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MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2: Differential Form of Maxwell’s Equati... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/1c014f0d8c71c1be1eb4fd3198b1d543_MIT6_641s09_lec02.pdf |
dV =
)
∫ ρ dV
S
V
V
∇ i ε0E = ρ
)
(
µ H i da = ∇ i µ H dV = 0
( 0 )
∫
V
0
S
(cid:118)∫
∇ i (
)
µ0H = 0
II. Stokes’ Theorem
1. Curl Operation
) i
A ds = ∫ Curl A
(
i
S
(cid:118)∫
C
da
(cid:118)∫
)n
Curl A = lim C
(
da
n →0
A ds
i
da
n
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Mark... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/1c014f0d8c71c1be1eb4fd3198b1d543_MIT6_641s09_lec02.pdf |
(
) - A x, y
)⎤
⎤
⎦
⎥
⎥
⎦
∆x
= da z ⎢
⎡ ∂Ay
⎣ ∂x
-
∂Ax ⎤
⎥
∂y ⎦
)z
(
Curl A
= (cid:118)∫
i
A ds
=
da
z
A
∂
y -
x
∂
A
∂
x
y
∂
By symmetry
(
Curl A
) = (cid:118)∫ A ds
i
y
da
y
∂A
= x
∂z
-
∂A
z
∂x
Curl A = (cid:118)∫ A ds
=
i
(
)x
dax
∂Az -
∂y
∂Ay
∂z
Curl A = i x ⎢
− ⎡ ∂A
⎣ ∂y
z -
−
⎥ ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/1c014f0d8c71c1be1eb4fd3198b1d543_MIT6_641s09_lec02.pdf |
3. Magnetic Field
∇ i ⎨∇ × E = - µ0
⎧
⎪
⎪
⎩
⎫
∂H⎪
⎬
∂t ⎪
⎭
0 = -
∂
t ⎣
∂
⎡∇ µ0
⎦ ⇒ ∇ i (µ0H) = 0
⎤
i H
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 8 of 10
4. Vector Identity
b
(
∫ i
E dl = Φ a
) − Φ ( ) b
a
if a=b
E dl
(cid:118)∫ i
C
= Φ ( )
a − Φ
(a) = 0
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/1c014f0d8c71c1be1eb4fd3198b1d543_MIT6_641s09_lec02.pdf |
⎡
(
)
µ0 J x ', y ', z ' dx dy dz
+ (y − y ')2
'
'
⎣(x − x ')2
+ (z − z ')2 ⎤
⎦
'
1
2
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 10 of 10 | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/1c014f0d8c71c1be1eb4fd3198b1d543_MIT6_641s09_lec02.pdf |
18.175: Lecture 5
More integration and expectation
Scott Sheffield
MIT
18.175 Lecture 5
1 Outline
Integration
Expectation
18.175 Lecture 5
2 Outline
Integration
Expectation
18.175 Lecture 5
3 Recall Lebesgue integration
� Lebesgue: If you can measure, you can integrate.
� In more words: if (Ω, F) is a m... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/1c016821763783c34c48816b1ac66969_MIT18_175S14_Lecture5.pdf |
<
<
If g ≤ f a.s. then
< fdµ.
< gdµ ≤
If g = f a.e. then gdµ =
fdµ.
<
|
|f |dµ.
When (Ω, F, µ) = (Rd , Rd , λ), write
fdµ| ≤
<
(cid:73)
�
(cid:73)
�
<
fdµ + b gdµ.
<
f (x)dx = 1E fdλ.
<
E
18.175 Lecture 5
5 Outline
Integration
Expectation
18.175 Lecture 5
6 Outline
Integration
Expectation
18.175 L... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/1c016821763783c34c48816b1ac66969_MIT18_175S14_Lecture5.pdf |
functions.
<
|f |pdµ)1/p for
H¨older’s inequality: Write lf lp = (
<
|fg |dµ ≤ lf lplg lq.
1 ≤ p < ∞. If 1/p + 1/q = 1, then
Main idea of proof: Rescale so that lf lplg lq = 1. Use
some basic calculus to check that for any positive x and y we
have xy ≤ x p/p + y q/p. Write x = |f |, y = |g | and integrate
to... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/1c016821763783c34c48816b1ac66969_MIT18_175S14_Lecture5.pdf |
If fn ≥ 0 then
lim inf
n→∞
fndµ ≥
lim inf fn)dµ.
n→∞
(cid:73)
�
(Counterexample for opposite-direction inequality using thin
and tall rectangles?)
Main idea of proof: first reduce to case that the fn are
increasing by writing gn(x) = infm≥n fm(x) and observing that
gn(x) ↑ g (x) = lim infn→∞ fn(x). Then truncat... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/1c016821763783c34c48816b1ac66969_MIT18_175S14_Lecture5.pdf |
18.445 Introduction to Stochastic Processes
Lecture 10: Hitting times
Hao Wu
MIT
16 March 2015
Hao Wu (MIT)
18.445
16 March 2015
1 / 8
Recall
Consider a network (G = (V , E), {c(e) : e ∈ E}). The effective
resistance is defined by
R(a ↔ z) = (W (a) − W (z))/||I||.
Consider a random walk on the network, the Green’s func... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/1c4e0f0bf28f0ec9f0fd6bc8f7b896f1_MIT18_445S15_lecture10.pdf |
/ 8
Transitive Markov chain
Roughly, a transitive Markov chain “looks the same" from any point in
the state space.
