multi-level paging used by a process that has following: a. Logical Address = 64 bits b. Page size = 1 M Byte c. Page table entry size = 4 Byte d. System is byte addressable. calculate the following. 1. Logical Address Space? 2. Maximum number of page table entries in a single page? 3. How many levels of page table entries will be used?
Q: Consider the following page table in a demand paging system. Assume the page size of 256 bytes.…
A: using demand paging system assume the page size of 256 bytes.
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: Assume we have a program that resides in the memory range of 80088000-820FOF00. The numbers are in…
A: a. PAGE SIZE(in bytes) = 2^offset = 2^8 = 256Bytes b. Number of entries in 1st level page table is =…
Q: Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a…
A: Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a…
Q: Consider a logical address of 128 pages of 1024 words each, mapped onto a physical memory of 64…
A: The Logical address will have 17 bits and the phycial address will have 16 bits. Explaination :…
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: 1. Consider a 32bit address space with 2K bytes page size, assuming that each entry consists of 4…
A: Solution : Address Space consist of 32 bits, program size is 2^32 bytes i.e. 4GB. Page Size = 2K…
Q: memory access time is 200 nanoseconds and average page-fault service time is 8 milliseconds, and you…
A: The Answer is
Q: Consider a system with 4-byte pages. A process has the following entries in its page table: logical…
A: An address of 32 corresponds to the byte that has the logical address of two. This is because the…
Q: a) A paging system with 512 pages of logical address space, a page size of 2³ and number of frames…
A: Here in this question we have given Page size = 256 No of frame = 1024 Page in logical address=…
Q: puter system with a 16-bit logical address and 2-KB page size. The system supports up to 1MB of…
A: Consider a computer system with a 16-bit logical address and 2-KB page size. The system supports up…
Q: 2. Consider the following assumptions: Size of virtual address: 64 bits Size of physical address: 29…
A: Find the required answer with calculation given as below :
Q: Assume a process containing 5 pages with 1024 bytes per page and physical memory with 10 page…
A:
Q: You have a byte-addressable virtual memory system with a two-entry TLB, a 2-way set-associative…
A: Finding the physical address: # of frames in main memory = 4 Frame number bits = log24 = 2 bit
Q: Q: Consider a system which has Logical Address = 4 GB, Physical Address PA = 64 MB, Page Size = 4…
A: An operating system (OS) is a system that acts as an interface between the user and the computer…
Q: 2. Consider the following assumptions: Size of virtual address: 32 bits Size of physical memory: 2…
A: Size of the Page table = Number of pages * Page table entry.
Q: Assuming a page size of 4 Kbytes and that a page table entry takes 4 bytes, how many levels of page…
A:
Q: Assume a 32-bit virtual address and 4 MBs of memory (i.e., DRAM). If the page size is 1 KB, then…
A: Assume a 32-bit virtual address and 4 MBs of memory (i.e., DRAM). If the page size is 1 KB, then…
Q: Define page fault and the reasons behind it. In the event of a page fault, what OS steps are…
A: According to the information given:- We have to define page fault and the reasons behind it. In the…
Q: 2. If the contents of the page table are as follows: VPN PPN Valid 021 1 31 20 3 11 4 5 01 10 6-0 7…
A: (a) 0x1AE in decimal is 430 which on modulus by 8 (since there are 8 VPN given) gives 6. Since at…
Q: Figure below refers to memory management using paging. The logical address space (left) and page…
A: ans is given below
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: Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but…
A: Below is the answer to the above question. I hope this will meet your requirement.
Q: Logical Address = 4 GB, Physical Address PA= 64 MB, Page Size = 4 KB, then calculate Number of…
A: Logical Address = 4 GB, Physical Address PA= 64 MB, Page Size = 4 KB, then Number of pages =…
Q: For the following problems assume 1 kilobyte (KB) = 1024 bytes and 1 megabyte (MB) = 1024 kilobytes.…
A: Below is the answer to above question. I hope this will be helpful for you...
Q: For the following problems assume 1 kilobyte (KB) 1024 kilobytes. 1024 bytes and 1 megabyte (MB) For…
A: a) Bits in offset = log2(page size)Given, every byte in the page have a unique address so we have…
Q: Consider a system with 36-bit addresses that employs both segmentation and paging. Assume each PTE…
A: This is Operating system related question.
