Операционные системы (база, состоящая из вопросов с РК1 и РК2)

Примечание: все вопросы, которые находятся в данном модуле являются частью ФИНАЛЬНОГО ЭКЗАМЕНА за исключением трёх задач.

Identify these registers:

Used to exchange data between I/O module and processor: I/OBR.
The MBR (memory buffer register) contains the data to be written into memory and receives the data read from memory.
Specifies the address in memory for the next read or write: MAR.
I/O address register to specify a particular I/O device: I/OAR.
Contains the most recently fetched instruction: IR (instruction register).
Holds the address of the instruction to be fetched next: PC (program counter).
Contains status information: PSW (program status word).
Bits set by the processor or hardware as the result of operations: condition codes.
Dedicated register that points to the top of the stack: SP (stack pointer).

Identify the process states described below:

The process cannot execute until some event occurs: Blocked.
The process is in the secondary memory and is available for execution (as soon as it is loaded into memory): Ready, suspend.
The process is in the secondary memory and awaiting an event: Blocked, suspend.
The process has been released from the pool of executable processes: Exit.
The process is currently being executed: Running.
The process is prepared to execute: Ready.
The process has been created, but has not yet been admitted to the pool of executable processes: New.

4th task (практические задания):

Задача №1: A computer has a cache, main memory and a disk used for virtual memory. An access to the cache takes 10 ns. An access to main memory takes 20 ns. An access to the disk takes 10 000 ns. Suppose the cache hit ratio is 0.9 and the main memory hit ratio is 0.8. What is the effective access time (EAT) in ns required to access a referenced word on this system? (Write formula and then show the solution):

Cache access time: 10 ns – T1.

Memory access time: 20 ns – T2.

Disk access time: 10 000 ns – T3.

Cache hit ratio: 0.9 – H1.

Memory hit ratio: 0.8 – H2.

Formula: T(avg) = (H1*T1) + (1 – H1)*(H2*(T1+T2) + (1 – H2)*(T1+T2+T3))

Solution: (0.9*10) + (1 – 0.9)*(0.8*(10+20) + (1 – 0.8)*(10+20+10 000)) = 212.

Задача №2: A computer has a cache, main memory and a disk used for virtual memory. An access to the cache takes 15 ns. An access to main memory takes 45 ns. An access to the disk takes 10 000 ns. Suppose the cache hit ratio is 0.8 and the main memory hit ratio is 0.7. What is the effective access time (EAT) in ns required to access a referenced word on this system? (Write formula and then show the solution):

Cache access time: 15 ns – T1.

Memory access time: 45 ns – T2.

Disk access time: 10 000 ns – T3.

Cache hit ratio: 0.8 – H1.

Memory hit ratio: 0.7 – H2.

Formula: T(avg) = (H1*T1) + (1 – H1)*(H2*(T1+T2) + (1 – H2)*(T1+T2+T3))

Solution: (0.8*15) + (1 – 0.8)*(0.7*(15+45) + (1 – 0.7)*(15+45+10 000)) = 624.

Задача №3: A computer has a cache, main memory and a disk used for virtual memory. An access to the cache takes 10 ns. An access to main memory takes 30 ns. An access to the disk takes 10 000 ns. Suppose the cache hit ratio is 0.9 and the main memory hit ratio is 0.7. What is the effective access time (EAT) in ns required to access a referenced word on this system? (Write formula and then show the solution):

Cache access time: 10 n – T1.

Memory access time: 30 ns – T2.

Disk access time: 10 000 ns – T3.

Cache hit ratio: 0.9 – H1.

Memory hit ratio: 0.7 – H2.

Formula: T(avg) = (H1*T1) + (1 – H1)*(H2*(T1+T2) + (1 – H2)*(T1+T2+T3))

Solution: (0.9*10) + (1 – 0.9)*(0.7*(10+30) + (1 – 0.7)*(10+30+10 000)) = 313.

Important extra information (from lectures of Mishina A. Y.):

Run through the four states:

Ready: The process is in main memory and available for execution.
Blocked: The process is in main memory and awaiting an event.
Blocked/Suspend: The process is in secondary memory and awaiting an event.
Ready/Suspend: The process is in secondary memory, but is available for execution as soon as it is loaded into main memory.

