How many bits are required to address a 4M X 16 main memory if a)Main memory is

Question

How many bits are required to address a 4M X 16 main memory if

a)Main memory is byte addressable?   ______

b)Main memory is word addressable? ______

A digital computer has a memory unit with 24 bits per word. The instruction 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 for memory? _________

Write, run and test a MARIE program to accept two accept two integer inputs and multiply them by repeated additions. You may assume that both inputs are positive. Provide a text file with your assembly code and machine code, and a screen shot of the simulator at the completion of the program.

Explanation / Answer

4M x 16 = 64Mb of memory = 67108864 bytes or 33554432 words (word is usually 2 bytes)

this means that last byte in memory will have address: 0x3FFFFFF (67108864-1) = 26 bits

last 2 byte word will have address: 0x1FFFFFF (67108864/2-1 = 33554432-1) = 25 bits

and if we consider 4 byte word - we have latest address 0xFFFFFF (67108864/4-1) = 24 bits

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