Q: Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory…
A: According to the Question below the Solution:
Q: Suppose the virtual address width is 32 bits. Also suppose that the virtual and physical page size…
A: We have , Virtual Address = 32-bits Virtual and Physical page size = 4KB…
Q: Let's assume a system has 64 bytes of physical memory, 4 byte pages, and 16-byte virtual address…
A: The information given are : Physical memory size = 64 bytes Size of page = 4 bytes Size of virtual…
Q: Suppose we have a virtual address of 26 bits in a byte addressable machine. Page size is 8K bytes.…
A: Suppose we have a virtual address of 26 bits in a byte addressable machine. Page size is 8K bytes.…
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 paged virtual memory system with 32-bit virtual addresses and 1K-byte pages. Each page…
A: a)
Q: Answer 4, 5 and 6 only !! ( NO PLAGARISM) Assume the following scenario for multi-level paging…
A: Given data, a. Logical Address = 64 bits b. Page size = 1 M Byte c. Page table entry size = 4 Byte…
Q: 9. In a paging system , the logical address is formed of 20 bits. the most significant 8 bits denote…
A: As per our guidelines, only 3 sub parts will be answered. So, please repost the remaining questions…
Q: Consider a virtual memory system with a 50-bit logical address and a 38-bit physical address.…
A: A) Total number of pages = virtual address size/ page size = 250/214 = 236 or 64 G pages
Q: Suppose you have a byte-addressable virtual address me system with 8 virtual pages of 64 bytes each,…
A: Solution: Considering that it is be available 64byte = 8 pages contains so, TNB(total number of bits…
Q: Consider a logical address space of 1024 pages with 2 KB page size, mapped onto a physical memory of…
A: As given, we need to find out, how many bits are required in the logical address. Given data -…
Q: Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB, and a PTE…
A: Data given in the question, 32-bit address system page size of 2 KB PTE (Page Table Entry) size of 1…
Q: A paging system has the following parameters 8GB of physical memory, page size of 4KB; 2 pages in…
A: Let's understand step by step : Given , Physical Address Space (Physical memory) = 8 GB…
Q: a) How many bits are in a virtual address? b) How many bits are in a physical address? c) What…
A:
Q: We use the concept "paging" to map logical address to physical address. If the process size is 9216…
A: As it is mentioned that paging is used to map the virtual address space to physical address space.…
Q: For the following problems assume 1 kilobyte (KB) 1024 kilobytes 1024 bytes and 1 megabyte (MB) For…
A: For solving this question, a user must know the meaning of the virtual address and the use of a…
Q: Given a page table and a logical address=0002022H, where is it physically (what is its physical…
A: As per our guidelines we are supposed to answer?️ only one question. Kindly repost other questions…
Q: In the following three questions, assume a 32-bit virtual address space and page size equal to 4096…
A: Given: Virtual address space = 32 bit page size = 4096 bytes To find: Number of page table entry
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: A system that uses a two-level page table has 212 bytes pages and 32-bit virtual addresses. Assume…
A: Data given, 2^12 bytes pages 32-bit virtual addresses 4-byte each entry 10 bits of address serve as…
Q: Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but…
A:
Assume the following scenario for multi-level paging used by a process that has following:
a. Logical Address = 64 bits
b. Page size = 1 M Byte
c. Page table entry size = 4 Byte
d. System is byte addressable.
calculate the following.
