A computer has a 256K word addressable memory Module with 16 bits per word. The instruction set consists of 166 different instructions. 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. Based on the above, answer the following questions: a. How many bits are there in the main memory? (Represent it in power of 2) b. How many bits are needed for the opcode? c. How many bits are left for the address part of the instruction?

Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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A computer has a 256K word addressable memory Module with 16 bits per word. The
instruction set consists of 166 different instructions. 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.
Based on the above, answer the following questions:
a. How many bits are there in the main memory? (Represent it in power of 2)
b. How many bits are needed for the opcode?
c. How many bits are left for the address part of the instruction?
d. How many additional instructions can be added to the existing 166 without affecting the
assigned size of the opcode part? Justify.
Transcribed Image Text:A computer has a 256K word addressable memory Module with 16 bits per word. The instruction set consists of 166 different instructions. 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. Based on the above, answer the following questions: a. How many bits are there in the main memory? (Represent it in power of 2) b. How many bits are needed for the opcode? c. How many bits are left for the address part of the instruction? d. How many additional instructions can be added to the existing 166 without affecting the assigned size of the opcode part? Justify.
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