Due to the fact that cache misses result in pipeline stalls, they increase a pipelined processor's CPI.True or False.
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Due to the fact that cache misses result in pipeline stalls, they increase a pipelined processor's CPI.True or False.
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- What is the impact of a cache miss on a pipelined processor's performance?Examine how pipelining affects the clock cycle time of the processor. Problems in this exercise assume that individual stages of the datapath have the following latencies: Also, assume that instructions executed by the processor are broken down as follows: What is the clock cycle time in a pipelined and non-pipelined processor? What is the total latency of an LW instruction in a pipelined and non-pipelined processor? If we can split one stage of the pipelined datapath into two new stages, each with half the latency of the original stage, which stage would you split and what is the new clock cycle time of the processor? Assuming there are no stalls or hazards, what is the utilization of the data memory? Assuming there are no stalls or hazards, what is the utilization of the write-register port of the “Registers” unit? f. Instead of a single-cycle organization, we can use a multi-cycle organization where each instruction takes multiple cycles but one instruction finishes…Assume the miss rate of an instruction cache is 4% and the miss rate of the data cache is 5%. If a processor has a CPI of 3 without any memory stalls, and the miss penalty is 50 cycles for all misses, determine how much faster a processor would run with a perfect cache that never missed. Assume the frequency of all loads and stores is 44%.
- Pipelining is used because it improves instruction throughput, increasing the level of pipelining cuts the amount of work performed at each pipeline stage, allowing more instructions to exist on the processor at the same time and individual instructions to complete at a more rapid rate. However, throughput will not improve as pipelining is increased indefinitely. Give two reasons for this.On the Motorola 68020 microprocessor, a cache access takes two clock cycles. Data access from main memory over the bus to the processor takes three clock cycles in the case of no wait state insertion; the data are delivered to the processor in parallel with delivery to the cache. a. Calculate the effective length of a memory cycle given a hit ratio of 0.9 and a clocking rate of 16.67 MHz. b. Repeat the calculations assuming insertion of two wait states of one cycle each per memory cycle. What conclusion can you draw from the results?How do processors use "pipeline replication" to handle multiple instruction streams simultaneously?
- A cache has a hit time Tc = 2 cycles and a miss rate Pmiss = 0.04. The main memory access time is Tmm = 36 cycles. The data-cache and instruction-cache have identical performance. A program has the following instruction distribution: probability of 0.3 for R-type instructions, 0.2 for load, 0.1 for store, and others for control instructions. Assume control instructions do not cause any loss. The processor is running at 1 GHz. Evaluate the average access time in nano-seconds of the memory system.True/False Cache performance gains are in part due to the Principle of Locality. This principle is applicable ONLY to pipelined machines and not to non-pipelined machines.What is the impact of pipeline flushes and stalls on CPU performance, and how are they managed?