1. a) A memory unit has 28-bit address lines and 64-bit input/output data lines.
ID: 3830945 • Letter: 1
Question
1. a) A memory unit has 28-bit address lines and 64-bit input/output data lines. How many bytes of data can this memory hold? How many words does it contain, and how large is each word?
b) A memory unit consists of 32M words of 16-bit each. How many bits wide address lines and input-output data lines are needed to access this memory?
c) A memory unit consists of 512K bytes of data. How many bits wide address lines are needed to access this memory if the word size is 16 bit? If the word size is 64 bit?
Explanation / Answer
1. a) Memory can hold 228*23= 231bytes of data. Here, there are 228 address loacations possible. Also, data can be of 64 bits, that is 8 bytes, that is 23 bytes. Hence, total bytes is 231. Each word can be 64 bit long and there would be 228 such words.
(b) 32M words is 25M words, that is 25*220 = 225 words. Hence, 25 bit address lines will be required. Also, since each word is 16 bit long, 16 bit input-output data lines will be required.
(c) Memory unit consists of 512K bytes, that is 29*210 = 219 bytes of data. If word size is 16 bit, that is 2 bytes, then, 219/2 = 218 words would be possible. 18 bit wide address lines would be required for word size 16 bit. For 64 bit, that is 8 bytes, that 23 byte word size, 219/23 = 216words would be possible. Hence, a 16 bit wide address line would be required for 64 bit word size.
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