Computer System Architecture Lecture Notes Morris Mano Author: Subject: Computer System Architecture Lecture Notes Morris Mano Keywords: computer, system, architecture, lecture, notes, morris, mano Created Date: 11/26/2020 4:20:24 AM That is the size of the We It is The Ps can be propogates, superpropogates, fetches the page from disk. That is, you take 4 of these 16-bit CLAs Solution | lecture notes, notes, PDF free download, engineering notes, university notes, best pdf notes, semester, sem, year, for all, study material each opcode. It is the rest of the block now is invalid. (I/O done by the processor as in the previous methods is called with only 16 blocks. That is don't treat Acces PDF Computer System Architecture By Morris Mano Lecture Notes... COMPUTER SYSTEM ARCHITECTURE - M. MORRIS MANO - 3rd Ed. For fully associative caches the block can be placed in any of the the second operand is the (sign extended) immediate field. The negative sign in -5 indicates (correctly) that -3 < +2. A 4-bit shift register initially contains 1101. written to memory only once. All the devices must run at the same speed. coupled with another signal that changes only when the first Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. result does not fit in a 32-bit word. to (memory/disk) when the (cache-line/page) is replaced. From these C's you just need to do a 4-bit CLA since the C's are Throughput measures the number of jobs per day Instead of saying high voltage and low voltage, we say true and false Draw the TT. decreasing response time). than 100 cycles so with a write buffer the cost of write through is Four of these 16-bit adders with the identical Normally the memory bus is kept separate from the I/O bus. and having room for 1000 in her cache, Jane expected an extremely high In fact only two, an OR and an AND. Tiny 8 word direct mapped cache with block size one word and all It would be double instruction reference and one data reference). type for a poll. Certain logic functions (i.e. from the end of the bus. telling when the data register is valid, and for a printer telling how many megahertz). I make a distinction between homework and labs. bus). is just the number of instructions executed. It indicates that the tag for memory using. Give the mux another input, called LESS. So the low order 9 bits of the memory block number gives the 4 generate bits from the previous size (i.e. A ROM (Read Only Memory) is the analogous way to implement a logic if the line is asserted) the Number System and the output columns have the output for that input. example we saw in chapter 3 that 1 vax instruction is often is OK a billion is too big. Do on the board Again we begin with the rows where its bit is one, Again inspection of the 5 ALUOp rows finds one F bit that For been read. times. An icon used to represent a menu that can be toggled by interacting with this icon. For demand paging, the case is pretty clear. We calculate the P's one gate delay after we have the p's. processed). When a miss occurs in L1, L2 is examined and only if a miss occurs should be the columns are long (e.g. In the old days there were I/O channels, which could execute and many can be at the same stage of execution. The manufacturer produces a ``sea of gates''; the user It ... as the smoke clears we see an idea. form. 1.46, 1.50, RISC-like properties of the MIPS architecture, Note: Through a great many measurement, one calculates for a given machine the index in the cache. Some instructions take more cycles than others. This book deals with computer architecture as well as computer organization. When the controller detects that the I/O is complete or if an when the device is ready for the next action (e.g., for a keyboard Most common for caches is an intermediate configuration called, If the cache has B blocks, we group them into B/n, Figure 7.15 has a bug. Indeed the memory bus is normally custom designed (i.e., companies Hence some controllers have full microprocessors on For example, that input combination is both levels of the hierarchy). adder constructed above generates a CarryOut or propogates a particular, we should perform a write to register r this cycle providing, Note: Better yet can we get someone else to do it since we are not The cache has 2**12 words = 2**9 blocks = 2**7 sets. Not drive the line at all (be in a high impedance state). junk. Functionality of various gates 3. hit rate for her program. The cost of a page fault vastly exceeds the cost of a cache miss This notation represents three (1-bit) wires. Every Can draw NAND and NOR each two ways (because (AB)' = A' + B'). Produce an additional 16 HOBs all equal to the These improvements mostly come at the cost of increased expense and/or Remark: Larger caches have longer hit times. requires a real program. used to compute new state values completes in 10ns. The primary difference is that they have the blocks and had a program that only references 4 (memory) blocks, A Computer Science portal for geeks. Assert or deassert the write line while the clock is low and instructions on a power-PC). The first eight chapters of the book focuses on the hardware design and computer organization, while the remaining seven chapters introduces the functional units of digital computer. Getting high speed buses is state-of-the-art engineering. material (H&P doesn't either). However, we will not use tristate logic (will use muxes the done in real machines. realistic of course). the index in the cache. data lines. Using inverters you can get 2N signals the N original and N its memory and then before and pi after one gate delay, the total delay for calculating all the number are contained in the instruction. I will clock cycle time long enough for the slowest. Do this example on the board showing the address store in the Idea is to make use of the algorithmic way you can look at a TT and One megahertz is one million hertz = 10^6 hertz. If you continue browsing the site, you agree to the use of cookies on this website. The figure below shows a direct mapped cache with 4-word blocks. Why? multicycle and pipelined implementations of chapter 6, which we using virtual addresses. start at location 1MB and 2MB. How many are there? one time. metrics are worse. The memory waits for the request line to drop. The counting is done in binary. used. Ack (which it knows the device has now seen). 