text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
distan
e
from
its
enter
of
mass
to
the
axle
enter
is
L
.
N
•
N
•
•
•
2.
The
axle
A
is
free
to
rotate,
and
we
an
ignore
any
fri
tional
for
es
(i e :
they
are
small) In
fa
t,
the
only
for
es
that
we
will
onsider
are
gravity,
and
the
torsional
for
es
indu
ed
on
the
axle
when
the
pendulums
are
not
all
aligned
3.
Any
d... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
Dis
rete
to
Continuum
Modeling.
9
MIT,
Mar
h,
2001
- Rosales.
Ea
h
one
of
the
pendulums
an
be
hara
terized
by
the
angle
e
=
e
(t)
that
its
suspending
n
n
rod
makes
with
the
verti
al
dire
tion.
Ea
h
pendulum
is
then
sub je
t
to
three
for
es:
(a)
Gravity,
for
whi
h
only
the
omponent
perpendi
ular
to
the
pendulum
rod
i... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
is
dire
tly
proportional
to
the
angle
of
twist,
and
inversely
proportional
to
the
distan
e
over
whi
h
the
twist
o
urs.
To
be
spe
if
:
in
the
problem
here,
imagine
that
a
se
tion
of
length
:£
of
the
axle
has
been
twisted
by
an
amount
(angle)
\.
Then,
if
is
the
torque
generated
by
this
twist,
one
an
write
T
(cid:20)
\
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
distan
e
from
the
respe
tive
ends
of
the
axle.
2(N
)
+ 1
The
tangential
for
e
(perpendi
ular
to
the
pendulum
rod)
due
to
gravity
o
n
ea
h
of
the
masses
is
1
F
=
N g
sin
e
,
where
n
= 1
, N
(3 3)
g
n
; : : :
:
-
N
su
essive
masses,
there
is
also
a
torque
whenever
e
=
e
.
This
is
generated
by
the
+1
n
n
+1
+
For
any
t
w... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
the
rod
itself,
whi
h
w
e
approximate
as
being
rigid.
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
N
£L
-
-
-
These
are
the
equations
for
N
torsion
oupled
equal
pendulums.
Remark
3.2
To
he
k
that
the
signs
for
the
torsion
for
es
sele
ted
in
these
equations
are
orre
t,
take
the
diferen
e
between
the
n
and
(n
)
equation.
Then
you
should
see
that
the
torsion
+ 1
th
th
for
e
(due
to
the
portion
of
the
axle
onne
ting... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
In
parti
ular,
note
that:
:x
=
x
x
=
:
(3 9)
+1
n
n
£
-
+ 1
N
1
Take
equation
(3 6),
and
multiply
it
by
N
/£
Then
we
obtain
2
d
e
N (N
)
(cid:20)
+ 1
n
p L
=
p g
sin
e
(
e
2e
e
)
,
+1
1
n
+
n
n
n-
+
2
2
dt
£
L
-
-
where
p
=
N
/£
is
the
mass
density
p
e
r
unit
length
in
the
N
limit Using
equation
(3 9),
this
an
b
e
wr... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
Dis
rete
to
Continuum
Modeling.
11
MIT,
Mar
h,
2001
- Rosales.
Thus,
fnally,
we
obtain
(for
the
ontinuum
limit)
the
nonlinear
wave
equation
(the
"Sine-Gordon"
equation):
2
2
e
e
=
w
sin ,
e
(3 11)
tt
xx
-
-
g
(cid:20)
where
w
=
is
the
pendulum
angular
frequen
y,
and
=
is
a
wave
propagation
speed
2
L
pL
f
I
(
he
k
th... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
0
end
(sin
e
the
axle
is
free
to
rotate
there).
Similarly,
one
obtains:
e
(£,
t) =
0
,
x
at
the
other
end
of
the
axle.
How
would
these
boundary
onditions
b
e
modifed
if
the
axle
where
fxed
at
one
(or
both)
ends?
Kinks
and
Breathers
for
the
Sine
Gordon
Equation.
Equation
(3 11),
whose
non-dimensional
form
is
e
e
=
sin
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
possesses
parti
le-like
solutions,
known
as
kinks,
anti-kinks,
and
breathers.
These
are
lo
alized
traveling
distur-
ban
es,
whi
h
preserve
their
identity
when
they
intera
t In
fa
t,
the
only
efe
t
of
an
intera
tion
is
a
phase
shift
in
the
parti
le
positions
after
the
intera
tion:
efe
tively,
the
"parti
les"
approa
h
ea... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
Dis
rete
to
Continuum
Modeling.
12
MIT,
Mar
h,
2001
- Rosales.
The
frst
step
is
to
present
analyti
al
expressions
for
the
various
parti
le-like
solutions
of
equation
(3 12) These
turn
out
to
be
relatively
simple
to
write
Example
3.1
Kinks
and
Anti-Kinks.
