Spring 04-05
ECOM 4416: Operating Systems Concepts
Final Exam
Duration: 2 hours
Student ID: __________________
Student Name: __________________
ECE Department
IUG- Faculty of Engineering
Dr. M. Mikki
Eng. Hasan Qannou'
Eng. Mohammed El-Shekh-Ali
Grade:__________
Notes to students:
1. There are 6 questions and 8 pages in total (including this cover sheet) for this
exam.
2. There are 100 points possible
3. Budget your time appropriately.
4. Write your name and student ID at the top of the first page.
5. In general, show your intermediate work and BRIEFLY state your assumptions
where appropriate.
6. Write clearly and legibly. If we can't read your answer, we can't grade it.
7. Read the questions carefully and answer the question that is being asked.
Question 1
1.
(7 pts) Circle “T” if the statement is always true. Otherwise circle “F”.
1
2
3
4
5
6
7
In paging systems, external fragmentation cannot occur
Race conditions cannot occur on a uni-processor
SJF can be implemented as a priority algorithm, where the
priority is determined by the arrival time of the job.
Hardware support is needed to synchronize more than two
processes
It is possible for a single process to deadlock with itself
Minimizing turnaround time is the most important criteria in
choosing a scheduling algorithm for an interactive system
With dynamic partitioning, all memory allocated for a process
must be contiguous
T/F
T/F
T/F
T/F
T/F
T/F
T/F
2. (5 pts) A DECSYSTEM-20 computer has multiple register sets. Describe the
actions of a context switch if the new context is already loaded into one of the
register sets. What else must happen if the new context is in memory rather
in a register set, and all of the register sets are in use?
Ans: (Pg. 103)
Changing pointers to the current register set.
If a new context is in memory:

Swapping

Copy register data first into a register set.
1.
Question 2
(6 pts) Explain the differences in the degree to which the following scheduling
algorithms discriminate in favor of short processes:
a. FCFS
Answer:
Neutral, does not consider burst time of jobs
b. RR
Answer:
Discriminates against long processes. If a process is longer than the time
slot of the CPU, it has to be preempted, and go to the end of the queue.
Number of \feedbacks a process has to make is directly proportional to
the length of the process.
c. Multilevel feedback queues
2.
(8 pts) Many CPU scheduling algorithms are parameterized. For example, the
RR algorithm requires a parameter to indicate the time slice. Multilevel
feedback queues require parameters to define the number of queues, the
scheduling algorithms for each queue, the criteria used to move processes
between queues, etc. These algorithms are thus really sets of algorithms (e.g.,
the set of RR algorithms for all time slices, etc.). One set of algorithms may
include another (in lecture, we talked about using Priority Scheduling to
implement SJF). What, if any, relation holds between the following pair of
algorithms?
a. Priority and SJF.
b. Multilevel feedback queues and FCFS.
c. Priority and FCFS.
d. RR and SJF.
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Question 3
1.
Consider a paging system with the page table stored in memory.
a. (3 pts) If a memory reference takes 200 nanoseconds, how long does a
paged memory reference take?
Solution
Teffective= fff
b. (3 pts) If we add TLBs, and 75 percent of all page-table references are
found in the TLBs, what is the effective memory reference time? (Assume
that finding a page-table entry in the TLBs takes zero time, if the entry is
there.)
2.
(4 pts) Explain why it is easier to share a reentrant module using
segmentation than it is to do so when pure paging is used.
3.
(10 pts) A system has a 32-bit virtual address space and uses a two-level page
table. 9 bits are used to index the top-level page table, and 11 bits are used to
index the second-level page table.
a) How large are the pages (in bytes)?
b) How many virtual pages can the process have?
c) What is the size (in bytes) of an entry in the second-level page table,
assuming the second-level page table is stored in a single physical page
frame?
d) If each page table entry includes valid, reference, and dirty bits in addition
to the page frame number, what is the maximum number of physical page
frames in the system?
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e) Assuming the page size is fixed, how can the address translation scheme be
altered to support more physical page frames?
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Question 4
1.
(8 pts) A file system is the data structure that holds files and directories.
a. Is it possible to have two file systems on a single physical disk drive? Why
or why not?
b. Is it possible to have a file system that spans two physical disk drives?
