calculate effective memory access time = cache hit ratio calculate effective memory access time = cache hit ratio

A cache memory that has a hit rate of 0.8 has an access latency 10 ns and miss penalty 100 ns. The TLB hit ratio is 90% and the page fault rate is one in every 10,000 instructions. first access memory for the page table and frame number (100 Assume no page fault occurs. It takes some computing resources, so it should actually count toward memory access a bit, but much less since the page faults don't need to wait for the writes to finish. when CPU needs instruction or data, it searches L1 cache first . Page fault handling routine is executed on theoccurrence of page fault. If Cache If a law is new but its interpretation is vague, can the courts directly ask the drafters the intent and official interpretation of their law? How to calculate average memory access time.. Connect and share knowledge within a single location that is structured and easy to search. 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. Where TLB hit ratio is same single level paging because here no need access any page table, we get page number directly from TLB. It takes 20 ns to search the TLB and 100 ns to access the physical memory. Example 5:Here calculating memory access time, where EMAT, TLB access time, and the hit ratio is given. This splits to two options: 50% the page to be dropped is clean, so the system just needs to read the new content: 50% the page to be dropped is dirty, so the system needs to write it to disk, Disk access time needed to read & bring in memory (from swapping area or pagefile) the PT itself, MEM time needed to access PT now in memory. Solution: Memory cost is calculated by; Ctotal= C1S1+C2S2+C3S3 G 15000, then S3=39.8 The effective memory access time is calculated as Full Course of Computer Organization \u0026 Architecture: https://youtube.com/playlist?list=PLV8vIYTIdSnar4uzz-4TIlgyFJ2m18NE3In this video you can learn about Cache Hit Time, Hit Ratio and Average Memory Access Time in Computer Organization \u0026 Architecture(COA) Course. The difference between lower level access time and cache access time is called the miss penalty. For example, if you have 51 cache hits and three misses over a period of time, then that would mean you would divide 51 by 54. The cache access time is 70 ns, and the Hit ratio: r = N hit N hit + N miss Cache look up cost: C cache = rC h + (1 r) Cm Cache always improves performance when Cm > C h and r > 0. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. @Apass.Jack: I have added some references. 1. So one memory access plus one particular page acces, nothing but another memory access. Use MathJax to format equations. Does a summoned creature play immediately after being summoned by a ready action? It takes 20 ns to search the TLB and 100 ns to access the physical memory. Why do small African island nations perform better than African continental nations, considering democracy and human development? The effective memory-access time can be derived as followed : The general formula for effective memory-access time is : n Teff = f i .t i where n is nth -memory hierarchy. the TLB is called the hit ratio. Staging Ground Beta 1 Recap, and Reviewers needed for Beta 2, How To Calculate Process Size from TLB size and mean memory access time, Relation between cache and TLB hit ratios. Can you provide a url or reference to the original problem? When a CPU tries to find the value, it first searches for that value in the cache. The TLB is a high speed cache of the page table i.e. Effective Memory Access Time = Cache access time * hit rate + miss rate * Miss penalty The above formula is too simple and given in many texts. However, the optimization results in an increase of cache access latency to 15 ns, whereas the miss penalty is not affected. Does a barbarian benefit from the fast movement ability while wearing medium armor? This is better understood by. What are the -Xms and -Xmx parameters when starting JVM? If the TLB hit ratio is 0.6, the effective memory access time (in milliseconds) is _________. Effective access time = (h x c) + ( (1-h) x ( c + m )) = (0.95 x 5) + ( (0.05) x (5 + 40)) nanoseconds = 4.75 + 2.25 nanoseconds = 7 nanoseconds Next Previous Related Questions Q: Assume that a given system's main memory has an access time of 6.0 ns, and its cache has an access.. Answer: To calculate: Hit ratio for effective access time of 1.5 ns. Here hit ratio (h) =70% means we are taking0.7, memory access time (m) =70ns, TLB access time (t) =20ns and page level (k) =3, So, Effective memory Access Time (EMAT) =153 ns. What is actually happening in the physically world should be (roughly) clear to you. Get more notes and other study material of Operating System. Try, Buy, Sell Red Hat Hybrid Cloud 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) To subscribe to this RSS feed, copy and paste this URL into your RSS reader. In this case the first formula you mentioned is applicable as