Monday, June 3, 2019
Equivalents In Little Man Computer
Equivalents In Little Man Computer-There argon two different registers .MAR keeps pass over and MDR takes the content and read from the memory which was indicated by the MAR. These to from each one one memory location has an address. These address are identifiers them and the data which is stored there, only as each mailbox in the LMC has both an address. Slip of paper including the data stuffed into the slot.9.19) Describe the steps that evanesce when a system receives multiple frustrates?-If we encounter with multiple cuts, first interrupt makes a suspension of the political program put to death at the time, memory of programs critical parameters and transfer of control to the program which are handles the determined interrupt. After first interrupt, the second interrupt occurs. Second interrupts priority is compared to that of the original interrupt. When its priority is bigger, it gets precedence and the original interrupt program is itself suspended. In the contrary, pr ocessing of the real (original) interrupt keeps going and the new interrupt is caught on till the original interrupt program is complete. If the higher priority interrupt process is completed, the lower interrupt is processed. When any interrupts do non occur and when any interrupts result do not occur in the suspension of all central processor processing, control eventually turn back to original program which then resumes processing. Generally, multiple interrupts cause in a queue of interrupt handler programs. These programs are penalise by priorities associated with each interrupt.11.19)Carefully explain the differences between a client-server net income and a peer-to-peer profits. Compare the Networks in terms of capability and performance.-There are many big diffarence between a client-server and a peer-to-peer network. Firstly a peer-to-peer network do not has a central server. Each workstation on the network shares its files equally with the others. There is no central s torage or authentication of users. On the other hand, there are allocate dedicated servers and clients in a client/server network. Via the client workstations, users can find lots of files that generally stored on the server. The server will determine that users can reach the files on the network. Peer-to-peer network is very usable because it can be installed in your homes or in a very small businesses. Employees can interact regularly there. On the other hand they are very expensive to set up and they gives almost no security. However, client-servers networks can become big to you need them. Millions of user support it and offer elaborate security measures but it is very expensive. When we compare two networks we can forgather that Peer-to-peer networks has any PC is an equal participant on the network, PCs are not reliant on one PC for resources like printer, access to the network is not centrally controlled, operate on a basic PC OS, generally simpler and cheaper. Client-server s A PC acts as the network controller, A PC controls access to network resources, network reach and security are centrally controlled, Need a special OS, generally more(prenominal) complicated but give the user more control.11.21)Clearly and carefully discuss each of the advantages of clustering.- foregather is two data processors are in interconnecting and can create a solution when a problem occurred. One of the most distinguished advantage of clustering is that if one of the computer has a fail, another computer can see the problem and automaticly recovery this problem. The users see no interrupt of access. Clustering computers for scalability include increased application performance and it has support of a greater number of users. It can cause ability to perform maintenance and upgrades with limited downtime and good scale up your cluster to a maximum of seven active Exchange Virtual Server.8.8)Carefully discuss what happens when a cache miss occurs.Does this result in a s tudy slowdown in execution of the instruction? If so, Why?Cache miss means that cache controller can not do true fill the cache via the data processor acculy needs next .Cache misses slow down programs because the program can not going on its executing till the requested page is fetched from the main memory. In other words, The first cache miss will recompute the data, another request will get a cache miss and also recompute. As a result, this situation added calculation might slow down the whole system leading you to a loop.7.14)As computer words get larger and larger,there is a law of diminishing returnsthe speed of execution of real application programs does not increase and may,infact,decrease.Why do you suppose that this is so?Firstly we collapse to know marginal utility for understand to deminishing returns. The law of diminishing marginal utility helps people to understand the law of demand and the detrimental sloping demand curve. If you have something less, the more satis faction you gain from each additional unit you consume. For instance when you eat a chocolate bar, it taste is sugariness and you were satisfied. However, when you continue to eat chocolate ,its taste started to disgusting for you and your pleasure will go decreasing. Another example can be classical System processors (CPUs). They are generally priced in an exponential manner the fastest CPU available at any given time is so expensive, and then prices decrease readily as you go down in speed yet, the increase in performance by getting a CPU thats a little smear blistering is very slight.7.16)Create the fetch-execute cycle for an instruction that moves a value from general purpose register1 to general purpose register2.Compare this cycle to the cycle for a LOAD instruction. What is the major advantage of the move over the LOAD?For movesREG1 - MARMDR - IRIR - MARMDR - REG2PC + 1 - PCFor loadPC - MARMDR - IRIR - MARMDR - APC +1 - PCThe move fetch-execute cycle is beter because it is faster than LOAD because it occures between two registers. Registers are always faster than main memory.8.11) a) Suppose we are trying to determine the speed op a computer that executes the Little Man instruction set. The load and store instructions each make up more or less 25% of the instructions in a typical program. Add, subtruct, in, and out take 10% each. The various appendagees each take about 5%. the halt instruction is almost never used (a maximum of once each program, of course). Determine the average number of instructions executed each second if the clock ticks at 100 MHz.TheloadandStoretakefive steps .The Addand Subtrack also require five steps, IN and outrequire four , SKIPs require four, and JUMPs require three. Then atypicalprogrammix requiresS = 0.25 (5+5) + 0.10 (5+5+ 4 + 4) + 0.05 (4 + 3) = 4.65 steps perinstructionon average.If the clock ticks at10MHz., the number ofinstructionsexecuted in a second,N = 10,000,000 / 4.65 = approximately 2.17instructionsper seco nd.b)Now suppose that the CPU is pipelined, so that each instruction is fetched while another instruction is executing. What is the avarage number of instructions that can be executed each second with the same clock in this circumstance?With pipelining,eachinstructionis reduced by the two steps required for the fetch. Then,N = 10,000,000 / ( 0.25 (2 + 2) + 0.10 (2 + 2 + 1 +1) + 0.05 (2 + 1) )= approx. 5.7 million IPS8.18) Some systems use a branch prediction method known as static branch prediction, so called because the prediction is made on the basis of the instructer, without regard to history. One possible scenario would have the system predict that all conditional backward branches are taken and all forward conditional branches are not taken. Recall your experience with this programing in the little man computer language. Would this algorithm be affective? Why or why not? What aspects of normal computer programming, in any programming language, support your conclusion.Little man algorithm can be affective for branch prediction method, because it is suitable for pipeling. Witout branch prediction,users have to wait till the conditional jump instruction has passed the execute stage before the next instruction can enter the fetch stage in the pipeline.You can avoid this flagellate of time via the branch predictor attempts.7.6) Most of the registers in the machine have two-way retroflex capability that is, you can copy to them from another register, and you can copy from them to another register. The MAR, on other hand, is always used as a destination register you only copy to the MAR. Explain clearly why this is so.-Addresses are always moved to the MAR there would never be a reason for an address transfer from the MAR to another register within the CPU, since the CPU controls memory transfers and is obviously aware of the memory address being used.
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