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T OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology.
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6.973 Communication System Design
Standard cell layout methodology
Celktructuru!hldden under interconnect layers
With limlted interconnect layers, dedicated routing channels
between rows of standard cells... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
Communication System Design, Spring 2006.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
17
Courtesy of Anantha Chandrakasan. Used with permission.
Clockina
Large Systems
Most large scale ASICs, and systems built... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
all flip-flops on chip
Clock
Distribution
Network length, metal width and
height, ctqdi/irg caps
jntmCe
Clock
Driver
Varrbtions in /ma/ clock
load, local power supply,
local gate length and --A
threshold: lccal
temprutwe
is "clock skew"
Local
Clock
Buffers
Courtesy of A ~ i n dand Krste Asanovic. Used ... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
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6.973 Communication System Design
23
Modern Interconnect Stack
Images removed due to copyright restrictions.
Tungsten local interconnect
IBM CMOS7 process
6 layers of copper wiring
1 layer o f tungsten ... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
Capacitance depends on geometry o f surrounding wires and
relative permittivity, &,,of
insulating dielectric
3-9
- silicon dioxide, SiOp E,
- silicon flouride, SiOF E, = 3.1
- SiLKTMpolymer, E, = 2.6
Can have different materials between wires and between
layers, and also different materials on higher layers
Ci... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
ed n model gives reasonable approximation
- Rw is lumped resistance o f wire
- Cw is lumped capacitance (put half at each end)
cw
Delay = Rdriver x -+ (Rdriver + Rw) x
2
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
M I T OpenCourseWare (http://ocw.mit.edu/), Ma... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
tracks improvement in gate delay
[Fnm Mark Homwitr, DAC ZdlXl]
Courtesy of Arvind and Krste Asanovic. Used with permission.
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
MIT Opencourseware (http://ocw.mit.edu/), Massachusetts Institute of Technology.
Downloaded ... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
Month YYYY].
6.973 Communication System Design
Courtesy of Arvind and Krste Asanovic. Used with permission. | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/44d1cff68d0f994cca25442de3e1ac78_lecture_5.pdf |
Topic 11 Notes
Jeremy Orloff
11 Argument Principle
11.1 Introduction
The argument principle (or principle of the argument) is a consequence of the residue theorem. It
connects the winding number of a curve with the number of zeros and poles inside the curve. This
is useful for applications (mathematical and otherwi... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
= = mult(
0).
1
11 ARGUMENT PRINCIPLE
2
Likewise, if
0 is a pole of order then the Laurent series for () near
0 is
() = ( −
0)−()
where () is analytic and never 0 on a small neighborhood of
0. Thus,
0)− ′()
′()
()
( −
= −
0)−−1() + ( −
0)−()
( −
′()
()
= −
−
0
+
Again we have that
0 is ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
We say maps to (cid:253).
We have done this frequently in the past, but it is important enough to us now, so that we will stop
here and give a few examples. This is a key concept in the argument principle and you should make
sure you are very comfortable with it.
Example 11.2. Let () = e with 0 ≤ ≤ 2 (the unit circle... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
′()
()
∫
= 2( , − , )
So we need to show is that the integral also equals the winding number given. This is simply the
change of variables = (). With this change of variables the countour = () becomes =
(cid:253)() and = ′() so
′()
()
∫
= ∫ (cid:253)
= 2 Ind( (cid:253), 0)
The last equality in the abo... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
′()
= ∫
1 + ()
Ind(1 + (cid:253), 0) = Ind( (cid:253), −1)
1+ , =
1+ , = ,
1+ ,
(because (1 + )′ = ′)
(1 + winds around 0 ⇔ winds around -1)
(same in both equations)
(poles of = poles of 1 + )
1 = circle of radius 2.
2 = circle of radius 1/2.
3 = circle of radius 1.
Example 11.5. Let () = 2 + Find the wi... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
, ℎ are analytic functions on and inside , except for some finite poles.
There are no poles of and ℎ on .
(cid:240)ℎ(cid:240) < (cid:240) (cid:240) everywhere on .
Then
That is,
Ind( (cid:253), 0) = Ind(( + ℎ)(cid:253), 0).
, − , = +ℎ, − +ℎ,
5
(4)
Proof. In class we gave a heuristic proof involving a person wa... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
:243)
(cid:243)
ℎ
(cid:243)
(cid:243)
Next, by assumption
(cid:243)
(cid:243)
(cid:243) (cid:243)
maps to the inside of the unit disk centered at 1. (You should draw a figure for this.) This implies
)
that
ℎ
(cid:253) is inside the unit circle. . This means that 1 +
( )
ℎ
< 1, so
+ ℎ
(6)
=
((
+ ℎ
)
(cid... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
.
Example 11.7. Show all 5 zeros of 5 + 3 + 1 are inside the curve
Solution: Let () = 5 and ℎ() = 3 + 1. Clearly all 5 roots of (really one root with multiplicity
5) are inside
2. The corollary to Rouchés theorem says all
5 roots of + ℎ = 5 + 3 + 1 must also be inside the curve.
