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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”) then output (G,0). Else, send a random bit b to P. • P: If b=0 then decommit to all commitments of message 1. If b=1 then open only the commitments of the edges in H. • V: Accept (output (G,1)) if all th...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/1d52d1e1de07e4fd8cf3ea9e3f689583_lecture3_4.pdf
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 realizes Fwzk . Proof sketch: Let A be an adversary that interacts with the protocol. Need to construct an ideal-process adversary S that fools all environments. There are four cases: ...
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ver, and send “cheat” on direct line. – Get output (G,b’) from TP. If b’=0 then output the ouptut of A from the run where decommitments failed. If b’=1 output the ouptut of A from the run where all decommitments succeed. If for both values of b some decommitments failed, give (G,-) to TP as the prover, send “no ch...
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). – Else output nothing (or alternatively output a default value, say (x0,1) for some x0 for which a w0 is known.) Analysis of the protocol Let A be an adversary that interacts with the protocol in the Fwzk R -hybrid 1. 2. R and fools all environments. There are four cases: model. Need to construct an ideal-pr...
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, and chose b=1. But this occurs with probability 2-k. Note: 1. 2. The protocol and analysis are purely combinatorial (“information theoretic”), no computational issues involved. All the “computational issues” are pushed to the realization Fwzk The same analysis works for a number of other ZK protocols with weak ...
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0 then c’ opens to a permutation of G. i This problem can be solved by using special commitments. But the general use of commitments failed. Parallel composition of Blum’s protocol • Solution 2 [Brassard-Crepeau-Yung]: Use “equivocable commitments” (given secret key, can open commitments both ways): VÆP: public c...
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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) ...
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) (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 equation of motion in the interaction ...
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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 and σy. Note that σx + iσy = (cid:18)0 1(cid:19) 1 0 (cid:19) (cid:18)0 −i i 0 i + = (cid:18) (cid:19) 0 2 0 0 . (9.14) This is why ...
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1 t. The frequency of ωR = (cid:12) (cid:12) (cid:12) (cid:12) geB 1 2m (cid:12) (cid:12) (cid:12) (cid:12) , (9.19) Lecture 9 8.321 Quantum Theory I, Fall 2017 50 is called the Rabi frequency. Note that the time-evolution operator, UI(t ) = e−iωRS t/(cid:126) x − = e (cid:16) ω tR 2 i (cid:17) x σ , does not oscillat...
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19)(cid:19)(cid:18)1 0 (cid:19) (cid:33)(cid:18)1 0 = = = (cid:18) cos (cid:18) ωRt 2 (cid:32) cos(cid:0) ωRt 2 −i sin(cid:0) ωRt 2 (cid:32) cos(cid:0) ωRt (cid:1) 2 −i sin(cid:0) ωRt 2 (cid:33) . (cid:1) Thus, after a time t = π ωR , we have |ψI(π/ωR)(cid:105) → (cid:18) (cid:19) 0 −i , and after a time t = 2π , we ha...
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ff + B (t)⊥ · S ) .⊥ ω γ , Beff := B0 + γ := ge 2m . (9.27) (9.28) (9.29) Now, let’s work in the interaction picture with these choices. Going to the interaction picture is going to a rotating frame, that rotates with H0. Previously, we went to a frame that rotated about the z-axis at frequency ω0; now we are going to a ...
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)a(cid:48)(cid:11) , ca(cid:48)(t) (cid:12) a(cid:48) where the |a(cid:48)(cid:105) form an energy eigenbasis, and ca(cid:48)(t) = e−iEa(cid:48) (t−t0)/(cid:126)ca(cid:48)(t0) . We can consider these particles in position space, and define ua(cid:48)(x) = (cid:10)x(cid:12) (cid:12)a(cid:48)(cid:11) (9.32) (9.33) (9.34) ...
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MIT OpenCourseWare https://ocw.mit.edu 8.321 Quantum Theory I Fall 2017 For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms.
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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] ...
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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) n )i1(i + 1 − n Here: A = $500/mo, i = 10/12 = 0.83%, n = 36 months ,5P = 000 + [500 = ,5 000 + ,15 505 = ,20$ .01( + 0083 ) 36 − ....
