Page Table Reference Time Calculation

The assignment content is about page-table lookups and synchronization in operating systems. It provides steps to convert a logical address into its decimal equivalent, calculate the page table, and determine the time taken for paged memory reference with and without a Translation Lookaside Buffer (TLB). Additionally, it presents pseudo-code for monitoring external sensors, highlighting potential issues when there is no synchronization, critical sections in the program, and modified pseudo-code using semaphores.

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 1
THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES
by (Student Name)
[Course name]
[Proffessor Name]
[University name]
[City name]
[Date]

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 2
Question 1.
void shuntTrucks() {
/* Code to shunt Trucks - do not touch */
}
void shuntWagons() {
/* Code to shunt Wagons - do not touch */
}
/* Shared Data between all processes */
Semaphore engineSleep=0;
Semaphore mutex=0;
int numWagons=0;
int numTrucks=0;
int buffer[N];
int input=0;
int output=0;
engine = new Semaphore(N);
track = new Semaphore(0);
wagons = new Semaphore(1);
EngineThread()
{
while (True) {
engine = shuntTrucks();
engine =shuntWagons();
wait(track);
wait(wagon);
buffer[input] = track;
buffer[input] = wagon;
input = (input + 1) % N;
signal(wagon);
signal(track);
consume(wagon);

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 3
consume (track);
}
}
WagonThread()
{
while (True) {
engine = shuntTrucks();
engine =shuntWagons();
wait(track);
buffer[input] = track;
input = (input + 1) % N;
signal(engine);
signal(track);
}
}
TruckThread()
{
while (True) {
engine = shuntTrucks();
engine =shuntWagons();
wait(wagon);
buffer[input] = wagon;
input = (input + 1) % N;
signal(engine);
signal(wagon);
}
}

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 4
Question 2
Question 2 a
First-fit:
Computer Processes Memory partitions
333 KiB 621 KiB
421 KiB 693 KiB
140 KiB 288 KiB (new partition from 621-333=288KiB)
503 KiB 693 KiB
- 132 KiB
Best-fit:
Computer Processes Memory partitions
333 KiB 417 KiB
421 KiB 621 KiB
140 KiB 337 KiB
503 KiB 693 KiB
- 132 KiB
Worst-fit:
Computer Processes Memory partitions
333 KiB 693 KiB
421 KiB 621 KiB
140 KiB 417 KiB
503 KiB Must wait for partitions.
-
Among the above algorithms the best is the first fit algorithm.

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 5
Question 2 b
i. Required page table pages for this process?
Four pages tables: There will be four page tables where each page data will be equal to p1 =
p2 = 210 = 1024 ,therefore either of the pages the pages will have size of 1024kbs. However for
the second page it will be pointing to 4MiB which is obtained as offset  2^12  4096kbs =
4Mbs).
Four data pages are required : since 2^10 x data page tables = 4Mbs therefore the number of
data pages is 4.
ii. Lookup steps within the page-tables of the logical address 0x00C0A005.
Steps
a. Converting 0x00C0A005 to decimal number we get 12623877
b. Calculating the page table we obtain 1262387/1024=1232Mibs.
c. Therefore we have 1232 page tables.
iii. Time taken for paged memory reference.
a. No TLB.
In this case the physical memory is required to be accessed for three times in the first and
second levels pages to the physical addresses too ,therefore the paged memory referencing will
require time as follows (William,2015,P. 230).
Total time 3 x 18ns=54nanoseconds.
b. There is a TLB, and .07 ns access speed and TLB does not contain information on the
required page.
For this case the total time will be different because the entry is not found in the
TLB ,therefore total time is given as follows:
Total time = 0.07 ns+3x18ns=54.07nanoseconds.
c. There is a TLB, and .07 ns access speed and TLB contain information on the required
page.

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 6
If there is entries in TLB the page frames will have direct access to it, therefore the time will be
as follows.
Total time: 0.07+20 = 20.07nanoseconds.
Question 3
Pseudo-code for monitor to access the external sensor.
SensorMonitor M;
int main() {
M.LockSensorAccess();
/* Access the sensor, reconfigure the sensor, ... */
M.FreeSensorAccess();
}
Monitor SensorMonitor {
boolean Busystatus; /* this variable test if the thread is engaged or not*/
condition w; /*this is a variable to set the time to wait before joining the thread*/
void LockSensorAccess()(int timetaken) {
if (Busystatus)
w.wait(timetaken);/* this is where time that the thread will have to wait before joining*/
}
void FreeSensorAccess(){
Busystatus = false;
w.signal()
}
initialization.code()
{
Busystatus = false;
}
}

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 7
Question 4
a. What potentially could go wrong
When there is no synchronization there will be possibility of having some problems
which could include:
i. Arising of Race conditions: This is where there will be competition of the frames that
need to display thus without synchronization the resources will not be well utilized.
ii. Occurrence of deadlock: This is a situation where there are many frames in the queues
and there is one frame that takes very long before it allow next frame to display
(Tom,2014,P. 130)
b. Critical section
The above pseudo code is used to get and display the frames from the camera however
there are some of critical sections in the program as described below:
The section that allows execution of one process since the variables being used are shared
which includes:
i. GetFrame(){ } and
ii. ShowFrame(){ }
However this can only be solved through solving of mutually exclusive problems by ensuring
the following is implemented (Gary, 2016,256)
i. The processes are locked before critical section is entered.
ii. Unlocking process after executions in the critical sections.
iii. Waiting to use critical sections in case it is locked.
iv. Avoid waiting forever before entering the critical sections.
c. Modified pseudo-code using semaphores.
/* in global memory area accessible by threads */
#define N 100
struct frame *emptyStack[N];
struct frame *displayQueue[N];
int processes;

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 8
int main() {
/*
** Allocate memory for N frames
** Place the frames' addresses into the empty Stack
*/
InitialiseStackMemory();
thread_t tid1, tid2;
thread_create(&tid1, GetFrame);
thread_create(&tid2, ShowFrame);
sleep(300);
}
GetFrame() {
struct frame *frame;
struct frame local;
while (processes>=1) {
CameraGrab(&local); /* get a frame from the camera */
frame = Pop(); /* pop an empty frame address
from the empty stack */
CopyFrame(&local, frame); /* copy data from the local frame
to the frame address*/
Enqueue(frame); /* push the frame address to the display queue */
wait(3000);
resume();
thread_t.join();
}( Charles,2015,P. 120).
}
ShowFrame() {
struct frame *frame;
struct frame local;
struct frame filtered;

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 9
while (processes>=1) {
frame=Dequeue(); /* pop the leading full frame from
the full queue */
CopyFrame(frame, &local); /* copy data to the local frame */
Push(frame); /* push the frame address to the
empty stack */
Solarise(&filtered, &local);
VideoDisplay(&filtered);
thread_t.start();
}
}(Avi , 2014,P.200)

THE OPERATING SYSTEM THREADS, MEMORY, PROCESSES 10
References
(Jones, 1985, p. 74)
Avi ,P.(2014) Operating System Concepts.2nd edn.Westminster :
willey publishers.
Charles, C.(2015) Operating System: A Design-oriented Approach.1st edn. New York:
Oxford University Press.
Gary,J.(2016) Operating Systems: A Modern Perspective.2nd edn.New Delhi:
Addison-Wesley
Tom, C.(2014) Microsoft Windows Operating System Essentials.3rd edn.
London: Sybex publishers
William,S(2015). Operating Systems: Internals and Design Principles .1st edn.
New Delhi: Prentice Hall.