Software-Engineering
November 30, 2023Question 7979 – Database-Management-System
November 30, 2023UGC NET CS 2018-DEC Paper-2
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Question 14
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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?
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7,7,18
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18,7,7
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7,6,18
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6,7,18
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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
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