It is given that on a system, on average, an instruction executes in 1 nano second. Suppose that it takes 20 micro seconds of processor time for page fault. While, it takes 300 micro seconds of disk time for reading and writing a single page. Suppose that on average 1/3 of the pages that need to be paged out are modified. What is the average number of instructions between page faults that would cause the disk to be busy doing page transfers all the time?

Q: Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If…

A: Question. Let the page fault service time be 10ms in a computer with average memory access time…

Q: Consider a system with 20 requests and out of 20 , 4 requests are already in cache memory. What will…

A: Introduction  Consider a system with 20 requests and out of 20 , 4 requests are already in cache…

Q: Consider a paging system with the page table stored in memory. a. If a memory reference takes 50…

A: The Answer is in step2

Q: Given a computer using a byte-addressable virtual memory system with a two-entry TLB, a 2-way set…

A:

Q: Imagine a virtual memory system with page size equal 256 bytes. Only the first 5 entries in the…

A: Here page size given is 256 bytes.  This means number of page offset bits is log 256 = 8 bits.  Thus…

Q: Consider a computer system that uses virtual memory with paging (no TLB). Suppose main access time…

A: Discontinuum Buffer for Localization: The CPU creates physical memory location in memory systems.…

Q: Consider a system using a 2-level page table tree for address translation, with 16 KiB pages, 8 byte…

A: The virtual memory address space is defined as space or the place that is used to store by the which…

Q: Consider a computer system that uses virtual memory with paging (no TLB). Suppose main access time…

A: Paging technique means dividing virtual memory into equal size partitions called pages and storing…

Q: Consider a system where the page fault service time is 200ms, and the main memory access time is…

A: To find Effective memory access time.

Q: b. Suppose we use two-level paging and arrange for all page tables to fit into a single page frame.…

A:

Q: Consider a system with 39-bit virtual addresses, 44-bit physical addresses, 4096 byte pages, and the…

A: The answer is

Q: Consider a computer system with a 24-bit logical address and a 28-bit physical address. Let's…

A: Question from Paging topic. We are given logical address, physical address, page size and page table…

Q: For an old computing system with 2K bytes physical memory, and the virtual address has 13 bits.…

A:

Q: a simple paging system with 224 bytes of physical memory, 256 pages of logical address space, and a…

A: Given : Physical memory = 224 bytes Logical address space = 256 pages Page size = 210 bytes

Q: Consider a computer with 128 MB of main memory, 64 KB of cache, and 4 bytes per memory block. Find…

A: Block size = 4B So block offset bits = log 4 = 2 bits.  Total number of cache block = 64/4 = 16…

Q: Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory…

A: The answer is as follows.

Q: Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses. (a) What is the…

A: Given  Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses.(a) What is…

Q: Unlike a byte that is always 8 bits, word size can vary from machine to machine. Let's say we have a…

A: since we have 4 address lines so no of unique addresses=2^4=16 so we can store 16 unique words

Q: ry split into pages. If each byte in the virtual memory has a virtual address, and the last 8 bits…

A: Pages are typically 512 to 8192 bytes in size, with 4096 being the most common value. For reasons…

Q: Consider a memory system with a 6-bit address space with a direct-mapped cache with two set bits and…

A: The memory address space is 6 bit and the cache is direct-mapped. We have to determine the offset…

Q: Consider the following scenario: we have a byte-addressable computer with fully associative mapping,…

A: Mapping in Associative Mode: In associative mapping, the associated memory is used to store…

Q: Consider a system with N bytes of physical RAM, and M bytes of virtual address and frames are K…

A: Here the virtual memory size = M bytes And page size is K bytes.  The virtual address is divided…

Q: Let's say in a system, it takes 8 milliseconds to service a page fault if an empty page is available…

A: Introduction :Given , Page fault service time = 8 ms service time if page is modified = 20 ms memory…

Q: Consider a system with a 7-bit virtual address space,32-byte pages and 512-byte physical memory. The…

A: In this question, we are given virtual address and physical address with page size. Virtual address…

Q: (a) Explain the use of TLBs to improve paging efficiency.  (b) Consider a paging system with the…

A: A). To overcome this problem a high-speed cache is set up for page table entries called a…

Q: Suppose a computer using set associative cache has 216 words of main memory and a cache of 128…

