Quiz: Paper 2 CS System Software & Operating System Part 1

In concurrent programming, concurrent accesses to shared resources can lead to unexpected or erroneous behavior, so parts of the program where the shared resource is accessed are protected. This protected section is the critical section .

Mutual Exclusion is a condition when one or more than one resource are non-sharable (Only one process can use at a time) i.e. processes claim exclusive control of the resources they require.

“fork() creates a new process by duplicating the calling process, The new process, referred to as child, is an exact duplicate of the calling process, referred to as parent, except for the following :
The child has its own unique process ID, and this PID does not match the ID of any existing process group.
The child’s parent process ID is the same as the parent process ID.
The child does not inherit its parent’s memory locks and semaphore
adjustments.
The child does not inherit outstanding asynchronous I/O operations from its parent nor does it inherit any asynchronous I/O contexts from its parent.”

“The values stored in registers, stack pointers and program counters are saved on context switch between the processes so as to resume the execution of the process.
There’s no need of saving the contents of TLB as it is invalidated after each context switch.”

“P: Wait operation decrements the value of the counting semaphore by 1.
V: Signal operation increments the value of counting semaphore by 1.

Current value of the counting semaphore = 10
a) after 3 P operations, value of semaphore = 10-3 = 7
d) after 2 v operations, and 5 operations value of semaphore = 10 + 2 – 5 = 7
“

“logical address space = 8 pages of 1024 words
number of bits in logical address space = p (page bits) + d (offset bits)
number of bits = log28 + log21024 = 3 + 10 = 13 bits”

“In Round Robin algorithm, it is very important to choose the the quantum carefully as smaller time quanta leads to more context switches thereby reducing the efficiency of the CPU and the larger time quanta makes the round robin algorithm regenerate into FCFS algorithm.
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“In Round Robin algorithm, it is very important to choose the the quantum carefully as smaller time quanta leads to more context switches thereby reducing the efficiency of the CPU and the larger time quanta makes the round robin algorithm regenerate into FCFS algorithm.
“

In Optimal Page replacement algorithm, pages are replaced which are not used for the longest duration of time in the future. This page replacement algorithm ensures the lowest page fault rate.

“Explanation: An operating system has three main functions:
(1) manage the computer’s resources, such as the central processing unit, memory, and other input – output sources
(2) establish a user interface, and
(3) execute and provide services for applications software.
OS provides an interface between the user and the hardware and thus making the computer easy to use for the user but the primary function of OS is to manage the hardware in the most efficient way.
“

“Scan is disk scheduling algorithm
In the round robin algorithm, each process will execute for particular time quantum,So it is the example for time sharing,
Stack is last in first out(LIFO)
In the batching processing, whatever process arrives first, it execute that process. So it First in First out(FIFO)”

A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process

Number of min resources required = (3-1) + (4-1) + (6-1) + 1 = 11

Only Single resource available never occur deadlock because it violates circular wait and hold & wait condition.

“Here, 20P operations means 20 wait operations. It decrement value by 1 every time.
xV operations means x increments operations. It increment value by 1 every time.
→ New value of semaphore is 5 after performing xV operations
= -13 + xV
= 5
xV = 5 + 13
= 18
→ After applying 20P operations in semaphore value is = 7-20 = -13
“

If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.

“→ Working set defines the amount of memory that a process requires in a given time interval.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
“

“Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32 – 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 – 5 = 15 bits
“

“→ Preemption is the act of temporarily interrupting a task being carried out by a computer system, without requiring its cooperation, and with the intention of resuming the task at a later time. Such changes of the executed task are known as context switches.
→ It is normally carried out by a privileged task or part of the system known as a preemptive scheduler, which has the power to preempt, or interrupt, and later resume, other tasks in the system.
→ Preemptive scheduling is the most suitable scheduling scheme for real time operating systems as it can always preempt other processes based on priority.
→ In real time operating systems, we have different tasks to perform simultaneously and also there are a few tasks which need to be performed first at priority basis.
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“→ Belady’s anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
→ This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
→ In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
“

“Spooling is a specialized form of multi-programming for the purpose of copying data between different devices.
A dedicated program, the spooler, maintains an orderly sequence of jobs for the peripheral and feeds it data at its own rate
Conversely, for slow input peripherals, such as a card reader, a spooler can maintain a sequence of computational jobs waiting for data, starting each job when all of the relevant input is available
“

“A process is the instance of a computer program that is being executed by one or many threads. It contains the program code and its activity.
Depending on the operating system (OS), a process may be made up of multiple threads of execution that execute instructions concurrently”

“→ While using a larger block size means that contains less number of blocks then that results better throughput.
→ This can be implemented in a fixed block size then the space utilization is not upto the mark. So the statement results better disk throughput but poorer disk space utilization
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“→ When a process is rolled back as a result of deadlock the difficulty which arises is starvation.
→ Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work.
→ Starvation may be caused by errors in a scheduling or mutual exclusion algorithm, but can also be caused by resource leaks, and can be intentionally caused via a denial-of-service attack such as a fork bomb
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“→Page size is total space taken up by page and Page table entry size is memory taken for indexing the Page in Page Table
→Size of logical address = 32 bits
→Page size = 4K =22210=212 Bytes
→Number of pages = logical address space/ size of each page = 232/ 212= 220
→Page table size = number of pages * size of a page table entry
= 220 * 22
= 222
“