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15,000 -400 -400 3,000 • The benefits from these alternatives are identical. We must select one. • Assume that i = 10%. CBA 3. Bases for Comparison of Alternatives 8 Total Investment Comparisons (2) 10(PW %) 3B −= ,12 000 − ,1 200 2 1.1 1 − 2 1.1x1.0 + ,1 500 3 1.1 = −= ,12 000 − ,1 200 736.1x + ,1 .1 500 331 −= ,12...
https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf
: F = 1,200(F/P, 9, 1) = 1,200x1.091 = $1,308 (cid:190) End of year 2: 1,200x1.092 = 1,426 End of Year 0 1 2 3 B3 -$12,000 -1,308 -1,426 1,943 B4 -$15,000 -436 -475 3,885 CBA 3. Bases for Comparison of Alternatives 11 Impact of Inflation (2) • Recalculate the PWs. 10(PW %) 3B −= ,12 000 − −= ,12$ 908 −< ,12$ 870 = −...
https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/1ddaa9c180aa7db04223045ffdc70817_cba3.pdf
1.1 −= $ 485 < 0 • B3 should be accepted, just as in slide 9. • PW- or AE-based results using total and incremental investments are identical. CBA 3. Bases for Comparison of Alternatives 14 Internal Rate of Return (IRR) • It is the interest rate for which the equivalent receipts of a cash flow equal the equivalent d...
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Flows with a Single IRR • The IRR is a useful concept when the shape of PW(i) is like the one on Fig. 6.3 (slide 18). Sufficient Conditions 1. F0 < 0 (The first nonzero cash flow is a disbursement) 2. The sequence F0, F1, F2, …, Fn, has one change in sign only. 3. PW(0) > 0 (sum of all receipts > sum of all disburse...
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alternative are all zero. CBA 3. Bases 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 differ...
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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 Alternatives 27 Example (5) To compare A3 to A1, we must solve 0 = -5,000 + 1,100(P/A, x31, 10) 0 −= ,5 000 + 100,1 10 )x1( 1 + − 31 10 )x1(x + 31 31 i ...
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(2) Similarly, i* A0 = %15 i* A1 = %25 i* A2 = %9.19 i* A3 = %4.21 • The “best” alternative (i.e., the one having the highest IRR) is A1, not A3 (slide 28). CBA 3. Bases for Comparison of Alternatives 31 IRR on Total Investment (3) A1 A2 A3 $16,000 $14,000 $12,000 $10,000 $8,000 $6,000 $4,000 $2,000 $0 ($2,000) 0 ($4...
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400 = ,2 026 > 0 10 + )15.01( − )15.01(15.0 + 1 10 = • Therefore, A1 becomes the current best alternative. CBA 3. Bases for Comparison of Alternatives 35 The Example Revisited (2) Similarly, PW(15)A2-A1 = -3,000 + 500(P/A, 15, 10) )15(PW −=− AA 2 1 ,3 000 + 500 10 + )15.01( − )15.01(15.0 + 1 10 = −= 490 < 0 • Therefor...
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3. Bases for Comparison of Alternatives 39 Example (1) Alternative A B C Initial Cost $4,000 $16,000 $20,000 Annual Cost $6,400 $1,400 $1,000 Lifetime 6 years 3 years 4 years A, B, and C all fulfill the same objective, but for a different number of years; select the least costly for i = 7% CBA 3. Bas...
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is (approximately) the same as the original value of 7.48K. • The AEs that we calculated in slide 41 represent the following extended cash flows. CBA 3. Bases for Comparison of Alternatives 43 t (years) 0 1 2 3 4 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...
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CBA 3. Bases for 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 ove...
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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...
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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, m is the mass, and a is the acceleration. III. 3rd Law - Forces that result from interactions of particles and such forces be­ tween two particles a...
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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: Acceleration is the time rate of change of its velocity. a = dv dt = d2 r dt2 Two coordinate systems: Cartesian and Polar Velocities and accelerations can be expressed using a variety of differe...
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ˆθ + 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 be related to changes in θ. Figure by MIT OCW. deˆr dt = lim Δ ˆer Δt→0 Δt |eˆr|Δθ =...
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­ 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 = mv = Linear Momentum) If � F = 0, p is constant. (Conservation of Linear Momentum) Angular Momentum Principle Define Angular Momentum about B in an ine...
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× mv. If τB = 0 and vB = 0 or vB � mv ⇒ h˙ B = 0 (Conservation of Angular Momentum) Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month ...
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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...