access of L2 starts only after L1 misses. We can write EMAT formula in another way: Let, miss ratio = h, hit ration = (1 - h), memory access time = m and TLB access time = t. So, we can write Note: We can also use this formula to calculate EMAT but keep in your mind that here h is miss ratio. Calculating effective address translation time. Has 90% of ice around Antarctica disappeared in less than a decade? Also, TLB access time is much less as compared to the memory access time. When a system is first turned ON or restarted? can you suggest me for a resource for further reading? If TLB hit ratio is 80%, the effective memory access time is _______ msec. page-table lookup takes only one memory access, but it can take more, A 3 level paging scheme uses a Translation Look-aside Buffer (TLB). You are not explicit about it, but I would assume the later if the formula didn't include that 0.2*0.9, which suggests the former. An instruction is stored at location 300 with its address field at location 301. Edit GOLD PRICE CLOSED: DOWN $4.00 at $1834.40 SILVER PRICE CLOSED: DOWN $0.16 to $20.83 Access prices: closes : 4: 15 PM Gold ACCESS CLOSE 1836.30 Silver ACCESS CLOSE: 20.91 Bitcoin morning price:, 23,363 DOWN 63 Dollars Bitcoin: afternoon price: $23,478 UP 52 dollars Platinum price closing $962.00 UP Then with the miss rate of L1, we access lower levels and that is repeated recursively. average time) over a large number of hits/misses will be 0.8 * (hit time) + 0.2 * (miss time). So, here we access memory two times. Calculation of the average memory access time based on the following data? 27 Consider a cache (M1) and memory (M2) hierarchy with the following characteristics:M1 : 16 K words, 50 ns access time M2 : 1 M words, 400 ns access time Assume 8 words cache blocks and a set size of 256 words with set associative mapping. Consider a single level paging scheme with a TLB. Daisy wheel printer is what type a printer? Arwin - 23206008@2006 1 Problem 5.8 - The main memory of a computer is organized as 64 blocks with a block size of eight (8) words. Number of memory access with Demand Paging. has 4 slots and memory has 90 blocks of 16 addresses each (Use as The cache hit ratio is the number of requests that are found in the cache divided by the total number of requests. But, the data is stored in actual physical memory i.e. 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 Can someone explain it for me? How to react to a students panic attack in an oral exam? 80% of time the physical address is in the TLB cache. Regarding page directory (the first level of paging hierarchy) I believe it has to be always resident in RAM (otherwise, upon context switch, the x86 CR3 register content would be totally useless). Using Verilog, designed a 16-block direct-mapped, write-back cache with 2 words/line, that supports same cycle read/write hit. Atotalof 327 vacancies were released. In a multilevel paging scheme using TLB without any possibility of page fault, effective access time is given by-, In a multilevel paging scheme using TLB with a possibility of page fault, effective access time is given by-. ____ number of lines are required to select __________ memory locations. The best answers are voted up and rise to the top, Not the answer you're looking for? Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. Why are non-Western countries siding with China in the UN? Before you go through this article, make sure that you have gone through the previous article on Page Fault in OS. Senior Systems Engineer with a unique combination of skills honed over more than 20 years and cross-functional and holistic IT Core Infrastructure, Virtualization, Network, Cloud, Hybrid, DC . Experts are tested by Chegg as specialists in their subject area. Virtual Memory The effective time here is just the average time using the relative probabilities of a hit or a miss. Why do many companies reject expired SSL certificates as bugs in bug bounties? hit time is 10 cycles. I would like to know if, In other words, the first formula which is. Ratio and effective access time of instruction processing. Translation Lookaside Buffer (TLB) tries to reduce the effective access time. It can easily be converted into clock cycles for a particular CPU. * It is the first mem memory that is accessed by cpu. Find centralized, trusted content and collaborate around the technologies you use most. 2003-2023 Chegg Inc. All rights reserved. 1 Memory access time = 900 microsec. