Suppose a byte-addressable memory with 4 frames of size 8 bytes each and a paged virtual memory using a three-entry TLB. Suppose a process P has 8 pages of virtual memory space. Assume the following TLB and page table for process P: TUB Page Framell 7. Frame Valid 1. 4. Page Table
Q: A paged memory is shared by two processes, p1 and p2. 16 24 15 1 3 -2 9 -3 17 -12 25 -3 18 -16 19 10…
A: Answer
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: Consider the main memory, which has a capacity of three frames. Assume that the process pages are…
A: Introduction Consider the main memory, which has a capacity of three frames. Assume that the process…
Q: Suppose a byte-addressable memory with 4 frames of size 32 bytes each and a paged virtual memory…
A: I will explain it in details,
Q: Given a 32-bit virtual address space featuring a 10-10-12 split and a 4-byte PTE size, suppose a…
A: The Answer is
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: 1. Suppose a process page table contains the entries shown below. Using the format shown in Figure…
A: Over here we have asked how virtual memory mapped to physical memory.
Q: A direct mapping cache memory of 46 line, main memory consists of 4K k of 128word . Show the format…
A: A) Total number of cache block = 46 So block offset bits = 6 bits Block size = 128 words =…
Q: For a machine with 4 GB virtual memory, 1 GB physical memory, 8 KB page size, 64 KB direct-mapping…
A:
Q: . Suppose we have 2 bytes of virtual memory and 28 bytes of physical main memory. Suppose the page…
A: Given: Virtual memory size=210 Physical memory size=28 Size of the page=24.
Q: Consider a system with 256Mbytes of main memory with page size of 4Kbytes. It has a logical address…
A: Answer: Given Main Memory 256Mbytes Page Size =4Kbytes and logical address=26 bits
Q: For an old computing system with 2K bytes physical memory, and the virtual address has 13 bits.…
A:
Q: How many bits are in a physical address?
A: There are 4 pages frames of size 64 each, or 22 x 26=28. So each physical address has 8 bits.
Q: Assume a process containing 5 pages with 1024 bytes per page and physical memory with 10 page…
A:
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: (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 byte-addressable memory with 4 frames of size 16 bytes each and a paged virtual memory…
A: In this, we are asked about the TLB hit and physical address corresponding to the given virtual…
Q: Consider a computer with byte-addressable main memory of size 8 GB and page size of 1 KB. Assume a…
A: Byte addressable memory upholds getting to information in units that are smaller than the transport.…
Q: Suppose the page table for a process A currently executing on the processor looks like the…
A: This is a multipart question, we are only allowed to solve 3 parts at a time, I am solving c, d, e…
Q: We have a paged virtual address system with the maximum size of virtual address space of 32 MB. The…
A: As per answering guidelines solving the first 3 sub question completely. A) Number of bits for…
Q: Suppose that an MMU has page table entries like the ones shown at right (formatted as in the slides…
A: The solution of the above question is:
Q: Consider the following Page tables: Process A : Page# Frame 0 3 1 4 2 8 3 10 Process B: Page…
A: Allocation of a given process (pages) in the main memory having 12 frames where A-process A,…
Q: b) Consider the following Page tables and show the process allocation in main memory having 19 frame…
A: Given:
Q: Consider a virtual memory system with a 50-bit logical address and a 38-bit physical address.…
A: A virtual memory system with a 50-bit logical address and a 38-bit physical address.Suppose that the…
Q: What is the primary advantage of using a two-level page table (instead of a one-level page table) in…
A: Two level paging means applying paging to page table. The 1st level page table includes information…
Q: Given a process with address space of size 32 bytes and page size of 8 bytes, if the CPU asks to…
A:
Q: Given a process with address space of size 32 bytes and page size of 8 bytes, if the CPU asks to…
A: As we know Physical address = page size*frame number + offset
Q: The PowerPC uses a hardware managed TLB with an inverted page table. Discuss its advantages and…
A: Given: The PowerPC uses a hardware managed TLB with an inverted page table. Discuss its advantages…
Q: Consider a rtual memory system with the following properties! • 42-bit virtual byte address • 32 KB…
A: Given Data : Virtual address bits = 42 Page size = 32KB Physical address bits = 36 Valid bit = 1…
Q: A machine has a memory of 64 frames, with each frame being 1K bytes. Current free-frame list is:…
A:
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: Consider a system with 4-byte pages. A process has the following entries in its page table: logical…
A: Answer (a): An address of 32 corresponds to the byte that has the logical address of two. This is…
Q: Consider a system consisting of thirty-two bits virtual address, page size is 16 KB and a 512 lines…
A: The virtual address is of 32 bits. Whenever CPU wants any data from the memory, it generates 32bit…
Q: Consider a system with 16-bit virtual addresses, 256 byte pages, and 4 byte page table entries. The…
A: PTE's per process = 216/28=28 PTE's per page = 28/22 = 26.
Q: Given a process with address space of size 32 bytes and page size of 8 bytes, if the CPU asks to…
A: Actually. OS is a system software which manages computer hardware and software.
Q: Consider a system with 256Mbytes of main memory with page size of 4Kbytes. It has a logical address…
A: Below is the answer to above questions.