Why or why not?
c. Is it possible to have a file that is not in a file system? Why or why not?
d. The mount operation allows you to graft the directory tree from one file
system onto the directory tree of another file system. Does this mean that
a file can span both file systems? Why or why not?
2.
Consider a dynamic partitioning system in which memory consists of the
following holes, sorted by increasing memory address (all sizes are in
Kilobytes):
10 4 18 9 20 7 12 15
Hole created by last allocation
(a) (8 pts) Suppose a new process requiring 11 kB arrives, followed by a
process needing 9 kB of memory. Show the list of holes after both these
processes are placed in memory for each of the following algorithms (start
with the original list of holes for each algorithm).
i) First Fit:
(uses 18 then 10)
4
9
20
7
12
ii) Worst Fit:
(uses 20 then 18)
10
4
9
7
12
iii) Best Fit:
(uses 12 then 9)
10
4
18
7
12
iv) Next Fit:
(uses 15 then 10)
4
18
9
20
7
15
15
15
15
(b) (2 pts) Keeping the hole list sorted by size can make some of the allocation
algorithms faster. State one advantage of keeping the list sorted by address
instead.
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Question 5
1.
( 8 pts) For a multi-threaded process, indicate whether each of the following
items should be part of the process descriptor (P), the thread descriptor (T),
or both (B):
a. Stack pointer
b. Priority
c. Page table
d. Resource usage accounting
P/T/B
P/T/B
P/T/B
P/T/B
2.
( 4 pts) What is the primary motivation for two-level page tables?
3.
(4 pts) Why is it better for disk scheduling to be performed by the disk
controller itself, rather than by the operating system?
4.
(4 pts) What is the difference between a race condition and deadlock?
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Question 6
Answer yes or no, or with a single term or phrase, as appropriate.
1.
(2 pts) Does the test-and-set instruction need to be a privileged instruction?
2.
(2 pts) A program containing a race condition will always/sometimes/never
result in data corruption or some other incorrect behavior?
3.
(2 pts) A system that meets the four deadlock conditions will
always/sometimes/never result in deadlock?
4.
(2 pts) A resource allocation graph that has a cycle in it
always/sometimes/never means that there is a deadlock?
5.
(2 pts) Can the system call instruction be a privileged instruction?
6.
(2 pts) Round-robin scheduling always/sometimes/never results in more
context switches than FCFS?
7.
(2 pts) What are the two components of a virtual address used in segmented
virtual memory?
8. (2 pts) The layout of virtual pages for a virtual address space is
always/sometimes/never contiguous in physical memory?
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1. (4 pts) Potpourri: Answer yes or no, or with a single term or short answer, as appropriate. You
should go through these quickly, answering with the first answer that comes to mind — it
probably is the correct one. Only dwell on ones you are unsure of after finishing the other
questions.
(a) A resource allocation graph that has a cycle in it always/sometimes/never means that there is a
deadlock?
sometimes (see the example in deadlock lecture)
(b) Can the system call instruction be a privileged instruction?
no – if it is, then user programs cannot invoke kernel operations
(c) In what Nachos file was the Semaphore class implemented?
synch.cc
(d) Round-robin scheduling always/sometimes/never results in more context switches than FCFS?
sometimes – if every job has an execution time less than the quantum, then it has the same
number as FCFS.
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Question
2. (15 pts) Give brief answers to each of the following questions.
(a) What is the purpose of system call instructions, and how do they work? Give two examples of
common system calls.
System call instructions enable user programs to invoke operations in the kernel. System call
instructions cause an exception, which causes the CPU to vector into the OS. The OS saves the
process state, detects that it is a system call exception, identifies the system call number from the
argument to the instruction, and invokes the system call routine. When the routine finishes, the
OS restores the process state and returns from the system call.
Two common examples are read() and write().
(b) What is the difference between interrupts and exceptions? Give two examples of each.
Interrupts are asynchronous events external the CPU (e.g., timer interrupt, device interrupt).
Exceptions are synchronous events that occur as the result of executing instructions (e.g., divide
by zero, system call).
(c) What is the difference between fork() and exec() on Unix?
fork() creates a new process and copies the address space from the parent into the new child
process. exec() stops executing the program in the process, overwrites it with a new program,
and starts executing the program at the beginning; exec() does not create a new process.
