08.705 RTOS (EL-III)

Module 1 University Questions and Solutions

Part A Questions and Solutions1. What are the functions of an OS? (Nov. 2012)Ans:The basic functions of an operating system:• OS provide services to programs and to users of those

programs. It provide environment for the execution ofprograms.• The common functions are:

1. Program execution: Loads the program into memory andexecutes the program. Also end its execution normally orabnormally.

2. I/O operation: Open or close any I/O device or any file.3. File system management: Read a file or write a file.4. Communications: Share the data between different

processes.5. Error detection: To detect and make appropriate action to

correct the errors for consistent computing.

2. What are the benefits of a multiuser OS as against a single user OS? (Nov. 2012)

Ans:

Advantages of multi-user OS:-

1. As the OS is handling multiple application at a time most of the CPU time is efficiently using.

2. Resources like memory, CPU are utilizing at the maximum.

3. Multiple users can use a system at a time.

4. Networked, distributed system can be implemented.

3. Describe how are system calls used while a file is copiedfrom a source location to destination location? (Oct.2011)

Ans: System call:• Provides an interface to the OS services.• Available as routines written in C and C++, also certain

low-level tasks may be in assembly language instructions.

• Types of system calls:1. Process control2. File manipulation3. Device manipulation4. Information maintenance5. Communication6. Protection

• An example to illustrate how system calls are used for writing a simple program to read data from one file and copy them to another file.

4. What is microkernel? (Oct. 2011)Ans:Kernel:• In one view, the kernel is the lowest level ofsoftware, immediately above the hardwarearchitectural level. In other view, it is the part of theOS that resides permanently in main memory. Byanother view, kernel is the part of the OS that runsin a protected mode.

Micro-kernel:• OS kernels can be quite small, consisting of most of

the two classes (Process and thread management,Interrupt and trap handling) and some limitedresource management, usually processor schedulingand perhaps some low-level virtual memoryallocation.

• The classes are:

1.Process and thread management: Processcreation, destruction, and basic interprocesscommunication and synchronization.

2.Interrupt and trap handling: Responding tosignals triggered by various system events;among these are the termination of aprocess, completion of an I/O operation, atimer signal indicating a timeout or clocktick, an error caused by a program, orhardware malfunctioning, etc.

5. List the criterion for considering the scheduler design(Oct. 2011)

Ans:

Scheduling Criteria:

• Different CPU-scheduling algorithms have differentproperties, and the choice of a particular algorithmmay favor one class of processes over another.

• The criteria include the following:

1.CPU utilization: To keep the CPU as busy as possible.

2.Throughput: The number of processes that arecompleted per time unit. For long processes, thisrate may be one process per hour; for shorttransactions, it may be ten processes per second

3. Turnaround time: It is the interval from thetime of submission of a process to the time ofcompletion. It is the sum of the periods spentwaiting to get into memory, waiting in the readyqueue, executing on the CPU, and doing I/O.

4. Waiting time: Waiting time is the sum of theperiods spent waiting in the ready queue.

5. Response time: This measure, called responsetime, is the time it takes to start responding, notthe time it takes to output the response. Theturnaround time is generally limited by thespeed of the output device

• It is desirable to maximize CPU utilization andthroughput and to minimize turnaround time,waiting time, and response time

6. What is shared data problem? (Nov. 2012)

Ans:

Shared data problems:

• ISRs and task code must share one or more variables that they can use to communicate with one another.

• The shared-data problem arises when the ISRs need to communicate with the rest of the code.

• A classical code with shared-data problem is shown below.

• The main code monitors two temperatures, whichare always supposed to be equal. If they differ, itindicates an alarm.

• The main code stays in an infinite loop and ISRvReadTemperatures happens periodically due tothe hardware interrupts if one or both of thetemperatures changes.

• Suppose both temperatures have been 73o for awhile and processor finished the execution of lineof code iTemp0 = iTemperatures[0];ie, iTemp0 = 73.

• Suppose an interrupt occurs and that bothtemperatures changed to 74o. The ISR writes thevalue into the elements of iTemperatures array.

• When ISR ends, processor continues the line of code iTemp1 = iTemperatures[1];

• This sets iTemp1 = 74. When the processor compares iTemp0 and iTemp1, they will differ and system set the alarm even though the actual values are same.

• This situation is called shared data problem.

7. What is critical section problem? (Oct. 2011)Ans:Atomic and Critical Section:• A part of the program is said to atomic if it cannot be

interrupted. By another way, we can say that the taskcode uses the shared data that is not atomic.

• When we disable interrupts around the lines of task codethat uses the shared data, we can make the collection oflines as atomic and can solve the problem.

• A part of the code is atomic means that it cannot beinterrupted at all, but it cannot be interrupted byanything that might mess up the data it is using.

