System Dynamics – 1ZM65/1ZS24
Lecture 6
October 7, 2014
Dr. Ir. N.P. Dellaert
Agenda
• Recap Bass model and Boiled Frog
• Delays
• Dynamic behavior of linear systems with 2 stocks
• Calculating the limits
• Oscillation or no oscillation?
• Calculating the behaviour
• Example Exam question
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PAGE 1
The Bass Diffusion Model
Potential
Adopters P
+
Adopters A
adoption rate AR
market saturation 1
+
word of mouth
Total Population N
+
adoption from
advertisting
adoption from
word of mouth
adoption fraction i
market saturation 2
+
contact rate c
+
advertising
effectiveness a
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number of problems per month
Oscillation and Growth of Problems
Staff Problems
Schedule Problems
Budget Problems
12
10
8
6
4
2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
month
phase 1:
staffing
problems
for
stage 1
phase 2:
schedule
problems
for
stage 1
phase 3:
budget
problems
for
stage 2
phase 4:
staffing
problems
for
stage 2
phase 5:
schedule
problems
for
stage 2
phase 6:
budget
problems
for
stage 3
Conclusion: The Boiled Frog Syndrome is not
just a fable
• Project management prefers short-term fixes over
fundamental solutions
• Short-term fixes create an illusion of control
• While in fact the project is on a slippery slope, where
an initially small problem is shifted around and
gradually becomes a big problem
• They should have seen it coming, but they didn't!
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Delays: examples
•
•
•
•
•
•
A company decides to build a new factory
A large mail delivery: 1000000 IKEA catalogues
Delivery time of a book
Ghost-driver
Being aware of an open fly
Time between price change and production change
of pigs ( pig-cycle)
• ...
General: 2 types of delays
material delays and information delays
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Delays and stocks
Delays always contain stocks
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Outflow patterns
different distribution with same mean
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Oscillation: structure and behavior
earlier example delays
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Pipeline delay
FIFO
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Higher-order delays
Higher-order delays are
formed by cascading firstorder delays together.
Non-fifo!
Memoryless!
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Example third-order delay
Figure 11-8 Pulse response of third-order delay by stage of processing
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Example linear second order systems
• Contain 2 stocks
• Linear relationship
between flows and
stocks
• Stocks contain groups in
different phases:
• In training/experienced
• Young/old
• WIP/Inventory
• Stocks describe delays
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First order inventory model
Figure 17-4 Structure for managing a stock when there
are no acquisition delays.
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Second order inventory model
Figure 17-6 The generic stock management structure
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General linear second order
model
Actual Value V is usually linear
combination of R and E
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Differential equations
dR
G V R
 inflow-outflow 

dt
A
B
dE
R E
 inflow-outflow  
dt
B C
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Steady state equilibrium
dR G  V R

 0
dt
A
B
dE R E
  0
dt B C
In equilibrium:
Inflow=outflow
For V=aR+bE, we have two equations, two unknowns
Easily solvable
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Steady state equilibrium calculation
dE R E
BE
   0 yields R 
dt B C
C
Using this and V=aR+bE to substitute in the other equation
dR G  V R G  bE R aR
G  bE
A  aB BE

 
( 
)
(
)
0
dt
A
B
A
B A
A
AB
C
G
A  aB  bC E
(
)  0 yields
A
A
C
CG
BG
E
and R 
A  aB  bC
A  aB  bC
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Exercise: calculate equilibrium
A, B, C  1
V  R  E, G  50
In equilibrium:
Inflow=outflow
CG
1  50
E
=
=16.67
A  aB  bC 1  1  1
BG
1  50
and R 

=16.67
A  aB  bC 1  1  1
Notice:
V and G are different!
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To reach the desired goal
E
CG
BG
and R 
A  aB  bC
A  aB  bC
aCG  bBG
aC  bB
V  aE  bR 
 G
A  aB  bC
A  aB  bC
If we really want to reach the desired goal, we could use a fictious goal:
A  aB  bC
G  G
aC  bB
*
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Steady state equilibrium
dR G  V R

 0
dt
A
B
dE R E
  0
dt B C
In equilibrium:
Inflow=outflow
These kind of calculations are possible
for any number of stock points, as long
as we have linear relationships
http://www.youtube.com/watch?v=Suugn-p5C1M
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Dynamic Behaviour
dR G  V R


dt
A
B
dE R E
 
dt B C
Now we consider the
process before we
reach equilibrium
To solve this, we use second order diff. equations
We also need initial values for R and E
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Second order differential equations
dR G  aR  bE R
R 


dt
A
B
'
dE R E
E 
 
dt B C
'
Taking again the derivative
d 2 E R' E '
E ''  2  
dt
B C
From the second equation we can express R in E and E’, from
the first equation we can express R’ in R(=E and E’) and E.
We then substitute in the third equation
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Second order differential equations
d 2 E R ' E ' G  aR  bE R E '
E ''  2   

 
dt
B C
AB
BB C
G  bE (aB  A) R E ' G  bE (aB  A)
E E'

 