(d) What is the difference between a race condition and deadlock?
A race condition is caused by more than one thread accessing shared state without
synchronization.
Threads continue to execute, but the program can have incorrect results. In a deadlock,
threads cannot make any progress. They are all waiting to acquire resources that each other
holds (circular wait) while the three other deadlock conditions hold.
(e) What is the difference between Thread::Yield() and Thread::Sleep()?
Yield() places the current thread on the end of the ready list, and context switches to another
thread. Sleep() halts the current thread and context switches to another thread. In particular,
Sleep() does not put a thread on a wait list (other code has to do that since Sleep() does not
know which list the thread should wait on).
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11
Question
3. (6 pts) Round-robin schedulers (e.g., the Nachos scheduler) maintain a ready list or run queue
of all runnable threads (or processes), with each thread listed at most once in the list. What can
happen if a thread is listed twice in the list? Briefly explain how this could cause programs that
use synchronization primitives to break on a uniprocessor.
The problem is that a thread on the ready list twice results in a spurious wakeup. Consider a
thread executing the PingPong procedure in the lecture notes. If a thread is on the ready list
twice, then the first time it runs on the CPU it will execute inside of PingPong and eventually
Wait on the condition variable. But, since it is on the ready list twice, it will run again, without
any other thread having called Signal on the condition variable. This could, for example, cause
that thread to print twice in a row, violating the problem constraints.
The most common error on this problem was to think of the thread on the list twice as two
different threads, with two different stacks, etc. Although this is not accurate, I still gave partial
credit when
an answer with this interpretation demonstrated a synchronization problem.
4
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Question
4. (15 pts) The table below lists five jobs, the time that they arrive in the system (when they are
created), how long they take to execute, and their priority (higher values correspond to higher
priorities). For example, job arrives at time “10”, requires “30” units of execution time to
complete, and has priority “1”.
_
Job Arrival Time Execution Time Priority
A 10 30 1
B 20 50 5
C 30 10 3
D 40 40 2
E 50 20 0
For all jobs in each of the following scheduling algorithms, calculate (a) the Start Time, the time
at which the job is first scheduled to run (it may have to wait when it arrives), and (b) the End
Time, the time when it finishes executing.
(a) First Come First Serve (FCFS)
Job Start Time End Time
A 10 40
B 40 90
C 90 100
D 100 140
E 140 160
(b) Shortest Job First (SJF)
Job Start Time End Time
A 10 50
B 110 160
C 30 40
D 70 110
E 50 70
(SRTF)
Job Start Time End Time
A 10 40
B 110 160
C 40 50
D 70 110
E 50 70
(c) Priority
Job Start Time End Time
A 10 140
B 20 70
C 70 80
D 80 120
E 140 160
5
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Question
5. (15 pts) Consider the following test program for an implementation of locks and condition
variables
in Nachos. It begins when the “main” Nachos thread calls ThreadTest(). Trace the execution of
this
program until it prints out the message “STOP HERE” and (a) write down the sequence of
context
switches that occurred up this point, (b) the output of the program, and (c) list the queues that the
threads are on at this point, and their relative order if more than one thread is on a queue (currentThread, the readyList, and any wait queues associated with synchronization primitives). For
example,
“A B” signifies that thread A context switches to thread B, and “readyList: A” signifies that A is
on the ready list.
Assume that the scheduler runs threads in FIFO order with no preemptive time-slicing (nonpreemptive
scheduling), all threads have the same priority, and threads are placed on wait queues in FIFO
order.
