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posted by  willydlw on 11/5/2009 2:33:59 AM  |  status: Closed  |  Earned Karma: 1563

Address, system memory, bits, byte locations

Course Textbook Chapter Problem Needs by
Computer Architecture N/A N/A N/A 11/10/2009 at 10:00:00 AM
Question Details:
Use the following K (Kilo) is 2^10 (2 to the power of 10), M (Mega) is 2^20, and G (Giga) is 2^30.

[Part A] Assume that a memory system has 16 address bits, how many byte locations are there?

[Part B] Assume that a memory system has 32Mbytes of storage, how many address bits does it
have?

[Part C] If a memory system has 2GB of RAM, how many address bits are there?
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posted by Martin's on 11/5/2009 6:16:26 AM  |  status: Live
Asker's Rating: Somewhat Helpful   
Response Details:
Dear User,
[Part A]:
Given,
The memory system has 16 address bits.
Then the number of byte locations=
                                                   =
                                                   =65536-1
                                                   =65535
[Part B]:
Given,
 A memory system has 32 Mbytes.
Then the address bits=
                                =
                                 =
                                 =28 address bits.
[Part C]:
Given,
A memory system has 2GB of RAM.
Then the address bits=
                                =
                                =
                               =34 address bits.
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posted by Mephisto on 11/5/2009 6:26:00 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
Part A:
16 address bits means 2^16 = 65536 possible addressable locations

Part B:
32MB means 32M = 2^5 * 2^20 addressable locations
This results in 5 + 20 = 25 address bits

Part C:
2GB means 2G = 2^1 * 2^30 addressable locations
This results in 1 + 31 = 31 address bits
If you have a question about my response, feel free to message me.
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