Question

You have 4 KBytes of byte addressable memory. For this memory, you consider the following organizations: 1. 64-bit words, with a 9-bit address line. 2. 32-bit words, with an 11-bit address line. 3.16-bit words, with a 10-bit address line. In which organization(s) is it the case that not all words in the memory can be addressed?

a. 1 b. 2 c. 3 d. 2 & 3

Answer 1

Memory is 4KB = 12 bits

A) word size = 64 bit which is 8B

9 bits address line therefore size is 2^9*8B = 2^12 B

Therefore memory size is 12 bits

B) 32 bit word size = 4B word size

11 bit line. Therefore total memory size is 2^11 * 2^2 = 2^13B

Therefore memory size is 13 bit

C) 16 bit word is 2B word size

10 bit address line. Therefore total memory size is 2^10* 2= 2^11B

Therefore memory is 11 bits.

Correct option is (c). The third organization cannot address all the word. Since it's size is smaller than required memory size.

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10 bit address line = 2^10*