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Memory Access Time for Read (MATRMATR) is, MATRMATR = HR1T1+(1−HR1)HR2T2HR1T1+(1−HR1)HR2T2 (b) Calculate the effective memory-access time with a cache hit ratio of h=0.95. Ideally, you will want the hit_ratio to be greater than 90%. Assume 8 words cache blocks and set size of 256 words with the set associative mapping (i) Show the mapping between M1 and M2 (ii) Calculate the effective access time with a cache hit ratio of h=0.95 (10 marks) 1 (b) What do you mean by fetch cycle, instruction cycle, machine cycle and interrupt acknowledgement cycle? effective-access-time = cache-access-time + miss-rate * miss-penalty Since "t1 means the time to access the L1 while t2 and t3 mean the (miss) penalty to access L2 and main memory, respectively", we should apply the second formula above, twice. The Note that the miss rate also equals 100 minus the hit rate. It also gives the effective access time (EAT) expression if it is not the case. access time due to . c. Find the effective memory access time. The fraction or percentage of accesses that result in a hit is called the hit rate. If Cache has 4 slots and memory has 90 blocks of 16 addresses each (Use as much required in question). Solution: Memory cost is calculated by; Ctotal= C1S1+C2S2+C3S3 G 15000, then S3=39.8 The effective memory access time is calculated as Primary cache " Focus on minimal hit time ! To improve the hit time for reads, • Overlap tag check with data access. 80% of the memory requests are for reading and others are for write. Calculate unknown specifications based on these . A computer has two levels of cache memories, which provide access times that are 0.01 and 0.1 times the access time of memory. Average memory access time ( AMAT) is the average time a processor must wait for memory per load or store instruction. This question was previously asked in. GATE 2015- Average Access Time - Cache Hit Miss - Computer Architecture When the time to access a page when not in associative memory. tc : cache access time. This equation will result in the average memory access time. Homework 3: Memory Management (10 pts) If the virtual address space supported is 2**64 bits (note bits not bytes), the page size is 1Kbyte, the size of the physical memory is 64Kbyte, the size of a PTE is two bytes, and the addressing is at the byte level, calculate the size of the page table required for both standard and inverted page tables. L-2 cache " Focus on low miss rate to avoid main memory access " Hit time has less overall impact ! Consider a system with 80% hit ratio, 50 nano-seconds time to search the associative registers , 750 nano-seconds time to access memory. Each processor is rated with 10 MIPS if a 100% cache hit ratio is assumed. We do now want our effective access time to increase much beyond 1 ns. A. b. Given a system with a memory access time of 250ns where it takes 12ms to load a page from disk to memory, update the page table and access the page.Calculate the Effective Access Time (EAT) when 100% of the pages are in memory.Calculate the Effective Access Time (EAT) with a page fault rate of 5%. 5.7 in text): APC (memory Access Per Cycle) is a new performance metric designed to measure concurrent memory system performance [5]. a. Assume 8 words cache blocks and set size of 256 words with the set associative mapping (i) Show the mapping between M1 and M2 (ii) Calculate the effective access time with a cache hit ratio of h=0.95 (10 marks) 1 (b) What do you mean by fetch cycle, instruction cycle, machine cycle and interrupt acknowledgement cycle? That is, Teff = t1 + (1-h1) [t2 + (1-h2)t3] = 32 Average Memory Access Time Look-through cache: main accessed after cache miss detected: T. C ,T. M = cache and main memory access times . • On a TLB or cache miss, the time required for access includes a TLB and/or cache update, but the access is not restarted. Don't forget that the cache requires an extra cycle for load and store hits on a unified cache because of the structural conflict. To improve the hit time for writes, Pipeline write hit stages Write 1 Write 2 Write 3 time TC W TC W TC W There are 10,000 memory references of which 10 causes L2 misses and 90 causes L1 misses. It follows that hit rate + miss rate = 1.0 (100%). The memory access times are 2 nanoseconds. Victim cache miss ratio = 0.26; and its transport time from L1 miss = 1 clock Given fixed L1 cache performance, it is fair to compare these head-to-head (but the comparison might not stay the same if L1 were changed): tea for L2 cache beyond the L1 access time is: 3 + 0.19 * (12+4+4+4) = 7.56 clocks in addition to L1 delay tea for L2 cache . How to calculate average memory access time.Computer Organization a. 33 nS c.) 24.6 nS d.) 27.0 nS e.) 2.4 nS f.) 3.0 nS a.) The miss penalty for either cache is 100 ns, and the CPU clock runs at 200 MHz. Calculate unknown specifications based on these conditions 22 Arwin - 23206008@2006 11 b. Reducing any of these factors reduces AMAT. Assume miss rates are as follows (Fig. Also, the total cost is upper-bounded by $1,500. • Av Access Time as function of hit ratio H: H * 0.01 