Question

There are 55 registers, and total 60 instructions available in a general-purpose computer. The computer allows only 2-address instructions, where one operand can be a register, and another can be a memory location. The memory is byte addressable with 64KB (Kilo bytes) in size. The minimum number of bits to encode the instruction will be……

Answer 1

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

$2^6 = 64$

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  

$2^6 = 64$

so here also  for represent instruction 6 bit is required

and

for size 64 kb

$2^6 * 2^{10} * 2^3$ = $2^{19}$

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