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6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 23. Networks — Links and Switches1 This handout presents the basic ideas for transmitting digital data over links, and for connecting links with switches so that data can pass from lots of sources to lots of destinations. You may ...
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copies of it in parallel. This goes by different names; ‘space division multiplexing’ and ‘striping’ are two of them. Common examples are: Parallel busses, as in the first four lines of the table. Switched networks: the telephone system and switched LANs. Multiple disks, each holding part of a data block, that can ...
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pair Broadcast µs 1.25 MB/s 100 + µs 6 MB/s 100 + µs 300 MB/s 5 6 MB/s 5 0.2 MB/s 5 KB/s 16 5 MB/s 3 3 ISDN CAP 16 memory packet 16 32/64 DMA block 1 1 1 16 1 1 packet packet ? 32 ? packet, 64-1500 B 1 1 1 µs/km 1 µs/km 1 µs/km 1 µs/km 1 µs/km 6 MHz byte or cell packet, 64-1500 B...
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and decreases it with each trouble-free send. This is called ‘exponential backoff'; when the 3 M. Sachs and A. Varman, Fibre channel and related standards. IEEE Communications 34, 8 (Aug. 1996), pp 40- 49. 4 G. Hoffman and D. Moore, IEEE 1394: A ubiquitous bus. Digest of Papers, IEEE COMPCON ’95, 1995, pp 334- 338. ...
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(backed off) and shortened depend on the properties of the channel. If the ‘trouble’ signal comes back very quickly and the cost of trouble is small, senders can shorten their waits aggressively; this happens in the Ethernet, where collisions are detected in at most 64 byte times and abort the transmission immediate...
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to control acquiring the channel, also uses backpressure, in the form of carrier sense, to keep a sender from interrupting another sender that has already acquired the channel; this is called ‘deference’. TCP uses credits to allow the receiver to control the flow. It also uses backoff to deal with congestion within ...
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and this is extra overhead that is serious if the messages are small (for instance, ATM cells are only 53 bytes, and only 48 bytes of this are payload). Stop sending messages and send a ‘resync’ request. When the receiver gets this it returns an absolute rather than an incremental credit. Once the sender gets this i...
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a label, framing is easy: You just label each X with the Y it is part of, and the position it occupies in that Y. For example, to frame (or encapsulate) an IP packet on the Ethernet, just make the ‘protocol type’ field of the packet be ‘IP’, and if the packet is too big to 10 F. Bonomi and K. Fendick, The rate-based...
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.13 If a ‘sync’ appears in the middle of a packet, that is taken as an error, and the next legal symbol is the start of a new packet. A simpler version of this idea requires at least one transition on every bit (in 10 Mb Ethernet) or byte (in RS-232); the absence of a transition for a bit or byte time is a sync. Ce...
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receiver makes a different guess. After at most 193 tries it will have guessed right. This takes a lot longer than the previous schemes to acquire sync, but it uses a constant amount of extra bandwidth (unlike escape schemes), and much less than fixed sync schemes: 1/193 for T-1 instead of 1/5 for FDDI, 1/2 for Ethe...
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at the other end. Of course the wide wire is more expensive than a single narrow one, and the multiplexers must also be paid for. • Traffic aggregated from several links may be more predictable than traffic from a single one. This happens when traffic is bursty (varies in bandwidth) but uncorrelated on the input li...
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and a sender has an n-bit ‘code’ (10010 in the figure) which it xor’s with its current data bit. The receiver xor’s the code in again and looks for either all zeros or all ones. If it sees something intermediate, that is interference from a sender with a different code. If the codes are sufficiently orthogonal (agre...
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sub-channel appears explicitly in the packet. If the packets are fixed size, they are often called ‘cells’, as in the Asynchronous Transfer Mode (ATM) networking standard. Fixed-size packets are used in other contexts, however, for instance to carry load and store messages on a programmed I/O bus. Variable sized pa...
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. Granting can be centralized, as in many I/O busses, or distributed, as in a daisy-chained bus or a token ring like FDDI. Flow control means buffering, as we saw earlier, and there are several ways to arrange buffering around a multiplexer, shown on the left side of figure 1. Having the buffers near the arbitratio...
