2 votes 2 votes Consider a processor which contains 32 time multiplexed pins used to carry address and data, and 8 address pins carry address only.Then calculate the size of memory and number of address bits needed to address memory? CO and Architecture co-and-architecture + – rahul sharma 5 asked Jan 24, 2018 rahul sharma 5 437 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Anu007 commented Jan 24, 2018 reply Follow Share is it 32 address bits and size = 32$\times$240 1 votes 1 votes rahul sharma 5 commented Jan 24, 2018 reply Follow Share Total pins for address =32+8 ,so address is 40 bits. Word size here =4B So if word addressable = $2^{40} W=>40$ If byte addressable then =$2^40 *2^2 B => 2^{42} Bytes=>42 $ I dont have answer.Please check if i am correct.Why have you taken address bits as 32 only? 0 votes 0 votes Anu007 commented Jan 25, 2018 reply Follow Share See Multiplexed pin used for a 1st clock for address and for remaining clock used to send data. so address bits will be 32 only, after that data bits will be 32+8 = 40 bits 1 votes 1 votes Please log in or register to add a comment.