Assignment
Contents:
Overview
Topic: Operating Systems
Related Reading: Ch. 10
Due:
Internet Requirements
You will not need an Internet connection for completing the assignment,
other than for submission.
Practice Problems
Problems to be Submitted (20 points)
-
(4 points)
Exercise 42 of Ch. 10 (p. 346)
You are asked to draw a Gantt chart. If you cannot easily mimic the
diagrams from the text, please feel free to report the same information
in whatever way you choose. For example, Exercise 41 of the text is a practice
problem with the answer in the back of the text. I might choose to typeset
that same answer in a table form, such as:
From To CPU Runs
------------------------
0 120 p1
120 180 p2
180 360 p3
360 410 p4
410 710 p5
-
(5 points)
Exercise 43 of Ch. 10 (p. 346)
-
(4 points)
If, in a dynamic-partition memory management system, the current value
of the base register is 58712 in decimal, and the current value
of the bounds register is 1587, compute the physical addresses
that correspond to the following logical addresses:
-
1422
-
2535
-
(3 points)
Assume that dynamic partitioning is used for memory management and
that currently there are six empty blocks with respective sizes 38, 85,
51, 99, 77, 27. A new job arrives requiring 52 blocks of main memory. Which
size block would be used be each of the following partition selection approaches:
-
first fit
-
best fit
-
worst fit
-
(4 points)
In a paged memory-management system, the frame and page sizes are 2048
and the following page map table applies to the currently executing process,
Page |
0 |
1 |
2 |
3 |
4 |
Frame |
64 |
22 |
8 |
71 |
55 |
Compute the physical addresses that correspond to the following logical
addresses:
-
4100
-
11000
Overall, please type your answers to all of the problems in a single
document to be submitted electronically. Please see details about the submission
process.
Extra Credit (2 points)
All of the CPU scheduling examples given in the text involve the simpler
case when all jobs "arrive" into the system at time 0. In this problem,
we wish to examine the more general case which is when jobs arrive at differing
times.
Consider the following situation:
Job |
Arrival Time |
Service Time |
p1 |
0 |
40 |
p2 |
0 |
20 |
p3 |
10 |
5 |
p4 |
11 |
1 |
As described in the book, the Shortest Job Next (SJN) CPU scheduling
algorithm is non-preemptive. Therefore, it produces the following Gantt
chart (for simplicity, I'm typesetting the "chart" as the following table):
from |
to |
CPU runs |
0 |
20 |
p2 |
20 |
21 |
p4 |
21 |
26 |
p3 |
26 |
60 |
p1 |
Notice that at time 0, processes p3 and p4 have not yet arrived, and so
of those processes which are ready, p2 is the shortest. It runs from time
0 to time 20. At time 20, when the scheduler decides on the next job, p1,
p3, and p4 are available, and so p4 is the next to run as it is the shortest.
After that job finishes, then p3 runs, and finally p1.
What is interesting to note is that SJN is non-preemptive. That is,
at time 10, when job p3 arrives, p3 is even shorter than the job which
is currently running. However, the SJN scheduling algorithm lets p2 complete.
Similarly when p4 arrives.
A common preemptive variant of this is one which is called Shorest Remaining
Processing Time (SRPT). With this scheme, if a new job arrives with a service
time even less than the currently running job's
remaining processing
time, then the current job is preempted in favor of the new job. When any
job completes, the scheduler will next give the CPU to the ready job with
the smallest remaining processing time. So on the above example, the SRPT
algorithm produces the following schedule:
from |
to |
CPU runs |
0 |
10 |
p2 |
10 |
11 |
p3 |
11 |
12 |
p4 |
12 |
16 |
p3 |
16 |
26 |
p2 |
26 |
60 |
p1 |
Your extra credit task is the following.
Please give the Gantt chart which results from running the SRPT scheduling
algorithm on the following set of jobs:
Job |
Arrival Time |
Service Time |
p1 |
0 |
15 |
p2 |
0 |
20 |
p3 |
10 |
1 |
p4 |
12 |
13 |
p5 |
14 |
5 |
p6 |
24 |
4 |
Solution to the last practice problem:
logical address 85 = 0*1024 + 85. Physical address
7*1024 + 85 = 7253
logical address 2359 = 2*1024 + 311. Physical address
99*1024 + 311 = 101687
logical address 6000 = 5*1024 + 880. Error:
out of bounds; no page number 5
Last modified: 29 April 2003