Equation
(3.12)
has
some
interesting
solutions,
that
orrespond
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
(kink,
or
antikink
speed),
and
let
z
= (
x
c t
x
)
b
e
a
moving
0
-
(cid:0)
-
(cid:0)
oordinate,
where
the
solution
is
steady
- the
"twist"
wil l
be
entered
at
x
=
t
+
x
,
where
x
is
0
0
the
position
at
time
t
= 0
.
Then
the
kink
solution
is
given
by
2
e
1
z
e
=
2
ar
os
=
4
ar
tan
exp
,
(3 13)
2
-
�
�
!
�
�
��
!!... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
"
lass"
of
models
proposed
for
nu
lear
inter
•
a
tions.
In
this
interpretation,
the
kinks
are
nu
lear
parti
les.
Sin
e
(in
the
nondimensional
version
(3.12))
the
speed
o
f
light
is
1,
the
restri
tion
1
< c <
1
is
the
relativisti
restri
tion,
and
the
fa
tor
(
in
orporates
the
usual
relativisti
ontra
tion.
-
The
ant... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
are
fol lowed
until
they
ol lide.
In
the
le
tures
the
results
of
numeri
al
experiments
of
this
type
wil l
be
shown
(the
numeri
al
method
used
in
the
experiments
is
is
a
"pseudospe
tral"
method).
11
(
(
x
(cid:0)
Solutions
of
the
form
=
(
,
where
is
a
onstant:
the
speed
of
propagation.
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
hand le
next.
A
breather
is
a
wave-pa
kage
kind
of
solution
(an
os
il latory
wave,
with
an
envelope
that
limits
the
wave
to
reside
in
a
bounded
r
e
gion
of
spa
e.
These
solutions
vanish
(exponential ly)
as
x
.
! (cid:6)1
This
last
property
al lows
for
easy
numeri
al
simulations
of
intera
tions
of
breathers
(and
kinks).... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
Q
=
Q
tanh(q )
:
(3 16)
p
q
-
The
breather
solution
(and
its
time
derivative)
is
then
given
by:
e
=
4
ar
tan(Q)
,
e
=
4
(1
+
Q
) (
C B Q
+
d B
V
Q
)
:
t
p
q
2
-
(3 17)
The
breather
solution
is
a
wavepa
kage
type
of
solution,
with
the
phase
ontrol led
by
p,
and
the
envelope
(
ausing
the
exponential
vanishing
of
the
so... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
0
)
we
have:
(group
speed)
=
1/(phase
speed)
- whi
h
is
exa
tly
tt
xx
-
the
relationship
satisfed
by
=
V
and
= 1
/V
for
a
breather.
This
is
be
ause,
for
x
large,
the
e
p
breathers
must
satisfy
the
linearized
equation.
Thus
the
envelope
must
move
at
the
group
velo
ity
(
(
orresponding
to
the
os
il lations
wavelength.
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
al
method
that
an
be
used
to
solve
the
SineGordon
equation.
This
remark
wil l
only
make
sense
to
you
if
you
have
some
familiarity
with
Fourier
Series
for
periodi
fun
tions.
The
basi
idea
in
spe
tral
methods
is
that
the
numeri
al
diferentiation
of
a
(smooth)
periodi
fun
tions
an
be
done
mu
h
more
eÆ
iently
(and
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
SineGordon
equation
would
require
periodi
in
spa
e,
solutions.
But
we
need
t
o
be
able
to
solve
for
solutions
that
are
mod-21
periodi
(su
h
as
the
kinks
and
antikinks),
sin
e
the
solutions
to
the
equation
are
angles.
Thus,
we
need
to
get
around
this
problem.
In
a
naive
implementation
of
a
spe
tral
method,
we
would
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
that
ignores
jumps
by
multiples
xx
of
21
in
u.
The
following
tri
k
works
in
doing
this:
Introdu
e
U
=
e
: Then
iu
gives
a
formula
for
u
that
ignores
21
jumps
in
u.
Warning:
In
the
a
tual
implementation
one
xx
2
(U )
U U
x
xx
u
=
i
xx
-
2
U
(3 21)
must
use
2
(U
)
U U
x
xx
u
=
imag
xx
-
2
-
U
�
!
�
to
avoid
smal l
imagi... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
ales.
1 Che
k
the
derivation
of
the
system
of
equations
(2 20)
2 Derive
the
ontinuum
equation
in
(2 21)
3 Look
at
the
end
of
se
tion
2,
under
the
title
"General
Motion:
String
and
Rods" Derive
ontinuum
equations
des
ribing
the
motion
(in
the
plane)
of
a
string
without
onstraints
4 Look
at
the
end
of
se
tion
2,
un... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
eri
al
ode
to
al
ulate
intera
tions
of
kinks,
breathers,
et
,
using
the
ideas
sket
hed
in
remark
3 5
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/6c889624a9bc707ffa544fb4b5fd41a5_MIT18_311S14_DiscreteTo.pdf |
MIT 2.853/2.854
Introduction to Manufacturing Systems
Inventory
Stanley B. Gershwin
Laboratory for Manufacturing and Productivity
Massachusetts Institute of Technology
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
1 Storage
Storage
Storage is fundamental!