Q: a. Design a two-level page table (Suppose we need to fit each page table into a physical frame). b.…
A:
Q: Suppose the main memory has 32 frames and memory size is 512 bytes. Now a process P of 192 bytes…
A: In this question, we are given main memory (physical address space) size with number of frames and…
Q: A system implements a paged virtual address space for each process using a one-level page table. The…
A: Note: - As per the guidelines we can only answer a maximum of three subparts. Please resubmit the…
Q: size is 1024 bytes and the maximum physical memory size of the machine is 2 megabytes. Assuming two…
A: 3. Number of bits in virtual address = log(virtual address size) = log(maximum size of process) =…
Q: 3
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Q: Suppose, a primary memory size is 56 bytes and frame size is 4 bytes. For a process with 20 logical…
A: Page size = Frame size = 4 bytes. So page offset bits = 2 bits. The least significant 2 bits will…
Q: Assume a memory subsystem with a cache of 8 blocks and a main memory of 64 blocks, where each block…
A: Given: Goal: We have to solve the above parts/questions given.
Q: 25. A system implements a paged virtual address space for each process using a one-level page table.…
A: Given Data : Page size = 1024 bytes Physical address space = 2MB Virtual address space = 16MB
Q: You are given the following data about a virtual memory system: (a) The TLB can hold 512 entries and…
A: The chance of a hit is 0.99 for the TLB and the access time for TLB is 1 nsec. So, the term will be…
Q: Consider a process with 4GB logical memory, 4KB page size, and a page table whose entry is 4B each.…
A: We are given logical memory size, page size and size of each page table entry. We need to find which…
Q: Consider a paged virtual memory system in which a process is allocated with 3 frames. Assume the…
A: Algorithms for replacing pages in virtual memory are critical components of virtual memory…
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- Suppose a computer using direct-mapped cache has 232 (that's 232)232) bytes of byte-addressable main memory, and a cache size of 512 bytes, and each cache block contains 64 bytes. How many blocks of main memory are there? 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? To which cache block will the memory address 0x13A4498A map?Suppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a cache of 1024 blocks, where each cache block contains 32 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields? c) To which cache block will the memory address 0x000063FA map?Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4
- Main memory size is 16 pages. Size of swap file on hard disk is 64 pages. Operating system locks down pages 0, 1, 2 and 7 in the main memory. Two programs are started and both are provided with a virtual address space of 10 pages. Initially operating system allocates 5 pages of main memory for each program and the rest from swap file. Virtual address space of each program starts from page 0. Construct page tables for both programs at program startup. A page table consists of P-bit which tells if the page is present in the main memory (1 = page is in main memory) and address of the page. Address tells either real address in main memory or the address of the page in swap file. b. A computer with a paged virtual memory system executes two load instructions that are adjacent to each other in memory. Describe the worst case scenario (in terms of performance) that can occur during the execution of the two instructions and explain under which conditions this case occurs. How does this affect…In a main memory-disk virtual storage system, the page size is 1KByte and the FIFO algorithm is used for page replacements. A given program has been allocated three page frames in the main memory and it makes the following 16 memory references when it starts executing (the addresses are given in decimal):500, 2000, 2500, 800, 4000, 1000, 5500, 1500, 2800, 400, 5000, 700, 2100, 3500, 900, 2400 Fill in the contents of the three page frames after each memory reference in a table and calculate the hit ratio. Hint: denote by 'a' the page consisting of locations 0 through 1023 in memory. Similarly, b: 1024-2047, c: 2048-3071, d: 3072-4095, e: 4096-5119 and f: 5120-6143. Round to three decimal places.Giventhe following assignment of some program’s virtual pages to physical pages in a system with 4 KiB byte pages, what physical memory address corresponds to virtual address 20000? (All values are given in decimal.)
- Consider a mini virtual memory system. It has a virtual address space of 2°. Each page is 2 bytes. There are 21 = 16 frames in the main memory.Initially, all the pages are on the hard disk. Assume the first two logical addresses generated by the CPU are 15 and 200. What are their physical addresses after their belonged pages are brought into the memory?Assume that each page is stored at the next available frames in the memory. no hand written8.14. In the S/370 architecture, a storage key is a control field associated with each page- sized frame of real memory. Two bits of that key that are relevant for page replace- ment are the reference bit and the change bit. The reference bit is set to 1 when any address within the frame is accessed for read or write, and is set to 0 when a new page is loaded into the frame. The change bit is set to 1 when a write operation is per- formed on any location within the frame. Suggest an approach for determining which page frames are least-recently-used, making use of only the reference bit.Consider a computer with byte-addressable main memory of size 8 GB and page size of 1 KB. Assume a process P has a virtual address space from 0 to 3653 (decimal). Determine the width of the virtual address , the width of the page # field and the width of the offset field (Note: enter only the numbers e.g. 8 for 8 bits etc.)
- The following data segment starts at memory address 1000h (hexadecimal) .data printString BYTE "ASSEMBLY IS FUN",0 moreBytes BYTE 25(DUP)0 dateIssued DWORD ? dueDate DWORD ? elapsedTime Word ? What is the hexadecimal address of dueDate ? a. 1045h b. 1029h c.1010h d. 102DhA disk has a capacity of two tera-byte size. Say the file system uses a multi-level inode structure for locating the data blocks of afile.The inode stores pointers to data blocks, including a single indirect block, a double indirect block, and several direct blocks in the 64 B of available space.Now, it is given that the disk has a block size of 512 B. The maximum file size that can be stored in such a file system in MB (round off upto 2 decimal places).Suppose a computer using fully associative cache has 4 GB of byte-addressable main memory and a cache of 256 blocks, where each block contains 256 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? c) To which cache block will the memory address 0X1A1B1C1D map?