Example 11.8. Show + 3 + 2e has on... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
240) ()(cid:240) > − 3 for large,
(cid:240)ℎ()(cid:240) = 2(cid:240)e+(cid:240) = 2e < 2 (since < 0).
So (cid:240)ℎ(cid:240) < (cid:240) (cid:240) on .
The only zero of is at = −3, which lies inside the contour.
Therefore, by the Corollary to Rouchés theorem, + ℎ has the same number of roots as inside
the contour,... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
:240) ()(cid:240) = , so we have shown (cid:240)ℎ(cid:240) < (cid:240) (cid:240) on the curve. Thus, the corollary to
Rouchés theorem says + ℎ and have the same number of zeros inside (cid:240)(cid:240) = . Since we know
has exactly zeros inside the curve the same is true for the polynomial + ℎ. Now let go to
infini... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
This assumption holds in many interesting cases. For example,
quite often () is a rational function ()∕ () ( and are polynomials).
We will be concerned with the stability of the system.
Definition. The system with system function () is called stable if all the poles of are in the left
half-plane. That is, if all the ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
visualize () using a pole-zero diagram. This is a diagram in the -plane where we put a
small cross at each pole and a small circle at each zero.
Example 11.12. Give zero-pole diagrams for each of the systems
1() =
( + 2)(2 + 4 + 5)
,
2() =
(2 − 4)(2 + 4 + 5)
,
3() =
( + 2)(2 + 4)
Solution: These are the same... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
If the system with system function () is unstable it can sometimes be stabilized by what is called a
negative feedback loop. The new system is called a closed loop system. Its system function is given
Re(s)Im(s)xxxG1(s)1iRe(s)Im(s)xxxxG2(s)1iRe(s)Im(s)xxxG3(s)1i
11 ARGUMENT PRINCIPLE
by Black’s formula
() =
()... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
( −
0) + …
0 it equals ∕() = 1∕.)
which is clearly analytic at
Example 11.13. Set the feedback factor = 1. Assume is real, for what values of is the open
loop system () =
stable? For what values of is the corresponding closed loop system
() stable?
(There is no particular reason that needs to be real in this ex... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
= −1∕3, so the closed loop system is stable. This is a case where feedback
stabilized an unstable system.
− 1
+ 1
Example 11.15. () =
when = 2.
Solution: The only pole of () is in the left half-plane, so the open loop system is stable. The closed
loop system function is
. Is the open loop system stable? Is th... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
(2i)G(−2i)G(∞)12w=G◦γ(ω)=1iω+1
11 ARGUMENT PRINCIPLE
11
Nyquist plot of () = 1∕( + 1), with = 1.
Example 11.17. Take () from the previous example. Describe the Nyquist plot with gain factor
= 2.
Solution: The Nyquist plot is the graph of (). The factor = 2 will scale the circle in the
previous example by 2. ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
number) of poles and zeros of in the right half-plane are all inside . Now we can
apply Equation 3 in the corollary to the argument principle to () and to get
−Ind((cid:253), −1) =
1+,
− ,
(The minus sign is because of the clockwise direction of the curve.) Thus, for all large
the system is stable ⇔
1+,
= 0 ⇔ I... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
the imaginary axis. Its image under () will trace out the
Nyquis plot.
Notice that when the yellow dot is at either end of the axis its image on the Nyquist plot is close to
0.
Example 11.20. Refresh the page, to put the zero and poles back to their original state. There are
two poles in the right half-plane, so the... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
1)
(−0.33)2+1.752
=
− 1
( − 0.33)2 + 1.752 + ( − 1)
11 ARGUMENT PRINCIPLE
13
So the poles are the roots of
( − 0.33)2 + 1.752 + ( − 1) = 2 + ( − 0.66) + 0.332 + 1.752 −
For a quadratic with positive coefficients the roots both have negative real part. This happens when
0.66 < < 0.332 + 1.752 ø 3.17.
Example ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
Given Equation 7, in 18.04 we can ask if there are any poles in the right half-plane without needing
any underlying physical model. Still, it’s nice to have some sense of where this fits into science and
engineering.
In a negative feedback loop the output of the system is looped back and subtracted from the input.
Exa... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/44f1db513a6a17d655abe0b6ff7748fc_MIT18_04S18_topic11.pdf |
18.404/6.840 Lecture 23
Last time:
- !"#$%↑ is EXPSPACE-complete
- Thus !"#$%↑ ∉ PSPACE
- Oracles and P versus NP
Today: (Sipser §10.2)
- Probabilistic computation
- The class BPP
- Branching programs
1
Probabilistic TMs
Defn: A probabilistic Turing machine (PTM) is a var... | https://ocw.mit.edu/courses/18-404j-theory-of-computation-fall-2020/44f3286933ae429ff3cd163e8181ee46_MIT18_404f20_lec23.pdf |
1. Run '% on 1 for 2 times and output the majority response.”
Details: Calculation to obtain 2 and the improved error probability.
Significance: Can make the error probability so small it is negligible.