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lexed to one BI bus, and 4 of these are multiplexed to one internal processor-memory bus. Demultiplexing uses the same physical mechanisms as multiplexing, since one is not much use without the other. There is no arbitration, however; instead, there is addressing, since the input channel must select the proper outp...
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it15. Instead, the task is distributed among all the senders. As with link arbitration in general, there are two ways to do it: scheduling and contention. 14 R. Rettberg et al.: The Monarch parallel processor hardware design. IEEE Computer 23, 18-30 (1990) 15 There are times when this religion is inappropriate. For ...
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However, it must keep this clock very close to the clock of the preceding node on the ring to keep from having to add sync marks or buffer a lot of data. • Maintaining the single token: with multiple tokens the broadcasting scheme fails. With no tokens, no one can send. So each node must monitor the ring. When it fi...
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is wasted whenever two packets overlap at the receiver; this is called a ‘collision’. How often does it happen? Fig. 3: A dual-attachment ring tolerates failure of one node • A practical way to solve this problem is to connect all the nodes to a single ‘hub’ in a so- called ‘star’ configuration, as shown in figure 4...
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one packet long. In this situation exponential backoff gives an efficiency of 1/e = .37 (see below). If a node that isn’t sending can tell when someone else is sending (‘carrier sense’), then a potential sender can ‘defer’ to a current sender. This means that once a sender’s signal has reached all the nodes without...
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0 Mbits/sec fast Ethernet has the same minimum packet size, and hence a maximum diameter of 5 µs, 10 times smaller. Gigabit Ethernet has a maximum diameter of .5 µs or 100 m. However, it normally operates in ‘full-duplex’ mode, in which a wire connects only two nodes and is used in only one direction, so that two wi...
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.7 at this maximum, including the successful transmission. The waste, also called the ‘contention interval’, is therefore 1.7r. For packets of length l the efficiency is l/(l + 1.7r)=1/(1 + 1.7r/l) ~ 1 - 1.7r/l when 1.7r/l is small. The biggest packet allowed on the Ethernet is 1.5 Kbytes = 20 r, and this yields an ...
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which of course is always changing as well). All versions of backoff arbitration have the problem that a selfish sender that doesn’t obey the rules can get more than its share. This isn’t a problem for Ethernet because there are very few sources of interface chips, and each one has been careful engineered to behave...
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ports. Fig. 5: An Ethernet with a hub can switch out failed nodes Handout 23. Networks — Links and Switches 13 Handout 23. Networks — Links and Switches 14 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 A drawback is that the hub is a single point of failure. Since it is ...
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view the links as simplex, but usually they are paired to form full-duplex links so that every input link has a corresponding output link which sends data in the reverse direction. Often the input and output links are connected to the same nodes, so that the switch allows any node to send to any other. A basic swit...
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0 µs 16 10-1000 KB/s 80 MB/s 125 1-100 s many µs 50K Handout 23. Networks — Links and Switches 15 Handout 23. Networks — Links and Switches 16 Storage can serve as a switch of the kind shown in figure 6(b). The storage device is the common channel, and queues keep track of the addresses that input and outpu...
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the slower one can deliver it. 3 3 2 concatenate 4 2 Fig. 7. Composing switches by concatenating. The only interesting thing about it is the rules for forwarding a single traffic unit: Can the unit start to be forwarded before it is completely received (‘wormholes’ or ‘cut­ through’)17, and Can parts of two uni...
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, and to watch out for alternating fast and slow links if you don’t have interleaving. However, a unit shouldn’t be too small on a variable TDM link because it must always carry the overhead of its address. Thus ATM cells, with 48 bytes of payload and 5 bytes of overhead, are about the smallest practical units (thou...
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an end node, and the Internet has 300 million nodes rather than 4. The new mechanism we need to make this work is routing, which converts an address into a ‘path’, a sequence of decisions about what output link to use at each switch. Routing is done with a map from addresses to output links at each switch. In addit...
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What spec does a mesh network satisfy? We saw earlier that a broadcast network provides unreliable FIFO delivery. In general, a mesh provides unreliable unordered delivery, because the routes can change, allowing one packet to overtake another, even if the links are FIFO. This is fine for IP on the Internet, which d...
