Recent questions tagged effective-memory-access

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If a page of a process is in secondary memory but not in main memory then what should be it's page table entry.Then how logical address is ... in calculating effective memory access time formula. This is problem of paging or mapping?
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Consider a 5-stage pipelined processor with stages - Instruction Fetch (IF), Instruction Decode (ID), Execute (EX), Memory Access (MA) and Write Back (WB). All stages except Memory ... : SUB R5, R3, R4 ; R5 ← R3 - R4 ans is 14 bt how?
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Assuming 3-level of paging is used by a paging system and a TLB for address translation. Calculate Average memory access time if each main memory access takes 100ns and ... given as 200ns and TLB hit ratio is 80% and page fault rate is 10%.
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The access time of a main memory is 800 ns and cache memory access time is 70 ns. The ratio of read and write requests is 7:3. The hit ratio for read ... e. both read and write requests) if a write-through policy is used for cache updation.
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Given values:1-level page tableMemory access time =10msPage fault service time=20ms Page fault rate=40% what is the effective memory access time?
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which one is correct to use in exam here (1-p) is as page faultEMAT=p(m+m)+(1-p)(pfst)https://gateoverflow.in/333178/gate-cse-2020-question- ... pfst)https://gateoverflow.in/401429/go-classes-test-series-2023-iiith-mock-test-2-question-94
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A demand paging system provides a TLB (15 ns access time), cache memory (25 ns access time), main memory (75 ns access time, NOT including the cache ... the time and in main memory 20% of the time. Calculate the effective access time.
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Could you please differentiate below. Memory access time.hit time. search time??Arey yar some one answer this.. why so dumb
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Consider a two-level memory hierarchy with separate instruction and data caches in level 1, and main memory in level 2. The clock cycle time in 1 ns. The ... cycle. The average access time of the memory hierarchy will be . nanoseconds
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Assume that a read request takes 50 nsec on a cache miss and 5 nsec on a cache hit. While running a program, it is observed that 80% of the processor’s read requests result in a cache hit. The average read access time is ………….. nsec.
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Consider the following memories with their miss rates and hit times Then the average memory access time is ______ (in ns)
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An operating system implements a single level paging. This system uses Translation Lookaside Buffer (TLB) and Physical Address Cache (PAC) as well. TLB has a hit ... are in main memory, the effective memory access time is ______ (ns).
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In a paged memory, the page hit ratio is 0.35. The page fault service time takes 100 ns. The time required to access a page in primary memory is 10 ns. The effective memory access time required to access a page is ____ ns
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How many memory accesses are there in this code?LOAD R1, a(RO)
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Can any one help me out with this question :This was asked in MadeEasy CBT held on 23rd jan
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I have a simple doubt, Given a question that says that memory access time is x and page fault service time is y. We applyT = h*x + (1-h)*yHere assuming ... access ? Reference : https://gateoverflow.in/85404/gate1990-7-b Help me out here :(
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A Computer uses two level Cache L1 and L2 and in 2000 memory references there are 320 misses in L1 and 150 misses in L2. If Miss penalty of L2 is 300 ... cycle .What is average memory access time ?3.6 cycles5.4 cycles25.06 cycles4.8 cycles
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Suppose: TLB lookup time = 20 nsTLB hit ratio = 80%Memory access time = 75 nsSwap page time = 500,000 ns50% of pages are dirtyOS uses a single level page ... toupdate the TLB, the page table, and the frame table (if needed) is negligible.
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Consider a paging system with a page table stored in memory where memory reference takes 200 nanoseconds and all pages are in memory. What is the effective memory ... access time is 10ns.The given answer is 299ns but i am getting 290ns.
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Suppose when there is cache “Miss” then memory Access is 30nsand when cache “Hit” then memory access time is 3nsif 80% is cache hit then effective memory access9ns8.4ns3ns9.10ns
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For page fault Some where I seeEMAT=page fault (page fault service time) +(1-page fault)(memory access time)BUT somewhere it's likeEMAT=page fault (page ... 1-page fault)(memory access time)Which one is correct and why this difference?
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Consider a single-level cache with an access time of 2.5 ns with a block size of 64 bytes. Main memory uses a block transfer capability that has a ... )which is correct?I find made easy accurate with answers for all subjects except Co.