s + (1-H)* 0.11 s • With H near 1 access time approaches 0.01 s . The hit ratio is clearly related to the number of associative registers. In Conclusion Calculating the hit and miss ratios in cache memories as described above can help you better understand how well your cache is performing. The hit rate and miss rate can measure reads, writes, or both, which means that the terms can be used to describe performance information in several ways. To calculate the average memory access time, add the hit time and the miss ratio, and multiply it by the miss penalty. Main memory t2 = unknown S2 = 32 Mbytes C2 = $0.02 Disk array t3 = 4 ms S3 = unknown C3 = $0.00002 Our aim is to achieve an effective memory-access time t=850 ns with a cache hit ratio h1 = 0.98 and a hit ratio h2 = 0.99 in the main memory. 1.26 b. of memory access leading to a cache miss to the total number of instructions - Miss penalty: time/cycles required for making a data item in the cache. In the typical computer system from Figure 8.3, the processor first looks for the data in the cache. Given a cache access time of 20ns, a main memory access time of 1000ns, and a cache hit ratio of 90 percent. Hit ratio of cache = 0.8 Cache Access Time = 30 ns Memory Access Time = 150 ns CPU Access Time = Cache Hit * Cache Access Time + (1 - Cache Hit)(Cache Access Time+ Memory Access Time) CPU Access Time= 0.8 * 30 + 0.2 * (30 + 150) = 60 ns Note : Effective Memory Access Time and Average Access Time both are same The effective memory-access time can be derived as followed : The general formula for effective memory-access time is : 1 . Results " L-1 cache usually smaller than a single cache " L-1 block size smaller than L-2 block size Multilevel Cache Considerations 45 The processor-memory system described in (a) is used to construct a bus-based shared-memory multiprocessor. We now add virtual memory to the system described in question 9. Which has the lower average memory access time? 4.7 A computer has a cache, main memory, and a disk used for virtual memory . Calculate the average memory access time. The average access time of the system for both read and write requests is. The formula for calculating a cache hit ratio is as follows: For example, if a CDN has 39 cache hits and 2 cache misses over a given timeframe, then the cache hit ratio is equal to 39 divided by 41, or 0.951. 1.68 c. 2.46 d. 4.52 Correct answer is (b). AMAT can be written as hit time + (miss rate x miss penalty). tm : main memory access time . Using the hit ratio, we can express the effective access time of a memory system using a cache as: teff = tcache + (1 - h ) tmain So, for a hit ratio of 90% (0.9), a cache access time of 10 ns, and a main memory access time of 60 ns, we have . 1 - h : miss ratio of the cache. For example, if the hit ratio is 90% and the memory access time is 12 nanoseconds, then, according to our simple model, the EAT would be calculated as (0.9)(12)+(0.1)(24) = 13.2 ns. The cache hit ratio can also be expressed as a percentage by multiplying this result by 100. —But we want speed to approach cache speed for all memory access —More cache is faster (up to a point) —Checking cache for data takes time The cache access time is 70 ns, and the time for transferring a main memory block to the cache is 3000 ns. Assume that the hit ratio and access times remain the same as in part (a). How do you calculate TLB? currently we have cache inserts=25, cache lookups=35, so i get ratio =0.28, is this a good number?-- When there is a miss in L1 cache and a hit in L2 cache, a block is transferred from L2 cache to L1 cache. When the page number is in associative memory. If h is the hit ratio, then we can also define (1 - h ) as the miss ratio . Compute the hit ratios of the L1 and L2 caches and the overall effective access time 2. Each access is either a hit or a miss, so average memory access time (AMAT) is: AMAT = time spent in hits + time spent in misses = hit rate * hit time + miss rate * miss time For example, if a hit takes 0.5ns and happens 90% of the time, and a miss takes 10ns and happens 10% of the time, on average you spend 0.4ns in hits and 1.0ns in misses . Use direct-mapped cache. The hit rate and miss rate can measure reads, writes, or both, which means that the terms can be used to describe performance information in several ways. check_circle. (ii) Given, t c = 50ns tm = 400 ns h= 0.95 Memory access time = ht c + (1- h) (t c +t m = 0.95 x 50 + ( 1-0.95) (50+400) = 47.5 + 22.5 = 70ns ) ( i ) Main Memory = 1 M words . Assuming fetches to main memory are started in parallel with look-ups in cache, calculate the effective (average) access time of this system. Block size = 8 words. Find the time to access a page a. CSE 471 Autumn 01 2 Improving Cache Performance • To improve cache performance: Calculate the Effective Access Time (EAT) by assuming the Hit ratio (?) b. Repeat the calculations assuming insertion of two wait states of one cycle each per memory cycle. The miss rate is similar in form: the total cache misses divided by the total number of memory requests expressed as a percentage over a time interval. If 80% of the processor's memory requests result in a cache "hit", what is the average memory access time? Notice that the on-chip access time is 1 ns. What conclusion can you draw from the results? Calculate the effect on CPI rather than the average memory access time. average time) over a large number of hits/misses will be 0.8 * (hit time) + 0.2 * (miss time). If a memory system consists of a single external cache with an access time of 20 ns and a hit rate of 0.92, and a main memory with an access time of 60 ns, what is the effective memory access time of this system? If the cache holds a copy of the memory location, the slow access to main memory can be skipped. • All . 