Operating Systems: Quiz 4

Semaphore is a/an integer variable to solve critical section problem.
An un-interruptible unit is known as: atomic.
A state is safe, if: the system can allocate resources to each process in some order and still avoid a deadlock.
A situation where several processes access and manipulate the same data concurrently and the outcome of the execution depends on the particular order in which access takes place is called: race condition.
The initial value of the semaphore, that allows only one of the many processes to enter their critical sections, is: 1.
Which of the following cannot be interrupted? Atomic operation.
Four necessary conditions for deadlock to exist are: mutual exclusion, no-preemption, circular wait and «hold and wait».
Deadlock inevitable region can be referred as: fatal region.
This matrix defines the amount of resources of type j requested by process i: request matrix.
This matrix defines the current allocation to process i of resource j: allocation matrix.
«Hold and wait» condition for deadlock is defined as: process holds a resource while awaiting for other resource.
Which of the following is deadlock avoidance algorithm? Banker’s algorithm.
The segment of code in which the process may change common variables, update tables, write into files is known as: critical section.
A problem encountered in multitasking when a process is perpetually denied necessary resources is called: starvation.
What is the reusable resource? A resource that can be used by one process at a time and is not depleted by that use.
This illustrates the progress of two processes competing for two resources: joint progress diagram.
This resource can safely be used by only one process at a time and is not depleted by that use: reusable resource.
Current allocation of resources to processes where there is at least one sequence that does not result in deadlock is called: safe state.
Current allocation of resources to processes where there is no sequence that does not result in deadlock is called: unsafe state.
The resource vector defines: total amount of each resource in the system.
The available vector defines: total amount of each resource NOT allocated to any process.
«No preemption» condition for deadlock is defined as: No resource can be forcibly removed from process holding it.
A minimum of two variables is/are required to be shared between processes to solve critical section problem.
For non-sharable resources like a printer, mutual exclusion: must exist.
For sharable resources, mutual exclusion: is not required.
A monitor is type of: high-level synchronization construct.
Part of program, where the shared memory is accessed and which should be executed invisibly, is called: critical section.
Which of the following is not one of the principles of dealing with deadlock? Initialization.
Deadlock inevitable region can be referred to as a: fatal region.
When each process in the set is blocked awaiting for resource seized by another blocked process: it is deadlock.
What is not an example of reusable resource? Interrupt.
This resource can be created and destroyed: consumable resource.
What is NOT an example of consumable resource? Information in I/O buffers.
A binary semaphore is a semaphore with integer values (multiple choices): 0 and 1.
If more than one user or program attempts to make use of a shared resource at the same time, it may cause the error of: failed mutual exclusion.
Mutual exclusion means that: if a process is executing in its critical section, then no other process must be executing in their critical sections.
The two kinds of semaphores are: counting, binary.
The degree of multi-programming is: the number of processes in memory.
The objective of multi-programming is to: maximize CPU utilization (or have process running at all times).
Listing the sequence of instructions that are executed is called: trace.
Fixed-length block of data in secondary memory is called: page.
A relationship between processes such that each has some part (critical section) which must not be executed while the critical section or another is being executed is known as: mutual exclusion.
What is a method of memory allocation by which the program is subdivided into equal portions, or pages and core is subdivided into equal portions or blocks? It is a method of allocating processor time? Paging.
The address of a page table in memory is pointed by: page table base register.

Operating Systems: Quiz 3

Choose the storage management responsibility for «Many application programs require means for storing information for extended periods of time, after the computer has been powered down»: long-term storage.
The principal tool available to system programmers in developing the early multiprogramming and multiuser interactive systems: Interrupt.
Utilities are: application programs.
Job control language was used to: provide instructions to the monitor to improve job setup time.
The hardware and software used in providing applications to a user can be viewed in a: layered (hierarchical) fashion.
The Process Control Block is: data structure.
Which one of the following is synchronization tool? Semaphore.
What controls the process in a single-threaded process model? Process Control Block (unknown).
In the non-blocking send: the sending process the message and resumes operation.
The degree of multi-programming is: the number of process in memory.
The first fit, best fit and worst fit are strategies to select a: free hole from a set of available holes.
For every process there is a: page table.
The memory allocation scheme subject to internal fragmentation is: fixed partitioning.
In memory systems, boundary registers: track the beginning and the ending of programs.
Any program, no matter how small, will occupy an entire partition results in: internal fragmentation.
This system handles multiple interactive jobs: multiprogramming batch system.
«More than one attempts to make use of a shared resource at the same time». This error is caused by: failed mutual exclusion.
«Two or more programs hung up waiting for each other». This error is caused by: deadlocks.
«Lost signals, duplicate signals received». This error is caused by: improper synchronization.
«The OS as a resource manager» corresponds to the following objective: Efficiency.
«The OS as a User/Computer Interface» corresponds to the following objective: Convenience.
Choose the storage management responsibility for «Programmers should be able to define program modules and to create, destroy and after the size of modules dynamically»: support of modular programming.
Choose the storage management responsibility for «The OS can achieve efficiency by assigning memory to jobs only as needed»: automatic allocation and management.
The context of a process in the PCB of a process DOESN’T contain: context switch time.
The link between two processes P and Q to send and receive messages is called: communication link.
Which of the following state transitions is not possible? Blocked to running.
If a page number is not found in the TLB, then it is known as a: TLB miss.
If a page table entry is not in a main memory, then it is known as a: page fault.
With paging there is no external fragmentation.
Fixed-length block of data in secondary memory is called: pages.
When the free memory is split into many unconnected pieces we call it: external fragmentation.
The address of a page table in memory is pointed by: page table base register.
Memory partitioning algorithm that uses binary tree as a data structure is called: Buddy system.
First fit is generally faster than best fit (and worst fit).
Main memory is broken into fixed-sized blocks called: frames.
Which of the following is not true about the memory management? Virtual memory is used only in multi-user systems.
The state of a process is defined by: the current activity of the process.
What is the ready state of a process? When process is scheduled to run after some execution.
What is a medium-term scheduler? It selects which process remove from memory by swapping.
What is a short-term scheduler? It selects which process has to be executed next and allocates CPU.
What is inter-process communication? Communication between two threads of same process.
Mutual exclusion can be provided by the: both.
The process index register: contains the index into the process list of process currently controlling the processor.
The end user views a computer system in terms of: a set of applications.
In operating system each process has its own: all of the mentioned.
There were no OS with: serial processing.
Inter-process communication: allows processes to communicate and synchronize their actions without using the same address space.
A sequence of instructions in a computer language to get the desired result is known as: program.
In contiguous memory allocation: each process is contained in a single contiguous section of memory.
A system clock generates an interrupts at a rate of approximately every 0.2 seconds. At each clock interrupt regained control and could assign processor to another user. Technique is known as: time slicing.
In internal fragmentation, memory is internal to a partition and: is not being used.
The operating system maintains a frame table that keeps track of how many frames have been allocated, how many are there and how many are available.
The algorithm is slow and may even tend to fill up memory with useless holes: Best fit.
Which of the following is false? Internal fragmentation is increased with small pages.
Short-term scheduler, or dispatcher, picks a process. Each process in the queue is given some time in turn. The strategy is called: round-robin technique.
The limit register: defines the size of the region of memory occupied by the process (in bytes or words).
Dispatcher is: small program which switches the processor from one process to another.
In operating system each process has its own: all of the mentioned.
Process spawning is: creation of child processes.
The code that changes the value of the semaphore is: critical section code.

