there are 35 signals out of which 3 signals can be active at a time
first lets consider 1st signal for that we will need 6 bit since it is vertical microprogrammed unit so we will need minimum 6 bit to represent 35 signal out of which only one is active for 1st signal then again same 35 signal out of which 1 will be active so 6 more bits then for 3rd signal out of 35 signal we need 6 more bits
total=6+6+6+5( 5 bits are for remaining 5 signals)= 23 bits
therefore bits in control word = 23 bits
now we are asked size of control memory
control memory size = size of one micro-instruction x no of microinstruction
we are given that addr of next micro instruction is 12 bit therefore
no of micro instruction = 2^12
and
size of microinstruction = control word size + mux select size + next microinstruction address
= 23 + 3 + 12 = 38 bits
hence control memory size = 38 x 2^12 = 152 kbits
