Respuesta :
Answer:
total bits = 6 + 6 + 19 = 31 bits
Explanation:
given data
total registers = 55
memory size = 64 KB
total instructions = 60
solution
here we have given 55 register so we get greater or equal power of 2 that is here 64
[tex]2^6 = 64[/tex]
so here for register operand 6 bit is required
and
when instruction 60 we get here greater or equal power of 2 that is here 64
[tex]2^6 = 64[/tex]
so here also for represent instruction 6 bit is required
and
for size 64 kb
[tex]2^6 * 2^{10} * 2^3[/tex] = [tex]2^{19}[/tex]
so 19 bits is required for memory location
and
as instruction in 2 parts are opcode and operand
and here given as 2 address instruction
they are memory operand and the register operand
so here
total bits will be = opcode + register operand + memory operand
total bits = 6 + 6 + 19 = 31 bits
total bits = 31 bits
