اسلام علیکم ورحمتہ اللہ و برکاتہ
Read Page No. 23 for detail or see the attachment named CS401_Q02_Ref
Best Wishes,
"...SUBHANALLAHI WABI HAMDIHI...SUBHANALLAH IL AZEEM..."
Muhammd Ishfaq
MCS 2nd Semester (PakPattan)
Muhammd Ishfaq
MCS 2nd Semester (PakPattan)
---------- 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>
16777216/8 = 2097152bytes
2097152/1024=2048KB
2048/1024 = 2MB
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
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,
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
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 = 1677721616777216/8 = 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=1024MB1024MB*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)
Muhammd Ishfaq
MCS 2nd Semester (PakPattan)
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