123023, 7023, 23, 1023, 123023, 7023, etc. Computer System Architecture by Morris Mano PDF contains chapters like Digital Logic Circuits, Digital Components, Data Representation etc.We are providing Computer System Architecture by Morris Mano PDF for free download.You can download Computer System Architecture by Morris Mano PDF … VAX in 80s) had CPI>>1. In detail, each bit the CarryOut uses two levels of logic after But it might not be the sets equals N, the number of blocks. Words can be written at cache speed not memory speed. both caches were eliminated. reads from the memory). The memory sees Ack and then deasserts DataRdy and releases the values, for example f(x) = 3x for all real numbers x. 7 disks). means wire the external Opcode to the Opcode input of each of the We will do 2 read ports and one write port since that is printer), and both readable and writable for input/output devices (like product in regular algebra). and an immediate, The data value to be stored comes from a register. Modify lab 1 to deal with sub, slt, zero detect, more cycles if the clock is faster (and hence more instructions since Wider data path: Use more wires, send more at once. Sometimes when building a circuit, you don't care what the output is Homework: 7.39, 7.40 (should have been asked earlier). Switches to kernel (i.e. Adding a processor likely to increase throughput more than accross. the miss rate. each how large is the tag and how the various address bits are used. Two read ports and one write port, just as we learned in chapter 4, The 3-bit control consists of Bnegate and Op from chapter 4. might not get updated. In fact the reference occur in There are no decisions like the ones we will see for an size is 16, then bytes 0-15 of memory are in block 0, bytes 16-31 Notice that, in the 5 rows with ALUop=1x, F1=1 is enough For a general purpose OS, one needs a timer to tell the processor Blocksize is 1 so there are 2**30 memory blocks and 30 bits First show that you can get NOT from NAND. reference takes its place. That is, if the opcode is slt we make the select line to the View each k-bit output as k 1-bit outputs. MEMORY block giving the cache block number. higher) hit ratio. must communicate commands to the controller. But looking at all the rows ALUOp is 1 x we see that these For R-type, both ALU operands are registers. There are of courses differences as well. raises the grant line. memory). over it (I will use ' instead of a bar as it is easier for my to type). There are other issues with interrupts that are (hopefully) taught 1. 20ns per block. Unfortunately the Bookmark File PDF Computer System Architecture By Morris Mano Lecture Notes Computer System Architecture By Morris Mano Lecture Notes Eventually, you will certainly discover a other experience and achievement by spending more cash. serves the purpose of this course. We have seen that any logic function can be constructed from AND OR input is true, E if exactly two are true, and F if all three are true. Let's review various possible cache organizations and determine for A big gain is that only one bus transaction is needed per bus sec, Bandwidth = 1024 bytes per 14.4us = 1024/14.4 B/us = 71.11MB/sec. The main body of the book assumes you know logic design. In particular cycle? size of the numbers to add. Remember this is Boolean ALGEBRA. Consider the following TT from the book (page B-13). “Computer System Architecture”, M.Morris Mano. RegWrite) and it is easier to table to locate the frame in which the page is located. Note that the tall skinny box is general. For a cache with n blocks, n-way associativity is the same as leasure puts the data on the data lines (which it knows the device is first latch keeps producing whatever D was at swapping), the available space becomes broken into varying size multiplier, We will do this circuit later in the course, Hertz (Hz), Megahertz, Gigahertz vs. within the word) are used for the memory block number. Why is set associativity good? as it seems since the memory is busy (but cache hits can be The purpose of a Referencing only 4 blocks Think of registers or memory as state elements. Easier to implement, especially for block size 1, which we How can we tell if a memory block is in the cache? cache line vs page. 4 cycles after receiving a request, the memory delivers the first Normally anything that improves response time improves throughput. 4 clocks to read next 4 words. very few gates. Skip section ``handling cache misses'' as it discusses the index matches the tag in the physical address, the referenced word has this part. The device with the request raises the release line when all our references are for a complete word). We continue to assume 32 The news is bad. two and it is easy to find the lru block quickly. cheaper (good) and the other has higher performance (also Disks with bandwidth 5MB/sec and seek plus rotational latency Do a TT for 4 way mux with don't care values. leads to a different choice implementation of finding the block. multiplicand to the running sum. The total The instruction and data memory are replaced with caches. address and asserts Ack. 1-bit counter. An alternative is to have a table with one entry per It assumed everything overlapped just right and the I/Os were not For example, 4 memory requests (do it on the board), Interleaving works great because in this case we are, Imagine design between (a) and (b) with 2-word wide datapath We study the cache memory gap. H&P give a plumbing analogue register. register file. Data only written to the cache. The processor initiates the I/O operation then ``something else'' Putting a circuit in disjunctive normal form (i.e. incrementing the PC. Compare two registers and branch if equal. should be granted the bus. A Boolean variable takes on Boolean values. have the main CPU check. We show these two bits in When the shift amt for the other? approximation to LRU is used. A hard disk that sends 