Definition
A Markov chain is called transitive if for each pair (x, y ) ∈ Ω × Ω, there
is a bijection ϕ : Ω → Ω such that
ϕ(x) = y ; P(ϕ(z), ϕ(w)) = P(z, w), ∀z, w.
Example : simple random walk on N-cycle,... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/1c4e0f0bf28f0ec9f0fd6bc8f7b896f1_MIT18_445S15_lecture10.pdf |
V → V such that
ϕ(x) = y ;
c(ϕ(z), ϕ(w)) = c(z, w), ∀z, w.
Remark : The random walk on a transitive network is a transitive
Markov chain.
Theorem
For the random walk on a transitive (connected) network, for any
vertices a and b, we have
Ea[τb] = Eb[τa].
Hao Wu (MIT)
18.445
16 March 2015
7 / 8
Summary
For random walk o... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/1c4e0f0bf28f0ec9f0fd6bc8f7b896f1_MIT18_445S15_lecture10.pdf |
Utility Theory
Week 4 Framing
Required Reading:
de Neufville, Richard, Applied Systems Analysis: Engineering Planning and Technology
Management, McGraw-Hill, New York, 1990. Chapters 18, 19, 20, 21.
McManus, H. L., and Ross, A. M., SSPARC Book Material for Lecture 4.
Gumbert, C. C., Violet, M. D., Hastings, D. E.... | https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1c5ef3f7a6f50076b2608123521822a9_04000lec4outlnv3.pdf |
onautics, Massachusetts Institute of Technology,
May 2001. (excerpt covering determination of utilities)
Spaulding, Timothy J., “Tools for Evolutionary Acquisition: A Study of Multi-Attributes
Tradespace Exploartion (MATE) Applied to the Space Based Radar (SBR), Master of
Science Thesis in Aeronautics and Astronaut... | https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1c5ef3f7a6f50076b2608123521822a9_04000lec4outlnv3.pdf |
1.4 Backward Kolmogorov equation
When mutations are less likely, genetic drift dominates and the steady state distributions are
peaked at x = 0 and 1. In the limit of µ1 = 0 (or µ2 = 0), Eq. (1.63) no longer corresponds
to a well-defined probability distribution, as the 1/x (or 1/(1 − x)) divergence close to x = 0
(or x... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
the next
following the transition rates, but irrespective of its previous history. This type of process
with no memory is called Markovian, after the Russian mathematician Andrey Andreyevich
Markov (1856-1922). We can use this probability to construct evolution equations for the
probability by focusing on the change of... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
1
2
(cid:18)Z
(cid:19)
15
Z
∂p(x, t|y)
∂y
(cid:19)
∂2p(x, t|y)
∂y2
+ · · · .
(1.66)
Using the normalization condition for R(δy, y) and the definitions of drift and diffusion
coefficients from Eqs. (1.36) and (1.37), we obtain
∂p(x, t|y)
∂t
= v(y)
∂p
∂y
+ D(y)
∂2p
∂y2 ,
(1.67)
which is known as the backward Kolmogorov equa... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
)
Z
= − ln (D(y)p∗(y)) .
(1.70)
The result of the above integration is related to an intermediate step in calculation of the
steady state solution p∗ of the forward Kolmogorov equation in (1.59). However, as we
noted already, in the context of absorbing states the function p∗ is not normalizable and
thus cannot be rega... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
if 2sN ≪ 1). If it is advantageous (2sN ≫ 1) it will be fixed with
probability Π1 = 1 − e−s irrespective of the population size! If it is deleterious (2sN ≪ −1)
it will have a very hard time getting fixed, with a probability that decays with population
size as Π1 = e−(2N −1)|s|. The probability of loss of the mutation is... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
mean fixation time is now computed from
hτ (y)ia =
∞
0
Z
dt t pa(t|y) =
1
Π∗(xa, y)
0
Z
∞
dt t
∂p(xa, t|y)
∂t
.
(1.74)
(1.75)
Following Kimura and Ohta (1968)3, we first examine the numerator of the above ex-
pression, defined as
(Writing limT →∞
equation by parts to get
T
0 rather than simply
R
0
Z
∞
0
Ta(y) = lim
T →∞
T... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
.67). Integrating the latter over time leads to
ByTa(y) = −p(xa, ∞|y) = −Π∗(xa, y) .
(1.80)
For example,
let us consider a population with no selection (s = 0), for which the
probability to lose a mutation is Π0 = (1 − y). In this case, Eq. (1.80) reduces to
y(1 − y)
4N
∂2T0
∂y2 = −(1 − y) ⇒
∂2T0
∂y2 = −
4N
y
.
(1.81)
... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
the mutation survives in the population.
The net probability that the mutation is still present at time t is
1−
S(t|y) =
dxp(x, t|y) ,
0+
Z
(1.85)
where the integrations exclude the absorbing points at 0 and 1. Conversely, the PDF that
the mutation disappears (by loss or fixation) at time t is
p×(t|y) = −
dS(t|y)
dt
= −... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/1c5ef58df27b3a6ae906ba0d5aaba9ed_MIT8_592JS11_lec5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
ESD.70J / 1.145J Engineering Economy Module
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ESD.70J Engineering Economy
Fall 2009
Session Two
Michel-Alexandre Cardin
Prof. Richard de Neufville
ESD.70J Engineering Eco... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
C4 and C5
ESD.70J Engineering Economy Module - Session 2
5
Random number generator
Follow the instructions, step by step
1. Go to tab “RAND”
2. Type “=Entries!C9*((1-
Entries!C25)+2*Entries!C25*RAND())” in cell C3
3. Type “=Entries!C10*((1-
Entries!C25)+2*Entries!C25*RAND())” in cell D3
4. Type “=Entries!C11*((1-
Entr... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
.70J Engineering Economy Module - Session 2
9
Monte Carlo Simulation
Generate many sets of random price for the
three-phase span
Calculate corresponding NPVs
Generate Distribution of NPVs
Statistical Analysis
ESD.70J Engineering Economy Module - Session 2
10
Setup simulation by Data Table
Follow these instructions, s... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
color
ESD.70J Engineering Economy Module - Session 2
13
Give it a try!