3
NP and BPP
NP
Computation trees
for ! on "
BPP
≥ 1 accepting
Many accepting
Few
rejecting
" ∈ $
" ∉ $
all rejec... | https://ocw.mit.edu/courses/18-404j-theory-of-computation-fall-2020/44f3286933ae429ff3cd163e8181ee46_MIT18_404f20_lec23.pdf |
1
5
Read-once Branching Programs
Defn: A BP is read-once if it never queries a variable more than once
on any path from the start node to an output.
Defn: !"ROBP = (), (+ () and (+ are equivalent read-once BPs}
Theorem: !"ROBP ∈ BPP
Check-in 23.2
Assuming (as we will show) that !"ROBP ∈ BPP,
can we use that to sho... | https://ocw.mit.edu/courses/18-404j-theory-of-computation-fall-2020/44f3286933ae429ff3cd163e8181ee46_MIT18_404f20_lec23.pdf |
.
7
*+
8+
1
0
*-
89
1
0
0
1
0
1
Boolean Labeling
Alternative way to view BP computation
1
!"
1
0
1
0
0
0
0
0
0
0
1
!#
1
0
!$
1
1
1
1
1
1
1
1
0
!#
0
0
!$
0
0
0
0
Show by example: Input is !" = 0, !# = 1, !$ = 1
The BP follows its execution path.
Label all nodes and edges on the execution path with 1
and off the executi... | https://ocw.mit.edu/courses/18-404j-theory-of-computation-fall-2020/44f3286933ae429ff3cd163e8181ee46_MIT18_404f20_lec23.pdf |
1 1 − 2
1 2 = 2
) (1 − !()
!(
0 1
)#
)"
)-
)" + )# + )-
) !(
−1
!#
0
1
1
2 = −1 1 − 3
−3 = −1 3
2
!#
0
2 1 − 3 = −4
2 3 = 6
8 = 2 + 6
0
−3 + −4 = −7
1
Check-in 23.3
Revised 4 for 56ROBP: “On input ;", ;#
What is the output for this branching program using
1. Pick a random non-Boolean input assignment.
the arithmetiz... | https://ocw.mit.edu/courses/18-404j-theory-of-computation-fall-2020/44f3286933ae429ff3cd163e8181ee46_MIT18_404f20_lec23.pdf |
Massachusetts Institute of Technology
Department of Materials Science and Engineering
77 Massachusetts Avenue, Cambridge MA 02139-4307
3.21 Kinetics of Materials—Spring 2006
February 22, 2006
Lecture 5: Effects of Capillarity and Stress on Diffusion.
References
1. Balluffi, Allen, and Carter, Kinetics of Material... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/44fe0b888d0d01732fa446f1df63383c_ls5.pdf |
3.74) and the resulting fluxes determine the rate of
surface smoothing.
i
◦
• Stresses in solids can affect diffusion in a variety of ways, including via the mobility, its effect on
diffusion potential, and via influencing the boundary conditions for diffusion.
• Even a uniform state of stress may break symmetry an... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/44fe0b888d0d01732fa446f1df63383c_ls5.pdf |
3.012 Fund of Mat Sci: Bonding – Lecture 1 bis
WAVE MECHANICS
Photo courtesy of Malene Thyssen, www.mtfoto.dk/malene/
3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
Last Time
1. Players: particles (protons and neutrons in
the nuclei, electrons) and electromagnetic
fields (photons)... | https://ocw.mit.edu/courses/3-012-fundamentals-of-materials-science-fall-2005/451781865b2544b82bfb2234d4808e31_lec01b_bis.pdf |
2000, p. 495, Figure 14.2.
3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
The total energy of the system
• Kinetic energy K
• Potential energy V
3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
Polar Representation
Diagram of the Argand plane remov... | https://ocw.mit.edu/courses/3-012-fundamentals-of-materials-science-fall-2005/451781865b2544b82bfb2234d4808e31_lec01b_bis.pdf |
926…)
3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
When is a particle like a wave ?
Wavelength • momentum = Planck
Image of the double-slit experiment removed for copyright reasons.
See the simulation at http://www.kfunigraz.ac.at/imawww/vqm/movies.html:
"Samples from Visual Qu... | https://ocw.mit.edu/courses/3-012-fundamentals-of-materials-science-fall-2005/451781865b2544b82bfb2234d4808e31_lec01b_bis.pdf |
012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
Stationary Schrödinger’s Equation (I)
−
2
h
2
m
2
Ψ∇
r
r
*
),(
),(
trVtr
+
r
),(
tr
Ψ
=
i
h
r
),(
tr
Ψ∂
t
∂
3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
Stationary Schrödinger’s Equation (II)
−
⎡
⎢
⎣... | https://ocw.mit.edu/courses/3-012-fundamentals-of-materials-science-fall-2005/451781865b2544b82bfb2234d4808e31_lec01b_bis.pdf |
Lecture 8
Understanding Transcription
RNA-seq analysis
Foundations of Computational Systems Biology
David K. Gifford
1 Lecture 8 – RNA-seq Analysis
• RNA-seq principles
– How can we characterize mRNA isoform
expression using high-throughput sequencing?