95%. Also suppose a TLB hit requires 7ns, the cache miss rate is 3%, the TLB hit rate is 95%, a cache hit requires 15 ns. We do now want our effective access time to increase much beyond 1 ns. What is the effective access time(in ns) if the TLB hit ratio is 90% and there is no page-fault? The fraction or percentage of accesses that result in a miss is called the miss rate. However, the effective memory access time will be different because every processor must now handle cache invalidation in addition to reads and writes. Assume eight0word cache blocks and a set size of 256 words with set-associative mapping: (a) show the mapping between M1 and M2. CalculatingAverage Access Time Let's say that we have two levels of cache, backed by DRAM: - L1 cache costs 1 cycle to access and has miss rate of 10% - L2 cache costs 10 cycles to access and has miss rate of 2% - DRAM costs 80 cycles to access (and has miss rate of 0%) Then the average memory access time (AMAT) would be: 1 + always access L1 . If the hit ratio to a TLB is 80%, and it takes 15 nanoseconds to search the TLB, and 150 nanoseconds to access the main memory, then what is the effective memory access time in nanoseconds? The access time of cache memory is 100 ns and that of the main memory is 1 μsec. Calculate unknown specifications based on these conditions 22 The block size in L1 cache is 4 words. = 2 20 words . A write of the procedure is used. Substituting values in the above formula, we get- Effective access time with page fault = 0.0001 x { 1 μsec + 10 msec } + 0.99999 x 1 μsec = 0.0001 μsec + 0.001 msec + 0.9999 μsec = 1 μsec + 0.001 msec = 1 μsec + 1 μsec Problem-04: Consider a single level paging scheme with a TLB. If the hit ratio in each cache is 0.9, the memory has an access time of 10 microseconds, and the time required to load a cache block is 5 times the access time of the slower memory, calculate the effective memory access time. To find the effective memory-access time, we weight the case by its probability: effective access time = 0.80 × 100 + 0.20 × 200 = 120 nanoseconds but in the 8th edition of the same book I'm confused with the effective access time HTH, Pierre xixi wrote: the formula for package cache hit ratio is 1 - (package cache inserts/package cache lookups), what the result would be a effective ratio? Note that the miss rate also equals 100 minus the hit rate. One of the main reasons why the cache hit rate for a table is low is that the database doesn't have enough space allocated in its internal cache . As both page table and page are in physical memory T (eff) = hit ratio * (TLB access time + Main memory access time) + (1 - hit ratio) * (TLB access time + 2 * main memory time) = 0.6* (10+80) + (1-0.6)* (10+2*80) = 0.6 * (90) + 0.4 * (170) = 122 This solution is contributed Nitika Bansal Quiz of this Question Next 185 B. (A) 54 (B) 60 (C) 65 (D) 75 Answer: (C) Explanation: Effective access time = hit ratio * time during hit + miss ratio * time during miss. Assume a memory access to main memory on a cache "miss" takes 30 ns and a memory access to the cache on a cache "hit" takes 3 ns. H. C = cache hit ratio . TLB time = 10ns, Memory time = 50ns Hit Ratio= 90% Nov. 2014 Computer Architecture, Memory System Design Slide 24 Cache, Hit/Miss Rate, and Effective Access Time One level of cache with hit rateh C eff = hC fast + (1 - h)(C slow + C fast) = C fast + (1 - h)C slow CPU Cache memory memory If most of the accesses are satisfied by the cache (a cache hit), then the cycles per instructions can be significantly reduced. Question 9. n eff i i i T f t = = ∑ where n is th n -memory hierarchy. (There is no virtual memory on this system -- no page table, no TLB). Cache memory example: L1 cache access in 10 ns; main memory access in 100 ns; 90% hit rate For every 10 memory accesses, 9 will be to cache and 1 to main memory Time without cache = 10 * 100ns = 1000 ns Time with cache = 9 * 10ns + 1 * 100ns = 190 ns Speedup = Speedup = 1000 190 »5.3 Time witho ut Enhancement Time with Enhancement Example . Cache Access Time (ε) is 20 microsecond and Memory Access Time (Τ) is 100 microsecond. Effective Memory Access Time • TLB Hit Rate (H): % time mapping found in TLB • Effective Access Time: [ (H)(TLB access time + mem access time) + (1-H)(TLB access + PT access + mem access)] • Example: mem access: 100ns, TLB: 20ns, hit rate: 80% Effective Access Time = (.8)(120) + (.2)(220) = 140 ns 40% slowdown in mem.

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