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March 3, 2015

March 3, 2015

Posted by **bode** on Saturday, March 26, 2011 at 11:40am.

a. The number of address lines needed for each memory chip

b. The number of data lines needed for each memory chip

c. The number of bytes that can be stored in each memory chip (note that there are 8 bits/byte)

128 KB words x 8 bits/word

512 KB words x 16 bits/word

512 MB words x 32 bits/word

8 GB words x 64 bits/word

- computer -
**MathMate**, Saturday, March 26, 2011 at 12:16pmAddress lines are required to address each word. For n lines, the number of possible address is 2^n addresses.

For 128K words, there are 2^7 addresses, so 7 address lines are required.

Each word has a width of 8 bits, so to address each bit individually (and internally), 8 lines are required.

The storage is

128K x 8bits

= 128K * 1 byte

= 128*2^10 bytes

= 131072 bytes

So answer for the first case is 7 address lines, 8 data lines and 131072 bytes.

The remaining cases are similar.

- computer -
**samrti**, Thursday, May 12, 2011 at 11:34amaddress lines is reqired in 512 k word memory

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