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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 MARKETING AND SALES SERVICE MARGIN POTENTIAL NEW ENTRANT FIRM INFRASTRUCTURE HUMAN RESOU...
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• “Kill or Cure” - Alternatives? Mfg Direct 60% (rising) (dropping) Chain Drug Store Distributors/ Wholesalers 40% { McKesson 20% Independent Drug Stores 9 RELATIONSHIP BETWEEN DRUGSTORE AND WHOLESALER • 3-5 Major Wholesalers, many small players • “Good-Old-Boy” Relationship • “Sales” calls to creat...
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ICE AND WAREHOUSE PRODUCTIVITY INCREASED • ROLE OF SALESPEOPLE CHANGED • Reduced by 50% • Sell “System” 13 LONG-TERM IMPACTS • MARKET SHARE? (mid-1970s=20%) < 30% 30-50% >50% • REVENUES? (mid-1970s=$720M) < $10B $10-20B >$20B 14 IMPACT ON DRUG WHOLESALING INDUSTRY STRUCTURE • RAISED “TABLE STAKES” • N...
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• Support cost and differentiation strategies • Alters industry structure • Spawn entirely new businesses • NOTE INTERDEPENDENCE OF STRATEGY, TECHNOLOGY AND ORGANIZATION 18
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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...
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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 ) > car length > 0 n n n +1 f g (cid:0) (identify ~x with the location of (say) the front end of the car). n Remark 1.1 We use tildes ove...
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1999 | Rosales. notation h = (1= ~(cid:26) ) = ~x ~x for the car separation. Typical shapes for the car velocity U and n n n n +1 ~ ~ the car (cid:13)ow Q = ~(cid:26) U (both functions of ~(cid:26)) are shown in (cid:12)gure 1.1. ~ ~ (cid:0) Then the model is given by the following set of coupled ODE’s d ~x n ~ d t = ~...
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) 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 nondimensional number | with the values above and with L a large J fraction of a mile, we get (cid:15) = O(10 ) . Note also that the nondimensional versions of the car (cid:0) 2 velocity and car (cid:13)ow functions have the...
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of a mile). Thus our assumption above (where we took L a large fraction of a mile) is quite reasonable. This, in addition to (cid:15) = O(10 ) , yields (in the nondimensionalization above in (1.3)) a time scale in the (cid:0) 2 L(cid:26) J q m order of a few minutes | 6 min for L a ful l mile and 3 min for half a mile....
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limit we are taking. In practice (cid:15) is (cid:12)xed. However, since (cid:15) is small, we expect the limit will give us useful information regarding the behavior of the model (1.4). 4 Think of how many points per wavelength are needed to have a reasonable drawing of a sine wave. Simple Tra(cid:14)c Flow Model. 5 ...
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We expect (cid:26) to be reasonably nice and (generally) have O(1) partial derivatives and . @(cid:26) @(cid:26) @ t @x (cid:26) = (cid:26)(x ; t) : (2.3) n n We now rewrite the equations for the model (1.4) in terms of the densities rather than the car positions. Thus we have d (cid:0) 1 2 d u u n n +1 (cid:26) = (cid...
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that (cid:26) above in (2.3) must satisfy the PDE 5 For example: we used = in remark 1.2 (with 20) to determine . (cid:26) (cid:26) (cid:3) m N L (cid:25) L min @(cid:26) @ q @(cid:26) @(cid:26) 0 = + = + c (2.5) @ t @x @ t @x Simple Tra(cid:14)c Flow Model. 6 MIT, Friday March 26, 1999 | Rosales. in the limit (...
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go beyond one, for as soon as (cid:26) reaches one, the n car will n th stop, while the (n + 1) car will be moving at a non-negative velocity. (iii) Thus, the condition th 0 < (cid:26) 1 will be preserved. (iv) This is enough to guarantee a solution for all times, for a solution n (cid:20) can cease to exist only if it...
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) behaves nicely, it does approximate quite well the behavior of the solution of (1.4). 2. The solution of (2.5) exhibits breakdown with formation of in(cid:12)nities in the derivatives in the regions where the density (cid:26) is increasing with x. In these regions, the solution of (1.4) also shows progressive steepen...
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function across a discontinuity. This condition is called the Rankine{Hugoniot jump condition. The (so called) entropy condition must hold across shocks the density increases. (2.8) In terms of the characteristic curves for equation (2.5), this means that the curves converge into the shock | and terminate there. Thus t...
