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Assignment No. 05 SOLUTION
Semester Spring 2011 Advance Computer Architecture - CS501 |
Total Marks: 20
Due Date: 01/07/2011
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Objective:
To learn and understand basic concepts of disk drives, and
error control
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
Note: You have to provide solution with all the calculations and
formulations involved, else no marks will be awarded.
For any query about the assignment, contact at cs501@vu.edu.pk
GOOD LUCK
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Question No 1
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10 Marks
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A hard drive has 4
surfaces, with 512 tracks per surface and constant 64 sectors per track.
Sector size is 2K byte. The average seek time is 3 ms, the drive runs at 3,600
rpm.
Solution:
Average access time= Seek time +
Rotational latency = 3 + 8.3 = 11.3 ms
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Question No 2
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10 Marks
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Use the hamming
code to encode the following 4-bit values (8 marks)
Also write down the
encoding scheme (2 mark)
Total (8+2 = 10 marks)
A.
0110
B.
1001
C.
0000
D.
1111
Note : Use odd bit parity to encode above values
Solution:
The encoding scheme is P1P2D3P4D5D6D7
, using odd parity
0110 → 0001110
1001 → 1110001
0000 → 1101000
1111 → 0010111
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Showing posts with label CS501. Show all posts
Showing posts with label CS501. Show all posts
Wednesday, 14 March 2012
CS501 Advance Computer Architecture ASSIGNMENT 5 SOLUTION
Advance Computer Architecture - CS501 ASSIGNMENT 5
|
|
Assignment No. 05
Semester Spring 2011 Advance Computer Architecture - CS501 |
Total Marks: 20
Due Date: 01/07/2011
|
|||
|
Objective:
To learn and understand basic concepts of disk drives, and
error control
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
Note: You have to provide solution with all the calculations and
formulations involved, else no marks will be awarded.
For any query about the assignment, contact at cs501@vu.edu.pk
GOOD LUCK
|
|||||
|
Question No 1
|
|
10 Marks
|
|||
|
A hard drive has 4
surfaces, with 512 tracks per surface and constant 64 sectors per track.
Sector size is 2K byte. The average seek time is 3 ms, the drive runs at 3,600
rpm.
|
|||||
|
Question No 2
|
10 Marks
|
||||
|
Use the hamming
code to encode the following 4-bit values (8 marks)
Also write down the
encoding scheme (2 mark)
Total (8+2 = 10 marks)
A.
0110
B.
1001
C.
0000
D.
1111
Note : Use odd bit parity to encode above values
|
|||||
Advance Computer Architecture - CS501 ASSIGNMENT 4
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Assignment No. 04
Semester Spring 2011 Advance Computer Architecture - CS501 |
Total Marks: 15
Due Date: 19/06/2011
|
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Objective:
To learn and understand basic concepts of I/O
subsystems, and DMA
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
Note: You have to provide solution with all the calculations and
formulations involved, else no marks will be awarded.
For any query about the assignment, contact at cs501@vu.edu.pk
GOOD LUCK
|
|||||
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Question No 1
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10 Marks
|
|||
|
An I/O device
transfers data at a rate of 5MB/s over a 100MB/s bus. The data is transferred
in 2KB blocks. If the processor operates at 500MHz, and it takes a total of
5000 cycles to handle each DMA request, find the fraction of CPU time
handling the data transfer with and without DMA.
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Question No 2
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5 Marks
|
||||
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A 4 K byte block of data is read
from a disk drive. What is the overall data transmission rate if the disk
drive has a latency of 7 ms, and a burst bandwidth of 2 MB per second?
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Advance Computer Architecture-CS501 ASSIGNMENT 3 SOLUTION
Assignment
No. 03 sOLUTION
Semester Spring 2011
Advance
Computer Architecture-CS501
|
Total Marks: 15
Due Date:
09/06/2011
|
|||
Objective:
To learn and understand basic concepts of I/O and polling loop.