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exists from every sender to every receiver. 19 C. Seitz: The cosmic cube. Communications of the ACM 28, 22-33 (1985) 20 W. Stallings, IPV6: The new Internet protocol. IEEE Communications 34, 7 (Jul 1996), pp 96-109. 21 Actually, this simple configuration can only deadlock if each node fills up with traffic going to th...
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other links up or down arbitrarily; the Autonet does this. Note that this kind of rule for preventing deadlock may conflict with an attempt to optimize the use of resources by sending traffic on the least busy links. Although figure 9 suggests that the resources being allocated are the links, this is a bit misleadi...
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assuming a global name space N for the nodes; this is usually coded with some kind of UID such as a LAN address, or by manually assigned addresses like IP addresses. If the nodes don’t have unique names, life becomes a lot more confusing. We name links with local names that are relative to the sending node, rather t...
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) -> Bool = RET p1.n = p2.n /\ p1.r <= p2.r FUNC Nodes(t, p) -> Ns = RET {p' | p' <= p | End(t, p')) 22 T. Anderson et al., High-speed switch scheduling for local area networks. ACM Transactions on Computer Systems 11, 4 (Nov. 1993), pp 319-352. 23 This is a complicated subject, and our treatment leaves out a lot. A...
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so that an entire subtree will be collapsed into a single node for the purposes of higher-level routing. n1 n2 n3 n0 ns = {n1, n2, n3} n0 IN ns TYPE L = (L + [n, ll]) FUNC Restrict(t, ns) -> T = RET (\ n | (\ l | (n IN ns /\ (t(n)(l) IN ns => t(n)(l)) )) FUNC IsConnected(t, ns) -> Bool = RET (ALL n1 :IN ns, n2...
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, do not do any switching of other people’s packets. The end-nodes don’t participate in the neighbor computation, since that would be an n2 process. Instead, only the routers on the LAN participate, and there is a separate scheme for the end-nodes. There are two mechanisms needed: 1. Routers need to know what end-n...
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each node how to map a destination node into a link24 on which to send traffic; you can think of it as the dual of a topology, which for each node maps a link to a destination node. Then a route is a path that is chosen by sw. TYPE SW = N -> N -> L PROC Route(t, sw, n1, n2) -> P = VAR p :IN Paths(t, n1, n2) | (ALL ...
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everyone knows the topology and the Cost function. For concreteness, we give a popular cost function: the length of the path. FUNC IsConsistent(t, sw) -> Bool = RET ( ALL n1, n2 | Route(t, sw, n1, n2) IN Paths(t, n1, n2) ) FUNC IsBest(t, sw) -> Bool = VAR best := {p :IN Paths(t,n1,n2) | | Cost(p)}.min | RET ( ALL n...
https://ocw.mit.edu/courses/6-826-principles-of-computer-systems-spring-2002/1686c03f438916c36cd791540d12985f_23.pdf
mesh network the sw function in every router has to map each address to a link that leads there. The structure of the address can make it easy or hard for the router to do the switching, and for all the nodes to learn the topology. Not surprisingly, there are tradeoffs. It’s useful to classify addressing schemes as...
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this method in connection with links, and the issues are the same in networks. One difference is that the round-trip time may be longer, so that more buffering is needed to support a given bandwidth. In addition, the round-trip time is usually much more variable, because traffic has to queue at each switch. Another ...
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An Overview of the Grammar of English Outline � Grammatical, Syntactic and Lexical Categories – Parts of Speech � Major Constituents – Noun Phrases – Verb Phrases – Sentences � Heads, Complements and Adjuncts Grammatical Categories � The dimensions – along with constituents can vary, and – to which the gramm...
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» let new words into them rather casually » and, therefore, tend to be very large. » Major ones are noun, verb, adjective and adverb. – Closed classes » change very little � Indeed, to a closed class is viewed as language change. » include “function” words, i.e., terms of high grammatical significance » Examples...
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the “number” grammatical category) » boy, boys, man, men – Count vs mass » “too many cats”, “too much water” » “Wine can be red or white.”, “Tigers have stripes.” Verbs � Types – auxiliary (closed) » List: do, have – modal (closed) » List: can, might, should, would, ought, must, may, need, will, shall (dare?) » c...