Check with your neighbors…
Check the solution sheet…
Ask me questions…
ESD.70J Engineering Economy Module - Session 2
14
Calculating descriptive statistics
• Useful to know mean, maximum, and
minimum values for the simulated
results
Follow step ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
0*F7”, and drag the formula down to G27
2. Set Cell H7
“=COUNTIF($B$9:$B$2008,"<="&G7)”, and drag
the formula down to H27
3. Set Cell I7 “=H7/2000”, and drag down to cell I27
4. Same is already done for Plan B
ESD.70J Engineering Economy Module - Session 2
19
Target curve
6. Right-click the chart on the right, selec... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
managers
and decision-makers!
ESD.70J Engineering Economy Module - Session 2
22
Question
• Why are high NPV values more cut off
for Plan B on the target curve and
histogram than for Plan A?
– A matter of constraints…
ESD.70J Engineering Economy Module - Session 2
23
Give it a try!
Check with your neighbors…
Check ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/1c610901c86e62fbe5cc1e555297b7ed_MITESD_70Jf09_lec02.pdf |
Turbulent Flow and Transport
4
Free Shear Flows I: Jets, Wakes, etc.−Solutions Based on
Simple Mean−Flow Closure Schemes
4.1 Mean−flow closure schemes for free shear flows.
4.2
4.3
4.4
for
4.5
Spreading of a velocity discontinuity with downstream distance in steady flow.
The nature of the laminar flow soluti... | https://ocw.mit.edu/courses/2-27-turbulent-flow-and-transport-spring-2002/1c80d3bfd54cb73fffc3a301657e8cfb_Free_shear_flows.pdf |
Mech., 74 (1976): 209−250.
Further reading:
Hinze. Chapter 6.
Abramovich. The Theory of Turbulent Jets:103−113, 120−125.
Rajaratnam. "Turbulent Jets." Elsevier, 1976.
Rodi. In "Studies in Convection" B. E. Launder. ed. Academic
Press, 1975: 79 ff.
Townsend. Chap.6 in The Structure of Turbulent Shear Flow. 2nd ed.... | https://ocw.mit.edu/courses/2-27-turbulent-flow-and-transport-spring-2002/1c80d3bfd54cb73fffc3a301657e8cfb_Free_shear_flows.pdf |
Introduction to Engineering
Introduction to Engineering
Systems, ESD.00
Lecture 6
Lecturers:
Professor Joseph Sussman
Dr Afreen Siddiqi
Dr. Afreen Siddiqi
TA: Regina Clewlow
Uncertainty Lecture 2 Outline
Uncertainty Lecture 2-- Outline
Global Climate Change
High-impact, Low-probability events
Decision-making Und... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1cb961bffd1d5e437be957c66ad547fd_MITESD_00S11_lec06.pdf |
ertainty: High-impact, Low
Probability Events
Probability Events
Ripped from the headlines
The Japan earthquake and tsunamis
5th biggest earthquake in recorded history,
biggest ever in Japan
biggest ever in Japan
Huge loss of life, injuries, property damage
Japan likely the most prepared nation in the
th
k di
wo... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1cb961bffd1d5e437be957c66ad547fd_MITESD_00S11_lec06.pdf |
Uncertainty: Annuities
Annuities
Buy an annuity for $X
You get $Y/ year for the rest of your life….
Why it is a [good, bad] deal for you?
Whyy it is a [g[g
you’re the annuity?
ood, bad]] deal for the comppanyy that sold
What might you do instead of buying an annuity?
What might you do instead of buying an annuity?
... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1cb961bffd1d5e437be957c66ad547fd_MITESD_00S11_lec06.pdf |
=
All M
Uncertainty: Bayes Theorem
Uncertainty: Bayes Theorem
’
Bayes’ Theorem
Conditional probabilities
P(event A happens)= [P(event A/given B occurs) for all
P(event A happens)= [P(event A/given B occurs) for all
possible outcomes of B] * P( each possible outcome of
B)]
Uncertainty: Bayes Theorem
Uncertainty... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1cb961bffd1d5e437be957c66ad547fd_MITESD_00S11_lec06.pdf |
D.00 Introduction to Engineering Systems
SSpriing 20112011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1cb961bffd1d5e437be957c66ad547fd_MITESD_00S11_lec06.pdf |
MIT OpenCourseWare
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6.080 / 6.089 Great Ideas in Theoretical Computer Science
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.080/6.089 GITCS
Mar 11, 2008
Lecturer: Scott Aaronson
Scribe: Yinmeng Zhang
Lecture 10
1 Administriv... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
SATISFIABILITY (or SAT for short). We saw Cook and Levin’s proof that SAT is NP-complete
by reduction from DUH. Today we will see some more reductions.
10-1
PNPNP-completeNP-hard 3 SAT reduces to 3SAT
Though SAT is NP-complete, specific instances of it can be easy to solve.
