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ser334_unit7_exercise_sample.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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\begin_body
\begin_layout Standard
\noindent
Arizona State University
\begin_inset space \hfill{}
\end_inset
SER334: Operating Systems & System Programming
\end_layout
\begin_layout Standard
\noindent
Lecturer Acuña (7), UGTA Lisonbee (2)
\begin_inset space \hfill{}
\end_inset
Revised 10/13/2021
\end_layout
\begin_layout Standard
\noindent
\begin_inset CommandInset line
LatexCommand rule
offset "0.5ex"
width "100col%"
height "1pt"
\end_inset
\end_layout
\begin_layout Standard
\size huge
Unit 7 Sample Problems - Process Synchronization I
\end_layout
\begin_layout Standard
\begin_inset VSpace 0.5cm
\end_inset
\end_layout
\begin_layout Standard
In this exercise, we will review the concepts of Process Synchronization.
\end_layout
\begin_layout Itemize
Length: 1:15 minutes with discussion.
\end_layout
\begin_layout Itemize
Questions: Q1-Q4, Q6-Q8.
(optional: Q9)
\end_layout
\begin_layout Standard
\begin_inset VSpace 0.5cm
\end_inset
\end_layout
\begin_layout Section
\series bold
\size larger
Background
\end_layout
\begin_layout Enumerate
[Acuña] Consider using an image editor that uses a background thread to
regularly save the file being edited to ensure the user doesn't lose data.
This image editor contains open, save, close, and apply filter functionality.
Does this program contain any possible race condition?
\series bold
Explain.
\series default
[2 points]
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\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
Yes.
There is a chance the image could auto-save, then since images are large
and take time to save, the user could make a small change and click save
manually.
This would result in two threads saving the image at once, with the result
that the contents of the file might be corrupted.
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña] Consider a program which creates three threads and uses each of
them to display "SER334" in the same console window.
No synchronization occurs.
Does this program contain a race condition?
\series bold
Explain.
\series default
Assume that print commands are atomic.
[2 points]
\end_layout
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\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
No.
At the end of the program, the state of the program and system will be
the same no matter which order the threads execute.
For example, the console window will display SER334 three times and since
each thread displays the exact same text, the window will look the same.
From the final point of view, it doesn't matter which thread printed which
SER334 because they are equivalent.
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Section
\series bold
\size larger
The Critical-Section Problem
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña]
\series bold
Explain
\series default
how an algorithm that solves the critical-section problem would help to
address the issues with the bounded buffer problem.
[2 points]
\end_layout
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\begin_layout Standard
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inverted 0
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\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
If we have a solution to the critical-section problem, that we could use
it to protect a critical-section of the bounded buffer problem that encapsulate
s all access to shared memory.
If we do this, then can be no conflict in that shared data.
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña]
\series bold
Explain
\series default
how it would be possible to have a situation where programs are making
progress but do not have bounded waiting time.
\end_layout
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\begin_layout Standard
\begin_inset Branch blank
inverted 0
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inverted 0
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\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
Progress only means that one program is able to do something (and will)
when no process is in the critical section.
This does not mean there could be another program running which never gets
to the critical section, or does not enter it enough times to complete.
(If a program doesn't complete, then its waiting time is unbounded.) This
can also be caused by the addition of many new processes which have a higher
\begin_inset Quotes eld
\end_inset
priority
\begin_inset Quotes erd
\end_inset
to enter the critical section.
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Section
Peterson's Solution
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña] Consider the code for Peterson's Solution:
\begin_inset listings
lstparams "language=C"
inline false
status open
\begin_layout Plain Layout
//shared memory
\end_layout
\begin_layout Plain Layout
int turn = 0;
\end_layout
\begin_layout Plain Layout
bool flag[2] = { false, false };
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
//for some process i
\end_layout
\begin_layout Plain Layout
do {
\end_layout
\begin_layout Plain Layout
flag[i] = true;
\end_layout
\begin_layout Plain Layout
turn = j;
\end_layout
\begin_layout Plain Layout
while (flag[j] /* && turn == j */);
\end_layout
\begin_layout Plain Layout
//critical section
\end_layout
\begin_layout Plain Layout
flag[i] = false;
\end_layout
\begin_layout Plain Layout
//remainder section
\end_layout
\begin_layout Plain Layout
} while (true);
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
Notice that part of the algorithm has been commented out.
\series bold
Explain
\series default
how this changes it's functionality.
Will it still solve the critical section problem?
\series bold
Explain.
\end_layout
\begin_layout Standard
\begin_inset Branch blank
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\begin_inset VSpace 4cm
\end_inset
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\end_inset
\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
This give the program a race condition since processes no longer have to
wait their term.
If both processes run their first line, then both flag indices will be
true and both processes will be allowed to entry (even though only one
should be going on the turn).
If only one process makes it to the while, then the opposite process will
start, and then first process will enter as well.
\end_layout
\begin_layout Standard
\begin_inset VSpace 1cm
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña]
\series bold
Explain
\series default
why Peterson's Solution needs to assume that loads/stores are atomic.
(
\series bold
Hint:
\series default
one reason is related to struct padding/alignment...)