• Storage is fundamental
in nature, management,
and en... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
orders.
? Long term: Total demand for a product next year.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
5 Manufacturing Inventory
Costs
Manufacturing Inventory
Motives/benefits
• Financial: raw materials are paid for, but no
revenue comes in until the item is sold.
• Demand risk: item loses value or is unsold... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
There is sometimes water in the tank when the flow is
variable.
Conclusions:
1. You can’t always replace random variables with their
averages.
2. Variability causes inventory!!
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
10 Variability and Inventory
Variability and Inventory
• To be more precise, non-synchro... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
= R =
if x ≤ w
rw + s(x − w)
if x > w
Profit = P =
(
r − c x
)
if
x ≤ w
rw +
s(x − w) − cx
= (r − s)w + (s − c)x if x > w
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
14 Newsvendor Problem Model
Newsvendor Problem
Model
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
r = 1.; c = .5... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
)x < wx > wwProfit=(r−c)xProfit=rw+s(x−w)−cxf(w) 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.417 Newsvendor Problem
Expected profit
Newsvendor Problem
Expected profit
Expected Profit = EP (x) =
(cid:82) x [(r − s)w + (s − c)x]f (w)dw+
−∞
(cid:82) ∞
x (r − c)xf (w)dw
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
18 Newsv... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
(x)x?21 Newsvendor Problem
Expected profit
Newsvendor Problem
Expected profit
EP (x) = (r − s) (cid:82) x wf (w)dw + (s
−∞
− c)xF (x) + (r − c)x(1 − F (x))
dEP
dx
= (r − s)xf (x) + (s − c)F (x) + (s − c)xf (x)
+(r − c)(1 − F (x)) − (r − c)xf (x)
= xf (x)(r − s + s − c − r + c)
+r − c + (s − c − r + c)F (x)
= r − c + (s −... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
∗) is the probability that W ≤ x∗.
So the equation above means
Buy enough stock to satisfy demand 100K% of the time, where
K =
r − c
r − s
.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.*
25 Newsvendor Problem
Expected profit
Newsvendor Problem
Expected profit
Inventory
Copyright (cid:13)c 2016 Stanley B. Gersh... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
. rr29 Newsvendor Problem
Example 2
Newsvendor Problem
Example 2:r = 1, c = .75, s = 0, µw = 100, σw = 10
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
0.70.91.11.31.51.71.9778185899397101105rx*x* vs. r30 Newsvendor Problem
Example 1
Newsvendor Problem
Example 1: r = 1, c = .25, s = 0, µw = 100, σw = 10
Inven... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
Problem
Example 2
Newsvendor Problem
Example 2: r = 1, c = .75, s = 0, µw = 100, σw = 10
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
020406080100120140160180200−103070110150190230x* vs. MeanMeanx*35 Newsvendor Problem
Example 1
Newsvendor Problem
Example 1: r = 1, c = .25, s = 0, µw = 100, σw = 10
Inventory... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
x∗ = µw + σwΦ−1
(cid:19)
(cid:18) r − c
r − s
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
39 Newsvendor Problem
Questions
Newsvendor Problem
Why does x∗ look linear in µw and σw?
Is it, really?
For what values of µw and σw is x∗ clearly nonlinear in
µw and σw?
Inventory
Copyright (cid:13)c 2016 Stanley B. G... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
Assumptions
Economic Order Quantity
Assumptions
• No randomness.
• No delay between ordering and arrival of goods.
• No backlogs.
• Goods are required at an annual rate of λ units per
year. Inventory is therefore depleted at the rate of
λ units per year.
• If the company orders Q units, it must pay s + cQ
for the order... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
a year is sλ/Q.
• The average inventory is Q/2. Therefore the
average inventory holding cost is hQ/2.
• Therefore, we must minimize the annual cost
C =
hQ
2
+
sλ
Q
over Q.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
*
49 EOQ
Formulation
Economic Order Quantity
Formulation
Then
or
dC
dQ
=
h
2
−
sλ
Q
2 = 0
Q∗... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
2016 Stanley B. Gershwin.
0 1 2 3 4 5 6 7 8 0 2 4 6 8 10Q*h55 More Issues
More Issues
• Random delivery times and amounts.
• Order lead time.
• Vanishing inventory.
• Combinations of these issues and random demand,
ordering/setup costs.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
56 Base Stock Policy Make-... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
the raw material buffer is never empty.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
Material FlowFromSupplierRawFactoryFG/DispatchProductionDemandMaterialFloorBacklogInformation Flow60 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue
• Whenever the factory floor buffer is not empty a... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
(t) does not change.
Q(t) + I(t) = S for all t.
• Therefore
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
QIMaterial FlowFromSupplierRawFactoryFG/DispatchProductionDemandMaterialFloorBacklogInformation Flow63 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue
• Also, Q(t) ≥ 0 and I(t)... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
Policy
Make-to-Stock Queue
• How much inventory will there be, on the average? (It
depends on what you mean by “inventory”.)