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gone for three weeks,} – Perfect participle (not quite the same thing) » He {has, will have, had} {cried, been, gone} » Always the same as the passive participle in English. Gerunds, BTW � Note that you can use the imperfective participle as a so-called “verbal noun”: Throwing stones at glass houses can be hazardou...
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: Jan threw up made up that story looked the word up put me down. � However, they are identical with the set of English prepositions. � So it is appealing to think of these as prepositions without complements. Adverbs � Types – manner (quickly, rarely, never) – directional/locative (here, home, downtown) – te...
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146. The Penn Treebank Tag Set tag description example coord. conjunction and, but, or cardinal number one, two, three determiner a, the existential there foreign word a propos preposition/sub-conj of, in, by, if adjective small adj., comparative smaller adj., superlative smallest CC CD DT EX...
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what possessive wh­ whose noun, plural cars WRB Wh-adverb how, where The Major Constituents � These syntactic categories are may be thought of as “bigger” versions of lexical categories: – Noun phrase (NP) – Verb phrase (VP, S) – Prepositional phrase (PP) – Adjective phrase (AP) – Adverbial phrase (ADVP) ...
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have said thus far: NP fi (PDT) (D) (Q) AP* NP* N � Note that – “X*” is just a shorthand for Xs fi Xs fi X Xs fi Xs X – “X fi (Y) Z” is an abbreviation for X fi Z X fi Y Z e An Approximate Grammar, Redux � However, most analyses have more embedded constituent structure. � So, a somewhat better set of rules might ...
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general, it is tempting to think that the grammar of noun phrases can be made simpler, and that at least some of these constraints can be explained semantically. – Exactly how to do so is not always clear. Preceding the Noun: Odds and Ends � Personal pronouns – can be NPs all by themselves. NPmin fi ProP – and ca...
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NP infC “the guy to go to in a pinch” Comments � Which “NP” are we talking about here? � Consider “most baguettes from the Cheese Board”, This should probably be analyzed as “[most [baguettes from the Cheese Board]]” � Also “a package from overseas delivery” is okay. � So, this looks like “NPint”. Followin...
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ositionals, prototype activity nouns…. � But our rules for NPs lose this distinction. Solutions? max fi fi fi N | NPint NPmin | PP NPmin (Q) AP* NPmin ((PDT) DP ) NPint � We can differentiate our grammar rules further. � E.g., instead of NPmin NPint NP we could have NPmin/scc NP NP NP NP NP Nscc | NPint NPmin/sc...
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that we can’t write down the grammar; it’s that we can’t write down a satisfying one. The Verb Phrase � Main clauses, e.g., “Pat baked Jan cookies” are typically analyzed as [S [NP Pat] [VP [V baked] [NP Jan] [NP cookies]]] as opposed to [S [NP Pat] [V baked] [NP Jan ] [NP cookies]] � I.e., the basic general str...
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baked and Jan iced a chocolate layer cake. which suggests that [Pat baked] and [Jan iced] are constituents. � But the other tests don’t bear this out: *What was done to the cake was Pat baked. *Pat baked a cake and so did frost. The Verb Phrase � Here are some common structures, and phrases that conform to them: ...
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E.g., “Jan verbed Pat a book.” « Pat.” (sometimes) “Jan verbed a book to Sentence Level Constructions � Sentences are generally regarded as a bigger form of VP, just as we had different forms of NP. � But, traditionally, we use the separate symbol “S” anyway. � Here are some common sentence types: S fi NP VP Ja...
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Adjuncts � For most constituents, there is a syntactically central part, and some less central parts. � For example, consider: “the conservative senator” – This is a noun phrase whose head is the noun phrase “conservative senator”. – This noun phrase in turn has the head “senator”. – We further say that “senator” ...
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were written, for a given lexical category X, X, X’, and X’’ (but pronounced “x bar” and “x double bar”). X-bar Theory N’’ Det that N’ N A’’ A’ A nice book In such theories: Complement is daughter of X’, sister of X. Adjunct is daughter of X’, sister of X’. Specifier is daughter of X’’, sister of X’ N’...