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
o The assignment is submitted after due date.
o The submitted assignment does not open or file corrupt.
o The assignment is copied (from other student or copy from
handouts or internet).
o Student ID is not mentioned in the assignment File or
name of file is other than student ID.
GOOD LUCK
|
||||
Question No 1
|
Marks = 10
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Let us have a 20MIPS processor, you have to calculate the
percentage of time it spends in busy wait loop of a 75-character line
printer, when it takes 2 msec to print a character and a total of 665
instructions require to be executed to print a 75 character line.
Suppose that two instructions
are executed in the polling loop.
Solution:
Out of
the total 665 instructions executed to print a line, 75x2=150 is required for
polling. For a 20MIPS processor, the execution of the remaining 515
instructions takes 515/ (20x106) =
25.75µsec. Since the printing of 75 characters
takes 150ms, (150-0.02575) =149.97msec is spent in the polling loop before
the next 75 characters can be printed. This is 149.97/150=99.98% of the total
time.
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Question
No 2
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Marks=5
|
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Let us assume that a program takes 400 seconds of elapsed
time to execute. Out of these 400
seconds, 320 seconds is the CPU time and the rest is I/O time. What
percentage of elapsed time is spent on I/O operations?
Solution:
Elapsed time = CPU time + I/O
time.
This
gives us the I/O time = 400 – 320 = 80 seconds at the beginning, which is 20
% of the elapsed time.
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Advance Computer Architecture-CS501 ASSIGNMEN 3
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Assignment
No. 03
Semester Spring 2011
Advance
Computer Architecture-CS501
|
Total Marks: 15
Due Date:
09/06/2011
|
||
|
Objective:
To learn and understand basic concepts of I/O and polling loop.
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
o The assignment is submitted after due date.
o The submitted assignment does not open or file corrupt.
o The assignment is copied (from other student or copy from
handouts or internet).
o Student ID is not mentioned in the assignment File or
name of file is other than student ID.
Note: You have to provide solution with all the calculations and
formulations involved, else no marks will be awarded.
GOOD LUCK
|
||||
|
Question No 1
|
Marks = 10
|
|||
|
Let us have a 20MIPS processor, you have to calculate the
percentage of time it spends in busy wait loop of a 75-character line
printer, when it takes 2 msec to print a character and a total of 665
instructions require to be executed to print a 75 character line.
Suppose that two instructions
are executed in the polling loop.
|
||||
|
Question
No 2
|
Marks=5
|
|||
|
Let us assume that a program takes 400 seconds of elapsed
time to execute. Out of these 400
seconds, 320 seconds is the CPU time and the rest is I/O time. What
percentage of elapsed time is spent on I/O operations?
|
||||
CS501 ASSIGNMENT 2 SOLUTION
ASSIGNMENT # 2(Solution)
CS501
QUESTION.1
Specify the condition for the branch instruction and the status of
the PC after the branch instruction executes with a true branch condition.
Solution:
|
Instruction
|
Condition
|
PC status
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Jump[20]
|
Always
|
PC← PC+20
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Jnz
r3,[33]
|
If
R[3] ≠ 0
|
PC← PC+33
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Jz r1,[16]
|
If
R[1] = = 0
|
PC← PC+16
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Jpl
r5,[37]
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If
R[5] ≥0
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PC← PC+37
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Jmi
r7,[40]
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If
R[7] < 0
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PC← PC+40
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Q.2
Identify the addressing modes and Register Transfer Language (RTL)
description (meaning) for the given Falcon-A instructions
Solution:
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Instruction
|
Addressing Mode
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RTL description(meaning)
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Load
r3,[r5+27]
|
Displacement
|
R[3]
← M[R[5] +27]
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Jnz
r5,[15]
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Relative
|
(If
R[5] ≠ 0): PC← PC+15
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Shiftl
r3,r2,5
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Immediate
|
Shift
r2 left 5 times and store in r3
|
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Add
r6,r5,r7
|
Register
|
R[6]
← R[5] + R[7]
|
|
Subi
r4,r3,17
|
Immediate
|
R[4]
← R[3] -17
|
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