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a Taylor series centered at x . That is u = u + u h + u h + : : :, n +1 n n n n +1 n n n (1) (2) 2 1 where h = x x and we use the notation u = (x ; t). Thus n n n n +1 n j (cid:0) @x j ( ) j @ u 2 u u @u 1 1 @u 1 1 n n +1 (2) (3) 2 (2) 2 (3) (cid:0) n n n = (x ; t) + u h + u h + : : : = (x ; t) + (cid:15)u + (cid:15) u...
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@(cid:26) 1 @ @ + c = (cid:15) (cid:23) ((cid:26)) (cid:26) ; (2.9) @ t @x 2 @x @x ! where (cid:23) = (notice that (cid:23) is a POSITIVE function of (cid:26)). Thus a (small) amount of dU (cid:0) d(cid:26) di(cid:11)usion is added to equation (2.5). As long as the derivatives are bounded, the e(cid:11)ects of this ...
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important for many reasons, some of which we will explain later on. In particular: in any numerical calculation we must make sure that all times scales are handled properly, even if they are not immediately apparent in the solution | the precise meaning of this last rather strange statement will be clari(cid:12)ed belo...
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this is the same type of solution used in the von Neumann stability analysis of numerical schemes. (cid:0) f (cid:0) g (cid:16) (cid:17) Simple Tra(cid:14)c Flow Model. 10 MIT, Friday March 26, 1999 | Rosales. where (cid:25) k (cid:25) (these solutions are periodic in the wavenumber k , since the exponential is (cid:0...
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) is precisely the time scale over m which rapid variations in the car separations are \wiped out" by the time evolution of the model. This is the process il lustrated by the MatLab script randCFSM. After these variations are eliminated, this time scale plays no role, except to the extent that it keeps eliminating any ...
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this need not be so with a numerical scheme if one is not careful. Precisely because the equations being approximated are so forceful about dissipating errors, naive numerical approximations can easily over do the e(cid:11)ect and end up amplifying the pertur- bations! A simple example of this is provided by the equati...
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in (3.3) have negative real parts, so al l the scales decay. Thus, if we wait long enough, not just the short wavelength (a few car distances long) variations wil l vanish, but the long ones as wel l. Although this conclusion is based on the linearized analysis in (3.1 { 3.3) and thus is valid only for smal l perturbat...
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28) . It is clear that m the model (1.4) does not al low accidents (car col lisions). These would require (at the very least) that (cid:26) > 1 somewhere, sometime. But we showed earlier (see the paragraph above the note 2.1) n that the equations wil l not let this happen. The time (cid:28) is closely associated with t...
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15) (cid:15) (cid:15) (cid:15) y y + = x x = = (cid:14) ; n n n n n +1 +1 (cid:0) (cid:0) n (cid:26) (cid:26) (cid:26) (cid:0) (cid:26) (cid:3) (cid:3) (cid:3) 2 where we used (3.1) and neglected quadratic terms in the perturbations. Thus 2 (cid:26) (cid:3) (cid:14) = (y y ) : n n n +1 (cid:0) (cid:15) (cid:0) Since u ...
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= 2 u . These numbers are m J m max m 1 2 q m 2 (cid:26) J compatible with the typical values given earlier above equation (1.3), except that the maximum car velocity seems a bit low (though not out of range). Then again, the typical values given are from measurements in the NYC Lincoln tunnel in the 1950’s (where, per...
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J ~(cid:26) ! (cid:0) where (cid:26) = ‘ . This yields J (cid:0) 1 u = u ; (cid:26) = and q = (cid:26) u = : m max m m m m v (cid:26) r J v (cid:26) u r J max v + u v + u r max r max With u = 50 mph, v = 10 mph and (cid:26) = 160 cpm this yields (cid:26) 27 cpm and q 1330 cph max r J m m (cid:25) (cid:25) | not altoget...
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1 (cid:11)) ! (cid:18) (cid:19) (cid:0) (cid:0) (cid:0) where 0 < (cid:11) = < 1. Note the strange feature of a piece-wise constant wave speed (cid:26) m (cid:26) J c. Thus, in the continuum limit, the parts of the density pro(cid:12)le with (cid:26) > (cid:11) move (backwards) at constant speed ((cid:11) 1) . Similarl...
