4 votes 4 votes The microinstructions stored in the control memory of a processor have a width of $26$ bits. Each microinstruction is divided into three fields. a micro operation field of $13$ bits, a next address field $\text{(X)},$ and a MUX select field $\text{(Y)}.$ There are $8$ status bits in the inputs of the MUX. How many bits are there in the $\text{X} $ and $\text{Y}$ fields, and what is the size of the control memory in number of words? $10, 3, 1024$ $8, 5, 256$ $5, 8, 2048$ $10, 3, 512$ CO and Architecture isro2009 co-and-architecture control-unit + – Desert_Warrior asked Jun 3, 2016 retagged Dec 4, 2022 by Lakshman Bhaiya Desert_Warrior 2.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes MUX has 8 states bits as input lines so we require 3 select inputs to select and input lines. No. of bits in control memory next address field=26-13-3 =10 10 bit addressing .we have 2^10=1024 memory size So X,Y size=10,3 So (A) is correct option. Pranabesh Ghosh 1 answered Jun 3, 2016 selected Aug 27, 2016 by Prashant. Pranabesh Ghosh 1 comment Share Follow See all 3 Comments See all 3 3 Comments reply Keval Kubavat commented Mar 5, 2017 reply Follow Share @Pranabesh Ghosh , it sounds odd but my doubt is "there is 8 status bits in the input of MUX" means there is 8 states can possible which requires 3 bits ??? please explain first line of your ans.. 0 votes 0 votes Rahul Jain25 commented Mar 5, 2017 reply Follow Share It means that we have 8 inputs to the MUX so it is 8:1 MUX and therefore it require 3 select inputs. Which makes Y = 3. Now remaining bits = 10 = X. Number of words in control memory is 210 = 1024. So option A) is answer. 0 votes 0 votes Keval Kubavat commented Mar 7, 2017 reply Follow Share Thanks Rahul Jain25.. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes The number of bits in Control memory =26. From the given data each instruction divided into op field (13)+X(next address field)+Y(MUX) 8(23) status bits in the inputs of the MUX then three bits in the MUX select field. No. of bits in control memory next address field=26-13-3 =10 size of the control memory in number of words is 210=1024 words topper98 answered Mar 23, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.