[discussion_vu] CS401_Q01_Clarification_with_Reference

---------- Forwarded message ----------
From: ..::ISHFAQ::.. <mc100402092@vu.edu.pk>
Date: Tue, Apr 19, 2011 at 2:32 AM
Subject: CS401_Q01_Clarification_with_Reference
To: zavia-lms <zavia-lms@googlegroups.com>, afaaq <afaaqtariq233@gmail.com>, Miss Kazmi <mc100402020@vu.edu.pk>, coool_vu_students <coool_vu_students@googlegroups.com>

اسلام علیکم ورحمتہ اللہ و برکاتہ

Dear All, i have found another solution of CS401 , Question n o 1 which totally wrong.

Question No. 1

a)       Suppose your computer has a processor with 24-bit address lines. What is maximum amount of memory that can be attached in your system? (Show the step(s) for calculation of maximum addressable memory) (2.5 marks)

2^24 = 16777216
16777216/8 = 2097152bytes
2097152/1024=2048KB
2048/1024 = 2MB

 b)       How many address bits are required for accessing 1GB RAM? (Show the step(s)for calculation of required address bits) (2.5 marks)

1GB*1024=1024MB
1024MB*1024=1048576KB
1048576KB*1024=1073741824bytes
1073741824Bytes*8=8589934592bits
2^33 = therefore 33 address lines exist
Correct is this Method:See the attachment for reference

However the maximum memory iAPX88 can access is 1MB which can be

accessed with 20 bits.

 

2^20=1048576 bytes

=1024 KB

=1 MB

See the attachment for Refrence which is taken from the handout at page number 12.Correct Solution are given Below.
2^24 = 16777216 Bytes
=16384KB
=16MB
Also
1GB*1024=1024MB
1024MB*1024=1048576KB
1048576KB*1024=1073741824bytes
2^30 = 1073741824bytestherefore 30 address lines exist

Best Wishes,

"...SUBHANALLAHI WABI HAMDIHI...SUBHANALLAH IL AZEEM..."
Muhammd Ishfaq
MCS 2nd Semester (PakPattan)

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