_
Lock *pumpkin;
Condition *cv;
void A(int arg) {
pumpkin->Acquire();
printf(‘‘giddy’’);
cv->Wait(pumpkin);
printf(‘‘gaddy’’);
currentThread->Yield();
cv->Signal(pumpkin);
pumpkin->Release();
}
void B(int arg) {
pumpkin->Acquire();
printf(‘‘goody’’);
cv->Signal(pumpkin);
currentThread->Yield();
printf(‘‘geddy’’);
cv->Wait(pumpkin);
pumpkin->Release();
}
void ThreadTest() {
Thread *t;
pumpkin = new Lock("l");
cv = new Condition("cv");
t = new Thread("A");
t->Fork(A, 0);
currentThread->Yield();
t = new Thread("B");
t->Fork(B, 0);
currentThread->Yield();
currentThread->Yield();
printf(‘‘STOP HERE\n’’);
}
(a) Context switches: main Amain _
_
_
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Bmain ABmain
(b) Output: giddy goody geddy
(c) Thread queues at “STOP HERE”:
currentThread: main
readyList: A
pumpkin:
cv: B
The one subtlety to this problem is at the end. When A is woken up from waiting on the condition
variable, it does not immediately return. Instead, it first tries to reacquire the lock (that is held by
B),
gets placed on the lock’s wait queue, and finally placed on the ready list when B finally calls wait
(but
A never does run again).
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6. (15 pts) Because his house is so popular on Halloween, Greg Ghoul likes to hand out
Halloween
candy in the following way. Each trick-or-treater coming to his door takes a single “ticket” with
a number from a ticket machine on his porch. The ticket numbers dispensed are guaranteed to be
unique and sequentially increasing. When Greg is ready to give candy to the next trick-or-treater,
he
calls out the “candycount”, the lowest unserved number previously dispensed from the ticket
machine.
Trick-or-treaters wait until the candycount reaches the number on their tickets. Greg waits until
the
trick-or-treater with the ticket he called comes forward with their bag held open before giving out
candy.
Show how to implement Greg() and trick-or-treater() procedures to solve this
problem
using locks and condition variables. Represent calling out the “candycount” by printing out its
value
in the appropriate place in Greg(). In addition, print out a message in trick-ortreater()
when the trick-or-treater gets candy and print out a message in Greg() when Greg gives out
candy.
_
_
_
_
// shared variables
Lock mutex;
Condition candy, greg;
int candycount = 0, ticketcount = 0;
void Greg() {
mutex->Acquire();
while (1) {
printf("Candy for %d!\n",
candycount);
candycount++;
candy->Broadcast(&mutex);
greg->wait(&mutex);
printf("Gave away the candy!\n");
}
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mutex->Release();
}
void trick-or-treater() {
int myTicket;
mutex->Acquire(&mutex);
myTicket = ticketcount++;
while (myTicket != candycount) {
candy->Wait(&mutex);
}
printf("Treater %d got candy!\n",
myTicket);
greg->Signal(&mutex);
mutex->Release();
}
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CSE 120 Principles of Computer Operating Systems
Fall Quarter, 2001
Midterm Exam
Instructor: Geoffrey M. Voelker
Name
Student ID
1. (10 pts) Potpourri: Answer yes or no, or with a single term or short answer, as appropriate. You
should go through these quickly, answering with the first answer that comes to mind — it
probably is
the correct one. Only dwell on ones you are unsure of after finishing the other questions.
(a) Does the test-and-set instruction need to be a privileged instruction?
No, it can be executed at user level.
(b) What approach to dealing with deadlock does the Banker’s algorithm implement?
Avoidance.
(c) Which of the following scheduling algorithms can lead to starvation? FIFO, Shortest Job First,
Priority, Round Robin
SJF, Priority
(d) A program containing a race condition will always/sometimes/never result in data corruption
or
some other incorrect behavior?
Sometimes...just because a race condition exists in code does not mean that a particular
execution
will encounter it.
(e) A system that meets the four deadlock conditions will always/sometimes/never result in
deadlock?
Sometimes.
2
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2. (6 pts) Categorize the following as one of the following: (I) interrupt, (E) exception, or (N)
neither.
(a) Timer – (I)
(b) Segmentation violation – (E)
(c) Keyboard input – (I)
(d) Divide by zero – (E)
(e) Procedure call – (N)
(f) System call – (E)
3
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3. (9 pts) Discuss the tradeoffs between user and kernel threads.
(a) What are the advantages and disadvantages of each?
(b) Assume we can make system calls as fast as procedure calls using some new hardware
mechanism.
Would this make one kind of thread clearly preferable over the other? Explain briefly.
(a) See slides 11-15 in Lecture 4.
(b) Kernel threads would be preferable. The primary disadvantage of kernel threads is the
overhead
of trapping to the kernel to manipulate them, context switch, etc. Some people mentioned that
kernel
threads still have the problem of having to be generic to all application needs, which is a good
observation and I gave points for this.
