30
-------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

-------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

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Page 1: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

-------------------------- ENGR 300

Dept. of Computer Science and EngineeringUniversity of Bridgeport, CT 06601

Page 2: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

NET PRESENT VALUE - NPV

Measures inflows vs. outflowsToday’s dollarsCradle to Grave

Page 3: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

NET PRESENT VALUE - NPV

Not only accounts for the Costs of Inflows and Outflows, but for their timing

Sales Revenues Loan Payments Development Costs Ramp Up and Production Costs Marketing and Support Costs Disposal Costs

Page 4: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

SENSITIVITY ANALYSIS

Answers “What If?” Questions

Helps in Making Project Tradeoffs

Cost is a Critical FactorCritical Factor in Design

??? ?

Page 5: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

SENSITIVITY ANALYSIS

Project Parameters can be Varied Development Time Project Loading Interest Rates Sales Price Product Quality

Page 6: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

QUALITATIVE ANALYSIS

Complex Factors

Risks

External Factors

Page 7: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

DEPRECIATION

Only a portion of the cost of an asset can be deducted for tax purposes in one year

Tangible Assets decrease in value over time Cars Equipment Buildings

Page 8: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

DEPRECIATION METHODS

An Asset has an Initial and a Salvage Value at the start and end of

its service life

The Book Value is the remaining undepreciated value of the asset

Straight Line Method (equal amounts) Accelerated Cost Recovery System

Modified Accelerated Cost Recovery System

Page 9: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

DECISIONS CAN BE INFLUENCED BY THE TIME VALUE OF MONEY

One dollar today

Will be worth more in the future

$ + time = $$$

Present

Future

Page 10: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

TIMING OF INFLOWS AND OUTFLOWS

PresentFuture cash outflows

Future cash inflows

time

Cash Flow Diagram graphically shows relationships

Page 11: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

BASIC TERMINOLOGY

P= Present value (NPV in Today's Dollars) F= Future value (Tomorrow’s Dollars) n = Number of compounding periods between

“present” and “future” A = uniform Amount received or paid out each

compounding period

Page 12: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

INTEREST RATE

The reward that investors demand for accepting delayed payment

Sometimes referred to as the Discount Rate n is the number of periods per year Must convert the yearly percentage rate to its decimal

equivalent rate

ii

nPeriod

Yearly

PerYear

% .01

Page 13: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

COMPOUNDING OF INTEREST

ANNUAL PERCENTAGE RATE (APR) INCREASES WITH SHORTER PERIODS OF COMPOUNDING

12% Yearly = 12% APR 3% Quarterly = 12.55% APR 1% Monthly = 12.68% APR Continuous = 12.75% APR

Page 14: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

COMPOUNDING FORMULAS

APR i

APR e

Periodn

i

per year

Yearly

( ) _

% .

1 1

101CONTINUOUS

FIXED PERIODS

Page 15: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT vs. FUTURE VALUE

Dollars today are worth more than the same amount of dollars in the future

$1000 today will grow to $3300.39 in 10 years at 12% compounded monthly

PF

iPeriodnTotal

( )1

Page 16: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT vs. FUTURE VALUE Find Present given the Future Value

$120 one year from now is worth $113.03 today n=12 periods or monthly Yearly interest rate is 6%, per month is .005

P

$120

( . )$113.

1 0050312

Page 17: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT vs. FUTURE VALUE How Many Periods?

How many years does it take to double your money if the APR=9%

PP

n years

n

2

1 09

2

1098 043

( . )

ln

ln( . ).

SOLVING

LOG FUNCTION WORKS TOO

Page 18: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT vs. FUTURE VALUEWhat interest rate is needed?

What interest rate is needed to make $200 grow to $1000 in ten years, if interest is paid yearly?

$200$1000

( )

. .ln( )

1

1 1746 17 46%

10

1000200

10

i

i e

SOLVING

Page 19: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601
Page 20: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT VALUE(P) OF A SERIES OF AMOUNTS

n = number of payments of amount A i=interest rate per period (decimal) A= amount of each payment

PA i

i

n

( ( ) )1 1

Page 21: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT VALUE OF EQUAL PAYMENTS

$10 monthly payments for one year interest rate is 6% per year = .005 per month

P

$10 ( ( . ) )

.$116.

1 1 005

00509

12

Present Value is greater than one single paymentof $120 after a year (in that case, P was $113.03)

Page 22: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

AMOUNT OF A LOAN PAYMENT

P=$100,000 i=9% per year = .0075 per month n=360 monthly payments

A

$100000 .

( ( . ) )$804.

0075

1 1007562360

Note in 30 years, $289,663 will be paid in payments

Page 23: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

MATHEMATICAL BASIS

A SERIES OF PAYMENTS IS BROKEN DOWN INTO SUMS OF INFINITE STRINGS OF PAYMENTS

etc.P1=A/iA

etc.A

A

P=P1-P2

Subtract

P2=A/(i(1+i) n)

Page 24: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

PRESENT VALUEOF INCREASING AMOUNTS

PA i

i

B i n i

i

P

nk

nk n

k

k

( ( ) ) ( ( ) ( ) ( ) )

$15 ( ( . ) )

.

$10 ( ( . ) ( . ) )

.$114.

1 1 1 1 1

1 1015

015

1015 3 1015

01591

1

1

41

34

A A+BA+2B A+3B

Present

A=$15 & B=$104A+6B=$120i=.06/4=.015n=4 (quarterly) Future

Page 25: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

ECONOMIC COMPARISON

Decisions often include comparisons of economic costs/benefits of alternative actions

Inflows/Outflows may occur at several different times

Time-value of money must be considered

Page 26: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

ALTERNATIVES WITH EQUAL LIVES

For each alternative, compute the Net Present Value (NPV)

Compute P for all inflows Compute P for all outflows NPV = P(inflows) - P(outflows)

Alternative with highest NPV is the best choice from an economic viewpoint

Page 27: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

ALTERNATIVES WITH UNEQUAL LIVES

For each alternative, compute the equivalent uniform cost per period (EUC/P)Assume identical replacement at end of life

Compute A for all inflows

Compute A for all outflows

EUC/P = A(outflows) - A(inflows)

Alternative with lowest EUC/P is the best choice from an economic viewpoint

Page 28: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

DEALING WITH RISK AND UNCERTAINTY

Use Expected Value (EV) for inflows and outflows with estimated uncertainties

EV = (p1)(V1) + (p2 )(V2) + .......+ (pn)(Vn)

pn is the probability that a value will be Vn

where p1 + p2 +.......+ pn = 1 Calculate NPV or EUC/P based on expected

values

Page 29: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

EXPECTED VALUE

Saturdays, the following probabilities exist .35 won’t study at all .15 will study for 4 hours .20 will study for 2 hours .30 will study for 1 hour

EV . . . . .35 0 15 4 20 2 30 1 131.3 IS THE EXPECTED NUMBER OF HOURS OF STUDY ON A TYPICAL SATURDAY

Page 30: -------------------------- ENGR 300 Dept. of Computer Science and Engineering University of Bridgeport, CT 06601

WHAT IF?

What If questions can be supported by doing a sensitivity analysis.– Take one variable at a time, holding others fixed,

make small changes in that variable observe effect on NPV or on EUC/P

– Spreadsheet program is useful for this purpose and doing time value of money calculations