6 votes 6 votes If there are 32 segments, each size 1 k bytes, then the logical address should have 13 bits 14 bits 15 bits 16 bits Operating System isro2015 operating-system virtual-memory segmentation + – go_editor asked Jun 17, 2016 go_editor 10.9k views answer comment Share Follow See 1 comment See all 1 1 comment reply Vishnu__ commented Jul 21, 2022 reply Follow Share In Segmentation, the Logical Address is divided into— Segment No. & Offset S d If no of segments is 32, we need 5bits to denote them in MSB-part (S). And given d=10 (PS=2^10) So Logical Address = (10+5) = 15bits. 0 votes 0 votes Please log in or register to add a comment.
Best answer 15 votes 15 votes Answer : 15 bits To specify a particular segment, 5 bits are required. How 2^5 = 32 ===> so 5 bits To select a particular byte after selecting a page, 10 more bits are required. Hence 15 bits are required. How 1 KByte ===> 2^10 So Total is 10+5 =15 shekhar chauhan answered Jun 17, 2016 • selected Jun 25, 2016 by Arjun shekhar chauhan comment Share Follow See all 3 Comments See all 3 3 Comments reply rameshbabu commented Jun 17, 2016 reply Follow Share with assumption of memory is byte addressable. 0 votes 0 votes shekhar chauhan commented Jun 17, 2016 reply Follow Share generally it is word addressable but here Byte. 0 votes 0 votes Arjun commented Jun 25, 2016 reply Follow Share Generally it is byte addressable - it is more common. 5 votes 5 votes Please log in or register to add a comment.
3 votes 3 votes Format of logical Address: segment number Block offset 32 segments, In order to represent these segments we require 5 bits Block offset = 1 Kbytes = 2^10 bytes =10 bits require for offset 5 + 10 =15 bits (C) will be answer akash.dinkar12 answered Apr 14, 2017 akash.dinkar12 comment Share Follow See all 0 reply Please log in or register to add a comment.