Segment table is stored in MMU. 5. If the number of bits required for page number...
Segment table is stored in MMU. 5. If the number of bits required for page number is 9 and the size of the page table entry is 4 bytes, then the page table size is 2KB. 6. In paging, external fragmentation reduces with increase in page size. 7. Multi-level page tables reduces the memory access time when compared to linear page table. 8. The number of levels in a multi-level page table increase with page size. 9. It is easier to allocate pages for page table in multi-level page table than in linear page table.
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Segment table is stored in MMU. 5. If the number of bits
required for page number...
A logical address is divided into 3 bits segment number, 4 bits page number, and 9...
A logical address is divided into 3 bits segment number, 4 bits page number, and 9 bits page offset. {(s,p,w) = (3,4,9)} Answer the following questions in relation to the address space. 1) The maximum number of segments per process is given by 2) The maximum number of pages per segment is given by 3) The maximum page size is given by 4) The maximum segment size is given by 5) The maximum size of the address space is given...
Number Name 3. Assuming no page fault on a page table access, what is the processor...
Number Name 3. Assuming no page fault on a page table access, what is the processor memory access time for the system depicted in the above figure, for a physical memory with 50ns read/write times? 4. Now, assume that the memory system has a translation look-aside buffer (TLB). The TLB requires 10 ns to determine a hit or mess. The physical memory system has an access time of 50ns. You may assume that page fault rate for the application is...
1. Consider a simple paging system with the following parameters: 232 bytes of physical memory; page...
1. Consider a simple paging system with the following parameters: 232 bytes of physical memory; page size of 210 bytes; 216 pages of logical address space. How many bits are in a logical address? How many bytes are in a frame! How many bits in the physical address specify the frame? How many entries are in the page table? How many bits are in each page table entry? Assume each page table entry contains a valid/invalid bit. 2. Consider a...
6) Paging [26 pts] Suppose you have a computer system with a 38-bit logical address, page size of 16K, and 4 bytes per...
6) Paging [26 pts] Suppose you have a computer system with a 38-bit logical address, page size of 16K, and 4 bytes per page table entry a) How many pages are there in the logical address space? Suppose we use two level paging and each page table can fit completely in a frame. [4 pts] How many pages? [2 pts] Show your calculations here: b) For the above-mentioned system, give the breakup of logical address bits clearly indicating number of...
I need help with this problem.I am currenlty struggelling with it. Consider a computer system using paging, where the address space of every process has a size of C = 2c bytes and the page size is S...
I need help with this problem.I am currenlty struggelling
with it.
Consider a computer system using paging, where the address space
of every process has a size of C = 2c bytes and the page size is S
= 2s bytes. Each entry in the page table uses E bytes.
Calculate the number of pages of a process, and the size of a
page table (in bytes).
Assume that the space wasted by a process in main memory is
defined...
Paging Questions 1. A page is 1 KB in size. How many bits are required to...
Paging Questions 1. A page is 1 KB in size. How many bits are required to store the page offset? 2. A page entry has 10 bits. What is the size of the page table? 3. A logical address is 32 bits long. The page size is 4 KB. Divide the address into its page number and offset. 4. The following hexadecimal addresses are used in a system with a 20-bit logical address where the page size is 256 bytes....
Please answer the following questions about paged memory... A) How much space needs to be allocated...
Please answer the following questions about paged memory... A) How much space needs to be allocated in the minimum and maximum cases for a two-level page table for a machine with a 32-bit virtual memory address, a 1K page size, and which has four times as many inner pages as outer pages? Assume any stored page table value requires 32 bits. B) For the two-level paging approach above, if a Translation Lookaside Buffer (TLB) is used and can cache both...
D | Question 1 7.14 pts 57 Minutes. 29 Secon What causes the VM subsystem to...
D | Question 1 7.14 pts 57 Minutes. 29 Secon What causes the VM subsystem to handle a page fault? 0 the M?U translates wrtual address and determines the page numbers beyond the page table length register's value the process attempts to exeoute this code 100. O the MMU translates a virtual address and whenit reads the page entry from the pare tables finds the vald/invalid bit to be imalid and thus trisgers a pae fault exception O the process...
36.-Assume a 32 bit memory space with 16[KB] pages/frames and a two-level page table. Answer the...
36.-Assume a 32 bit memory space with 16[KB] pages/frames and a two-level page table. Answer the following questions: a).-what is the size of the page table in bytes needed for a process that has only 64[MB] of code at the start of its virtual address range? b).-how many bits are used for address (frame) offset, and how many for page table indexing (total). Assume each page table entry has 32 bits. a).-Size of page table b).-Offset bits b).-Page table indexing...
A system implements a paged virtual address space for each process using a one-level page table....
A system implements a paged virtual address space for each process using a one-level page table. The maximum size of virtual address space is 8MB. The page table for the running process includes the following valid entries (the -> notation indicates that a virtual page maps to the givenpage frame, that is, it is located in that frame): Virtual page 2 -> Page frame4 Virtual page 4 -> Page frame 9 Virtual page 1 -> Page frame2 Virtual page 3...
A logical address is divided into 3 bits segment number, 4 bits page number, and 9 bits page offset. {(s,p,w) = (3,4,9)} Answer the following questions in relation to the address space. 1) The maximum number of segments per process is given by 2) The maximum number of pages per segment is given by 3) The maximum page size is given by 4) The maximum segment size is given by 5) The maximum size of the address space is given...
Number Name 3. Assuming no page fault on a page table access, what is the processor memory access time for the system depicted in the above figure, for a physical memory with 50ns read/write times? 4. Now, assume that the memory system has a translation look-aside buffer (TLB). The TLB requires 10 ns to determine a hit or mess. The physical memory system has an access time of 50ns. You may assume that page fault rate for the application is...
1. Consider a simple paging system with the following parameters: 232 bytes of physical memory; page size of 210 bytes; 216 pages of logical address space. How many bits are in a logical address? How many bytes are in a frame! How many bits in the physical address specify the frame? How many entries are in the page table? How many bits are in each page table entry? Assume each page table entry contains a valid/invalid bit. 2. Consider a...
6) Paging [26 pts] Suppose you have a computer system with a 38-bit logical address, page size of 16K, and 4 bytes per page table entry a) How many pages are there in the logical address space? Suppose we use two level paging and each page table can fit completely in a frame. [4 pts] How many pages? [2 pts] Show your calculations here: b) For the above-mentioned system, give the breakup of logical address bits clearly indicating number of...
I need help with this problem.I am currenlty struggelling
with it.
Consider a computer system using paging, where the address space
of every process has a size of C = 2c bytes and the page size is S
= 2s bytes. Each entry in the page table uses E bytes.
Calculate the number of pages of a process, and the size of a
page table (in bytes).
Assume that the space wasted by a process in main memory is
defined...
D | Question 1 7.14 pts 57 Minutes. 29 Secon What causes the VM subsystem to handle a page fault? 0 the M?U translates wrtual address and determines the page numbers beyond the page table length register's value the process attempts to exeoute this code 100. O the MMU translates a virtual address and whenit reads the page entry from the pare tables finds the vald/invalid bit to be imalid and thus trisgers a pae fault exception O the process...
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