• Differential expression and PCA
– What genes are di... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
expression
Sox2
6 RNA-seq reads map to exons and across exons
Reads over exons
Junction reads (split between exons)
Smug1
7 Two major approaches to RNA-seq analysis
Short sequencing reads,
randomly sampled from a
transcript
exon 1
exon 2
exon 3
1. Assemble reads into
transcripts. Typical
issues with cov... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
Tj ) = 0
Ri
Tj
Intron in Tj
Courtesy of Cole Trapnell. Used with permission.
Slide courtesy Cole Trapnell
12
P(Ri | T=Tj) – Single end reads
Cufflinks assumes that fragmentation is roughly uniform. The probability
of observing a fragment starting at a specific position Si in a transcript of
length lj is:""
P... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
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Tracked 8,134 transfrags
across all time points, 5,845
complete matches to UCSC/
Ensembl/VEGA
Tracked 643 new isoforms of
known genes across all points
Slide courtesy Cole Trapnell
16
Case study: myogenesis
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0.01
0.1
1
10
100
1000
10000
0.01
0.1
1
10
100
1000
10000
Reads per bp
Courtesy of Cole Trapnell. Used with permission.
•
•
~25% of transcripts have
light sequence coverage,
and are fragments of full
transcripts
Intronic reads, repeats, and
other artifacts are
numerous, but account for
less than 5% of the
ass... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
js
ijµ = ip( j )
q
js
2s
2σ = ijµ + j
ij
pv
q(
)
ip( j )
ijK ~ NB ijµ ,
2σ(
ij
)
25
2
σij = µij + s jv p(qip( j ))
2
Orange Line – DESeq
Dashed Orange – edgeR
Purple - Poission
Courtesy of the authors. License: CC-BY.
Source: Anders, Simon, and Wolfgang Huber. "Differential Expression Analysis for Sequence Count... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
3
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$
& N − n1
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%
"
$
!
&
#
%
"
N
n2
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P x ≥ k
) =
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0.017
0.020
29
30 31 32 33 34 35 Lecture 8 – RNA-seq Analysis
• RNA-seq principles
– How can we characterize mRNA isoform
expression using high-throughput sequencing?
• Differential expre... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
038/nature12172
39 Analysis of co-variation in single-cell mRNA expression levels
reveals distinct maturity states and an antiviral cell circuit.
Courtesy of Macmillan Publishers Limited. Used with permission.
Source: Shalek, Alex K., Rahul Satija, et al. "Single-cell Transcriptomics Reveals Bimodality in Expression... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/45a80bf19f88d0cabfb091b4edbc0f36_MIT7_91JS14_Lecture8.pdf |
Course Organization
Spirit of the Undertaking
6.871: Knowledge-Based Systems
Spring 2005
Randall Davis
Logistics
• Info sheet, syllabus
• Personnel:
– Lecturers: Davis (and friends)
• Course notes:
– 1st installment ready now
• You are responsible for what happens in lecture.
• No open laptops.
2
Course Charac... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/45d32f86dec5685e9c9fcbf8b61d804b_lect01_intro1.pdf |
– Manufacturer’s Hanover: Inspector
– The clever (?) paper clip
8
In 1995, in Singapore…
Crime Case Closed Infamous Crimes
Nick Leeson and Barings Bank
The week before Nick Leeson disappeared he had kept
throwing up at work.
Colleagues did not know why but were soon to find out.
The ego of a 28-year-old trader on ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/45d32f86dec5685e9c9fcbf8b61d804b_lect01_intro1.pdf |
Applied AI leads to advances in basic
science
– Knowledge acquisition/learning
– Explanation
– Knowledge sharing
12
Character of the problems
attacked
• Balancing your checkbook vs.
Getting out of the supermarket
• Telling it what to do vs.
Telling it what to know
– Write down some relevant knowledge
– Advice... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/45d32f86dec5685e9c9fcbf8b61d804b_lect01_intro1.pdf |
of
thought”
19
Intellectual Origins
• 19th century: Boole’s logic and The Laws
of Thought
To see this image, please visit:
http://images.google.com/images?q=cgboole.gif
20
Intellectual Origins
• 19th century: Babbage and the Analytical Engine.
Lady Lovelace conjectured that it “would weave
algebraic patterns... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/45d32f86dec5685e9c9fcbf8b61d804b_lect01_intro1.pdf |
on
expressions to produce other expressions:
Create, Modify, Reproduce and Destroy
28
The Physical Symbol System
Hypothesis
The Knowledge Level Hypothesis: There exists a
distinct computer systems level which is
characterized by knowledge as the medium and
the principle of rationality as the law of behavior.
Pr... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/45d32f86dec5685e9c9fcbf8b61d804b_lect01_intro1.pdf |
Chapter 7
Scattering
c(cid:13) B. Zwiebach
In high energy physics experiments a beam of particles hits a target composed of par-
ticles. By detecting the by-products one aims to study the interactions that occur during
the collision.