Some useful terminology: a clause is a sin... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
straightforward – every
in the formula becomes a NOT, AND, or OR gate in the circuit. One detail we need to take care
of is that when we say multiple literals ∧ed or ∨ed together, we first need to specify an order in
which to take the ∧s or ∨s. For example, if we saw (x1 ∨ x2 ∨ x3) in our formula, we should parse
it... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
two inputs and one output for AND
and OR, and one input and one output for NOT. Let’s define a variable for every gate. We want
to know if there is a setting of these variables such that for every gate, the output has the right
relationship to the input/inputs, and the final output is set to true.
So let’s look at th... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
And similarly for the AND gate. The big idea here is that we’re taking small pieces of a problem
we know to be hard (gates in a circuit) and translating them into small pieces of a problem we’re
trying to show is hard (CNF formulas). These small pieces are called “gadgets”, and they’re a
recurring theme in reduction... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
triangle. So for a given good 3-coloring, whatever color the first vertex gets assigned,
we’re going to call True; the second is going to be False; and the third is going to be Neither.
Here is a gadget for the NOT gate.
Because x and ¬x are connected to each other, any good coloring will assign them different
colors... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
, we can turn any circuit into a graph so that assignment of T/Fs
to the wires translates to colorings of the graph. If we then connect the final output vertex to the
False vertex in the palette, then the graph we’ve constructed will have a good 3-coloring iff the
circuit had a satisfying assignment.
So we’ve just pr... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
∧y132a2b 5
In conclusion
In 1972, Karp showed 21 problems were NP-complete using the above techniques and a good dose
of cleverness. We could spend a month looking at clever gadgets, but we’re not going to. The
takeaway lesson is that NP-completeness is ubiquitous. Today, thousands of practical problems
have been ... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
will be solvable efficiently, and every problem in
NP will be NP-complete: the computational universe will look pretty monotonous. If P=� NP,
then what do we know? All the problems in P will be easy, and all NP-complete problems will
be hard. Would there be anything in between? Would there be problems in NP which are ... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/1cce8117881c1d1e5a21935bdf76d377_lec10.pdf |
14. Instability of Superposed Fluids
Figure 14.1: Wind over water: A layer of fluid of density ρ+ moving with relative velocity V over a layer
of fluid of density ρ− .
Define interface: h(x, y, z) = z
The unit normal is given by
−
η(x, y) = 0 so that ∇h = (
ηx,
−
ηy, 1).
nˆ =
∇h
∇h
|
|
=
−
ηy, 1)
1/2
ηx,
(
−
... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
±
∂y
yˆ +
∂φ±
∂z
zˆ
from which
∂η
∂t
=
1
V +
2
∂φ±
∂x
)
(
−
(
∓
ηx) +
∂φ±
(
∂y
−
ηy) +
∂φ±
∂z
(14.1)
(14.2)
(14.3)
(14.4)
Linearize: assume perturbation fields η, φ± and their derivatives are small and therefore can neglect
their products.
Thus ηˆ
ηy, 1) and ∂η =
ηx,
(
1 V ηx +
2
∂φ±
∂z
∂t... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
1
+ ρ±
2
V
(
∓
∂φ±
∂x
)
+ p± + ρ±gη = G(t)
(14.7)
so
p−
−
p+ = (ρ+
−
ρ−)gη + (ρ+
∂φ±
∂t
−
ρ−
∂φ−
∂t
) +
V
2
(ρ−
∂φ−
∂x
+ ρ+
∂φ+
∂x
) =
−
σ(ηxx + ηyy)
(14.8)
is the linearized normal stress BC. Seek normal mode (wave) solutions of the form
η = η0e
iαx+iβy+ωt
φ± = φ0±e
∓kz iαx+iβy+ωt
e ... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
= ω ρ+(ω
[
1
2
−
iαV ) + ρ−(ω +
1
2
iαV ) + gk(ρ−
]
ρ+) +
−
1
2
iαV ρ+(ω
[
−
1
2
iαV ) + ρ−(ω +
1
2
iαV )
]
−
so
ω2 + iαV
ρ−
ρ+
−
ρ− + ρ+
(
ω
)
−
1
α2V + k2C0
2
4
2 = 0
where C 2
k.
0
Dispersion relation: we now have the relation between ω and k
σ
ρ−+ρ+
ρ−−ρ+
ρ−+ρ+
+
≡
g
k
(
)
ω =
1
2
i
... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
We proceed by considering two important special cases, Rayleigh-Taylor and Kelvin-Helmholtz instability.
MIT OCW: 18.357 Interfacial Phenomena
56
Prof. John W. M. Bush
14.1. Rayleigh-Taylor Instability
Chapter 14. Instability of Superposed Fluids
14.1 Rayleigh-Taylor Inst... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
= ΔES + ΔEG < 0, i.e. when ρg > σk2 ,
so λ > 2πlc.
The system is thus unstable to long λ.
Note:
h2 dx =
)
−
−
0
Figure 14.2: The base state and the per-
turbed state of the Rayleigh-Taylor system,
heavy fluid over light.
1. The system is stabilized to small λ disturbances by
σ
2. The system is always unstable for su... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
gravitationally stable, but destabilized by the shear.
Take k parallel
instability
(V · k)
to V ,
k2V
and
the
so
=
2
2
ρ+ so the
≥
criterion becomes:
2
ρ−ρ+V > (ρ−
ρ+)
−
g
k
+ σk
(14.17)
Equivalently,
2
ρ−ρ+V > (ρ−
ρ+) g
−
λ
2π
+ σ
2π
λ
(14.18)
Note:
1. System stabilized to short λ disturban... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
the min
Figure 14.5: Fluid speed V (k) required for
the growth of a wave with wavenumber k.