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ow function (4.1) in example 4.1. A (cid:12)nite number of cars N is used, with x < x < : : : < x and the density 0 < (cid:26) < 1 at the leading car given and constant . The ini- 1 2 N N 9 tial conditions are such that (see (cid:12)gure 5.1) x (0) = 0, x (0) < (cid:25) and N 1 (cid:0) (cid:26) (0) = (cid:26) + (1 (cid...
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MIT, Friday March 26, 1999 | Rosales. This equation determines the value of (cid:15) in terms of the number of cars in the hump and the densities given by (5.1). Note also the relationship (cid:0) N 1 (cid:15)(N 1) = (cid:26) (x x ) : h n n n +1 (cid:0) (cid:0) n p = X As the number of cars increases (continuum limit) ...
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equation @ c @ 1 2 0 = + ( c ) ; with c(x; 0) = c C (x) ; (5.3) N @ t @x 2 (cid:0) where 4 < c = 4 8(cid:26) < 4 and C = (4 + c )r(x). Thus the initial pro(cid:12)le for c has a \dip" N N N (cid:0) (cid:0) instead of a \hump". In terms of c the shock condition (2.7) states: the shock speed is the average of the charact...
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Then the shock starts at 1 t = and x = (cid:16) + (c C ((cid:16) ))t : S S m N m S (cid:0) m S (cid:0) Note that (cid:16) must correspond to a location on the back end of the initial hump. m Typical solution after shock forms. r r S r N Shock x S c t N Areas under shock and multiple valued curve are equal. Ty...
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and terminate there. The only part that remains is the very stretched out front. Because the stretching is linear in (cid:26), this part becomes a straight line, joining the front edge of the shock with the position of the leading characteristic starting at the front edge of the initial hump (i.e. x = c t). Thus the wa...
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0 A = C (x)dx = (4 + c )A 8((cid:15)(N 1) (cid:25)(cid:26) ) ; N r h N (cid:0) (cid:25) Z (cid:25) (cid:0) (cid:0) where we have used (5.2) and the fact that 4 + c = 8(1 (cid:26) ) to write the last (approximate) N N (cid:0) equality. The formula for x follows because this area must be conserved. Speci(cid:12)cally, no...
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Thus, in order to do the graphical comparisons of the continuum limit with the actual solutions of the (1.4), the script quadCFSM uses an improved approximation, which we describe next (the idea is actually very simple). As stated earlier, after a while the details of the solution are - p x = x L Area = A c » c N ...
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the (cid:25) t (cid:25) (cid:0) x front part of the sawtooth, by the solution that follows from initial conditions as in (5.6). Other than this, we use the same ideas that lead to (5.5), to obtain the improved approximation: 1. There is a shock at : : : : : : : : : : : : : : : : : : : : : : : : : x = c t . S N 2A(1 + B...
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> > > > > > > > > > > > > > Just one issue remains now and it is how to best choose B . For a given target time around which one ; desires the approximation to be good, one can use (5.5) to get an estimate of what is the range of characteristics that are making up the front of the saw-tooth. That is, one can determine ...
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X MIT OpenCourseWare http://ocw.mit.edu 18.306 Advanced Partial Differential Equations with Applications Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Simple Tra(cid:14)c Flow Model. 21 MIT, Friday March 26, 1999 | Rosales. Thus, we obtain (cid:15)(N ...
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Strategic Architectural Approaches at NASA Gary Martin November 14, 2004 MIT Overview • Decadal Planning Team (DPT) /NASA Exploration Team (NEXT) • Space Architect Team/New Vision for Space Exploration • Advanced Planning and Integration Office • A Few Points to Remember 2 Decadal Planning Team (DPT) /NASA Explorat...
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bottom). • Transition to next chart: The technology needed to overcome these hurdles and enable new missions is determined in a systematic way. The NEXT is structured to conduct the analysis and drive technology investment 7 Progressive Exploration Capabilities Sustainable Planetary Surface Capability Accessible ...
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InIn--space propulsion, Isp>3000 sec, high thrust Isp>3000 sec, high thrust Sustainable power •• Sustainable power systems systems Intelligent systems, orbital Intelligent systems, orbital and planetary and planetary Crew countermeasures for •• Crew countermeasures for indefinite duration indefinite duration...
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2.52.5 Exploration Exploration and and Expeditions Expeditions 2.62.6 Space Space Transportation Transportation 9 An Agency-Wide Approach THREADS THREADS Enterprises... Enterprises... 1.0 Systems Integration, Analysis, Concepts, Modeling 2.0 Enabling Advanced Research and Technology 3.0 Technology Flight Demons...
