ISE218_Unit06_Practice_Problems_Answers_v1
docx
keyboard_arrow_up
School
Stony Brook University *
*We aren’t endorsed by this school
Course
218
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
docx
Pages
3
Uploaded by AmbassadorWolfMaster380
ISE 218: Fundamentals of Information Technology
Practice Problems for Unit 6: Input/Output Systems
Directions: Below is a set of exercises and problems you should do for practice. Answering this set of questions
should be just one part of your preparation for quizzes and exams. In addition to solving these problems you
should:
•
Review your homework assignments.
•
Review the lecture notes and read the relevant sections of the course textbook.
•
Do additional practice problems from the textbook; see the problems at the end of each chapter.
•
Form a study group and test each other on the material.
1.
Calculate the overall speedup of a system that spends 65% of its time on I/O with a disk upgrade that
provides for 50% greater throughput.
1.28 or 28% (
S
= 1
.
2766;
f
= 0
.
65;
k
= 1
.
5
) rewrite this solution to match
the quality of the homework 6 solutions
2.
Suppose your company has decided that it needs to make certain busy servers 300% faster. Processes in
the workload spend 80% of their time using the CPU and 20% on I/O. In order to achieve an overall system
speedup of 300%, how much faster does the CPU need to be? Your final answer should be a whole
number.
k
= 0
.
8
/
0
.
05 = 16
times as fast (1500% faster)
3.
Suppose your company has decided that it needs to make certain busy servers 400% faster. Processes in
the workload spend 90% of their time using the CPU and 10% on I/O. In order to achieve an overall system
speedup of 400%, how much faster does the CPU need to be? Your final answer should be a whole
number.
ISE 218 – Fall 2023
Unit 6 Practice Problems
Page 1 of 3
k
= 0
.
9
/
0
.
1 = 9
times as fast (800% faster)
4.
Using Amdahl’s Law, show which is better: making 20% of the instructions in a program 80% faster, or
making 80% of the instructions 20% faster.
Making 20% of instructions 80% faster:
Making 80% of instructions 20% faster:
Making 80% of instructions 20% faster is the better option.
5.
Suppose the daytime processing load of some computing server consists of 70% CPU activity and 30% disk
activity. Customers are complaining that the system is slow. After doing some research, you have learned
that you can upgrade your disks for $2,500 to make them 3 times as fast as they are currently. You have
also learned that you can upgrade your CPU to make it 1.5 times as fast as it currently is for $3,000.
a.
Which would you choose to yield the best performance improvement for the least amount of money?
Justify your answer through appropriate application of Amdahl’s Law and a cost-benefit analysis
similar to the one we did in lecture.
or 30.43%
or
25% Choose
the CPU upgrade:
CPU = $3000/30.43% = $98.59 per 1% increase in performance Disk
= $2500/25% = $100 per 1% increase in performance
b.
What is the break-even point for the upgrades? That is, what price would we need to charge for the
CPU (or the disk – change only one) so the result was the same cost per 1% increase for both?
ISE 218 – Fall 2023
Unit 6 Practice Problems
Page 2 of 3
We want the price per 1% to be the same. A correct response must include
at least one
of the
following analyses:
•
If we change the CPU price, we have
X/
30
.
43 = 100
, and
X
= $3043
. This means if the CPU
costs $3043, we break even on cost per 1% improvement.
•
If we change the price of the disk, we have
Y/
25 = 98
.
59
, and
Y
= 2464
.
75
. This means if the
disk costs $2464.75, we break even on cost per 1% improvement.
6.
Suppose a disk drive has the following characteristics:
•
8 surfaces
•
100,000 tracks per surface
•
1500 sectors per track
•
1024 bytes/sector
•
Track-to-track seek time of 6 milliseconds
•
Rotational speed of 10,000 RPM.
a.
What is the capacity of the drive in GB? Use base-2 units of measure, i.e., 1 GB =
2
30
bytes.
8 surfaces
×
100,000 tracks / surface
×
1500 sectors / track
×
1024 bytes/ sector = 1,228,800,000,000
bytes or 1144.4 GB
b.
What is the access time in ms?
Access time = seek time + rotational delay
60 seconds
×
1000 ms
Rotational delay =
60 seconds
×
1000 ms
Access time = 8.5 ms + 3 ms = 11.5 ms
ISE 218 – Fall 2023
Unit 6 Practice Problems
Page 3 of 3
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help