4
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4. (15 pts) The Coronado Bridge is undergoing repairs and only one lane is open for traffic. To
prevent
accidents, traffic lights have been installed at either end of the bridge to synchronize the traffic
going
in different directions. A car can only cross the bridge if there are no cars going the opposite
direction
on the bridge. Sensors at either end of the bridge detect when cars arrive and depart from the
bridge,
and these sensors control the traffic lights.
Below is a skeleton implementation of two routines, Arrive() and Depart(). You may assume that
each
car is represented by a thread, and threads call Arrive() when they arrive at the bridge and
Depart()
when the leave the bridge. Threads pass in their direction of travel as input to the routines.
int num_cars = 0;
enum dir = {open, north, south};
dir cur_dir = open;
void
Arrive (dir my_dir) {
A.-1 lock->Acquire();
A.0 while (cur_dir != my_dir ||
cur_dir != open) {
cv->Wait(lock);
}
A.1 num_cars++;
A.2 cur_dir = my_dir;
A.3 lock->Release();
}
Lock *lock = new Lock();
Condition *cv = new Condition();
void
Depart (dir my_dir) {
D.-1 lock->Acquire();
D.0 num_cars--;
D.1 if (num_cars == 0) {
D.2 cur_dir = open;
cv->Broadcast(lock);
D.4 }
D.5 lock->Release();
}
(a) The code above does not have any synchronization statements. Outline an execution sequence
(referring to the numbered statements) where two threads can cause two cars to travel on the
bridge in opposite directions at the same time.
Thread 1 (north): A.0, A.1 (context switch)
Thread 2 (south): A.0, A.1
and now we have two threads on the bridge in opposite directions.
(b) Show how a condition variable and lock could be used to correctly synchronize the cars by
annotating the above routines with calls to Condition and Lock operations.
See above.
(c) Draw brackets around the sections of code that are critical sections.
Lines A.-1 – A.3, D.-1 – D.5.
(d) Can this solution lead to starvation? Briefly explain.
Yes. Once cars start moving in one direction, they will starve out cars waiting to go in the other
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direction indefinitely (as long as more cars continue to arrive in the first direction).
(e) Can this solution lead to deadlock? Briefly explain.
No, not all of the deadlock conditions are met (e.g., no circular wait). In specific terms, for
example, there is no way for cars going in both directions to wait at the same time.
5
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5. (15 pts) Consider the following test program for an implementation of Join in Nachos. It begins
when
the “main” Nachos thread calls ThreadTest(). You do not need to know the details of how Join is
implemented. All you need to know is that when a “parent” thread calls Join on a “child” thread,
the
“parent” does one of two things: (1) if the “child” is still running, the “parent” blocks until the
“child”
finishes (at which point the “parent” is placed on the ready queue); (2) if the “child” has finished,
the
“parent” continues to execute without blocking. The semantics of Fork are exactly those of
Nachos in
Project 1.
void ThreadTest() {
Thread *t;
t = new Thread("A", 1);
t->setPriority(10);
printf(‘‘fee ’’);
t->Fork(A, 0);
printf(‘‘foe ’’);
t->Join();
printf(‘‘fun\n’’);
}
void A(int arg) {
Thread *t2;
t2 = new Thread(‘‘B’’, 1);
t2->setPriority(20);
printf(‘‘foo ’’);
t2->Fork(B, 0);
printf(‘‘far ’’);
t2->Join();
printf(‘‘fum ’’);
}
void B(int arg) {
printf(‘‘fie ’’);
}
(a) Assume that the scheduler runs threads in round-robin order with no implicit time-slicing (i.e.,
non-preemptive scheduling), priorities are ignored, and threads are placed on queues in FIFO
order. What will this test program print out?
fee foe foo far fie fum fun
(b) Now assume that the scheduler runs threads according to priority. You should not need to
worry
about how priority is implemented. It is enough to know that (1) the highest priority thread on
the ready queue runs first, and (2) when a thread is added to the ready queue, it will preempt the
current thread if the new thread has higher priority. Assume that there is no priority donation,
and that the priority of the “main” thread is 0. What will the test program print out now?
fee foo fie far fum foe fun
6
GOOD LUCK
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