Question 4878 – UGC NET CS 2018-DEC Paper-2
November 30, 2023Question 7979 – Database-Management-System
November 30, 2023Question 4881 – UGC NET CS 2018-DEC Paper-2
A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?
Correct Answer: C
Question 14 Explanation:
An instruction size is given as 32-bits.
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(2 6 ) so to identify each register uniquely 6-bits are needed.
Address part : 256K(2 18 ) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Operation code:
Size of Operation code = Complete instruction size – (size of indirect bit + size of register code + size of address part)
Size of Operation code= 7-bits
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(2 6 ) so to identify each register uniquely 6-bits are needed.
Address part : 256K(2 18 ) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Operation code:
Size of Operation code = Complete instruction size – (size of indirect bit + size of register code + size of address part)
Size of Operation code= 7-bits
7,7,18
18,7,7
7,6,18
6,7,18
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