CS 170:
Computing for the Sciences and
Mathematics
Error: Finding Creative ways to Screw Up
Administrivia
 Last time
 Basics of modeling
 Assigned HW 1
 Today
 HW1 due!
 Assign HW 2
 Monday 9/13– NO CLASS
Error
 What is the value of a model…
 That is completely wrong

That is a perfect match to a physical system

That may be off by as much as 5%
Types of Error
 Input Data Errors
 Faulty/inaccurate sensors, poorly calibrated, mis-read results..
 Modeling Errors
 Poor assumptions, bad math, mis-understanding of system
 Implementation Errors
 Bug in computer program, poor programming
 Precision

The limits of finite number representation
Input Data Errors
 NSIDC
 Sensor drift led to their real-time sea ice estimates to be off by
over 500,000 km2
Still only off by 4%
 http://nsidc.org/arcticseaicenews/2009/022609.html

Modeling Errors
 Obvious type: math formulation errors, mis-writing
a formula, etc.
 By virtue of making assumptions and simplifications,
models will have “error” versus reality.

This isn’t necessarily a bad thing, as long as we manage it well
 Lord Kelvin used his knowledge of temperature
dissipation to estimate that the earth was between
20-40 million years old.


Actual: ~12 billion
Assumed there was no heat source but the sun
Implementation Errors
 Incorrect programming (a “bug”)
 Didn’t implement the model correctly
Wrong equations
 Mis-defined inputs/outputs
 Implemented the solution to a different problem

 Precision Errors
 Computers only have so much space to store numbers
 This limits the range and precision of values!
Precision Errors
 In a computer, a number is stored in a number of
bits (binary digits)

IEEE 754 standard floating point
Variation on standard normalized scientific notation
 single-precision is 32 bits
 double-precision is 64 bits


Stored in 3 parts
sign (1 bit) – is it positive or negative?
 magnitude – what is the exponent?
 mantissa/significand – what is the number?

i.e. 6.0221415 × 1023
Exponential notation
 Example: 698.043990  103
 Fractional part or significand?

698043990
 Exponent?

3
 Normalized?

6.98043990  105
Significant digits
 Significant digits of floating point number

All digits except leading zeros

Number of significant digits in

9 significant digits
 Precision

Number of significant digits
698.043990  103?
Round-off error
 Problem of not having enough bits to store entire
floating point number
 Example: 0.698043990  105 if only can store 6
significant digits, rounded?

0.698044  105
 Do not test directly for equality of floating point
variables

Note that many numbers that seem “safe” really aren’t!

i.e. 0.2 is an infinite repeating series when expressed in binary
Absolute error
 |correct – result|
 Example: correct = 0.698043990  105 and result =
0.698043  105
 Absolute error = ?

|0.698043990  105 - 0.698043  105| = 0.00000990  105 =
0.990
Relative error
 |(correct - result) / correct|
 Example: (correct - result) = 0.990 and correct =
0.698043990  105

0.990/(0.698043990  105) = 1.4182487  10-5
Relative error
 Why consider relative errors?

1,000,000 vs. 1,000,001

1 vs. 2
 Absolute error for these is the same!
 If exact answer is 0 or close to 0, use absolute error

Why?
Addition and subtraction errors
 Beware if there is a big difference in the magnitude
of numbers!
 Example: (0.65  105) + (0.98  10-5) = ?

65000 + 0.0000098 = 65000.0000098
 Suppose we can only store 6 significant digits?

65000.0 = (0.650000  105)
Associative property
 Does not necessarily hold!
 Sum of many small numbers + large number may
not equal adding each small number to large number
 Similarly, distributive property does not necessarily
hold
To reduce numerical errors
 Round-off errors
 Use maximum number of significant digits
 If big difference in magnitude of numbers
 Add from smallest to largest numbers
Error Propagation (Accumulated Error)
 Example: repeatedly executing:
t = t + dt
 Better to repeatedly increment i and calculate:
t = i * dt
Overflow/underflow
 Overflow - error condition that occurs when not
enough bits to express value in computer
 Underflow - error condition that occurs when
result of computation is too small for computer to
represent

Truncation error
2
3
4
n
x
x
x
x
e x  1 x 


  
1 2 1 2 3 1 2 3 4
n!
1
1
1
1
e 11


 
1 2 1 2 3 1 2 3 4
20!
 Truncation error
 Error that occurs when truncated, or finite, sum is used
as approximation for sum of infinite series
1
1
1



21! 22! 23!
 2.051020
Error is Not Inherently Bad
 Almost all of these issues can be managed and
controlled!
 A certain amount of error is normal
 What’s important is that we:
 Know how much error there might be
 Keep it within a bound that allows the results to still be valid
Rates: The Basics of Calculus
Calculus
 Mathematics of change
 Two parts
 Differential calculus
 Integral calculus
Rates
 “Rate of Change”
 Often depends on current amounts
 Rates are important in a lot of simulations
 growth/decay of…
populations
 materials/concentrations
 radioactivity
 money




motion
force
pressure
Height (y) in m vs time (t) in sec of ball
thrown up from bridge
change in position s(b) s(a)
average velocity

change in time
b a
Time (t) in
seconds
Height (y) in
meters
0.00
11.0000
0.25
14.4438
0.50
17.2750
0.75
19.4938
1.00
21.1000
1.25
22.0938
1.50
22.4750
1.75
22.2438
2.00
21.4000
2.25
19.9437
2.50
17.8750
2.75
15.1937
3.00
11.9000
3.25
7.9938
3.50
3.4750
3.75
-1.6563
Average velocity
 between 0 sec & 1 sec?
 between 1 sec & 2 sec?
 between 0.75 sec & 1.25 sec?
 Estimate of instantaneous
velocity at t = 1 sec?
Derivative
s(1 t) s(1)
instantaneous velocity at 1 sec
 lim
t 0
t
 Derivative of y = s(t) with respect to t at t = a is the
instantaneous rate of change of s with respect to t at
a (provided limit exists):
dy
s'(a) 
dt
s(a  t) s(a)
 lim
t 0
t
t a
Derivative at a point is the slope of the curve
at that point
Differential Equation
 Equation that contains a derivative
 Velocity function
 v(t) = dy/dt = s'(t) = -9.8t + 15
 Initial condition
 y0 = s(0) = 11
 Solution:
 function y = s(t) that satisfies equation and initial condition(s)
 in this case:

s(t) = -4.9t2 + 15t + 11
Second Derivative
 Acceleration - rate of change of velocity with respect
to time
 Second derivative of function y = s(t) is the
derivative of the derivative of y with respect to
independent variable t
 Notation
 s''(t)
 d2y/dt2
Systems Dynamics
 Software package that makes working with rates much
easier.
 Components include:




“Stocks” – collections of things (noun)
“Flow” – activity that changes a stock (verb)
“Variables” – constants or equations – converter
“Connector” – denotes input/information being trasmitted
HOMEWORK!
 READ pages 17-48 in the textbook
 On your own
 Work through the Vensim PLE tutorial
 Turn-in the final result files on W:
 Vensim is being deployed tonight. If there is a
problem, I will notify everyone ASAP.
 NO CLASS on Monday