Вопросы с Рубежного Контроля

The process is in the secondary memory and awaiting an event: Blocked, suspend.
In the Ready/Suspend state: the process is available for execution as soon as it is loaded into memory.
In the Running state: the process is currently being executed.
In the Blocked state: the process cannot execute until some event occurs.
In which of these states the process is in secondary memory: blocked/suspend & ready/suspend.
In which of these states the process is in main memory: Running, blocked, ready.
Identify which state the process will transit: P1 executes a command to read from disk unit 3: Blocked.
Identify which state the process will transit: An interrupt occurs from disk unit 1 — P2’s read is complete: Ready.
Identify which state the process will transit: P1 time slice expires: Ready.
Identify which state the process will transit: P4 terminates: Exit.
Identify which state the process will transit: P3 swapped out: Ready/Suspend.
Identify which state the process will transit: P2 is chosen by dispatcher: Running.
Which of the following is not the state of the process? Old.
Which of the following state transitions is not possible? Blocked to running.
Suppose that the process is in “Blocked” state waiting for some I/O device. When the service is completed, it goes to the: Ready state.
With the use of swapping one other state must be added to the 5-state model: Suspend state.
We can characterize the behavior of an individual process by listing the sequence of instructions that execute for that process. Such a listing is referred to as: trace.
The state of a process is defined by: the current activity of process.
The Process Control Block is: Data Structure.
A reason for swapping: processor is so much faster than I/O it will be common for all processes in memory to be waiting for I/O.
These tables are used to keep track of both main and secondary memory: Memory tables.
A task in a blocked state: is waiting for same temporarily unavailable resources.
A program: is a device that performs a sequence of operations specified by instructions in memory.
A memory: is a service where information is stored.
A processor: is a device where information is stored.
The process is in main memory and available for execution: Ready.
The OS uses this table to know the location in main memory being used as the source of destination of the input/output transfer: I/O tables.
the banker's algorithm is reffered to as: resource allocation denial
The code that changes the value of the semaphore is: critical section code
Two kinds of semaphores: binary, counting
Which of the following condition is required for deadlock to be possible: all of the above
A state is safe if: the system can allocate resources to each process in some order and avoid a deadlock
When memory is divided into several fixed sized partitions, each partition may contain: · exactly one process
Which of the following is not a principle of storage management responsibilities: process termination
A solution to the problem of external fragmentation is: compaction
In fixed sized partition, the degree of multiprogramming is bounded by: the number of partitions
The memory allocation scheme subject to "external" fragmentation is? · segmentation
The principal tool available to system programmers in developing the early multiprogramming and multiuser interactive jobs: Interrupt.
What is not example of consumable resource? Main memory.
A system is in safe state, if: the system can allocate resources to each process in some order and still avoid deadlock.
A set of processes is deadlock if: each process is blocked and will remain so forever.
With multiprogramming only one process can execute at a time; meanwhile all other processes are waiting for the processor. With multiprocessing more than one process can be running simultaneously each on a different processor.
A process can be terminated due to: all of the mentioned (fatal error, normal exit, killed by another process).
In a multiprogramming environment: more than one process resides in memory.
The principal objective of time-sharing system is to: minimize response time.
The process that can be created and destroyed: consumable resource.

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