( E ' ) 
AB
AB B C
AB
AB
C
C
aBC  AC  AB
bC  aB  A
G
E '' E '(
)  E(
)
ABC
ABC
AB
In equilibrium: first and
second derivative are 0
We could use this to calculate
equilibrium
This is NOT a general expression, but only valid for this
particular way of control
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General approach Second order DE
First consider homogeneous DE:
E '' xE ' yE  0
suppose E  ezt
E '' xE ' yE  ( z 2  xz  y)  e zt  0
 x  x2  4 y
z1,2 
2
for x 2  4 y imaginary roots
for x 2  4 y real roots
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General approach Second order DE
First consider homogeneous DE:
E '' xE ' yE  0
suppose E  ezt
E '' xE ' yE  ( z 2  xz  y)  e zt  0
for x2  4 y imaginary roots
General solution
j
e  cos  j sin 
E (t )  C0  C1e kt cos(t )  C2e kt sin(t )
with k   x / 2
 x2  4 y
and  
2
Constants are
determined by
initial values
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Determining constants in oscillation case
General solution
E (t )  C0  C1e kt cos(t )  C2e kt sin(t )
E ' (t )  C1ke kt cos(t )  C2 ke kt sin(t )  C1e kt sin(t )  C2e kt cos(t )
E ()  C0
E (0)  C0  C1
E '(0)  C2  kC1
Constants are
determined by
initial values
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Example: calculate dynamic behaviour
A, B, C  1
R(0)  10, E (0)  20
V  R  E, G  50
aBC  AC  AB
bC  aB  A
G
)  E(
)
ABC
ABC
AB
E '' 3E ' 3E  50 gives oscillation
E '' E '(
E (t )  C0  C1e kt cos(t )  C2e kt sin(t )
with k   x / 2  3 / 2
 x2  4 y 1
and  

3
2
2
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Example determining constants in oscillation case
E (t )  C0  C1e kt cos(t )  C2e kt sin(t )
E ()  C0  50 / 3
E (0)  C0  C1  20, so C1  10 / 3
R(0) E (0)

 10  20  10
B
C
1
3 10
10
C2  kC1 
3C2 
 10, so C2  
3
2
2 3
3
E '(0)  C2  kC1 
E (t )  50  10 e1.5t cos( 12 t 3)  10 3e1.5t sin( 12 t 3)
3
3
3
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General approach Second order DE
First consider homogeneous DE:
E '' xE ' yE  0
suppose E  ezt
E '' xE ' yE  ( z 2  xz  y)  e zt  0
for x2  4 y real roots
General solution
E (t )  C0  C1e k1t  C2 e k2t
with k1,2   x / 2 
x  4y
2
2
Constants are
determined by
initial values
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Determining constants in non-oscillation case
General solution
E (t )  C0  C1e k1t  C2 e k2t
E ' (t )  C1k1e k1t  C2 k 2e k2t
E (  )  C0
E (0)  C0  C1  C2
E '(0)  k1C1  k 2C2
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General approach Second order DE
First consider homogeneous DE:
E '' xE ' yE  0
suppose E  ezt
E '' xE ' yE  ( z 2  xz  y)  e zt  0
for x2  4 y a double real root, so-called critical dampening
General solution
E (t )  C0  C1ek1t  C2tek1t
with k1   x / 2
Constants are
determined by
initial values
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Determining constants in critical case
General solution
E (t )  C0  C1e k1t  C2te k1t
E ' (t )  C1k1e k1t  C2 k1te k1t  C2 e k1t
E ()  C0
E (0)  C0  C1
E '(0)  k1C1  C2
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Example Vensim
Duration B
E
R
inflow
Adjustment
Time A
Duration C
move rate
outflow
Actual Value V
Desired Value D
V=E, B=1, C=5
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Vensim example
Critical dampening for A=6.25
E
60
45
30
15
0
0
E : a=10
E : a=2
1
2
3
4
5
6
Time (Month)
7
8
9
10
E : a=1
E : a=6
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Example Vensim
Duration B
E
R
inflow
Adjustment
Time A
Duration C
move rate
outflow
Actual Value V
Desired Value D
V=R+E, B=1, C=1
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Vensim example
Critical dampening for A=0.25
E
60
45
30
15
0
0
E : 100a=1
E : a=3
1
2
3
4
5
6
Time (Month)
7
8
9
10
E : a=1
E : 4a=1
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Example Exam Question
Duration B
E
R
inflow
Adjustment
Time A
Duration C
move rate
outflow
Actual Value V
Desired Value D
Consider the standard second order linear feedback system as shown
above, with actual value V= E, B=1, C=3, Desired Value 50. Determine
the equilibrium state for stock E as a function of parameter A. Which
statement is true
a)
b)
c)
d)
For all A values, stock E will grow to 50
For A=1, E will grow to 40; for A=3 the limit of E=33.33
For A=1, E will grow to 40; for A=3 the limit of E=36.67
For all A values, stock E will grow to 33.33
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Example Exam Question
• Same picture
Consider the standard second order linear feedback system as shown
above, with actual value V= E, B=1, C=3, Desired Value 50. Which
picture does NOT describe the behaviour of this system for various Avalues
E
R
R
20
40
17
30
14
20
60
30
10
11
0
0
2
4
6
Time (Month)
E : a=3
8
10
0
8
0
E : a=1
E
1
2
R : a=3
3
4
5
6
Time (Month)
R : a=1
7
8
9
10
0
1
2
3
4 5 6 7
Time (Month)
8
9 10
R : 4a=1
60
45
30
15
0
0
E : a=3
1
2
3
4
5
6
Time (Month)
E : a=2
7
8
9
10
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Conclusions
• Second order linear feedback systems can be solved
analytically
• Three possible reactions
• Critical dampening
• Oscillation
• Dampening
• Parameters are important for behaviour
• Be aware of names/relations between parameters
• Handout describes one example and some other
typical elements
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Instruction
Week 6
10-Oct
15:45-17:30
Study Hub 2 (PAV)
1. Phase Space of the Bass Diffusion Model (pp.
333 & 334)
2. Extending the Bass Diffusion Model (pp. 335 &
339)
3. Modelling Fads (pp. 341 & 342)-Modeling the
Life Cycle of Durable Products (pp. 345 & 346)
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