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, EVA) √ Regenerative Life Support Systems √ Surface Science & Mobility (Human-Involved) √ Materials and Structures (Manufacturing Validation) √ Advanced Habitation Systems x Space Nuclear Power x In Situ Resource Utilization x In Situ Manufacturing √ Cryogenic Propellant Depots x Space Medicine and Health Care x Fl...
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3 Min Energy Tether / Chemical SEP/ Chem HPEP/NEP VaSImR Gas Core NT Trip time is for the crew departing from HEO. 0.5 1.0 1.5 2.0 Round Trip Mission Duration (years) 2.5 3.0 3.5 12 Agency Investments Prioritized In-Space Propulsion Technologies Process •Requirements/Goals Established by NASA Enterprises •...
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. Report Available - Space Radiation Cancer Risk Projections 14 Hurdles: Crew Health & Safety Artificial Gravity NEP Vehicle System Concepts • Objective – Develop and assess integrated NEP and artificial gravity (AG) vehicle systems concepts as a means to mitigate the deleterious effects of zero gravity on humans ...
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) (45%) Aerobraking (42%) Consum- ables HW s s a M S S I o t d e z i l a m r o N s g n v a S s s a M i 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 Space Architect Team/New Vision for Space Exploration 19 Architecture Studies • Architectures are used to: – Understand requirements for exploration in the context of ...
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exploration spacecraft Artificial-g Mars Transfer Vehicle GEO SAR 22 Architecture Study #1 Exploration Metro Map Sun, Mercury, Venus Sun-Earth L1 , L2 High Earth Orbit Earth-Moon L1, L2 Earth Low Earth Orbit Moon d o o h r o b h g i e N s ’ h t r a E s e c a f r u S y r a t e n a l P e l b i s s e c c A Mars Outer ...
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h t r a E s e c a f r u S y r a t e n a l P e l b i s s e c c A Mars Outer Planets and beyond 25 National Vision for Space Exploration THE FUNDAMENTAL GOAL OF THIS VISION IS TO ADVANCE U.S. SCIENTIFIC, SECURITY, AND ECONOMIC INTEREST THROUGH A ROBUST SPACE EXPLORATION PROGRAM Implement a sustained and afforda...
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Mars Scout Mars Robotic Missions Outer Moons Cassini Saturn Arrival Cassini Titan Landing Jupiter Icy Moons Orbiter Earth-Like Planets and Life Hubble Space Telescope Kepler Mission Webb Space Telescope Terrestrial Planet Finder Spitzer Space Telescope Space Interferometry Mission Optical Comm Demo Nuclear Power / Pro...
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search for evidence of life, to understand the history of the solar system, and to prepare for future human exploration. (1.5) NASA shall conduct human expeditions to Mars to extend the search for life and to expand the frontiers of human exploration after successfully demonstrating human exploration mission to the ...
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liquid water as candidate site for evidence of extant life/pre-biotic chemistry Explore Global Evolution of Mars: Examine initial conditions and investigate why terrestrial planets evolved differently, much more so than we had thought, if no evidence of past or present liquid water has been found 30 Notional Archit...
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conjunction opposition (See chart ? for details) crew/cargo* separately predeploy assets conjunction (See chart ? for details) deploy together opposition Aeroassisted Orbit Insertion Direct Entry from Earth Propulsive Orbit Insertion Mars Arrival, Descent/Ascent & Departure Separate Lander & Habitat Lander as Habitat M...
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O Existing LV GEO GEO Update existing infrastructure Build new NASA Infrastructure Use new Industry infrastructure Nuclear Nuclear predeploy assets crew & cargo together as required for specific mission in campaign Advanced Chemical Advanced Chemical deploy together Crew Separate/ Split Mission Transfer to Moon Chem/...
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Develop Capabilities through infusion of new technologies as they become available consistent with flight hardware schedule Develop Capabilities through infusion of breakthrough technologies when they become available Agency Programmatic Approach Current Structure & new management methods Smaller NASA Infrastruct...
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unar Arrival Lunar Ascent/Descent Lunar Departure Earth Arrival Earth Surface Landing Surface Landing Infrastructure Mars Robotic Precursors Lunar Robotic Precursors Bioastronautics Management Structure Agency Infrastructure Civil Service Role International Partnerships Acquisition Strategy Budget/Cost Strategy Capabil...