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Jan left. Oddly, Jan sang folks songs. So we might add S fi AA* S AA fi PP | AdvP � You can also get these at the end, but then they are best analyzed as part of the VP: Jan left in the morning/quickly. Jan sang folks songs oddly. Jan quickly left the meeting So one might add VP fi AA* VP AA* An Approximate Grammar, ...
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iPaq Intro, Python, and Connectivity Feb 13, 2006 Larry Rudolph 1 Pervasive Computing MIT 6.883 SMA 5508 Spring 2006 Larry Rudolph Administration • iPaq’s and Mobile Phones very similar • both use python, bluetooth, internet • This week: • Ipaq comments, Python, Network • Problem set, due in one week • On your own, w...
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aq, your ipaq? • Anonymous vs Personal handheld • Telephone example • Landline telephones are anonymous • Cell/Mobile phones are personal • Tradeoffs • private state • can be lost or stolen; should be protected • setup overhead on user • daily underhead on user: setup once & • less dependent on connectivity • public/...
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Client Code import sys import socket if len(sys.argv) < 3: print "usage: socketclient <address> <port>" sys.exit(2) s = socket.socket(socket.AF_INET, socket.SOCK_STREAM) s.connect( (sys.argv[1], int(sys.argv[2]) ) ) print "connected. type stuff." while True: data = sys.stdin.readline() if len(data) == 0: print "closin...
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Lecture 14 Interlude: Problem Solving Supplemental reading in CLRS: None This lecture was originally given as a pep talk before the take-home exam. In note form, this chapter will be light reading, a break in which we look back at the course material as veterans. 14.1 What to Bring to the Table In most technical underg...
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even need to tell you this. 14.2 How to Attack a Problem We’ll now discuss several tips that might help you to crack a problem open. We’ll keep the following concrete example in the back of our mind: Problem 14.1 (Bipartite Matching). In a group of n heterosexual people, each woman has a list of the men she is interes...
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the general problem!) – Whatever your solution is, its running time must of course fall within the lower and upper bounds you find. • Is there a useful diagram? 1 A comparison-based sorting algorithm is a sorting algorithm which makes no assumptions about the items being sorted except that there is a well-defined notion ...
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Is it similar to a known problem? Can we modify the known algorithm to suit it to this version? • Can the problem be solved with convolution? If so, the fast Fourier transform will improve efficiency. • Is there any further way to take advantage of the given information? • Can randomization help? It is often better to fi...
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, here is a solution: Algorithm: Let M be the set of men and let W be the set of women; let s and t be objects representing a source and a sink. Create a bipartite directed graph∗ G with vertex set M ∪ W ∪ {s, t}. Draw an edge from s to each man and an edge from each woman to t. Next, for each man m and each woman w, d...
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O (cid:161)n3(cid:162). Actually, assuming each person has a relatively small list of acceptable spouses, we ought to be more granular about this bound: we have (cid:195) E = O n + (cid:181)size of the ith person’s list (cid:182)(cid:33) , n (cid:88) i=1 and the running time is at most (cid:195) (cid:195) T = O n · n +...
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.3 Recommended Reading There are many great references and exercise books for algorithmic problem solving and mathemat- ical problem solving in general. Two particular selections are: • George Pólya, How to Solve It • The Princeton Companion to Mathematics, VIII.1: “The Art of Problem Solving.” Another good source of a...
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1 Pipeline Hazards Arvind Computer Science and Artificial Intelligence Laboratory M.I.T. Based on the material prepared by Arvind and Krste Asanovic 6.823 L6- 2 Arvind Technology Assumptions • A small amount of very fast memory (caches) backed up by a large, slower memory • Fast ALU (at least for integers) • Multipo...
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Data Memory wdata I-Fetch (IF) Decode, Reg. Fetch (ID) Execute (EX) Memory (MA) s e c r u o s e R time IF ID EX MA WB t0 I1 t1 I2 I1 t2 I3 I2 I1 t3 I4 I3 I2 I1 t4 I5 I4 I3 I2 I1 t5 t6 t7 . . . . I5 I4 I3 I2 I5 I4 I3 I5 I4 I5 September 28, 2005 6.823 L6- 4 Arvind Write -Back (WB) Pipelined Execution: ALU Instruc...