• A set of instructions that must be atomic for the systemto work properly is called critical section.

• Critical section problem is share-data problem. Theshared-data problem arises when the ISRs need tocommunicate with the rest of the code.

Part B Questions and Solutions1. Define process. Explain the operations on

process. (Oct. 2011)

Ans:

Process:

• OS executes variety of programs and is called jobs, tasks etc.

• Process is a program in execution and process execution must progress in sequential fashion.

• The structure of process in memory is:

• A process is more than the program code, which issometimes known as the text section.

• It includes the current activity, as represented bythe value of the program counter (PC) and thecontents of the processor's registers.

• A process generally also includes the process stack,which contains temporary data (such as functionparameters, return addresses, and local variables),and a data section, which contains global variables.

• A process may also include a heap, which is memory that is dynamically allocated during process run time.

• Process is an active entity, but a program is passive entity, which is also referred as executable file.

Operation on process:

• The processes in most system can execute concurrently, and they may be created and deleted dynamically. Therefore two mechanisms provided by OS for this are:

1. Process creation

2. Process termination

Process creation:• A process may create several new processes using

create-process system call during the execution ofprocess. The creating process is called parent processand the new processes are called children of thatprocesses, forming a tree of processes.

• Most OS identify the processes according to a uniqueprocess identifier (pid), which is typically an integernumber

• Parent and children can share all resources. Children share subset of parent’s resources, but parent and child share no resources.

• When a process creates a new process, two possibilities exist in terms of execution:– Parent and children execute concurrently.– Parent waits until children terminate.

• There are also two possibilities in terms of theaddress space of the new process:

– Child duplicate of parent.

– Child has a program loaded into it.

• A UNIX example:

fork() is a system call and create a new process. This allows the parent process to communicate with child process. Both process continue execution after system call fork(), but one difference: the return code for the fork()is zero for child process, whereas the a nonzero pid of the child is returned to the parent.

exec() is a system call after fork(). Parent can create more children or it has nothing else to while the child runs, it can issue a wait() system call to move itself in the ready queue until the termination of child.

Process Termination:

• Process executes last statement and asks the OS to delete it by using the exit system call.

• Output data from child to parent via wait system call.

• Process’s resources are deallocated by OS

• Parent may terminate execution of children processes via abort system call for a variety of reasons, such as:

– Child has exceeded allocated resources.

– Task assigned to child is no longer required.

– Parent is exiting, and the operating system does not allow a child to continue if its parent terminates.

• A UNIX example to terminate a process is system call exit().

2. Explain various memory management schemes. (Nov 2012)

Ans:

Memory Allocation Schemes:

• The main memory must accommodate both the OS and the various user processes.

• There is a need to allocate main memory in the most efficient way possible.

• Two types of memory allocation scheme:

1. Contiguous Allocation

2. Non-contiguous Allocation

Contiguous Allocation:

• Main memory usually into two partitions:

1. Resident OS, usually held in low memory with interrupt vector.

2. User processes held in high memory

• Multiple-partition allocation: fixed-sized partitions and variable-sized partition

• A simplest method for allocating memory is to divide memory into several fixed-sized partitions.

• When a partition is free, a process is selected from the input queue and is loaded into the free partition.

• When the process terminates, the partition becomes available for another process

• In variable-sized partition scheme, OS keeps a table indicating which parts of memory are available and which are occupied.

• Initially, all memory is available for userprocesses and is considered one large block ofavailable memory called hole.

• When a process arrives, it is allocated memoryfrom a hole large enough to accommodate it.

• The memory blocks available comprise a set ofholes of various sizes scattered throughoutmemory.

• OS maintains information about the allocated partitions and free partitions (hole).

• Dynamic Storage-Allocation Problem; which concerns how to satisfy a request of size n from a list of free holes.

• Many solutions to this problem. The first-fit, best-fit and worst-fit strategies are the common, used to select a free hole from the set of available holes.

1. First-fit: Allocate the first hole that is big enough

2. Best-fit: Allocate the smallest hole that isbig enough; must search entire list, unlessordered by size. Produces the smallest leftoverhole.

3. Worst-fit: Allocate the largest hole; mustalso search entire list. Produces the largestleftover hole.

• This type allocation causes externalfragmentation: Due to the gaps betweenallocated contiguous memory. The totalmemory space exists to satisfy a request, butit is not contiguous.

Non-Contiguous Allocation:Paging:• Memory-management scheme that permits the

physical address space of a process to benoncontiguous.

• Process is allocated physical memory whenever itis available.

• Paging avoids external fragmentation and theneed for compaction.

• Divide physical memory into fixed-sized blockscalled frames (size is power of 2, between 512bytes and 8192 bytes).