√
103
2
1.2·10−3
E.g. Air blowing over water: (cgs)
2
Vc ∼
V =
1
c
·
mum wind speed required to generate waves.
These waves have wavenumber kc =
waves.
1·103
70
70
J
⇒
·
≈
650cm/s is the mini-
3.8 cm , so λc = ... | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/1d1f0dbdfd935d542ea0242a4649fada_MIT18_357F10_Lecture14.pdf |
6.897: Selected Topics in Cryptography
Lecturer: Ran Canetti
Lectures 3 and 4:
ZK as function evaluation and
sequential composition of ZK
• Review the definition of ZK and PoK
• Give SFE-style definition of ZK and show
equivalence to the standard one.
• The Blum protocol for Hamiltonicity:
– Commitment schemes
– ... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
(x,0). (But this is easy to achieve, by
having P send (x,”reject”) to V.)
Review: Zero-Knowledge
[Goldwasser-Micali-Rackoff 85]
• Zero-Knowledge:
For any verifier V* there exists a machine S
such that for all x,w,z, S(x,R(x,w),z) ≈ V*
P(x,w)(z).
“Whatever V* can gather from interacting with P, it could have
comp... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
in the spirit of the “enhanced PoK” of [Lindell 03].)
ZK PoK as an SFE task
Let R(x,w) be a binary relation.
Consider the 2-party function:
F
R ((x,w), - , - ) = ( - , (x,R(x,w)) , (x,R(x,w)))
zk
Theorem: A two-party protocol securely
R (with respect to
realizes Fzk
non-adaptive adversaries) if and only if
it is ... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
um protocol
Common relation:
HC(G,H)=1 if H is a Hamiltonian cycle in graph G.
Common input: k-node graph G
Secret witness for P: Hamiltonian cycle H in G.
• P(G,H): If HC(G,H)=0 then send (G,”reject”) to V. Else, choose a
random permutation p on [1..k] and send (G, C(p(G)), C(p)) to V.
• V: If received (G,”reject... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
½. Let R(x,y) be a binary relation. Recall:
Fzk
R ((x,w), -,- ) = ( - ,(x,R(x,w)),(x,R(x,w)))
• Define:
Fwzk
R ((x,w), -,c) = If R(x,w)=1 then output (-,(,x,1),(x,1).
Else, if c=“no cheat” output (,(,x,0),(x,0)).
If c=“cheat” output (,(,x,b),(x,b)) for bÅ R {0,1}.
H
Claim: The basic Blum protocol securely realize... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
value to F and on output:
•
•
•
If all decommitments succeed, then construct the Hamilt. Cycle. (if
fail then some commitment was broken). Then give (G,H) to the
TP (as the prover), and hand Z the output of A from either the first
or the second run (chosen at random).
If all decomm. for one value of b succeed, t... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
that S does not know.)
From Fwzk
R to Fzk
R
A protocol for realizing Fzk
• P(x,w): Run k copies of Fwzk
R in the Fwzk
R -hybrid model:
R , sequentially. Send
(x,w) to each copy.
• V: Run k copies of Fwzk
R , sequentially. Receive
i
(x ,b ) from the i-th copy. Then:
– If all x’s are the same and all b’s are the... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
,w) to to Fzk
R
Else S gives (x0 ,w0) (the default value) to Fzk .
i
R . (If didn’t find such w then
R , where w is an invalid witness.
i
i
i
–
Finally, S outputs whatever A outputs.
Analysis of S:
– When the verifier is corrupted, the views of Z from both interactions
are identically distributed.
– When the pr... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
)),C(p1)…C(pk(G)),C(pk)
VÆP: D(b1...bk)
PÆV: D(p1(G)),D(p1)…D(pk(G)),D(pk)
V: accept if all copies accept.
Parallel composition of Blum’s protocol
This solves the ZK problem. But now soundness fails:
Assume P,V use the same commitment C, and C is “malleable”,
say given C(b) can generate C(1-b) Then, P* can:
• Recei... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf |
Lecture 9
8.321 Quantum Theory I, Fall 2017
48
Lecture 9 (Oct. 4, 2017)
9.1 Spin- 1
2 in an AC Field
Consider a spin- 1 system in a time-dependent magnetic field. The Hamiltonian is
2
H = −
ge
2
m
S
· B(t .
)
(9.1)
We will consider a particular class of time-dependent magnetic fields, which we can write in the
form
B(t) ... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
(ωt)xˆ + sin(ωt)yˆ) .
⊥
ω =
0
(cid:12)
(cid:12)
(cid:12)
(cid:12)
geB
0
2m
(cid:12)
(cid:12)
(cid:12)
(cid:12)
,
H0 = ω0Sz
.
(9.4)
(9.5)
(9.6)
(9.7)
(9.8)
(Note that we are taking e < 0 so that the signs work out here.) The time-evolution operator due
to this part of the Hamiltonian is
(cid:126)
U0(t) = e−iH0t/ .
The e... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
)
cos
φ
2
− iσz sin
(cid:19)
φ
2
(cid:18)
cos
= S+
− iσz sin
(cid:19)
φ 2
φ
2
2
= S+(cos φ
− iσz sin φ)
= S+eiφ .
