the AMAT (in number of clock pulses)?
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Suppose a two layer memory hierarchy has a 4 clock pulse hit time, a 35 clock pulse miss penalty, and the miss ratio is 20%. What is the AMAT (in number of clock pulses)?
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- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?Question 18 Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag hield isConsider a demand-paging system with a paging disk that has an average access/transfer time of 50 ms. Addresses are translated through a page table in main memory, with an access time of 500 ns per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added a TLB that reduces access time to one memory reference if the page-table entry is in the TLB. Assume that 80% of the accesses are in the TLB and that, of those remaining, 15% (or 3% of the total) cause page faults. We assume that the TLB access time is 20 ns. What is the effective access time?
- (b) Consider a paging system with the page table stored in memory. If a memory referencetakes 200 nano seconds, how long does a paged memory reference take? If we add a TLB,and 75% of all page references are found in the TLB, what is the effective memory referencetime? (Assume that it takes zero time to find an entry in the TLB if it is already present).Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, set, and offset fields?In a certain computer, the virtual addresses are 32 bits long and the physical addresses are 48 bits long. The memory is word addressable. The page size is 16 kB and the word size is 2 bytes. The Translation Look-aside Buffer (TLB) in the address translation path has 64 valid entries. Hit ratio of TLB is 100% then maximum number of distinct virtual addresses that can be translated is K.
- Suppose a computer using set associative cache has 216 words of main memory and a cache of 128 blocks, and each cache block contains 8 words. If this cache is 2-way set associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields?Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 bits, the blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, and the size of the tag field is bits. size of the set field isSuppose a computer using direct mapped cache has 2^32 bytes of byte-addressable main memory and a cache size of 512 bytes, and each cache block contains 64 bytes. a. How many blocks of main memory are there? b. What is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields?
- In a computer with a 32-bit data-bus, how many 4-bit wide memory components are used? the answer to this part is 32/4 = 8 components (2-bit wide) I need the answer to part two, please If the size of each 4-bit memory component is 4 x n cells where n = 1G (i.e., 4 x n uniquely addressable locations—n : row, 4 : 2 column/width), what is the total capacity of the memory system? Show your answer in power of 2. (hint: 1000 ~ 210)Suppose we have a byte-addressable computer using fully associative mapping with 16-bit main memory addresses and 32 blocks of cache. If each block contains 16 bytes: (a) Determine the size of the offset field. (b) Determine the size of the tag field.Suppose we have a byte-addressable computer using fully associative mapping with 16-bit main memory addresses and 32 blocks of cache. Supposed also that each block contains 16 bytes. The size of the offset field is 19 bits and the size of the block field is 0.625 bytes.