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Store instructions - Write back September 28, 2005 Pipelined MIPS Datapath without jumps E IR M IR 6.823 L6- 7 Arvind W IR 31 RegDst RegWrite we rs1 rs2 rd1 ws wd rd2 GPRs Imm Ext OpSel ALU Y A B MD1 MD2 MemWrite WBSrc we addr rdata Data Memory wdata wdata R ExtSel BSrc F D 0x4 Add PC addr inst IR Inst Memory Septemb...
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3 L6- 10 Arvind Freeze earlier pipeline stages until the data becomes available ⇒ interlocks If data is available somewhere in the datapath provide a bypass to get it to the right stage Speculate about the hazard resolution and kill the instruction later if the speculation is wrong. September 28, 2005 Feedback to Re...
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IF3 IF3 stalled stages t5 t6 t7 . . . . ID2 EX2 MA2 WB2 IF3 ID3 EX3 MA3 WB3 IF4 ID4 EX4 MA4 WB4 IF5 ID5 EX5 MA5 WB5 time t0 I1 IF ID EX MA WB t1 I2 I1 t2 I3 I2 I1 Resource Usage September 28, 2005 . . . . t4 I3 I2 t5 I3 I2 t6 t3 I4 I3 I3 I2 nop nop nop I2 I1 t7 I5 I4 I3 nop nop nop I2 I1 I5 I4 I3 nop nop nop I2 I5 I...
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IR we rs1 rs2 rd1 ws wd rd2 GPRs Imm Ext A B ALU Y MD1 MD2 we addr rdata Data Memory wdata wdata Should we always stall if the rs field matches some rd? not every instruction writes a register ⇒ we not every instruction reads a register ⇒ re September 28, 2005 6.823 L6- 15 Arvind IR 31 Cdest R Source & Destination R...
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ALR ... ⇒ on ⇒ off Cstall re1 = Case opcode ALU, ALUi, LW, SW, BZ, JR, JALR J, JAL re2 = Case opcode ALU, SW ... ⇒ on ⇒ off ⇒ on ⇒ off stall = ((rsD =wsE).weE + (rsD =wsM).weM + (rsD =wsW).weW) . re1D + ((rtD =wsE).weE + (rtD =wsM).weM + (rtD =wsW).weW) . re2D th e full sto ry ! T his is n ot September 28, 200...
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28, 2005 20 Five-minute break to stretch your legs Complications due to Jumps PCSrc (pc+4 / jabs / rind/ br) stall 6.823 L6- 21 Arvind 0x4 Add Add nop Jump? E IR I1 M IR PC 104 addr inst Inst Memory IR I2 Note fetching the next instruction before decode is speculation ⇒ kill I1 I2 I3 I4 096 100 104 304 ADD J 200 ...
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304: ADD t5 t6 t7 . . . . t2 t3 t4 t1 ID1 EX1 MA1 WB1 IF2 ID2 EX2 MA2 WB2 IF3 nop nop nop nop IF4 ID4 EX4 MA4 WB4 time t0 I1 IF ID EX MA WB t1 I2 I1 t2 I3 I2 I1 Resource Usage September 28, 2005 t5 t6 t7 . . . . t4 t3 I4 I5 nop I4 I2 I1 I5 nop I4 I2 I1 I5 nop I4 I2 I5 nop I4 I5 nop ⇒ pipeline bubble Pipelining Cond...
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? zero? M IR I1 E IR I2 A ALU Y I1 I2 I3 I4 096 100 104 304 ADD BEQZ r1 200 ADD ADD September 28, 2005 If the branch is taken - kill the two following instructions - the instruction at the decode stage is not valid ⇒ stall signal is not valid Pipelining Conditional Branches stall PCSrc (pc+4/jabs/rind/br) 6.823 L6- ...
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5 Control Equations for PC and IR Muxes 6.823 L6- 28 Arvind PCSrc = Case opcodeE BEQZ.z, BNEZ.!z ⇒ br ... ⇒ Case opcodeD ⇒ jabs J, JAL JR, JALR ⇒ rind ... ⇒ pc+4 IRSrcD = Case opcodeE BEQZ.z, BNEZ.!z ⇒ nop ... ⇒ Case opcodeD J, JAL, JR, JALR ⇒ nop ⇒ IM ... Give priority to the older instruction, i.e., execute ...