Collisions can be rather intricate. For example, the particles involved may be not
ele... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
26)
∇2 + V (r)
ψ(r) = Eψ(r)
(cid:20)
−
(cid:126)2
2M
(7.1.1)
(7.1.2)
(7.1.3)
Figure 7.1: Potential is finite range, or vanishes faster than 1
r as r → ∞
Will be consider solutions with positive energy. The energy is the energy of the particle
far away from the potential, E = (cid:126)2k2
2M . The Schr¨odinger equation t... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
will look like
ϕ(r) = eikz
(7.1.5)
If we assume V (r) has a finite range a, this ϕ(r) satisfies (7.1.4) for any r > a. For r < a,
however, it does not satisfy the equation; ϕ(r) is a solution everywhere only if the potential
vanishes.
Given an incident wave we will also have a scattered wave. Could it be an ψ(r) = eikr ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
r
,
r (cid:29) a .
(7.1.9)
136
CHAPTER 7. SCATTERING
As indicated this expression is only true far away from the scattering center. We physically
expect fk(θ, φ) to be determined by V (r). fk(θ, φ) is called the scattering amplitude.
We now relate fk(θ, φ) to cross section!!
dσ =
(cid:104)# particles scattered per uni... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
)
m all these particles in the little volume will go cross out in time dt = dr
v ,
With velocity v = (cid:126)k
therefore the number of particles per unit time reads
(cid:126)k
dn
m
dt
= |fk(θ, φ)|2 dΩ dr
=
dr
v
|fk(θ, φ)|2 dΩ
Back in the formula for the cross section we get
(cid:19)(cid:126)k
m |fk(θ, φ)|2 dΩ
dσ = (ci... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
E(cid:96)(r) =
(cid:126)2k2
2m
uE(cid:96)(r)
(cid:19)
uE(cid:96)(r) = k2uE(cid:96)(r) .
(7.2.1)
(7.2.2)
Now take ρ = kr, then (7.2.2) reads
(cid:18)
−
d2
dρ2 +
(cid:96)((cid:96) + 1)
ρ2
(cid:19)
uE(cid:96)(ρ) = uE(cid:96)(ρ)
(7.2.3)
Since k2 disappeared from the equation, the energy is not quantized. The solution to (7... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
8)
(cid:96)=0
a(cid:96)P(cid:96)(cos θ)j(cid:96)(kr)
(7.2.7)
for some coefficients a(cid:96). Using
Y(cid:96),0(θ) =
(cid:114)
2(cid:96) + 1
4π
P(cid:96)(cos θ)
and
j(cid:96)(x) =
1
2i(cid:96)
(cid:90) 1
−1
eixuP(cid:96)(u)du
(7.2.8)
138
it can be shown that
CHAPTER 7. SCATTERING
eikz =
√
4π
∞
(cid:88)
√
(cid:96)=0
2(ci... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
)
Thus
eikz =
√
4π
k
∞
(cid:88)
√
(cid:96)=0
2(cid:96) + 1 i(cid:96) Y(cid:96),0(θ)
1
2i
(cid:104) ei(kr− (cid:96)π
2 )
r
(cid:123)(cid:122)
outgoing
(cid:124)
(cid:125)
−
e−i(kr− (cid:96)π
2 )
r
(cid:123)(cid:122)
incoming
(cid:124)
(cid:125)
(cid:105)
,
r (cid:29) a
(7.2.11)
7.2.1 Calculating the scattering amplitude... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
39) eikz + fk(θ)
eikr
r
,
r (cid:29) a .
(7.2.15)
7.2. PHASE SHIFTS
139
The incoming partial waves in the left-hand side must be equal to the incoming partial
waves in eikz since the scattered wave is outgoing.
Introducing the phase shifts on the
outgoing waves of the left-hand side we get
ψ(r) =
√
4π
k
∞
(cid:88)
√
(... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
4π
k
=
∞
(cid:88)
√
(cid:96)=0
2(cid:96) + 1 Y(cid:96),0(θ)eiδ(cid:96) sin δ(cid:96)
eikr
r
,
(7.2.16)
(7.2.17)
where we noted that e− i(cid:96)π
2 = (−i)(cid:96) and i(cid:96)(−i)(cid:96) = 1. Therefore we get
fk(θ) =
√
4π
k
∞
(cid:88)
√
(cid:96)=0
2(cid:96) + 1 Y(cid:96),0(θ)eiδ(cid:96) sin δ(cid:96) .
(7.2.18)
This ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
∗
(cid:96),0(Ω)Y(cid:96)(cid:48),0(Ω)
(cid:125)
(cid:123)(cid:122)
δ(cid:96)(cid:96)(cid:48)
(7.2.20)
Now let us explore the form of f (θ) in the forward direction θ = 0. Given that
Y(cid:96),0(θ) =
(cid:114)
2(cid:96) + 1
4π
P(cid:96)(cos θ) =⇒ Y(cid:96),0(θ = 0) =
(cid:114)
2(cid:96) + 1
4π
CHAPTER 7. SCATTERING
fk(... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
to a fixed (cid:96)
ψ(x)
(cid:12)
(cid:12)
(cid:12)(cid:96)
= (A(cid:96)j(cid:96)(kr) + B(cid:96)n(cid:96)(kr)) Y(cid:96),0(θ)
x > a
(7.2.22)
If B (cid:54)= 0 then V (cid:54)= 0. As a matter of fact, if V = 0 the solution should be valid everywhere
and n(cid:96) is singular at the origin, thus B(cid:96) = 0.