(9.13)
Here, in the first line we have expanded the exponentials, using the fact that the Pauli matrices
square to the identity. In the second line, we have used the fact that σz anticommutes with both
σx an... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
(9.16) simplifies to
which is time-independent. The interaction picture time-evolution operator is then
VI = −
geB1
4m
(cid:0)S+ + S−(cid:1) = −
geB1 Sx ,
4m
UI(t) = ei
geB
2m(cid:126)
1 Sxt
.
(9.17)
(9.18)
This operator rotates states about the x-axis of spin space by an angle
rotation,
geB
2m(cid:126)
1 t. The frequen... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
Rabi frequency.
Imagine that we start in a spin-up state, i.e.,
|ψI(0)(cid:105) →
(cid:19)
(cid:18)
1
0
,
(9.23)
where the arrow indicates that we are representing the state as a vector. At time t, we then have
(cid:18)1
0
(cid:19)
|ψI(t)(cid:105) → e−iωRtσx/2
(cid:19)
(cid:19)
− iσx sin
(cid:18) ωRt
2
(cid:1)
(cid:1) ... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
Resonant Drive
Up until now, we have been working in the case of resonant drive. What happens if we’re off
resonance, i.e., ω (cid:54)= −ω0? In this case, even though we went to the interaction picture, we still
have a time-dependent Hamiltonian that we don’t know how to solve, because VI(t) is still time-
dependent. It... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
9.30)
first term in brackets is unaffected by the time evolution, because Sz commutes with the
The
exponentials outside the brackets. The second term in brackets, along with the time-evolution
exponentials, looks exactly like the calculation we did in Eq. (9.16), but with the frequencies in all
of the exponentials exactl... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
, t) =
(cid:88)
(cid:126)
e−iEa(cid:48) (t−t0)/ ca(cid:48)(t0)ua(cid:48)(x) .
a(cid:48)
∞
ˆ
−∞
ψ(x, t) =
ddx(cid:48) K(cid:0)x, t; x(cid:48), t0
(cid:1)ψ(cid:0)x(cid:48), t0
(cid:1) ,
with
K(cid:0)x, t; x(cid:48), t0
(cid:1) =
(cid:88)(cid:10)x(cid:12)
(cid:12)
(cid:12)
a(cid:48)(cid:11)e−iEa(cid:48) (t−t0)/ (cid:10)a(... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/1d8cd3d97c8c26f16bea95051466574c_MIT8_321F17_lec9.pdf |
Engineering Risk Benefit Analysis
1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82, ESD.72
CBA 3.
Bases for Comparison of Alternatives
George E. Apostolakis
Massachusetts Institute of Technology
Spring 2007
CBA 3. Bases for Comparison of Alternatives
1
Overview
•(Net) Present Worth or Value [(N)PW or
(N)PV] ... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
CBA 3. Bases for Comparison of Alternatives
4
Present Worth: Example
• You buy a car and you put down $5,000. Your payments
will be $500 per month for 3 years at a nominal interest rate
of 10%. Assuming monthly compounding, what is the
present price you are paying?
From CBA 2, Slide 14, we get
)n,i,A/P(
=
(1
+
i)
... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
Ak
if
if
PW(i)Aj > PW(i)Ak
AE(i)Aj > AE(i)Ak
• These criteria are using the total investment.
CBA 3. Bases for Comparison of Alternatives
7
Example of Total Investment Comparisons
End of Year
0
1
2
3
B3
-$12,000
-1,200
-1,200
1,500
B4
-$15,000
-400
-400
3,000
• The benefits from these alternatives are
identical. We... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
,1
200
+
,2
700
1.0
3
1.1
=
1
−
−=
,1
200
+
,2
302.0x700
,5$
209
10(AE
%)
−=
,15
000
402.0x
−
400
+
,3
400
.0x
302
=
−=
,5$
403
,5$
209
4B
−<
Thus, B3 should be preferred over B4, just as before.
CBA 3. Bases for Comparison of Alternatives
10
Impact of Inflation (1)
• Suppose that the inflation rate is 9% and that the... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
1.1
%)
4B
+
,1
943
3
1.1
=
• Therefore, B4 is now preferred, while, in the case
without inflation, B3 was preferred (slide 9).
CBA 3. Bases for Comparison of Alternatives
12
PW and AE on Incremental Investment
• Derive the difference between the two alternatives.
Year
0
1
2
3
B3
-$12,000
-1,200
-1,200
1,500
B4
-$15,... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
*
)i1(F
+
t
−
t
• The IRR represents the percentage or rate earned on the
unrecovered balance of an investment such that the
payment schedule makes the unrecovered investment equal
to zero at the end of investment life.
CBA 3. Bases for Comparison of Alternatives
15
End of year t
0
1
2
3
4
5
Example
Ft
-1,000
-800
... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
sequence F0, F1, F2, …, Fn, has one change in
sign only.
3. PW(0) > 0 (sum of all receipts > sum of all
disbursements)
CBA 3. Bases for Comparison of Alternatives
19
Examples
• Both B3 and B4 of slide 8 fail the third condition.
For B3, PW(0) = -$12,900 <0.
• For the cash flow on slide 16, PW(0) = 900 > 0.
• Consid... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
ases for Comparison of Alternatives
22
IRR on Incremental Investment
1. Make sure the cash flows satisfy the conditions on
slide 19.
2. List alternatives in ascending order based on
initial cost.
3. The “current best” alternative can be the “Do
Nothing” one.
4. Determine the differences between the
“challenging” a... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
)x1(x
+
21
21
i
*
A
2
≡−
A
1
x
21
=
%15%5.10
<
Therefore, A1 remains the current best alternative.