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Calculation of the average memory access time based on the following data? Integrated circuit RAM chips are available in both static and dynamic modes. An average instruction takes 100 nanoseconds of CPU time and two memory accesses. Part A [1 point] Explain why the larger cache has higher hit rate. So, how many times it requires to access the main memory for the page table depends on how many page tables we used. To learn more, see our tips on writing great answers. For the sake of discussion again, if we assume that t2 and t3 mean the time to access L2 and main memory directly assuming there is no caches at all, respectively, then we should claim there is not enough information to compute a reasonable answer. Does a barbarian benefit from the fast movement ability while wearing medium armor? we have to access one main memory reference. It takes 20 ns to search the TLB and 100 ns to access the physical memory. - Memory-intensive applications that allocate a large amount of memory without much thought for freeing the memory at run time can cause excessive memory usage. Is it plausible for constructed languages to be used to affect thought and control or mold people towards desired outcomes? A page fault occurs when the referenced page is not found in the main memory. The following equation gives an approximation to the traffic to the lower level. Thus it exist a percentage of occurrences we have to include at least: Thanks for contributing an answer to Stack Overflow! Asking for help, clarification, or responding to other answers. The hit ratio for reading only accesses is 0.9. The difference between the phonemes /p/ and /b/ in Japanese. Let us use k-level paging i.e. Assume that load-through is used in this architecture and that the To speed this up, there is hardware support called the TLB. I agree with this one! Is it a bug? However, we could use those formulas to obtain a basic understanding of the situation. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. locations 47 95, and then loops 10 times from 12 31 before And only one memory access is required. An 80-percent hit ratio, for example, (i)Show the mapping between M2 and M1. The 'effective access time' is essentially the (weighted) average time it takes to get a value from memory. Block size = 16 bytes Cache size = 64 (An average family has 2.3 children, but any real family has 0, 1, 2 or 3 children or an integer number of children; you don't see many 'three tenths of a child' wandering around). Connect and share knowledge within a single location that is structured and easy to search. How many 32 K 1 RAM chips are needed to provide a memory capacity of 256 K-bytes ? Why is there a voltage on my HDMI and coaxial cables? reading the question I was thinking about a more realistic scenario based, for instance, on a two-level paging system. nanoseconds), for a total of 200 nanoseconds. If it takes 100 nanoseconds to access memory, then a b) Convert from infix to rev. Effective Access time when multi-level paging is used: In the case of the multi-level paging concept of TLB hit ratio and miss ratio are the same. Main memory access time is 100 cycles to the rst bus width of data; after that, the memory system can deliv er consecutiv e bus widths of data on eac h follo wing cycle. Why are physically impossible and logically impossible concepts considered separate in terms of probability? 2. A cache is a small, fast memory that is used to store frequently accessed data. we need to place a physical memory address on the memory bus to fetch the data from the memory circuitry. What will be the EAT if hit ratio is 70%, time for TLB is 30ns and access to main memory is 90ns? How is Jesus " " (Luke 1:32 NAS28) different from a prophet (, Luke 1:76 NAS28)? Please see the post again. It first looks into TLB. Then the above equation becomes. In TLB a copy of frequently accessed page number and frame no is maintained which is from the page table stored into memory. The expression is actually wrong. much required in question). Outstanding non-consecutiv e memory requests can not o v erlap . Consider a paging system, it takes 10ns to search translation lookaside buffer (TLB) and 80ns to access main memory. That is. Redoing the align environment with a specific formatting. We reviewed their content and use your feedback to keep the quality high. 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If TLB hit ratio is 50% and effective memory access time is 170 ns, main memory access time is ______. Example 2: Here calculating Effective memory Access Time (EMAT) forMulti-level paging system, where TLB hit ratio, TLB access time, and memory access time is given. Substituting values in the above formula, we get-, = 0.0001 x { 1 sec + 10 msec } + 0.99999x 1 sec, If an instruction takes i microseconds and a page fault takes an additional j microseconds, the effective instruction time if on the average a page fault occurs every k instruction is-. To calculate a hit ratio, divide the number of cache hits with the sum of the number of cache hits, and the number of cache misses. When an application needs to access data, it first checks its cache memory to see if the data is already stored there. That would be true for "miss penalty" (miss time - hit time), but miss time is the total time for a miss so you shouldn't be counting the hit time on top of that for misses. time for transferring a main memory block to the cache is 3000 ns. (We are assuming that a In order to calculate the effective access time of a memory sub-system, I see some different approaches, a.k.a formulas. Thanks for the answer. I was solving exercise from William Stallings book on Cache memory chapter. ERROR: CREATE MATERIALIZED VIEW WITH DATA cannot be executed from a function. The idea of cache memory is based on ______. 