Consider the following system: Byte addressable 16-bit addresses 256B pages Single level Page Table System The...
Question 31 supus Given a computer using a byte-addressable virtual memory system with a two-entry TLB, a 2-way set associative cache, and a page table for a process P. Assume cache blocks of size 16 bytes. Assume pages of size 32 bytes and a main memory of 4 frames. Assume the following TLB and page table for Process P: TLB 03 4 هما 0 1 2 3 4 5 6 7 Page Table f Vali d 1 1 0 2...
3. Virtual Memory (20 points) An ISA supports an 8 bit, byte-addressable virtual address space. The corresponding physical memory has only 256 bytes. Each page contains 32 bytes. A simple, one-level translation scheme is used and the page table resides in physical memory. The initial contents of the frames of physical memory are shown below. VALUE address size 8 bit byte addressable each byte of addressing type memory has its own address 32 B page size physical memory size 256...
A certain byte-addressable computer system has 32-bit words, a virtual address space of 4GB, and a physical address space of 1GB. The page size for this system is 4 KB. Assume each entry in the page table is rounded up to 4 bytes. a) Compute the size of the page table in bytes. b) Assume this virtual memory system is implemented with a 4-way set associative TLB (Translation Lookaside Buffer) with a total of 256 address translations. Compute the size...
A computer uses a byte-addressable virtual memory system with a four-entry TLB and a page table for a process P. Pages are 16 bytes in size. Main memory contains 8 frames and the page table contains 16 entries. a. How many bits are required for a virtual address? b. How many bits are required for a physical address?
18. You have a byte-addressable virtual memory system with a two-entry TLB, a 2-way set associative cache, and a page table for a process P. Assume cache blocks of 8 bytes and page size of 16 bytes. In the system below, main memory is divided into blocks, where each block is represented by a letter. Two blocks equal one frame. Given the system state as depicted above, answer the following questions: a) How many bits are in a virtual address...
Question # 3 Consider a demand-paged system where the page table for each process resides in main memory. In addition, there is a fast associative memory (also known as TLB which stands for Translation Look-aside Buffer) to speed up the translation process. Each single memory access takes 1 microsecond while each TLB access takes 0.2 microseconds. Assume that 2% of the page requests lead to page faults, while 98% are hits. On the average, page fault time is 20 milliseconds...
3. (6 pts) Consider a new processor. The memory system is 32-bit byte- addressable. The on-chip cache memory is 128 KByte 4-way set-associative, with a 64 byte block size. (a) Draw a diagram showing how the cache controller will split the memory address: for each field. show its name and number of bits. (b) The design team decided to change the cache architecture to a direct mapped one. For each of the parameters in the following table, indicate the impact...
As described in 5.7, virtual memory uses a page table to track the mapping of virtual addresses to the physical addresses. This exercise shows how this table must be updated as addresses are accessed. The following data constitutes a stream of virtual addresses as seen on a system. Assume 4 KiB pages, a 4-entry fully associative TLB, and true LRU replacement. If pages must be brought in from disk, increment the next largest page number. 4669, 2227, 13916, 34587, 48870,...
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...
Let's start with a memory system that is byte addressable and has 14-bit addresses. Now let's add a direct mapped cache that can hold 16 blocks (or has 16 lines) and each block is 4 bytes. The table shoun below details the contents of the cache at a particular point in time. if the valid bit is 1 there is a cache block loaded into the cache with Tag as detailed in the tag column. The columns B0-D3 then show...