CBA 3. Bases for Comparison of Alternatives
26
$12,000
$10,000
$8,000
$6,000
$4,000
$2,000
$0
0
($2,000)
($4,000)
Example (4)
A1 A2 A2-A1
0.1987
0.2499
0.1
0.2
0.3
0.4
0.5
0.1056
CBA 3. Bases for Comparison of Alternativ... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
6,000
$4,000
$2,000
$0
0
($2,000)
($4,000)
($6,000)
IRR on Total Investment (1)
• We can calculate the IRRs for the total cash flows
on slide 24.
x ≡
• Let then
2
i*
A2
10
)x1(
1
+
−
2
10
)x1(x
+
2
2
0
−=
,8
000
+
,1
900
x
2
≡
*
A2 =
i
%9.19
CBA 3. Bases for Comparison of Alternatives
30
IRR on Total Investment (2)
... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
AE-)based results using total and
incremental investments are identical.
Approach
1. Make sure the cash flows satisfy the conditions on
slide 19.
2. List alternatives in ascending order based on
initial cost.
CBA 3. Bases for Comparison of Alternatives
33
PW on Incremental Investment Revisited
(2)
3. The “current b... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
atives
36
The Example Revisited (3)
• Finally,
)15(PW
−=−
AA
3
1
,5
000
+
100,1
10
+
)15.01(
−
)15.01(15.0
+
1
10
=
=
521
>
0
• Therefore, A3 becomes the current and final best
alternative. This is the same as in slide 28, but not
the result on slide 31 (best: A1).
CBA 3. Bases for Comparison of Alternatives
37
Sum... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
atives
40
Example (2)
• AE(A) = 6.4K + 4K (A/P, 7%, 6) 6.4K + 4K (0.21)
≈
= $7,240
• AE(B) = 1.4K + 16K(A/P, 7%, 3) 1.4K+16K (0.38)
≈
= $7,480
• AE(C) = 1K + 20K (A/P, 7%, 4) 1K+ 20K (0.3) =
≈
= $7,000
CBA 3. Bases for Comparison of Alternatives
41
Repeating Cash Flows (1)
• When sequences... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
5
6
7
8
9
10
11
12
Extended Cash Flows
A
(in $000)
4
6.4
6.4
6.4
6.4
6.4
10.4
6.4
6.4
6.4
6.4
6.4
6.4
B
(in $000)
16
1.4
1.4
17.4
1.4
1.4
17.4
1.4
1.4
17.4
1.4
1.4
1.4
C
(in $000)
20
1
1
1
21
1
1
1
21
1
1
1
1
CBA 3. Bases for Comparison of Alternatives
44
Present ... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
Comparison of Alternatives
45
Present Worth of Original Alternatives
• PW(A, 6 yrs) = 4 + 6.4(P/A, 7%, 6) = 4 +(6.4)(4.76) =
= $34.46K
• PW(B, 3 yrs) = 16 + 1.4 (P/A, 7%, 3) =
= 16 +(1.4)(2.62) = $19.68K
• PW(C, 4 yrs) = 20 + 1 (P/A, 7%, 4) = 20 + 1(3.4) =
= $23.4K
• These are the PWs of costs over the actual life... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf |
Big Picture
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
2/7/2007
Lecture 1
Newton’s Laws, Cartesian and Polar
Coordinates, Dynamics of a Single Particle
Big Picture
First Half of the Course → Momentum Principles (Force, Vectors) Newtonian
Dynamics
Second Half of the Course → Lagra... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
or continues to move in a straight line
with constant velocity if there is no resultant force acting on it.
II. 2nd Law - A particle acted upon by a resultant force moves in such a manner
that the time rate of change of its linear momentum is equal to the force.
�
F = ma for a single particle.
where F is the force... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
Two coordinate systems: Cartesian and Polar
3
r is position, and t is time. The direction of v is in the direction of Δr as Δt → 0.
The acceleration:
Ac... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
by MIT OCW.
=
eˆr + r
dr
dt
deˆr = r˙eˆr + rθ˙eˆθ
dr
dt
dt
= ¨reˆr + ˙rθ˙eˆθ + ˙rθ˙eˆθ + rθ ¨ eˆθ − rθ˙2eˆr
v =
a =
dv
dt
a = (¨r − rθ˙2) ˆer + (rθ ¨ + 2 ˙rθ˙) ˆeθ
d
dt
eˆθ = −θ˙eˆr
Proof that dt = θeˆθ:
deˆr
˙
Figure 3: Differentiation of unit vectors. Changes in the direction of unit vector
eˆr can... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
Single Particle (Review)
Figure 4: A point mass m moves from r(t0) to r(t). Figure by MIT OCW.
Consequences of Newton’s Second Law: Linear and Angular Mo
mentum Conservation
Using an inertial frame of reference, here is the expression of Newton II:
�
F = ma
.
Linear Momentum Principle
�
F = (mv) = p˙
d
dt
(p =... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
cross product of two
parallel vectors is zero.
Define Resultant Torque around B. τ B = r × �
F is the moment of the total
force about B. Combining this definition with Equation (2) yields
τ B = h˙
B + vB × mv.