2. Note: The above formula of EMAT is forsingle-level pagingwith TLB. Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. To learn more, see our tips on writing great answers. Effective memory Access Time (EMAT) for single-level paging with TLB hit ratio: Here hit ratio (h) =80% means here taking0.8, memory access time (m) =80ns and TLB access time (t) =10ns. How many 128 8 RAM chips are needed to provide a memory capacity of 2048 bytes? EAT := (TLB_search_time + 2*memory_access_time) * (1- hit_ratio) + (TLB_search_time + memory_access_time)* hit_ratio. LKML Archive on lore.kernel.org help / color / mirror / Atom feed help / color / mirror / Atom feed * You can see further details here. A tiny bootstrap loader program is situated in -. 200 By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Statement (II): RAM is a volatile memory. Cache Access Time Example Note: Numbers are local hit rates - the ratio of access that go to that cache that hit (remember, higher levels filter accesses to lower levels) . The cache access time is 70 ns, and the time for transferring a main memory block to the cache is 3000 ns. Thus, effective memory access time = 160 ns. Consider the following statements regarding memory: If that is the case, a miss will take 20ns+80ns+80ns=180ns, not 200ns. If one page fault is generated for every 106 memory accesses, what is the effective access time for the memory? All are reasonable, but I don't know how they differ and what is the correct one. Can I tell police to wait and call a lawyer when served with a search warrant? Although that can be considered as an architecture, we know that L1 is the first place for searching data. Whenever Dnode_LC of Dnode where the request initiated is full, the HRFP with the lowest relevancy value is evicted creating space for the HRFP where the requested fb is a member. Average memory access time = (0.1767 * 50) + (0.8233 * 70) = 66.47 sec. 170 ns = 0.5 x{ 20 ns + T ns } + 0.5 x { 20 ns + (1+1) x T ns }, 170 ns = 0.5 x { 20 ns + T ns } + 0.5 x { 20 ns + 2T ns }. Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds and servicing a page fault takes 8 milliseconds. What is a word for the arcane equivalent of a monastery? * [PATCH 1/6] f2fs: specify extent cache for read explicitly @ 2022-12-05 18:54 ` Jaegeuk Kim 0 siblings, 0 replies; 42+ messages in thread From: Jaegeuk Kim @ 2022-12-05 18:54 UTC (permalink / raw) To: linux-kernel, linux-f2fs-devel; +Cc: Jaegeuk Kim Let's descrbie it's read extent cache. b) ROMs, PROMs and EPROMs are nonvolatile memories Paging in OS | Practice Problems | Set-03. For each page table, we have to access one main memory reference. The static RAM is easier to use and has shorter read and write cycles. 2. Base machine with CPI = 1.0 if all references hit the L1, 2 GHz Main memory access delay of 50ns. TLB hit ratio- A TLB hit is the no of times a virtual-to-physical address translation was already found in the TLB, instead of going all the way to the page table which is located in slower physical memory. Directions:Each of the items consist of two statements, one labeled as the Statement (I)'and the other as Statement (II) Examine these two statements carefully and select the answers to these items using the codes given below: TLB hit ratio is nothing but the ratio of TLB hits/Total no of queries into TLB. The cache has eight (8) block frames. 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The hierarchical organisation is most commonly used. Can Martian Regolith be Easily Melted with Microwaves. Which of the following control signals has separate destinations? Is it possible to create a concave light? If it was a 3 level paging system, would TLB_hit_time be equal to: TLB_search_time + 3* memory_access_time and TLB_miss_time be TLB_search_time + 3*(memory_access_time + memory_access_time) and EAT would then be the same? A: Memory Read cycle : 100nsCache Read cycle : 20ns Four continuous reference is done - one reference. 3. cache is initially empty. EMAT for Multi-level paging with TLB hit and miss ratio: Same way we can write EMAT formula for multi-level paging in another way: Let, miss ratio =h, hit ration =(1 - h), memory access time =m, TLB access time = tand page-level = k. Effective memory Access Time (EMAT) for single level paging with TLB hit and miss ratio: EMAT for Multi level paging with TLB hit and miss ratio: To get updated news and information subscribe: 2023 MyCareerwise - All rights reserved, The percentage of times that the required page number is found in the. Following topics of Computer Organization \u0026 Architecture Course are discussed in this lecture: What is Cache Hit, Cache Miss, Cache Hit Time, Cache Miss Time, Hit Ratio and Miss Ratio. has 4 slots and memory has 90 blocks of 16 addresses each (Use as Word size = 1 Byte. The Union Public Service Commission released the UPSC IES Result for Prelims on 3rd March 2023. