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 $(X),$ and a MUX select field $(Y).$ There are $8$ status bits in the input of the MUX.
How many bits are there in the $X$ and $Y$ fields, and what is the size of the control memory in number of words?
I am not getting how the size of control memory$=1024$?
I am ok with calculating $X$ and $Y$.But i think
control memory size=number of different microinstruction * size of each micro instruction
=$1024 *26 \,\,bits=26624bits$=$3328 Bytes$=$3328 words$ (if i consider word size = 1 Byte)
If the solution is correct then this solution is wrong
@sourav, Control memory is nothing but the collection of control words, in this question next address field is 10 which means there are 1024 control words are there.