If τB = 0 and vB = 0 or vB � mv ⇒ h˙ B = 0 (Conservation of Angular
Momentum)
Cite as: Thomas Peacock an... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1e1173e9515ba00437cf6685d160de3e_lec01.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
SLOAN SCHOOL OF MANAGEMENT
15.565 Integrating Information Systems:
Technology, Strategy, and Organizational Factors
15.578 Global Information Systems:
Communications & Connectivity Among Information Systems
Spring 2002
Lecture 4
INTER- AND INTRA-
ORGANIZATIONAL SYSTEMS
1
Diff... | https://ocw.mit.edu/courses/15-565j-integrating-esystems-global-information-systems-spring-2002/1e11f1d7504656e102fec6aa1f2283f7_lecture04.pdf |
. SUPPORTING COST AND DIFFERENTIATION
STRATEGIES
2. ALTERING INDUSTRY STRUCTURE
3. SPAWNING ENTIRELY NEW BUSINESSES
EFFECTIVE STRATEGIES DO ALL THREE!
6
EXAMPLE
FIRM INFRASTRUCTURE
HUMAN RESOURCE MANAGEMENT
TECHNOLOGY DEVELOPMENT
MARGIN
PROCUREMENT
INBOUND
LOGISTICS
OPERATIONS
OUTBOUND
LOGISTICS
MARKETI... | https://ocw.mit.edu/courses/15-565j-integrating-esystems-global-information-systems-spring-2002/1e11f1d7504656e102fec6aa1f2283f7_lecture04.pdf |
6% / yr
• Profit growth 2% / yr
8
MCKESSON - CASE (continued)
MCKESSON - CASE (continued)
• DRUG WHOLESALING
• Industry ($9B)
• Problem Area
• “Kill or Cure” - Alternatives?
Mfg
Direct
60%
(rising)
(dropping)
Chain Drug Store
Distributors/
Wholesalers
40%
{ McKesson
20%
Independent
Drug Stores
9
... | https://ocw.mit.edu/courses/15-565j-integrating-esystems-global-information-systems-spring-2002/1e11f1d7504656e102fec6aa1f2283f7_lecture04.pdf |
•
• Customer reduced by 25% but average order size increased
• Increase “Switching Costs”
IMPROVEMENTS IN OPERATIONS
• Telephone Order -taker Clerks: Reduced from 700 to 15
• Purchasing Staff: Reduced from 140 to 12
• Warehouses: Decrease 50% (“mother trucks”)
• OFFICE AND WAREHOUSE PRODUCTIVITY INCREASED
• ROL... | https://ocw.mit.edu/courses/15-565j-integrating-esystems-global-information-systems-spring-2002/1e11f1d7504656e102fec6aa1f2283f7_lecture04.pdf |
00 M/yr prescription)
16
OTHER LESSONS LEARNED
• Changing customer behavior difficult
• Changing corporate “culture” difficult
• Needs lots of top management patience
• Attempts to use these approaches not always
successful
17
SUMMARY
• NEED TO EXPLORE NEW STRATEGIES AND PROCESSES --
MIGHT TRANSFORM COMPANY A... | https://ocw.mit.edu/courses/15-565j-integrating-esystems-global-information-systems-spring-2002/1e11f1d7504656e102fec6aa1f2283f7_lecture04.pdf |
Simplest Car Following Tra(cid:14)c Flow Model.
Rodolfo R. Rosales
.
(cid:3)
MIT, Friday March 26, 1999.
Abstract
These notes describe in some detail the continuum limit behavior of a very simple car following
tra(cid:14)c (cid:13)ow model. The formation and behavior of shock waves is described. This model is the
one s... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/1e6a824d45c144cecd4c0154ea3006a0_MIT18_306f09_lec24_CF_Simple_Model.pdf |
ondimensionalization.
Consider a line of cars on a road, with car n located at ~x
= ~x
(t), moving at speed ~u
=
.
n
n
n
dt
d ~x
n
Measure distance ~x along the road in the same direction the cars move (so the car velocities ~u
are all
n
non-negative). Number the cars so that
~x
is an increasing sequence ( ~x
~x
) > ca... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/1e6a824d45c144cecd4c0154ea3006a0_MIT18_306f09_lec24_CF_Simple_Model.pdf |
ow functions.
As long as the situation is not changing too rapidly, this is not unreasonable. Note that in this model we will,
implicitly, deal with all the cars as if they were equal copies of each other | all the cars obey exactly the same rules.
r
r
Simple Tra(cid:14)c Flow Model.
3
MIT, Friday March 26, 1999 | Ros... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/1e6a824d45c144cecd4c0154ea3006a0_MIT18_306f09_lec24_CF_Simple_Model.pdf |
=
u
;
U ( ~(cid:26)) =
U ((cid:26)) ;
Q( ~(cid:26)) = q
Q((cid:26)) and
t =
t :
n
n
n
J
n
n
n
m
~
~
~
q
m
q
m
L(cid:26)
J
(cid:26)
J
(cid:26)
J
q
m
(1.3)
Then the equations take the form
dx
n
(cid:15)
= u
= U ((cid:26)
)
and (cid:26)
=
;
(1.4)
n
n
n
dt
x
x
n
n
+1
(cid:0)
where (cid:15) = 1=(L(cid:26)
) is a small nondi... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/1e6a824d45c144cecd4c0154ea3006a0_MIT18_306f09_lec24_CF_Simple_Model.pdf |
spaced, so that
=
.
L
1
Simple Tra(cid:14)c Flow Model.
4
MIT, Friday March 26, 1999 | Rosales.
(cid:26) along the road | say, a hump or a sinusoidal up and down. This perturbation is \marked" by
a discrete set of points (the car positions) and needs a minimum number of them before it can be
clearly identi(cid:12)ed |... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/1e6a824d45c144cecd4c0154ea3006a0_MIT18_306f09_lec24_CF_Simple_Model.pdf |
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