1. Logical Address Space?
2. Maximum number of page table entries in a single page?
3. How many levels of page table entries will be used?
4. Compute the size of page tables at each level?
5. Compute the number of entries in a single page?
6. Compute the number of pages
Step by step
Solved in 2 steps
- Suppose we have a computer system with 44-bit logical addresses, page size of 64KiB, and 4 bytes per page table entry. b. Suppose we use two-level paging and arrange for all page tables to fit into a single page frame. How will the bits of the address be divided up? (show your work) c. Suppose we have a program with a 4GiB address space. Counting the program and all page tables, using the two-level page table scheme from the previous question, how much memory, in number of page frames, is used? (show your work)Computer Science 9. In a paging system , the logical address is formed of 20 bits. the most significant 8 bits denote the page number, the least significant 12 bits denote the offset. Memory size is 256K bits. a. What is the page size (in bits)? b. What is the maximum number of pages per process? c. How many frames does this system have? d . Give the structure of the page table of a process executing in this system. Assume 2 bits are reserved for attributes. e. How many bits are required for each page table entry? f. If physical memory is upgraded to 1024 K bits, how will your answer to c and e change?Consider a paged virtual memory system with 32-bit virtual addresses and 1K-byte pages. Each page table entry requires 32 bits. It is desired to limit the page table size to one page. a. How many levels of page tables are required? b. What is the size of the page table at cach level? Hint: One page table size is smaller. c. The smaller page size could be used at the top level or the bottom level of the page table hierarchy. Which strategy consumes the least number of pages?
- Consider a paging system with the following: Physical memory= 32 bytes. Page size=4 bytes. Page Table: Page Frame 0 5 1 6 2 1 3 2 What are the physical addresses for the following logical addresses? A) 15 B) 8 C) 4Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? 1024 Bytes Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…9. Consider a system that uses 32-bit addresses and page table structures as discussed in class. If the address space of a process contains exactly 256 pages, what is the minimum number of pages that might be needed for its page table structure? What is the maximum number that might be needed? Briefly explain your answer.
- Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a multi-level page table design to store all the page table entries of a process, and each page table entry and index entry are 4 bytes in size. What is the total number of page that are required to store the page table entries of a process, across all levels of the page table? You may follow the hint below or finish from scratch to fill the blanks. Please show your calculations to get partial points like 2^10/2^4=2^6. 1. We need to calculate the total number of page table entries needed for a process (i.e., the total number of pages for a process) 2. We need to calculate how many entries each page can store 3. With 1 and 2, we can calculate how many pages needed for the lowest (innermost) level 4. Each page from 3 requires an entry (pointer) in the upper (next) level. We need to calculate how many pages are required to store this next level entries (please note the entry size is always 4…On a simple paging system with 224 bytes of physical memory, 256 pages of logical address space, and a page size of 210 bytes. 1. How many bits are needed to store an entry in the page table (how wide is the page table)? Assume a valid/invalid 1-bit is included in each entry. 2. If the page table is stored in the main memory with 250nsec access time, how long does a paged memory reference take? 3. If the page table is implemented using associative registers that takes 95nsec. and main memory that takes 200nsec, what is the total access time if 75% of all memory references find their entries in the associative registers?Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up to 512 MB of physical memory. How many entries are there in each of the following? Assume that each page table entry is 4 Bytes. c. A conventional single-level page table?d. An inverted page table?e. A two-level hierarchical page table? Consider a virtual memory system with a 50-bit logical address and a 38-bit physical address. Suppose that the page/frame size is 16K bytes. Assume that each page table entry is 4 Bytes. a. How many frames are in the systems? How many pages in the virtual address space for a process? b. If a single-level page table is deployed, calculate the size of the page table for each process. c. Design a multilevel page table structure for this system to ensure that each page table can fit into one frame. How many levels do you need? Draw a figure to show your page systems
- Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up to 512 MB of physical memory. How many entries are there in each of the following? Assume that each page table entry is 4 Bytes. A conventional single-level page table? An inverted pagetable? A two-level hierarchical page table?Suppose a computer using 8-way set associative cache has 1 M words of main memory, and a cache of 16 K words, where each cache block contains 8 words. What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, set, and ?word fields Tag = 9-bit, Set = 8-bit, Word = 3-bit Tag = 9-bit, Set = 7-bit, Word =4-bit Tag = 9-bit, Set = 6-bit, Word = 5-bit Tag = 9-bit, Set = 5-bit, Word = 6-bitConsider a simple paging system with the following parameters: Size of the program = 256 bytes Page size of 16 bytes 2-Byte addressable memory Main memory is divided into 32 frames Page table for the program is as follows: Page No. 0 1 2 3 4 5 6 7 8 9 10 11 a) 49 b) 22 12 13 14 15 Frame No. 14 22 10 17 12 9 31 Convert the following logical addresses into physical addresses using the mathematical procedure. You are expected to show each and every step of your solution: