Question # 1 (4 marks) A digital computer has a memory unit of 16 bits per word.

Question

Question # 1 (4 marks) A digital computer has a memory unit of 16 bits per word. The instruction set consists of 30 different operation stt an operaion code part (opeode) and one address field (one for the memory address). Ea a) How many bits are needed for the opcode? (l marks) b) How many bits are left for the memory address part of the instruction? (1 marks) c) What is the maximum allowable size for memory? (1 marks) d) What is the largest unsigned binary number that can be accommodated in one memory word? (1 mark

Explanation / Answer

Answer:

Given - computer has a memory unit of 16 bits per word

instruction set consist of 30 different operation .

a)

no of bits needed for opcode = ?

we need the number of bits with which we can represent 30 operations

so lets try

So 5 bits are needed for opcode to represent 30 operation.

b)

We know each instrcution take one word to store itself . which means we have 16 bits for an instruction.

and one instruction have two parts one for opcode and other for memory address

so now that we use 5 bits for opcode.

bits left for memory = 16 bits - 5 bits =11 bits.

c) maximum allowable size of memory is 16 bits

d) size of one memory word = 16 bits

unsigned binary number means all bits can be used to represent a number while in signed one (left-most bit alone is used to represent the sign(+/-) )

so all the 16 bits can be used -

so the largest unsigned binary number is = 1111111111111111 = 131071 (in decimal)

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