Question

Can someone clarify how I find the lenght in bits of the OFFSET given this problem. The problem is solved here completely from 1to6 i just wanted a quick run down on how the numbers were obtained, how the word size, total frame size, program size and op code was used

PROBLEM 1 decimal value of o codes 16 bits word size: load fetch 2 total frame size: 64 bytes. store 3 total size of program 256 words add The CPU gets the next instruction which is a "fetch, and is given the following bit sequence of which the first eight bits are the 'op' code 0 0 0 0 0 0 1 0 0 1 0 1 1 1 0 0 logical page table physical memory frame space in memory 20AAA 20AB 20AF 2 20AC 20AD 20AE page of data 20AF 20B0 20B1 1. mark all known age numbers in the logical space 2. fill in the page table with whatever exact data is known binary decimal 3. logical page number 010 2 11100 28 4. offset 5. total logical address space in bytes 512 6. total page size in bytes 64 15 points

Can someome

Answer 1

Answver GIVen: word size-16bits total framesize 64 bytes total size of program-256words opcode-8bits total number of frames total programsizeI framesize - (256*2bytes) /64 bytes (because a word-16 bits -2bytes) Therefore we need 3 bits to represent the frames (23 8) Total address bits 16 bits Offset- 16 - (opcode frame bits) -16 (8+3) -5 bits In the address, the first 8 bits represent the opcode, the next 3 bits represent the logical page number (page number on the frame), remaining 5 bits for offset Hence opcode 00000010 Logical page number-010 Offset 11100