?
EQ =
ρ
1 − ρ
• EI = S − EQ is not the expected finished goods inventory.
It is the expected inventory/backlog.
• We want to know
I if I > 0
+E(I ) = E
0
otherwise
= E(I|I > 0)prob(
I > 0)
... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
1 − ρ)
=
S
−
q=0
S−1
(cid:88)
q=0
ρq
S−1
(cid:88)
qρq
q=0
S−1
(cid:88)
q=0
ρq
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
69 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue
Therefore
E(I|I > 0) = S −
S−1
(cid:88)
qρq
q=0
S−1
(cid:88)
q=0
ρq
and
+E(I ) = S(1 − ρ)
S−1
(cid:88)
q=0
... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
(cid:88)
q=S
=
(S
− q) prob(Q = q)
∞
(cid:88)
s=S
prob(Q = s)
where, because Q = S − I, we substitute q = S − i and s = S − j.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
72 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue
∞
(cid:88)
q=S
=
S prob(Q = q)
−
∞
(cid:88)
q=S
q prob(Q =... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
0.7 0.8 0.9 1 prob I pos75 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue
S
=
10, µ = 1
Inventory
λ
Copyright (cid:13)c 2016 Stanley B. Gershwin.
-40-30-20-10 0 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1EIposEIneg76 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Make-to-Stock Queue... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
for one time
unit and b is the cost of having one unit of backlog for one time
unit.
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
79 Base Stock Policy Make-to-Stock Queue
Base Stock Policy
Optimization of Stock
Inventory
Copyright (cid:13)c 2016 Stanley B. Gershwin.
costhI−b80 Base Stock Policy Make-to-Stock... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
Supplier Lead Time
Issues
Issue: Supplies are not delivered instantaneously. The time
between order and delivery, the lead time L > 0.
• If everything were deterministic, this would not be a problem.
Just order earlier.
Issue:
... but demand is random.
• Problem: The demand between when the order arrives and
and when t... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
approximately R + Q/2. Therefore the
average inventory holding cost is approximately h(R + Q/2).
• Therefore, we choose R and Q to minimize the approximate expected
annual cost
such that
(cid:18)
C = h
R +
(cid:19)
Q
2
+
sλ
Q
prob{total demand during a period of length L > R} ≤ K 0
for some K 0.
Inventory
Copyright c
(... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6c8fec8f99bcf7059b73b82e96d43901_MIT2_854F16_Inventory.pdf |
Lecture 7
Contents
1 Wavepackets and Uncertainty
2 Wavepacket Shape Changes
3 Time evolution of a free wave packet
1 Wavepackets and Uncertainty
B. Zwiebach
February 28, 2016
1
4
6
A wavepacket is a superposition of plane waves eikx with various wavelengths. Let us work with
wavepackets at t = 0. Such a wavepacket is o... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
Note that we are plotting there the absolute value
j|Ψ(x, 0)j| of the wave packet. Since Ψ(x, 0) is complex, the other option would have been to plot the
real and imaginary parts of Ψ(x, 0).
1
Figure 2: The Ψ(x, 0) corresponding to Φ(k) shown in Fig. 1, centered around x = 0 with width ∆x.
Indeed, in our case Ψ(x, 0) ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
−1
−∞
∗
Φ(cid:3)(−k)eikx
(cid:90) ∞
Z
1
1
dk = p
√
2π −1
−∞
Φ(k)eikxdk .
This is equivalent to
1
√
p
2π
(cid:90) ∞
Z
1
∗
(cid:31)
Φ(cid:3)( k)
− −
|
{z
(cid:123)(cid:122)
(cid:124)
object o
ver the
−1
−∞
the
Φ(k) eikxdk = 0 .
(cid:1)
(cid:1)
}
(cid:125)
brace must vanish.
This equation actually means that
Indeed, the i... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
Z
1
(cid:101)
Ψ(x, 0) = p eik0x
1
√
2π
(cid:16)
(cid:16)
Φ
k
0 +
(cid:17)
(cid:17)
ek
(cid:101)
e (cid:101)kx dek.
i
(cid:101)
(1.9)
−∞
−1
2
(1.5)
(1.6)
(1.7)
Figure 3: The real and imaginary parts of Ψ(x, 0).
As we integrate over ek, the most relevant region is
(cid:101)
(cid:2)
(cid:2)
k 2 − ∆k , ∆k
e
(cid:101)
2
∈
... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
clearly unreliable in this argument, we simply record
≈
∈
∆x ∆k (cid:25) 1 .
