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ser334_exam2_practice.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
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\begin_body
\begin_layout Standard
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SER 334
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Practice
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Exam 2
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\begin_layout Standard
\align center
Updated 4/12/2022
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inverted 0
status open
\begin_layout Standard
NOT FOR STUDENT DISTRIBUTION
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\begin_inset Formula $ $
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Last Name:
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status open
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$
\backslash
rule{6cm}{0.15mm}$
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First Name:
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status open
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$
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rule{5.94cm}{0.15mm}$
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\begin_layout Standard
Last 4 digits of ASU ID:
\begin_inset ERT
status open
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$
\backslash
rule{4.05cm}{0.15mm}$
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\begin_layout Standard
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\align center
Exam Instructions
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\begin_layout Standard
The exam is open one note card (3x5 inches).
\bar under
No electronic items are allowed.
Write legibly.
\bar default
Please use a pen (instead of a pencil) if you have one.
There are 90 points available and the exam must be completed in 37.5 minutes.
This exam has two types of questions:
\end_layout
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\begin_inset Formula $ $
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\begin_layout Standard
\series bold
Multiple choice questions:
\series default
There are 30 points of multiple choice questions.
An answer is selecting one option among the choices given.
Each multiple choice is worth 2 to 5 points.
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\begin_inset Formula $ $
\end_inset
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\begin_layout Standard
\series bold
Short answer questions:
\series default
There are 40 points of short answer questions.
A typical answer is two or three sentences.
Each short answer question is worth ~10 points.
\end_layout
\begin_layout Standard
\begin_inset Formula $ $
\end_inset
\end_layout
\begin_layout Standard
\series bold
Scenario questions:
\series default
The programming questions are given near the end of the paper.
They must be answered on the question paper.
There are 0 points of write-in programming questions.
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\begin_layout Standard
\align right
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<lyxtabular version="3" rows="10" columns="3">
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Topic
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\begin_inset Formula $\;\;$
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\series bold
Possible
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<row>
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\begin_inset Text
\begin_layout Plain Layout
MC: Processes
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</cell>
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<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
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5
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SA: Processes
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20
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</cell>
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<row>
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\begin_inset Text
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MC: Threads
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10
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</cell>
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<row>
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\begin_inset Text
\begin_layout Plain Layout
MC: Process Synchronization I
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5
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SA: Process Synchronization I
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20
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MC: Process Synchronization II
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10
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Scen: Threads
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0
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Scen: Process Synchronization
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Total:
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70
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Processes
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\begin_layout Enumerate
Imagine that you are trying to write a program that performs a data-intensive
computation, and which needs to support some level of recoverability.
Would you be better off using shared data or message passing? Explain.
[5 points] [Acuña, Pilcher]
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\begin_deeper
\begin_layout Enumerate
Shared data – because every thread has individual access to the memory
\end_layout
\begin_layout Enumerate
Message passing – because there will be a record of each message sent which
can be saved
\series bold
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
ANS
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\begin_layout Enumerate
Shared data – because it would require less memory to be used than message
passing
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\begin_layout Enumerate
Message passing – because if a message gets corrupted, it only affects one
other process
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\begin_layout Standard
\begin_inset VSpace 0.5cm
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\begin_layout Enumerate
Trace the following program.
As pids come into existence, assume they come from the following list:
2600, 2601, 2602, 2603, and so on.
[Acuña, Silberschatz]
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inline false
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int main() {
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pid_t pid, pid1;
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\end_layout
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pid = fork();
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if (pid == 0) {
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pid1 = getpid();
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printf("child: pid = %d", pid); /* A */
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printf("child: pid1 = %d", pid1); /* B */
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\begin_layout Plain Layout
}
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\begin_layout Plain Layout
else {
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pid1 = getpid();
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printf("parent: pid = %d", pid); /* C */
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printf("parent: pid1 = %d", pid1); /* D */
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wait(NULL);
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}
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return 0;
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}
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\begin_deeper
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From the program above, identify the values of the process ids at lines
A, B, C, and D.
[10 points]
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inverted 1
status open
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\end_inset
\end_layout
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inverted 0
status open
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\series bold
Ans:
\end_layout
\begin_layout Standard
A: 0
\end_layout
\begin_layout Standard
B: 2601
\end_layout
\begin_layout Standard
C: 2601
\end_layout
\begin_layout Standard
D: 2600
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Trace the previous program and then list all of the possible outputs (just
write the A/B/C/D text; no need for pids) that could be generated.
(Assume that fork will always succeed.) [10 points] [Acuña]
\begin_inset Newline newline
\end_inset
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
ANS:
\series default
ABCD, ACDB, ACBD, CABD, CDAB, CADB
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset VSpace 4cm
\end_inset
\end_layout
\begin_layout Standard
\series bold
Threads
\end_layout
\end_deeper
\begin_layout Enumerate
Consider the following program.
\begin_inset listings
lstparams "language=C"
inline false
status open
\begin_layout Plain Layout
int main() {
\end_layout
\begin_layout Plain Layout
pid_t pid;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pid = fork();
\end_layout
\begin_layout Plain Layout
if (pid == 0) {
\end_layout
\begin_layout Plain Layout
fork();
\end_layout
\begin_layout Plain Layout
thread_create(...);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
thread_create(...);
\end_layout
\begin_layout Plain Layout
return 0;
\end_layout
\begin_layout Plain Layout
}
\end_layout
\end_inset
How many unique threads are created in the program above? [5 points] [Acuña,
Silberschatz]
\end_layout
\begin_deeper
\begin_layout Enumerate
2
\end_layout
\begin_layout Enumerate
3
\end_layout
\begin_layout Enumerate
4
\end_layout
\begin_layout Enumerate
5
\series bold
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
ANS
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Is it faster to switch between processes or threads within those processes?
Explain.
[5 points] [Acuña]
\end_layout
\begin_deeper
\begin_layout Enumerate
Processes - they are permanently assigned to a specific CPU so switching
never occurs.
\end_layout
\begin_layout Enumerate
Processes - they have kernel level privileges to access the PCB handling
functions.
\end_layout
\begin_layout Enumerate
Threads - they don't require the call stack to be saved/restored since the
memory is shared among all threads for a given process.
\end_layout
\begin_layout Enumerate
Threads - they are light weight and don't require saving/restoring many
resources.
\series bold
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
ANS
\end_layout