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nop nop I5 I2 I1 nop nop I5 I2 nop nop I5 nop ⇒ pipeline bubble Reducing Branch Penalty (resolve in decode stage) 6.823 L6- 30 Arvind • One pipeline bubble can be removed if an extra comparator is used in the Decode stage PCSrc (pc+4 / jabs / rind/ br) 0x4 Add Add nop E IR PC addr inst nop Inst Memory IR D we rs1 rs2...
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WB2 ID3 EX3 MA3 IF3 ID4 EX4 IF4 ID5 IF5 Each stall or kill introduces a bubble in the pipeline ⇒ CPI > 1 A new datapath, i.e., a bypass, can get the data from the output of the ALU to its input t0 t1 IF1 time (I1) r1 ← r0 + 10 (I2) r4 ← r1 + 17 (I3) (I4) (I5) September 28, 2005 t3 t4 t5 t2 ID1 EX1 MA1 WB1 IF2 t6 t7 . ...
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34 Arvind stall = ( ((rsD =wsE).weE + (rsD =wsM).weM + (rsD =wsW).weW).re1D +((rtD =wsE).weE + (rtD =wsM).weM + (rtD =wsW).weW).re2D ) ws = Case opcode we = Case opcode ALU ⇒ rd ALUi, LW ⇒ rt JAL, JALR ⇒ R31 ALU, ALUi, LW ⇒(ws ≠ 0) JAL, JALR ⇒ on ⇒ off ... ASrc = (rsD=wsE).weE.re1D Is this correct? No because only A...
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E IR M IR W IR 31 PC addr D inst IR Inst Memory we rs1 rs2 rd1 ws wd rd2 GPRs Imm Ext A B ALU Y BSrc MD1 MD2 we addr rdata Data Memory wdata wdata R Is there still a need for the stall signal ? September 28, 2005 stall = (rsD=wsE). (opcodeE=LWE).(wsE≠0 ).re1D + (rtD=wsE). (opcodeE=LWE).(wsE≠0 ).re2D Why an Instructio...
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18.03 Class 4, Feb 10, 2010 First order linear equations: integrating factors [1] First order homogeneous linear equations [2] Newtonian cooling [3] Integrating factor (IF) [4] Particular solution, transient, initial condition [5] General formula for IF Definition: A "linear ODE" is one that can be put in the "s...
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|x| = - int p(t) dt + c ln|x| = - t^2 + c Exponentiate: |x| = e^c e^{ - int p(t) dt } |x| = e^c e^{-t^2} Eliminate the absolute value and reintroduce the lost solution: x = C e^{- int p(t) dt} x = C e^{-t^2} In the example, we chose a particular anti-derivative of k ,...
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the process for so long that the insulation of the cooler starts to break down! Systems and signals analysis: The system is the cooler. The output signal = system response is x(t) , the temperature in the cooler. The input signal is the external temperature T_ext(t) . Note that the right-hand side is k times the ...
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standard form, but not reduced standard form.) Here's a *trick*. Multiply both sides by t : t^2 x' + 2t x = t^2 The left hand side is now the derivative of a product: (d/dt) (t^2 x) = t^2 We can solve this by integrating: t^2 x = t^3/3 + c so x = t/3 + c t^{-2} [In the first lecture, I posed this (with a dif...
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20 + 2t) e^{t/3} Integrate: e^{t/3} x = 60 e^{t/3} + \int 2t e^{t/3} dt Um. Parts: \int u dv = uv - \int v du u = 2t , dv = e^{t/3} dt du = 2dt , v = 3 e^{t/3} e^{t/3} x = 60 e^{t/3} + 6 t e^{t/3} - 18 e^{t/3} + c = ( 42 + 6 t ) e^{t/3} + c Solve for x: x = ( 42 + 6t ) + c e^{-t/3} That's the general so...
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does. It is then called a TRANSIENT. All solutions come to look more and more alike as time goes on. This is a funnel! I graphed the solutions 42 + 6t and x , and some others along with T_ext . If the temperature in the cooler is more than 60 degrees at the start, then it declines at first, crosses the nullcline x...