≈
(1.12)
This is what we wanted to show. The product of the uncertainty in the momentum distribution and
the uncertainty in the position is a constant of order one. This uncertainty product is not quantum
mechanical; as you have seen, it follo... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
12)
(cid:12)
(cid:12)
∆k
2
− ∆k
2
(1.16)
=
√
p
1
2π∆k
√
= p
1
2π∆k
We display Ψ(x, 0) in Fig. 5. We estimate
i ∆kx
e 2
e−i kx
∆
2
−
ix
2
x
sin ∆kx =
2
(cid:114)
r
∆
k
2π
sin
∆kx
2
∆kx
2
.
Figure 5: The Ψ(x, 0) corresponding to Φ(k).
≈
∆x (cid:25)
2π
∆k
→
! ∆x∆k (cid:25) 2π .
≈
(1.17)
2 Wavepacket Shape Changes
In order... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
:12)
(cid:12)
(cid:12)
(cid:12)
.
k0
(2.19)
The second term played a role in the determination
y and the next term, with
second derivatives of ω is responsible for the shape distorsion that occurs as time goes by. The
derivatives are promptly evaluated,
of the group velocit
dω
dk
=
(cid:126)k
dE
p
dp m m
=
=
,
(cid:126... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
t (cid:28) (∆x)2 .
(cid:126)
(cid:28)
∆p t
m
(cid:28)
(cid:28)
(cid:126)
∆p
,
∆p
m
t
(cid:28) ∆x .
(cid:28)
(2.23)
(2.24)
(2.25)
(2.26)
(2.27)
This inequality has a clear interpretation. First note that ∆p/m represents the uncertainty in the
velocity of the packet. There will be shape change when this velocity uncertai... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
(k) = p
2π (cid:0)1
−∞
(cid:90) ∞
Z
1
dx Ψ(x, 0)e(cid:0)ikx .
−
2. Use Φ(k) to rewrite Ψ(x, 0) as a superposition of plane waves:
1
√
Ψ(x, 0) = p
2π
(cid:90) ∞
Z 1
−∞
(cid:0)1
Φ(k)eikxdk .
(3.1)
(3.2)
This is useful because we know how plane waves evolve in time. The above is called the Fourier
representation of Ψ(x, 0... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
packet at t = 0. The constan
t a has units of length and ∆x (cid:25) a. The state
|
ψa is properly normalized, as you can check that
dxjψa(x, 0)j2 = 1.
≈
(cid:82)
R
|
6
We will not do the calculations here, but we can imagine that this packet will change shape as
time evolves. What is the time scale τ for shape change... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/6ca5bd79a2121967796f7c0b8f6f48c9_MIT8_04S16_LecNotes7.pdf |
2.853/2.854
KKT Examples
Stanley B. Gershwin(cid:3)
Massachusetts Institute of Technology
October 1, 2007
The purpose of this note is to supplement the slides that describe the Karush-Kuhn-Tucker
conditions. Neither these notes nor the slides are a complete description of these conditions; they
are only intended to pro... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
4
2 (cid:0) (cid:21) 1 0
B
C
B
C
B
C
@
A
=
x1 = x2 = x3 = x4 =
1
C
C
C
A
0
0
0
0
(cid:21)
2
(2)
x1 + x2 + x3 + x4 = 4
(cid:21)
2 !
= 1 or (cid:21) =
1
2
x1 = x2 = x3 = x4 =
1
4
and J =
1
4
x4
x =1 x =4
l
2
(
1
4
,
1
4
,
1
4
,
)
1
4
x1
equality constraint
J=constant
Figure 1: Example 1, represented in two dimensions
Com... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
4-dimensional sphere
that intersects with the equality constraint (a 3-dimensional plane in 4-dimensional space).
Equation (2) essentially tells us that the solution point is on a line that intersects with that
plane.
(cid:15) If we asked for the maximum rather than the minimum of J, the same necessary conditions
would... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
which J
and g are arbitrary functions that are di(cid:11)erentiable, whose derivatives are continuous, and where
J has a minimum:
subject to
min J(x)
x
g(x) (cid:20) 0
(5)
There are two possibilities: the solution x? satis(cid:12)es g(x?) < 0 (ie, where the solution is strictly in the
interior of the inequality conditi... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
dx
(x?)(cid:14)x
dg
dx
(x?)(cid:14)x (cid:20) 0
(8)
We now seek conditions on
(x?) and
that (8) is a linear programming problem.
dJ
dx
dg
dx
(x?) such that the solution to (8) is (cid:14)x = 0. Note
To be really exhaustive about it, we must now consider the four cases in the table below1. Case
1, for example, reduces t... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
only cases in which the solution is (cid:14)x = 0 are Cases 2 and 3, the cases in which
(x?) are opposite. Therefore, there is some positive number (cid:22) such
(x?) and
the signs of
that
dJ
dx
dg
dx
or
dJ
dx
(x?) = (cid:0)(cid:22)
dg
dx
(x?)
dJ
dx
(x?) + (cid:22)
dg
dx
(x?) = 0
Equation (6) is implied by this if we r... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
= x2 = x3 =
x1 + x2 + x3 + x4 = 4
(cid:21)
2 !
