Consider a system running ten I/O-bound tasks and one CPU-bound task. Assume that the I/O-bound tasks issue an I/O operation onc
1 answer:
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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
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
so here also for represent instruction 6 bit is required
and
for size 64 kb
=
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