pay off interest (It)
and principal (Bt) (the ‘t’s should be subscripts)
At = It + Bt
• It = A*(P/A, i, n-t+1)*i
• Bt = A – It
where A is the loan payment amount
‘t’ is the tth payment against the loan
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Loan payoff early
• If you increase the loan payment by E,
how much faster do you pay it off?
• We had P/A = (P/A, i, n-t+1) after ‘t’
payments
• Now find the number of payments ‘j’ for
which P/(A+E) = (P/A, i, j) is true
• Again, interpolation is often needed
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Interpolation review
• Linear interpolation lets you find the
intermediate value between table entries
• The example on page 83 needs to
interpolate between 18 and 19% interest
entries for the value of F/P:
18%
Z%
19%
F/P=3.759
F/P=3.852
F/P=4.021
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Interpolation
• To solve this, these ratios hold
• (Z-18)/(19-18) =
(3.852-3.759)/(4.021-3.759)
• Solve for Z
• (Z-18)/1 = 0.093/0.262 = 0.35496
• Z = 18 + 1*0.35496 = 18.355%
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Other tricks
• Notice that solving the problems in this
chapter often involve converting the given
interest rates into the effective annual rate
• Then apply the right F, A, or P formula to
solve the problem
• Also note that interpolation is often helpful
to solve for interest rates or remaining loan
durations
INFO630 Week 6
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Other tricks
• In other real world situations, look out for
sneaky deals where interest is calculated
based on the total principal of the loan
– The real or effective interest rate could be
much higher than it seems, since you’re
paying interest on principal which has already
been paid off 
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