(cid:0)
(cid:22)
2
= 1
4(cid:21) (cid:0) (cid:22) = 2 or (cid:21) =
2 + (cid:22)
4
x1 = x2 = x3 =
2 + (cid:22)
8
=
1
4
+
(cid:22)
8
; x4 =
2 + (cid:22)
8
(cid:0)
(cid:22)
2
=
1
4
(cid:0)
3(cid:22)
8
1
4
(cid:0)
3(cid:22)
8
3(cid:22)
8
(cid:21)
1
4
(cid:20) A
(cid:0) A
From... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
=4 This behaves like Case 1. The unconstrained optimum lies on the boundary.
Therefore, if we ignored the inequality constraint, we would get the same x(cid:3).
Case 3: A < 1=4 If x4 were strictly less than A, then (13) would require that (cid:22) = 0. But then
(14) would imply x = 1=4, which violates (11).
Therefore x... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
KKT conditions was modeled on that of Bryson and Ho (1975).
There are plenty of other references, but this discussion is especially intuitive.
References
Bryson, A. E. and Y.-C. Ho (1975). Applied Optimal Control: Optimization, Estimation, and
Control. Hemisphere Publishing Corporation.
9
MIT OpenCourseWare
https://oc... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/6cbf08435a0e531bcfeb396636e216d5_MIT2_854F16_KktExample.pdf |
9.3 Perturbative Results
dσ
dM 2d
M
2
(cid:90)
= σ0H(Q, µ)
dl+dl− Jn(M 2 − Ql+
,µ) Jn¯(
9.3 Perturbative Results
9.4 Results with Resummation
10 SCET II
10 SCET II
(9.8)
2
M − Ql,µ
− ) S(l+, l−)
(ROUGH) When soft gluons interact with collinear particles, the resulting particle has momentum
Q(λ, 1, λ) and is there... | https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/6cc1695ed771c13f14fecb38b45dc258_MIT8_851S13_SCETII.pdf |
(10.4)
The fact that our standard current J = ξn,pWnΓµhv is not gauge invariant under the soft transformations
suggests that it is an incomplete description of the physics of this process. We can make this current soft
gauge invariant by including the soft Wilson line. The soft Wilson line Sn transforming as
Sn → U... | https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/6cc1695ed771c13f14fecb38b45dc258_MIT8_851S13_SCETII.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
8.512 Theory of Solids II
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 7: The Scaling Theory of Conductance
Part II
Last time, we studied localization and conductance in one dimension using arguments a... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
al system. A two dimensional system can be realized by evaporating thin films to a
thickness of order Lφ or less.
We also defined a localization length ξ as the length at which the Ohm’s law 1/L conductance
becomes of order 1. Ordinarily, there are two length scales in a system – the mean free path
� = vF τel, and th... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
scale. Thus we will consider the logarithmic derivative
d ln g
d ln L
= β [g(L)]
(7.4)
where β is some function of its argument.
Notice that the right hand side of equation (7.4) depends on L only through the function g(L).
This is exactly what we mean by our assumption of scaling behavior. To determine the form ... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
.9)
(7.10)
(7.11)
(7.12)
Scaling Theory of Conductance: Results in 1, 2, and 3 Dimensions
3
7.2 Scaling Theory of Conductance: Results in 1, 2, and 3 Dimensions
Based on the limiting behavior calculated above, we can draw a qualitative picture of how we
expect β to behave in the βg plane. We analyze the behav... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
g as a function of L. For
large g and d = 2,
d ln g
d ln L
dg
d ln L
= −
c
g
= − c
g = g0 − c ln
� �
L
L0
(7.13)
(7.14)
(7.15)
where we have used the initial conditions L0 = �, the mean free path characterizing the small
length scale over which we began the scaling procedure, and g0 = σBoltz is the Bo... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
stable Fixed Point in Three Dimensions
4
Three Dimensions: In three dimensions, the situation is a little more interesting. The
large g limit has β[g] positive ((d − 2) = 1), while the small g limit crashes to −∞ like ln g.
By continuity, this means that the function β[g] must cross the β = 0 at some point gc on th... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
Thus the fixed point at gc is an
unstable fixed point.
With such interesting mathematical behavior in the vicinity of gc, you can be sure that there
is some interesting physics involved. In fact, this point describes a metalinsulator transition:
finite scale systems prepared with conductance less than gc become asymp... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
our condition for the validity of the powerlaw scaling region, we can define
the localization length ξ through the relation
�
�
�
�
g(ξ) − gc
�
�
�
�
gc
= 1
(7.26)
Why is it reasonable to associate ξ with L? In the scaling region, the correlation length
of the system diverges (as with crucial opalescence) a... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
Ohm’s Law type scaling behavior. Thus we can define a large lengthscale limiting
conductivity σ(L → ∞)
σ
(L → ∞) ≈ σ
(L = ξ) =
g˜
−2
ξd
(7.29)
This definition makes sense because σ is essentially constant for length scales beyond the
localization length ξ. Substituting in equation (7.28) for ξ, we get
σ(L → ∞) ∝
... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
of stationary phase, it can be shown that the primary contribution
to this amplitude comes from paths very close to the path satisfying the classical EulerLagrange
equations. For a free particle, this means that the amplitude is dominated by paths confined to
a tube of width proportional to k−1 where k is the wave v... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
|
2
|
(7.34)
2 that we got previously assuming cancellation
which should be compared with the result 2|Ai
|
of all cross terms. This shows that quantum interference results in an enhanced probability of a
particle coming back to where it started from in a system with timereversal symmetry.