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The first part of this course will cover the foundational material of homogeneous big bang cosmology. There are three basic topics: 1. General Relativity 2. Cosmological Models with Idealized Matter 3. Cosmological Models with Understood Matter 1 General Relativity References: • Landau and Lifshitz, Volume 2: Th...
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0 This will be a terse introduction to general relativity. It will be logically complete, and adequate for out later purposes, but a lot of good stuff is left out (astrophysical applications, tests, black holes, gravitational radiation, . . . ). 1.1 Transformations and Metrics We want equations that are independent ...
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∂x�β ∂xλ ) We can write the transformation law for gµν in matrix form: G� = R−1G(R−1)T From linear algebra, we can insure G� is diagonal with ±1 (or 0) entries. The signature, e.g. � 1 � −1 −1 −1 , is determined. There are residual transformations that leave this form of gµν intact. They are the Lorentz transfor...
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(x�) = φ(x) • Vector A� µ(x�) = ∂xα ∂x�µ Aα(x) • Operator ∂xα ∂ � ≡ ∂x�ν = ∂x�ν ∂xα ∂ν ∂ Is there an invariant derivative? � (R−1)α � µAα � A� ∂ν µ = = � Aβ � ∂xα ∂ ∂xβ ∂x�ν ∂xα ∂x�µ ∂xα ∂xβ ∂x�ν ∂x�µ �� � good ∂2xβ Aβ ∂αAβ + ∂x�ν ∂x�µ � �� � � bad (hard to use ­ not a tensor) Add correct...
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xσ ∂x�ν∂x�µ Note that the inhomogeneous part is symmetric in µ ↔ ν. So we can assume Γλ = Γλ βα αβ consistently. (The antisymmetric “torsion” part is a tensor on its own!) Given Γ, we can take covariant derivatives as �αTµ1...µm ν1...νn = ∂αT µ1...µm ν1...νn −Γλ T αµ1 λµ2...µm ν1...νn −. . .−Γλ T αµm µ1...
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µ − ∂λgµν Conversely, this works! Extra term in Γ�: ∂xσ � 1 λβ �� � � ������ µ ∂x�λ gσν + ∂µ g �∂� 2 � � ∂xσ ∂x�ν gλσ � � ������ ∂xσ ν ∂x�λ gσµ + � + �∂� � � ∂ν � ∂xσ ∂x�µ gλσ � ∂x� � � − �∂� λ ∂x�µ � � � � �� σ���� ∂� ∂xσ gσν − ��λ���� gσµ ∂x�ν The boxed terms give the desired inhomogeneous ...
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� · This Rα βµν automatically transforms as a proper tensor; the “hard” part is to show that the derivatives on Aα all cancel so we get this form. �µAβ = ∂µAβ − Γλ νβ Aλ �µ (�ν Aβ ) = ∂µ µβ �ν Aσ (�ν Aβ ) − ����� −Γσ σ �σAβ Γµν � � �� ⇒ symm etric drop it νλAσ) − Γσ (Γσ − ∂µ νλAσ − Γσ Aσ − Γσ ρ − Γα ...
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∂µ∂σgβν − ∂ν ∂β gµσ − ∂ν ∂σgβµ = g ασRσβγδ 1 � 2 1 2 with Rαβγδ = 2 � � 1 ∂α∂δgβγ + ∂β∂γ gαδ − ∂α∂γ gβδ − ∂β∂δgαγ ⇒ Rαβγδ = −Rβαγδ Rαβγδ = −Rαβδγ Rαβγδ = Rγδαβ Rαβγδ + Rαγδβ + Rαδβγ = 0 (e.g. Look at the coefficient of ∂α∂δgβγ : +1 in Rαβγδ −1 in Rαγδβ 0 in Rαδβγ ) Since these are tensor identities, they h...
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γ Aν = αβγ �ν Aµ + Rν (1) −�αRν µβγ �αAν (4) (3) µβγ Aν − Rν (2) µβγ �αAν (2) cancels against (4), (3) will go away by the symmetry of Rν generates the Bianchi identity. This identity is the gravity analogue of αβγ + Rν βγα + Rν γαβ = 0, so (1) in electromagnetism, or � · B = 0, � × E = − ∂B (existence of vect...
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