A digital computer has a memory unit with 24 bits per word. the instructions set

Question

A digital computer has a memory unit with 24 bits per word. the instructions set consists of 150 different operations. all instructions have an operation code part(opcode) and an address part (allowing for only one address) Each instruction is stored in one word of memory.
a. how many bits are needed for the opcode
b. how many bits are left for the address part of the instruction.
c. What is the maximum allowable size of memory.
d. what is the largest unsigned binary number that can be accommodated in one word of memory?

Explanation / Answer

Answer: a) As we know that: 2^8 = 256 As it can seen that it is larger than 150 => we will take lower => 2^7 = 128 bits Result: The the operation code needs 8bits b) Since: 24-8=16 Hence there are 16 bits available for the address part of the instruction c)Since: 2^16 = 65536 =>Max number is XXXXX = 65535 d) As 2^15 = 32768-1 = 32767 Ans Please rate

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