When this result is taken... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
to equation (7.34), paths which close on themselves acquire an enhanced proba
bility. Thus one expects a reduction of conductivity proportional to P0. In a previous lecture,
we derived the Einstein relation for the conductivity, which yields g ∝ D. This suggests that in
two dimensions
σ = σ0(1 −
ln
)
g = g0(1 − ... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
random phases and the interference is
destroyed. In this case, everything goes back to the normal case without the enhanced return
probability. From this observation, we can define a magnetic length scale
�
hc 1
2e B
LB =
(7.45)
Time Reversal Symmetry and the Probability of Return
8
When LB < Lφ, we should rep... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/6cc5023caca0b6c23e7c88c6e36b38b0_MIT8_512s09_lec07_rev.pdf |
Introduction to Algorithms: 6.006
Massachusetts Institute of Technology
Instructors: Erik Demaine, Jason Ku, and Justin Solomon
Lecture 3: Sorting
Set Interface (L03-L08)
Container build(X)
len()
find(k)
Order
Static
Dynamic insert(x)
delete(k)
iter ord()
find min()
find max()
find next(k)
find prev(k) ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
∈ {1, . . . , n}
• Example: [8, 2, 4, 9, 3] → [2, 3, 4, 8, 9]
• A sort is destructive if it overwrites A (instead of making a new array B that is a sorted
version of A)
• A sort is in place if it uses O(1) extra space (implies destructive: in place ⊆ destructive)
Permutation Sort
• There are n! permutations of A,... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
9], [3, 2, 4, 8, 9], [2, 3, 4, 8, 9]
1 def selection_sort(A, i = None):
’’’Sort A[:i + 1]’’’
2
if i is None: i = len(A) - 1
if i > 0:
3
4
5
6
7
j = prefix_max(A, i)
A[i], A[j] = A[j], A[i]
selection_sort(A, i - 1)
8
9 def prefix_max(A, i):
10
’’’Return index of maximum in A[:i + 1]’’’
if i > 0:
11
1... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
)
P n−1
i=0 1 = Θ(n)
• selection sort analysis:
– Base case: for i = 0, array has one element so is sorted
– Induction: assume correct for i, last number of a sorted output is a largest number of
the array, and the algorithm puts one there; then A[:i] is sorted by induction
– T (1) = Θ(1), T (n) = T (n − 1) + Θ(n)... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
0 and A[i] < A[i - 1]:
A[i], A[i - 1] = A[i - 1], A[i]
insert_last(A, i - 1)
10
11
12
# T(i)
# O(1)
# O(1)
# T(i - 1)
# S(i)
# S(i)
# O(1)
# O(1)
# S(i - 1)
• insert last analysis:
– Base case: for i = 0, array has one element so is sorted
– Induction: assume correct for i, if A[i] >= A[i - 1], array i... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
3, 4, 5, 6, 7, 9]
1 def merge_sort(A, a = 0, b = None):
2
’’’Sort A[a:b]’’’
if b is None: b = len(A)
if 1 < b - a:
c = (a + b + 1) // 2
merge_sort(A, a, c)
merge_sort(A, c, b)
L, R = A[a:c], A[c:b]
merge(L, R, A, len(L), len(R), a, b)
10
11 def merge(L, R, A, i, j, a, b):
12
# T(b - a = n)
# O(1)
# O(1)... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
n = 0, arrays are empty, so vacuously correct
– Induction: assume correct for n, item in A[r] must be a largest number from remaining
prefixes of L and R, and since they are sorted, taking largest of last items suffices;
remainder is merged by induction
– S(0) = Θ(1), S(n) = S(n − 1) + Θ(1) =⇒ S(n) = Θ(n)
• merge so... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/6d1ae5278d02bbecb5c4428928b24194_MIT6_006S20_lec3.pdf |
Electricity and Magnetism
• Electric field continued
Feb 15 2002
Electric Field
• New concept – Electric Field E
• Charge Q gives rise to a Vector Field
• E is defined by strength and direction of
force on small test charge q
Feb 15 2002
Electric Field
• For a single charge
• Visualize using Field Lines
– Cartoon!
–... | https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/6d95f57de45b1b7062e2fb644db8490f_2_15_2002_edited.pdf |
Total Charge Q = 2 λ L
dQ = λ dy
r
x
x0
dE
y
+L
0
-L
Feb 15 2002
More on Fields and Field Lines
• What’s wrong with this picture?
• Magnitude and direction of field have to be
unique at each point!