Now, let’s ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
6)
cos δ(cid:96)
(cid:18)
cos
kr −
(cid:19)(cid:21)
(cid:96)π
2
Y(cid:96),0(θ)
(cid:18)
sin
kr −
(cid:96)π
2
(cid:19)
+ δ(cid:96)
Y(cid:96),0(θ)
(7.2.25)
=
1
1
kr
2i
(cid:39) e−iδ 1
kr
(cid:104)
ei(kr− (cid:96)π
2 +δ(cid:96)) − e−i(kr− (cid:96)π
2 +δ(cid:96))(cid:105)
Y(cid:96),0(θ)
(cid:104)
ei(kr− (cid:96)π
2 +2δ(cid... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
96)
ψ(x)
(cid:12)
(cid:12)
(cid:12)(cid:96)
(cid:20)
(cid:18)
∼
i sin
kr −
(cid:19)
(cid:96)π
2
(cid:18)
+ cos
kr −
(cid:96)π
2
(cid:19)(cid:21) Y(cid:96),0(θ)
kr
∼
ei(kr− (cid:96)π
2 )
kr
Y(cid:96),0(θ) (7.2.28)
• If A(cid:96) = iB(cid:96)
ψ(x)
(cid:12)
(cid:12)
(cid:12)(cid:96)
∼
e−i(kr− (cid:96)π
2 )
kr
Y(cid:96),0(... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
are determined
tan δ(cid:96) =
j(cid:96)(ka)
n(cid:96)(ka)
(7.2.30)
(7.2.31)
(7.2.32)
(7.2.33)
(7.2.34)
(7.2.35)
142
CHAPTER 7. SCATTERING
We can now easily compute the cross section σ, which, recalling (7.2.20) is proportional to
sin2 δ(cid:96)
sin2 δ(cid:96) =
tan2 δ(cid:96)
1 + tan2 δ(cid:96)
=
j2
(cid:96) (ka)
(ci... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
πa2
Full area of the sphere!
Not the cross section!
(7.2.39)
One more calculation!
i tan δ =
eiδ − e−iδ
eiδ + eiδ =
e2iδ − 1
e2iδ + 1
=⇒
e2iδ =
1 + i tan δ
1 − i tan δ
Therefore
e2iδ(cid:96) =
1 + i j(cid:96)(ka)
n(cid:96)(ka)
1 − i j(cid:96)(ka)
n(cid:96)(ka)
=
n + ij
n − ij
=
i(j − in)
−i(j + in)
(7.2.40)
(7.2.41)
e2... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
:17)
(cid:96)(ka)
143
(7.2.44)
Let’s form the ratio and define β(cid:96) as the logarithmic derivative of the radial solution
β(cid:96) ≡
aR(cid:48)
(cid:96)(a)
R(cid:96)(a)
= ka
A(cid:96)j(cid:96)(ka) + B(cid:96)η(cid:48)
(cid:96)(ka)
A(cid:96)j(cid:96)(ka) + B(cid:96)n(cid:96)(ka)
= ka
j(cid:96)(ka) + B(cid:96)
A(cid:... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
:16) j(cid:96)−iη(cid:48)
(cid:96)
j(cid:96)−in(cid:96)
(cid:16) j(cid:96)+iη(cid:48)
(cid:96)
j(cid:96)+in(cid:96)
which we can also write as
e2iδ(cid:96) = e2iξ(cid:96)
(cid:124)(cid:123)(cid:122)(cid:125)
Hard sphere
phase shift
β(cid:96) − ka
β(cid:96) − ka
(cid:17)
(cid:17)
(cid:16) j(cid:96)−iη(cid:48)
(c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
Confirm the impact parameter intuition from partial wave expression. Consider the
free partial wave ∼ j(cid:96)(kr)Y(cid:96),0. The impact parameter b(cid:96) of such a wave would be
estimated to be (cid:96)/k.