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Robotic Science continues af Human Departure Robotic Science concluded at departure Mission Design Approach Convenient Opportunities Any Opportunity SFL extant life (Hydrothermal deposits) SFL Global Evolution 91-500 Days (Long) (See Chart ? for inferred capabilities) Sustained Outpost (McMurdo model) > 500 Days Whe...
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LEO 28.5 HEO Different LV Existing LV Other Lunar Vicinity Update existing infrastructure Build new NASA Infrastructure Use new Industry infrastructure Heliocentric Transfer to Mars Crew Propulsive transit Nuclear (See chart ? for propulsion metrics) Advanced Chemical Cargo Propulsive Transit Nuclear Advanced Chemical...
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ISS Higher autonomy than ISS Full autonomy Operations Infrastructure for Moon Navigation/Communication Space Weather Use existing US assets, no additional investment Use existing US assets, no additional investment Emplace additional evolved infrastructure Emplace additional evolved infrastructure Cargo to Earth Orbi...
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biter in 2008/Lander 2009 only Orbiter 2008/Lander by 2012 on (See Chart ? for lunar robotic optio Precursors Start in 2011 - multiple missions (See Chart ? for Mars robotic options) Start in 2016 or later Multiple orbiters, landers & sample returnsMultiple orbiters and landers Bioastronautics Use ground research only...
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Location for Assembly** Orbit Location for final Crew Rendezvous Launch Infrastructure Crew Propulsive transit Cargo Propulsive Transit Crew/Cargo HLLV 51.6 LEO New LV LEO 28.5 HEO Different LV Existing LV Other Lunar Vicinity HEO Lunar Vicinity Update existing infrastructure Build new NASA Infrastructure Use new Indu...
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Operations Infrastructure for Moon Navigation/Communication Space Weather Use existing US assets, no additional investment Use existing US assets, no additional investment Emplace additional evolved infrastructure Emplace additional evolved infrastructure Earth Orbit Approach Crew to Orbit approach Orbit Location for...
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in 2016 or later Orbiter in 2008/Lander 2009 only Orbiter 2008/Lander by 2012 onlyMultiple orbiters, landers & sample returnsMultiple orbiters and landers Use ground research only to certify crews for deep space Use ISS and ground research t certify crews for deep space Use ISS, ground research and Moon to certify crew...
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Different LV Existing LV Other Lunar Vicinity Update existing infrastructure Build new NASA Infrastructure Use new Industry infrastructure Heliocentric Transfer to Mars (In-Space Transportation) Crew Propulsive transit Nuclear (See chart ? for propulsion metrics) Advanced Chemical Cargo Propulsive Transit Nuclear Adva...
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) Campaign Different Site Campaign same sites Sustained presence with robotics Mission Control Same autonomy as ISS Higher autonomy than ISS Full autonomy Operations Infrastructure for Moon Navigation/Communication Space Weather Use existing US assets, no additional investment Use existing US assets, no additional inv...
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2011 - 1 mission only Lunar Robotic Precursors Orbiter in 2008/Lander 2009 only Orbiter 2008/Lander by 2012 onl (See Chart ? for lunar robotic optio Precursors Start in 2011 - multiple missions (S Chart ? for Mars robotic options) Start in 2016 or later Multiple orbiters, landers & sample returnsMultiple orbiters and l...
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ate Separate Lander Lander Crew Launch EARTH EARTH 33 Notional Architecture Example Short-Stay Missions • Typically referred to as opposition class missions • Characterized by – Only 1 Hohman transfer (short leg) – High-propulsive requirements for other leg (long leg) • Venus swing-by or deep-space Maneuvers • Cl...
https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1eb80d3e0594fe46d8a6cc8a1321d6dc_architecture_mit.pdf
2/12/33 ) s / m k ( V D n o i s s i M 24 20 16 12 8 4 0 Sun Arrive Earth 9/18/33 γ Arrive Mars 9/16/31 01-Jan-15 31-Dec-18 30-Dec-22 29-Dec-26 28-Dec-30 Earth Departure Date 36 Notional Architecture Exploration Research Testbeds Exploration Research Needs Integrated Mission Validation Ground ISS Lunar Mars Robo...
https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1eb80d3e0594fe46d8a6cc8a1321d6dc_architecture_mit.pdf