A: Given Data:   Size of Main Memory = 216 words Size of Cache = 128 block The Cache is 4-way Set…

Q: Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns.…

A: Correct answer is 30 ns

Q: A process contains 8 virtual pages on the disk, and is assigned a fixed allocation of 4 page frames…

A:

Q: Given a 32-bit virtual address space featuring a 10-10-12 split and a 4-byte PTE size, suppose a…

A: The memory is split as  10 - 10 - 12 This means page size is 4KB. Now process size is 9MB Therefore…

Q: (b) Consider a paging system with the page table stored in memory. If a memory reference takes 200…

A: Ans:-  a) If memory access takes 200 nanoseconds, how long does a paged memory reference take?200…

Q: Consider a logical address space of 512 pages with a 4-KB page size, mapped onto a physical memory…

A:

Q: What is the effective memory access time if the memory access time is 450 ns and a page fault takes…

A: Effective access time The 'effective access time' is essentially the (weighted) average time it…

Q: Consider a machine where total addressable physical memory size is 32 Kbytes. If we would like to…

A: total addressable physical memory size=32 Kbytes. we like to run 2 processes P and R as 8KB using…

Q: Consider a computer system with 32-bit virtual addressing and a page size of sixty-four kilobytes.…

A: In this question, we are given virtual address and physical memory size and also page size. Page…

Q: Suppose on-demand paging is employed in addition to TLB caching. The time for a TLB hit is T= 1 ns,…

A: Probability of TLB miss = 1 - probability of TLB Hit = 1- 0.9 = 0.1   Probability of No page fault =…

Q: Consider a system with 36-bit virtual addresses, 32-bit physical addresses, and 4KB pages. The…

A: Given,The virtual Address space = 236 bytesPage size = 212 bytesPages = 236 / 212 = 224  Pages

Q: Suppose a computer using direct mapped cache has 232 words of main memory, and a cache of 1024…

A: Given: Suppose a computer using a direct-mapped cache has 232 words of main memory and a cache of…

Q: a memory reference takes 10 ns (nanoseconds), how long does a paged m mke? we add TLBS, and if 90…

A: A) Paged memory reference takes 20 ns Total 20 nanoseconds 10 nanoseconds to access the page table…

Q: Suppose a computer using fully associative cache has 224 words of main memory and a cache of 512…

A: As per the policy and  guidelines of Bartleby we are supposed to answer only first question or the…

Q: Consider a paging system with the page table stored in memory. a. If a memory reference takes 400…

A: Since, number of memory accesses required by a paged memory reference = 2   Time taken by a memory…

Q: Suppose a computer using 8-way set associative cache has 1 M words of main memory, and a cache of 16…

A: Given Data :  8 way set associative Number of words in MM = 1M Number of words in cache = 16K Block…

Q: Suppose we have a byte-addressable computer with a cache that holds 8 blocks of 16 bytes each.…

A: Below is the answer to above question. I hope this will be helpful for you...

Q: Suppose a computer using fully associative cache has 220220 words of main memory and a cache of 128…

A: solution:

Q: Given a virtual memory of size 4 GiB, physical memory of size 1 GiB, and page size equal to 256 KiB.…

A: Here in this question we have given Virtual memory size = 4GiB Physical memory size = 1GiB Page size…

Q: Given a computer using a byte-addressable virtual memory system with a two-entry TLB, a 2-way set…

A: Below is the answer with explanation and calculation:

Q: Consider a processor that uses a 48-bit virtual memory address. The main memory installed on the…

A: virtual memory address=48bits Main memory=4 GB Page Size=32KB Determine the following:I. Address…

Q: Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up…

A: (1) it is given that: logical memory space = 30 bit = 2^30 bytes   given page size is 4KB = 2^12…

Q: Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up…

A: Actually, 1 byte = 8 bits.

Q: Consider a logical address space of 1024 pages of 2048 words each, mapped onto a physical memory of…

A: Question :-

Fundamentals of Input and Output Performance

The I/O subsystem is a component of computer system design that uses buses to assist communication between the Central Processing Unit, Computer Memory/Storage, and Input and Output Devices (that is known as peripheral devices). The I/O subsystem includes disc controllers, tape drives, discs, system buses, and other devices. The I/O database application reads and writes drives using the operating system windows I/O.