• Divide logical memory into blocks of same sizecalled pages.

• Keep track of all free frames.

• To run a program of size n pages, need to find n free frames and load program.

• Set up a page table to translate logical to physical addresses.

• Cause internal fragmentation: Allocated memory may be slightly larger than requested memory; this size difference is memory internal to a partition, but not being used.

• Hardware support for paging:

• Physical address generated by CPU is dividedinto:

Page number (p) – used as an index into apage table, which contains base address ofeach page in physical memory

Page offset (d) – combined with base addressto define the physical memory address that issent to the memory unit

3. What is paging? How paging is implemented with TLB. (Oct. 2011)

Ans: Paging:

• Non-Contiguous memory allocation scheme.

• Memory-management scheme that permits the physical address space of a process to be noncontiguous.

• Process is allocated physical memory whenever it is available.

• Paging avoids external fragmentation and the need for compaction.

• Divide physical memory into fixed-sized blocks called frames (size is power of 2, between 512 bytes and 8192 bytes).

• Divide logical memory into blocks of same size called pages.

• Keep track of all free frames.

• To run a program of size n pages, need to find n free frames and load program.

• Set up a page table to translate logical to physical addresses.

• Cause internal fragmentation.

• Hardware support for paging:

• Address generated by CPU is divided into:

1. Page number (p) – used as an index into a page table, which contains base address of each page in physical memory

2. Page offset (d) – combined with base address to define the physical memory address that is sent to the memory unit

• The paging model example:

• If the size of the logical address space is 2m, and a page size is 2n addressing units (bytes or words) then the high-order m- n bits of a logical address designate the page number p, and the n low-order bits designate the page offset d.

• The logical address is as follows:

Where p is an index into the page table and d is the displacement within the page.

Implementation of Page Table

• Page table is to improve the speed of page access with large number of pages.

• Hardware implementation of the page table can be done in several ways.

• A simple way is the page table is implementedas a set of dedicated registers. If the addressconsists of 16 bits, and the page size is 8KB.The page table thus consists of eight entriesthat are kept in fast registers. The use ofregisters for the page table is satisfactory ifthe page table is reasonably small (forexample, 256 entries). This is not suitable forlarge page table, because fastest registers withlarge size in CPU is not feasible.

• Another method is keeping page table in main memory. There is a Page-Table Base Register (PTBR) points to the page table and a Page-Table Length Register (PTLR) indicates size of the page table.

• In this scheme every data/instruction access requires two memory accesses. One for the page table and one for the data/instruction. Thus, memory access is slowed by a factor of 2.

• The two memory access problem can be solved by the use of a special fast-lookup hardware cache called associative memory or Translation Look-aside Buffers (TLBs).

• The TLB is associative, high-speed memory. Each entry in the TLB consists of two parts: a key (or tag) and a value. When the associative memory is presented with an item, the item is compared with all keys simultaneously. If the item is found, the corresponding value field is returned.

• The search is fast; the hardware, however, is expensive. Typically, the number of entries in a TLB is small, often numbering between 64 and 1,024.

• A paging hardware with TLB is shown below.

• The TLB contains only a few of the page-tableentries.

• When a logical address is generated by theCPU, its page number is presented to the TLB.If the page number is found, its frame numberis immediately available and is used to accessmemory. If the page number is not in the TLB(known as a TLB miss) a memory reference tothe page table must be made.

• If the TLB is already full of entries, the OSmust select one for replacement.Replacement policies range from leastrecently used (LRU) to random.

4. Consider the following set of processes with CPUburst given in milliseconds.

a)Draw the Gantt chart that illustrates theexecution of these process using the followingscheduling algorithms: FCFS, SJF and non-preemptive priority (a smaller priority numberimplies a higher priority)

b) What is the turnaround time and waiting time ofeach process for the above algorithms?(Oct. 2011)

Ans:

5. a) Explain the performance measures in RTOS. (Nov. 2012)Ans: (Refer file: C M Krishna, Kang G. Shin- Real-Time Systems:Chapter 2, page 12 to 18)b) Briefly discuss atomic and critical section. (Nov. 2012)

Ans:Atomic and Critical Section:• A part of the program is said to atomic if it cannot be

interrupted. By another way, we can say that the task codeuses the shared data that is not atomic.

• When we disable interrupts around the lines of task code thatuses the shared data, we can make the collection of lines asatomic and can solve the problem.

• A part of the code is atomic means that it cannot beinterrupted at all, but it cannot be interrupted by anythingthat might mess up the data it is using.

• A set of instructions that must be atomic for the system towork properly is called critical section.

6. Describe the steps followed after an interrupt is caused by an I/O call. (Nov 2012)

Ans:

Interrupt driven I/O cycle is shown below.