Recall that j(cid:96)(kr) is a solution of the V = 0 radial equation with angular momentum
(cid:96) and energy... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
ψ(r) ,
(7.3.1)
7.3. INTEGRAL SCATTERING EQUATION
145
set the energy equal to that of a plane wave of wavenumber k and rewrite the potential in
terms of a rescaled version U (r) that simplifies the units:
E =
(cid:126)2k2
2M
and V (r) =
(cid:126)2
2M
U (r)
then the Schr¨odinger equation reads
rewrite as
(cid:2)−∇2 + U (... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
δ(3)(r − r(cid:48))U (r(cid:48))ψ(r(cid:48))
= U (r)ψ(r)
(cid:88)
To find G we first recall that
(7.3.2)
(7.3.3)
(7.3.4)
(7.3.5)
(7.3.6)
(7.3.7)
(7.3.8)
Ce±ikr
r
− 1
4πr
for r (cid:54)= 0
for r → 0
G(r) ∼
Thus try
as a matter of fact
as a matter of fact ∇2 (cid:0)− 1
4πr
(cid:0)∇2
r + k2(cid:1) e±ikr
r = 0 ∀ r (cid... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
2
−
1
4πr
(7.3.11)
Recall that
Therefore
then
∇2 = ∇ · ∇ ,
∇r =
r
r
,
∇ · (f A) = ∇f · A + f ∇ · A
(7.3.12)
∇e±ikr = ±ik
r
r
e±ikr ,
∇2e±ikr =
(cid:18)
−k2 ±
(cid:19)
2ik
r
e±ikr
(7.3.13)
(cid:16)
±ike±ikr r
r
(cid:17)
·
(cid:16) r
(cid:17)
4πr3
∇2G±(r) =
(cid:18)
−k2 ±
(cid:19)
2ik
r
(cid:19)
(cid:18)
1
4πr
−
e±ikr
(c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
|
|r − r(cid:48)|
(7.3.14)
(7.3.15)
(7.3.16)
(7.3.17)
We now want to show that this is consistent with our asymptotic expansion for the energy
eigenstates. For that we can make the following approximations
For the G+ denominator:
For the G+ numerator:
|r − r(cid:48)| (cid:39) r
|r − r(cid:48)| (cid:39) r − n · r(cid:48... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
form of eiki·r with ki the incident
wave number, |ki| = k. For the outgoing wave we define ks ≡ nk (in the direction of n),
the scattered wave vector. The expression for ψ(r) then becomes
ψ(r) = eiki·r +
(cid:90)
(cid:20)
−
1
4π
d3r(cid:48)e−iks·r(cid:48)
U (r(cid:48))ψ(r(cid:48))
(cid:21) eikr
r
(7.3.23)
We will do bet... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
eikir +
(cid:90)
(cid:90)
G U
(cid:90)
G U
G U eikir + . . .
(7.3.28)
The approximation in which we keep the first integral in this series and set to zero all others is
called the first Born approximation. The preparatory work was done in (7.3.23), so we
have now
ψBorn(r) = eiki·r −
(cid:18)(cid:90)
1
4π
d3r(cid:48)e−i... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
(cid:126)2
d3re−iK·rV (r) .
(7.3.33)
By spherical symmetry this integral just depends on the norm K of the vector K. This is
why we have a result that only depends on θ: while K is a vector that depends on both θ
and φ, its magnitude only depends on θ. To do the integral think of K fixed and let Θ be
7.3. INTEGRAL SCAT... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
26)2
(cid:90) ∞
0
dr e−µr sin(Kr) = −
2mβ
(cid:126)2(µ2 + K2)
.
(7.3.37)
We can give a graphical representation of the Born series. Two waves reach the desired
point r. The first is the direct incident wave. The second is a secondary wave originating at
the scattering “material” at a point r(cid:48). The amplitude of a ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/45dbd7038c3e969491eceeae86c44d42_MIT8_06S18ch7.pdf |
MIT OpenCourseWare
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18.917 Topics in Algebraic Topology: The Sullivan Conjecture
Fall 2007
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Introduction (Lecture 1)
Let X be an algebraic variety defined over a field k. If k is the field C of complex... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
et
for every integer p. In other words, ´etale cohomology provides a purely algebraic recipe for extracting the
cohomology groups H∗(X(C); Z/pZ). This raises the question: to what extent can we recover the topological
space X(C) itself in purely algebraic terms?
Of course, algebro-topological invariants like cohomol... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
(M ∨; Z) exhibits Hn(M ∨; Z) as the p-adic completion
Remark 1. The p-adic completion of a space M is, in some sense, determined up to homotopy equivalence
by the mod-p cohomology H∗(M ; Z/pZ). Of course, for this to be true one must consider H∗(M ; Z/pZ) as
endowed with more structure than just a graded vector spa... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
ic methods.
Suppose that the algebraic variety X is defined instead over the field R of real numbers. In this case,
one has an underlying topological space of real points X(R). To what extent can this topological space be
reconstructed in purely algebraic terms? We note that there is a canonical inclusion
X(R) ⊆ X(C)... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
M . The fixed point set M G can be identified with space
of G-equivariant maps MapG(∗, M ); here ∗ denotes a point with G acting trivially. As noted above, the
functor M �→ M G need not preserve homotopy equivalences. One explanation for this is that the G-space ∗
is badly behaved. To get a better functor, we need to ... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
construction: every G-fixed point on a space M determines a (constant) G-equivariant
map EG M . This construction yields a natural transformation
→
M G M hG .
→
Let us now return to our algebraic variety X, defined over the field of real numbers R. We have a sequence
of maps
X(R) � X(C)Z/2Z → X(C)hZ/2Z → (X(C)∨)hZ/2... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
omorphic to a circle, which is definitely not simply connected.
