A computer system has a word addressable memory. Each word is 32 bits (4 bytes)

Question

A computer system has a word addressable memory. Each word is 32 bits (4 bytes) wide, and system uses 24-bit addressing. How many memory locations can be addressed? a. 32 b. 24k c. 16M d. 24 e. 32G How many data lines are required for the system in the previous question? a. 8 b. 16 c. 24 d. 26 e. 32 If the system in the previous questions were instead byte-addressable: and were to retain the same number of bytes of memory, how many address lines would it require? a. 8 b. 16 c. 24 d. 26 e. 32 Given a memory that is addressed by 22 bits and is 6-bit addressable, how many bits of storage does the memory contain? a. 132 bits b. 64*22 = 1408 bits c. 12 Mbits d. 24Mbits e. 64Mbits

Explanation / Answer

7) It is given that system uses 24-bit addressing. Hence there are total 2^24 address are possible in the memory which is equal to 16M (2^20==1M)

8)In the previous question, it is given that each word is of 32 bits(4 bytes). Hence in each address it stores 32 bits. Hence data lines are 32. Remember data lines are required for each memory location and address lines are required to assign address all the memory locations.

9) If the memory is byte-addressable, then 24+2 =26 address lines are required. 24 is required to address the particular word and 2 bit is required to address the byte in the word (as each word contains 4 bytes. Hence to find particular byte in the word 2 bits are required.)

10) It is given that memory is addressed by 22 bits. Hence, there are total 2^22 memory locations are possible in the memory and it is also given that each memory is 6-bit addressable. Hence each memory location contains 6 bits. Hence total bits in the storage should be 2^22 *6 which is equal to 4M*6=24M, But I don't know about the your answer.

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