\end_layout
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inverted 0
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\end_inset
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\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
Because instances of the algorithm load/store data in the same turn and
flag variables.
This is problematic in the case that those values are not word aligned,
causing the compiler to generate more than one load/store operation (meaning
that store could save the first part of a new value, then a load could
read the entire value, and then store would save the remaining bits of
the value; potentially producing inconsistent state).
\end_layout
\begin_layout Standard
\begin_inset VSpace 1cm
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Section
\series bold
\size larger
Synchronization Hardware
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña] Why would it be appropriate to say that an assignment operation
(a = 5;) is typically an atomic operation while an increment operation
(a++;) is not? Will your answer be the same for all possible hardware?
\series bold
Explain.
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Branch blank
inverted 0
status collapsed
\begin_layout Standard
\begin_inset VSpace 3.5cm
\end_inset
\end_layout
\end_inset
\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
\series bold
Ans: [Acuña]
\end_layout
\begin_layout Standard
An assignment operation has only an value on the right side, so it will
compile down to just an assignment without any preconditions.
Since it's just a single operation, it's naturally atomic.
For the increment, we need to emit code that loads, adds, and stores the
value.
Since we have multiple operations, it cannot be atomic.
\end_layout
\begin_layout Standard
\begin_inset VSpace 1cm
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Section
Mutex Locks
\end_layout
\end_deeper
\begin_layout Enumerate
[Lisonbee] The following code creates 4 new threads and increments and prints
the contents of a global variable.
\begin_inset listings
lstparams "language=C"
inline false
status open
\begin_layout Plain Layout
#include <stdlib.h>
\end_layout
\begin_layout Plain Layout
#include <stdio.h>
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
#define NUM_THREADS 4
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_mutex_t lock;
\end_layout
\begin_layout Plain Layout
int data = 0;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
void* runner(void* arg) {
\end_layout
\begin_layout Plain Layout
pthread_mutex_lock(&lock); //ACQUIRE LOCK
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
data++;
\end_layout
\begin_layout Plain Layout
printf("Thread %d: %d
\backslash
n", *((int*)arg), data);
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_mutex_unlock(&lock); //RELEASE LOCK
\end_layout
\begin_layout Plain Layout
pthread_exit(0);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
int main() {
\end_layout
\begin_layout Plain Layout
pthread_t threads[NUM_THREADS];
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_mutex_init(&lock, NULL);
\end_layout
\begin_layout Plain Layout
for (int i = 0; i < NUM_THREADS; i++) {
\end_layout
\begin_layout Plain Layout
pthread_create(threads[i], NULL, runner, (i + 1));
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
for (int i = 0; i < NUM_THREADS; i++) {
\end_layout
\begin_layout Plain Layout
pthread_join(threads[i], NULL);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_mutex_destroy(&lock);
\end_layout
\begin_layout Plain Layout
return 0;
\end_layout
\begin_layout Plain Layout
}
\end_layout
\end_inset
\begin_inset Newpage pagebreak
\end_inset
What would happen to the output of the program if the commented lines were
removed (ACQUIRE LOCK & RELEASE LOCK)? What impact do these lines have
on the program's final state?
\series bold
Justify
\series default
your answer.
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Branch blank
inverted 0
status open
\begin_layout Standard
\begin_inset VSpace 4cm
\end_inset
\end_layout
\end_inset
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
Ans: [Lisonbee]
\end_layout
\begin_layout Standard
If these lines of code were removed then the value of data could change
before a thread has had the chance to print the value.
This could mean that the output could look like this:
\end_layout
\begin_layout Standard
\begin_inset listings
lstparams "language=C"
inline false
status open
\begin_layout Plain Layout
Thread #: 2
\end_layout
\begin_layout Plain Layout
Thread #: 2
\end_layout
\begin_layout Plain Layout
Thread #: 4
\end_layout
\begin_layout Plain Layout
Thread #: 4
\end_layout
\end_inset
\end_layout
\begin_layout Standard
In this instance, 'data' was incremented twice by two different threads
before it was printed out, so the same value shows up twice.
This happened again for the second two threads, causing the output to be
2, 2, 4, 4.
The commented lines of code ensure that the value contained in 'data' is
always incremented once and then printed.
The order of threads that print may change from run to run, but since the
data variable is incremented and printed out within the critical section
of code (where the thread has the mutex lock) the value of data will always
be incremented by 1 each time it's printed.
\end_layout
\begin_layout Standard
The program's final state is deterministic in terms of the data value that
is printed out by all of the threads.
Keeping the commented lines of code means that the output will always look
like:
\end_layout
\begin_layout Standard
\begin_inset listings
lstparams "language=C"
inline false
status open
\begin_layout Plain Layout
Thread #: 1
\end_layout
\begin_layout Plain Layout
Thread #: 2
\end_layout
\begin_layout Plain Layout
Thread #: 3
\end_layout
\begin_layout Plain Layout
Thread #: 4
\end_layout
\end_inset
\end_layout
\begin_layout Standard
The '#' can change between runs because it isn't guarenteed which thread
will execute the runner function first, however the data value will always
be output in the order 1, 2, 3, 4.
\end_layout
\begin_layout Standard
\begin_inset VSpace 1cm
\end_inset
\end_layout
\end_inset