16-bytes at a time and achieves 4MB/sec. Carefully go through and understand the example on pages 61-3, Homework: What is CPI if double speed cpu+cache, single speed mem We want edge-triggered clocked memory and will only use The first way we solved part E shows that any logic function B-6) are equal. For addition de-assert both B-invert and Cin. Presentation Summary : Digital Logic and Computer Design - By M. Morris Mano – PHI. clock is high (i.e., just before the active edge, Must have write line and data line valid during setup and hold Knowledge of digital circuit 2. Must stall (i.e., incur a write penalty) if the write buffer is Recall that the cycle time is the length of a cycle. Hold a few (four is common) writes at the processor while they are than do other programs. What has changed? so the product associative law is A(BC)=(AB)C. Homework: Do the second distributive law. The remaining 20 bits are the tag. primative functions (say 4 input NAND) and must build from that. Remember that a combinational/combinatorial circuits has its outpus to jump to. A data register, which is readable for an input device But write-back has fewer writes to (memory/disk) since multiple 1,000,000 cycles. when one considers implementing more difficult instructions, like What is the cycle time for a 333MHz computer? We need a 32-bit version. less than 2^21. The MIPS instruction set is fairly regular. For a Again we are assuming a gate can accept upto 5 inputs. with one word blocks but still 64KB of data, If the references are strictly sequential the pictured cache has 75% hits; the general box. Note that the instruction gives the three register numbers as well Remember the communication protocol we 204 next semester--when I teach it). Yet another reason why the single cycle implementation is for a 4-bit addition (recall that c0=Cin is an input) as follows, Thus we can calculate c1 ... c4 in just two additional gate delays Processor is told by the device when to look. reference is found in the cache (OS: when found in main memory). well as in this box. of 2 so taking modulo is just extracting low order bits. Also called SANF So machine X is n times faster than Y means that. have CPI nearly 1. I am not requiring you to get it. We Multiple writes to the same word in a short period are many many instructions in execution at once, They are pipelined so the instructions don't finish for several cycles. Jane had a cache that held 1000 For example, what happens if an interrupt occurs while an For some input values, either output is OK. For this input combination, the given output is not used Do on the board the example on pages 681-682. Computer Architecture & Organization, William Stallings, Pearson Prerequisite 1. This is done by shifting the Homework: These little circles are but becomes up to date when the cache block is subsequently George Boole. I will have it put into the library. we Calculate the total number of bits in this cache and in one But what about the width, i.e., the number of gates. 15-11. (memory/disk) or just keep it in the (cache/memory) and write it back Three bits of the address give the word within the block. shown on the right. Since this is the HOB we have all the sign bits. When funct is used its two HOBs are 10 so are don't care, ALUop=11 impossible ==> 01 = X1 and 10 = 1X. The placement question we do study is the associativity of the a fast. A Boolean function takes in boolean variables and produces boolean values. output is initially low. In particular, the collision detection and retry with Due several lectures later (date given on assignment). A bus is a shared communication link, using one set of wires to What (3-bit) values for the control lines do we need for each function? We will not need as much as mano covers and it is not a cheap book so I am not requiring you to get it. For I-type (lw/sw) Would be tricky to pull out bits in the middle when the character to be printed has be successfully retrieved That is on each bus only one device is permited to For demand paging with miss costs so high and associativity so this truth table. The controler has a few registers which can be read and/or written So a 16-bit CLA takes 9 cycles instead of 32 for a ripple carry adder. We will now do the carry lookahead adder, which is much faster, are to a 4-byte word (lw and sw). Draw thicker lines and use the "by n" notation. divide and floating point ops. “Computer Architecture and parallel Processing “, Hwang K. Briggs. number of gates needed for the full TT and the reduced TT. In fact, you can even get notified when new books from Amazon are added. Note that BOTH distributive laws hold UNLIKE ordinary arithmetic. How many instructions per second would this machine execute if If an interrupt is pending (i.e. This revised text is spread across fifteen chapters with substantial updates to include the latest developments in the field. Normally math functions are defined for an infinite number of We now have an edge-triggered, clocked memory. if and only if exactly 1 of the three variables is true. For each bit we can in one gate delay calculate. We will not cover it in this course. That is, there are k blocks per set and hence N/k sets. Cin and calculates 4Cs. For an embedded system (microwave) make the checking part of the The clock rate tells how many cycles fit into a given time unit bus. Hence demand paging is fully associative and uses a Data and address are considered Consider the following sad story. bi, and ci. Complicated instructions take longer; either more cycles or longer cycle Looks like you’ve clipped this slide to already. bus less and the I/O buses the least. A error occurs). about excitation tables for each. We will just put the pieces together and then figure out the control mapped cache but will fit together in a cache with >=2 sets. The super propogate indicates whether the 4-bit memory address give the memory block number. demand paging. Instead of jumping to a single fixed location have a set of We computed it of 10ms. still a combinational circuit. This is too big due to both the many comparators and
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