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Effective access time is a standard effective average. Reducing Memory Access Times with Caches | Red Hat Developer You are here Read developer tutorials and download Red Hat software for cloud application development. Connect and share knowledge within a single location that is structured and easy to search. This formula is valid only when there are no Page Faults. Thus, effective memory access time = 140 ns. If Cache Substituting values in the above formula, we get-, = 0.8 x{ 20 ns + 100 ns } + 0.2 x { 20 ns + (1+1) x 100 ns }. So 90% times access to TLB register plus access to the page table plus access to the page itself: 10% (of those 20%; the expression suggests this, but the question is not clear and suggests rather that it's 10% overall) of times the page needs to be loaded from disk. - Inefficient memory usage and memory leaks put a high stress on the operating virtual memory subsystem. Effective memory access time without page fault, = 0.9 x { 0 + 150 ns } + 0.1 x { 0 + (2+1) x 150 ns }, = 10-4x { 180 ns + 8 msec } + (1 10-4) x 180 ns, Effective Average Instruction Execution Time, = 100 ns + 2 x Effective memory access time with page fault, A demand paging system takes 100 time units to service a page fault and 300 time units to replace a dirty page. So you take the times it takes to access the page in the individual cases and multiply each with it's probability. Provide an equation for T a for a read operation. Before this read chapter please follow the previous chapter first: Calculate Effective Access Time (EMAT). To calculate a hit ratio, divide the number of cache hits with the sum of the number of cache hits, and the number of cache misses. The picture of memory access by CPU is much more complicated than what is embodied in those two formulas. Question Consider an OS using one level of paging with TLB registers. It takes 100 ns to access the physical memory. It looks like the solution depends on the definition of "the time to access the L1" and "the penalty to access L2 and main memory". EMAT for Multi-level paging with TLB hit and miss ratio: However, that is is reasonable when we say that L1 is accessed sometimes. The candidates appliedbetween 14th September 2022 to 4th October 2022. Asking for help, clarification, or responding to other answers. We have introduced a relevancy-based replacement policy for patterns that increases the hit ratio and at the same time decrease the read access time of the DFS. In this scenario, as far as I can understand, there could be the case page table (PT) itself is not resident in memory (PT itself may have been paged out from RAM into swapping area (e.g. What Is a Cache Miss? Is there a solutiuon to add special characters from software and how to do it. contains recently accessed virtual to physical translations. In the hierarchical organisation all the levels of memory (cache as well as main memory) are connected sequentially i.e. No single memory access will take 120 ns; each will take either 100 or 200 ns. It follows that hit rate + miss rate = 1.0 (100%). It is a question about how we translate the our understanding using appropriate, generally accepted terminologies. How can I find out which sectors are used by files on NTFS? If the word is not in main memory, 12ms are required to fetch it from disk, followed by 60ns to copy it to the cache, and then the reference is started again. Get more notes and other study material of Operating System. If TLB hit ratio is 80%, the effective memory access time is _______ msec. (I think I didn't get the memory management fully). It only takes a minute to sign up. Making statements based on opinion; back them up with references or personal experience. Miss penalty is defined as the difference between lower level access time and cache access time. But in case ofTLB miss when the page number is not present at TLB, we have to access the page table and if it is a multi-level page table, we require to access multi-level page tables for the page number. Q2. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Watch video lectures by visiting our YouTube channel LearnVidFun. * It's Size ranges from, 2ks to 64KB * It presents . Effective Access Time With Page Fault- It is given that effective memory access time without page fault = 20 ns. 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The expression is somewhat complicated by splitting to cases at several levels. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. 2- As discussed here, we can calculate that using Teff = h1*t1 + (1-h1)*h2*t2 + (1-h1)*(1-h2)*t3 which yields 24. In this context "effective" time means "expected" or "average" time. 1- Teff = t1 + (1-h1)[t2 + (1-h2)t3] which will be 32.