What if p = 2? In this case, the algebraic answer (X(C)∨)hZ/2Z turns out to be a reasonably close
approximation to the topological space X(R). This is a consequence of the following conjecture of Sullivan:
Conjecture 6 (Sullivan). Let p be a prime number... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
). The
ultimate goal of this course is to give proofs of Conjecture 6 (and Theorem 7). Let us outline our strategy.
For simplicity, let us consider the proof of Theorem 7 in the special case where G = V � (Z/pZ)n is an
elementary abelian p-group and M is simply connected. To prove that the mapping space Map(BG, M ) ... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
ology operation (of degree k) is a collection of
maps
Hm(X, Y ; Z/pZ)
→
Hm+k(X, Y ; Z/pZ),
defined for every pair of spaces (X, Y ) and every integer m. We require that these operations depend
functorially on the pair (X, Y ), and behave well with respect the the boundary maps in long exact sequences.
The collectio... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
H∗(Y ; Z/pZ)
→
H∗(Map(BV, Y ), Z/pZ).
Moreover, this map is an isomorphism if Y is a sufficiently nice p-complete space.
The functor TV is called Lannes’ T -functor. Because of its many good properties, Lannes’ T -functor
is an extremely useful tool. In particular, it can be used to give very elegant proofs of Conject... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
(a more detailed
outline, which may turn out to be grossly inaccurate, is given in the syllabus). We will begin by giving a
construction of the Steenrod algebra A, and establishing its basic properties. Though our ultimate goal is to
4
∞
prove Sullivan’s conjecture, we will take the scenic route: for example, we w... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
assume that the prime p is equal to 2 whenever it is convenient to do so. Generally
speaking, we will do all calculations at the prime 2, though more conceptual arguments will usually work for
any prime p.
5 | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/45dd9ddfde563d3f01efed5cc60f13fe_lecture1.pdf |
MIT 2.852
Manufacturing Systems Analysis
Lectures 15–16: Assembly/Disassembly Systems
Stanley B. Gershwin
http://web.mit.edu/manuf-sys
Massachusetts Institute of Technology
Spring, 2010
2.852 Manufacturing Systems Analysis
1/41
Copyright �2010 Stanley B. Gershwin.
c
Assembly-Disassembly Systems
Assembly System ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
buffer downstream.
◮ Each buffer has exactly one machine upstream and one machine
downstream.
◮ Discrete material systems: when a machine does an operation, it
removes one part from each upstream buffer and inserts one part into
each downstream buffer.
◮ Continuous material systems: when machine Mi operates during
[... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
Manufacturing Systems Analysis
7/41
Copyright c�2010 Stanley B. Gershwin.
Assembly-Disassembly Systems
Models and Analysis
◮ Systems with loops are not ergodic. That is, the steady-state
distribution is a function of the initial conditions.
◮ Example: if the system below has K pallets at time 0, it will have K ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
(0)
.
�
�
�
�
�
2.852 Manufacturing Systems Analysis
9/41
s
�
Copyright c�2010 Stanley B. Gershwin.
¯nb(s(0)) =
nb prob (s|s(0)).
Assembly-Disassembly Systems
Decomposition
B (j, i)
B (i, m)
Mi
B (n, i)
B (i, q)
Mj
•
•
•
Mn
Mm
•
•
•
Mq
Part of Original Network
2.852 Manufacturing System... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
Case 1:
ri = .1, pi =
.1, i = 1, ..., 8;
Ni = 10, i =
1, ..., 7.
2.852 Manufacturing Systems Analysis
13/41
Copyright c�2010 Stanley B. Gershwin.
Numerical examples
Eight-Machine Systems
7.9444
7.9444
7.9444
7.9444
7.0529
7.0529
2.0555
Case 2:
Same as Case
1 except
p7 = .2
2.852 Manufacturing Syst... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
Assembly Line Designs
A product is made of three
subassemblies (blue, yellow,
and red). Each subassembly
can be assembled
independently of the others.
We consider four possible
production system
structures.
Machine 6 (the first machine
of the yellow process) is the
bottleneck — the slowest
operation of all.
2.852 Manufa... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
�2010 Stanley B. Gershwin.
c
Equivalence
Assembly System State Space
N = 2
1
µ
2
N = 3
2
µ
1
µ
3
µ
1
µ
1
20
10
µ
2
µ
3
µ
2
µ
3
µ
1
µ
1
11
µ
2
µ
3
µ
2
µ
1
12
µ
1
µ
2
µ
3
µ
2
21
22
µ
3
µ
3
00
µ
3
01
µ
3
02
µ
3
µ
1
03
µ
1
13
23
2.852
Manufacturing Systems Analysis
2... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
the two systems
are the same except for the labels of the
states.
◮ Therefore, the steady-state probability
distributions of the two systems are the
same, except for the labels of the states.
◮ The relationship between the labels of the
states is:
(n1
A , n2
A) ⇐⇒ (n1
T )
T , N2 − n2
◮ Therefore, in steady st... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/460acb49eb2c6f02fa3a7ba41481f026_MIT2_852S10_a_d_systems.pdf |
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