• Field lines can’t cross!
E
Feb 15 2002
More on Fields and Field Lines
• Very close to surface of charged object
• Fie... | https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/6d95f57de45b1b7062e2fb644db8490f_2_15_2002_edited.pdf |
Tell It What to Know
6.871 - Lecture 2
A Reminder
• Checkbook balancing vs.
getting out of the supermarket
• Character of task
• Character of solution
• Go past image to technical ideas and concepts
6.871 - Lecture 2
2
Purposes of This Lecture
• Explain the mindset of knowledge engineering
• Change your mind a... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
simple,
consistent
– Algorithms for use are simple
6.871 - Lecture 2
7
What’s a Good Representation?
• Consider: 1996 vs. 11111001100
• Which would the computer rather use in
arithmetic? Why?
– Algorithms for use are simple
– And simplicity is in the eye of the
interpreter
6.871 - Lecture 2
8
The Power of A ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
numerator of which is the holding period of the first party
multiplied by the capital contributed by the first party, and the denominator of
which is a sum, the first term of which is the holding period of the first party, the
second term of which is the holding period of the second party; and the second
term of wh... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
+ 4 x + 5x + 7
2
is entered by typing
3 4 5 7
PROCEDURE POLY-DIFF (REAL ARRAY PROBLEM)
FOR I = DEGREE TO 1 STEP –1 DO
ANSWER [I-1] = I * PROBLEM [I]
6.871 - Lecture 2
15
Version 2
PROCEDURE POLY-DIFF (REAL ARRAY PROBLEM)
FOR I = DEGREE TO 1 STEP –1 DO
ANSWER [COEFF, I] = PROBLEM [EXPON, I] *
PROBLEM [COEFF, I]
ANS... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
?
6.871 - Lecture 2
21
Observations about the knowledge
• It’s organized around the operators.
• It’s organized around nested sub-expressions
• Top-down tree descent is the natural approach
• The representation should reflect that.
• The representation should facilitate that.
6.871 - Lecture 2
22
Use a Natura... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
is style and pragmatics, not theory
• A program can be much more that just code.
It can be a repository for knowledge, an environment
for the development of knowledge
• Embody the reasoning, not (just) the calculation.
• Don’t tell it what to do, tell it what to know.
– Task changes from writing a program to specif... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
.99
$25.00
$72.54
$105.00
$14.00
$24.00
6.871 - Lecture 2
31
A Spreadsheet is Almost Right
• The right mindset: focus on the
knowledge
But:
– They are numeric and we want more
– They have only one inference engine
• KBS as “conceptual spreadsheets”
6.871 - Lecture 2
32
Search Basics
• Lecture 2, Part 2.... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
Figure by MIT OCW.
39
Pruning
• Throw away unpromising nodes
• Some risk that the answer is still there
• Great savings in time and space
• Breadth limited search, beam search
KEY
A Node
An Operator
b
100
-5
d
6.871 - Lecture 2
Figure by MIT OCW.
40
Optimum Often isn’t Optimum
In the real world things... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
pairs of integers,
coefficients and exponents, putting the coefficients
in the COEFF row of P and the exponents in the EXPON
row of P. Example:
3x 3 + 4 x + 5x + 7
2
results in EXPON row: 3 2 1 0
COEFF row: 3 4 5 7
6.871 - Lecture 2
45
Version 2
PROCEDURE POLY-DIFF (REAL ARRAY PROBLEM)
FOR I = DEGREE TO 1 STEP –1... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/6dccb18e39e0eb7b43ec990caa3fc294_lect02_know.pdf |
Electronics B
Joel Voldman
Massachusetts Institute of Technology
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
JV: 2.372J... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/6de08ad7d4ee670c283ce5ec0e6a653b_07lecture07e.pdf |
, 1 = 0)
s orbital 2
allowed levels
Image by MIT OpenCourseWare.
s 2.1 and 2.6 in Razeghi, M. Fundamentals
Adapted from Figure
of Solid State Engineering. 2nd ed. New York, NY: Springer,
2006, pp. 48 and 59. ISBN: 9780387281520.
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Micr... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/6de08ad7d4ee670c283ce5ec0e6a653b_07lecture07e.pdf |
.
Image by MIT OpenCourseWare.
Figure 1 on p. 559 in: Chelikowsky, J. R., and M. L. Cohen. "Nonlocal
Pseudopotential Calculations for the Electronic Structure of Eleven
Diamond and Zinc-blende Semiconductors." Physical Review B 14, no.
2 (July 1976): 556-582.
Razeghi, Fundamentals of Solid State Engineering
Cite as: ... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/6de08ad7d4ee670c283ce5ec0e6a653b_07lecture07e.pdf |
valence
band
Cit
Op
e as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Device
enCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
s, Spring 2007. MIT
JV: 2.372J/6.777J Spring 2007, Lecture 7E - 5
Elements of... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/6de08ad7d4ee670c283ce5ec0e6a653b_07lecture07e.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.