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.
Need to watch both lectures for complete understanding.