At least six killed as bomb hits Pakistan passenger train: Officials

QUETTA, Pakistan: A bomb hit a passenger train in Pakistan's restive southwest on Monday, killing at least six people and wounding more than 17 others, officials said.

The device, apparently planted on the railway track, exploded when the train approached a station in Naseer Abad district of the insurgency hit southwestern Baluchistan province.

"It was a bomb blast, the target was the passenger train. At least six people have been killed," Asad Gilani, provincial home secretary said, adding that more than 17 others were wounded in the blast.

Now, move videos from one gadget to another by touch

WASHINGTON: Now you can move images and videos from a smartphone to another tablet just by touch.

Scientists from Finland have developed a new technology whereby a ring, nail or wristband acts as a user interface allowing files to be transferred directly from one screen to another simply by touch.

The new InTouch user interfaces allows you to touch a file on one screen, for example, with a structure nail and transfer it through the nail to your friend's smartphone or even to your car.

From the end user viewpoint, the data transfer between devices currently happens using memory sticks, short-range point-to-point connections (for example, Bluetooth) or the cloud (sharing services).

The solution, developed by VTT Technical Research Centre of Finland, allows files to be received, sent and shared much more easily than before.

The InTouch user interface operates with touchscreens and allows file transfer either directly or via a cloud service.

The new concept also enables entirely new kinds of interactive devices and product categories, and opens up new business possibilities, the researchers claim. Areas of application for the technology might include digital devices, cars, the manufacturing industry, logistics and health care. There is a patent pending for the technology.

Nokia leaks 6-inch Lumia 1520's specifications

NEW DELHI: Just two days ahead of the NokiaWorld Event, the company has accidentally revealed the specifications of its upcoming top-end phablet. Nokia China has listed the Lumia 1520, which will be biggest handset in the smartphone maker's portfolio after launch, on its official Tmall store, an online marketplace.

According to the listing by Nokia China, the upcoming Lumia 1520 will have a 6-inch display, 20MP PureView camera and quad-core Snapdragon 800 chipset. This phablet will cost 4,999 yuan ($819), according to the listing. These specifications are in line with previous speculation about the device.

Other rumoured specifications of the Lumia 1520 are Full HD screen (1920x1080p resolution), 2GB RAM, 32 and 64GB storage options, 3,400mAh battery and NFC. Several leaked images of this handset have emerged on the internet, showing a big screen and new layout of homescreen tiles.

The Lumia 2520 will compete against the likes of Samsung Galaxy Note 3, HTC One Max and Sony Xperia Z Ultra. Apple is also said to be testing a handset with 6-inch display.

Nokia has teased its October 22 event with an image of a smartphone, a tablet and a laptop positioned alongside each other. It is said that Nokia will launch a total of six products at the event, including a phablet, a tablet and four accessories. Leaked images of the Nokia tablet, currently called Lumia 2520, have appeared on the internet.

Apple is also holding an event on October 22, where it is expected to launch the new iPads.

Basic concepts of OPERATING SYSTEM

INTRODUCTION

          An operating system is a program that is an intermediary between a user and the computer hardware. Different levels in a computer system are shown below.

   An operating system provides an environment in which a user can execute programs. The purpose of an operating system is two fold:

·         To make the computer system convenient to use.

·         To use the computer hardware in an efficient manner.

OPERATING SYSTEM AS A RESOURCE MANAGER

   A computer system consists of hardware, the operating system, application programs and users .

          Hardware includes the central processing unit (CPU), memory and input / output (I/O) devices. These provide the basic computing resources. Application programs like compilers, database management systems, editors and many more allow users to use the hardware to solve their problems. The operating system controls and co-ordinates the use of the hardware among the various application programs of the various users. It provides an environment where software and data make use of the hardware in the computer system.

The operating system is a resource allocator. Many hardware and software resources are needed to solve a problem. These include CPU time, memory space, file storage, I/O devices and so on. The operating system as a manager of these resources, allocates resources to programs of users as required by the tasks. Since there is always a possibility of conflicts for resources because of more than one user requesting for resources, the operating system resolves the conflicts for efficient operation of the computer system.

The operating system as a control program controls the execution of user programs to avoid errors and improve use of the computer system.

A computer system has been developed to execute user programs there by making work easier for the user. But the bare hardware alone is not easy to use. Hence application programs are developed. These require certain common operations such as I/O device handling. The common functions of controlling and allocating resources are present in one piece of software – the operating system.

The primary goal of an operating system, that is convenience for the user is seen in small computers where as the secondary goal of efficient operation is important in large shared multi-user systems.

STORAGE STRUCTURE

Programs to be executed need to reside in the main memory. This is the memory that the CPU directly accesses. The main memory can be visualized as an array of bytes / words, each having its own address. Load / Store instructions allow movement of a word from the main memory to a CPU register and vice versa.

The Von-Neumann architecture brought in the concept of a stored program. The following is the sequence of steps in one instruction execute cycle:

 Instruction execute cycle

Program and data are to be in the main memory for execution. Ideally they should reside there permanently, which is not possible because main memory

·         is not all that big to hold all needed programs and data permanently.

·         is volatile (contents are lost when power is turned off).

·         cost is very high.

Secondary storage is an answer to the above listed problems. The main characteristics of secondary storage include

·         large storage space

·         Non-volatile or permanent storage

·         Low cost

Thus program and data need to be loaded from secondary storage to main memory during execution. Common secondary storage devices include disks and tapes.

STORAGE HIERARCHY

The various storage devices differ in speed, cost, size and volatility (permanence of storage). They can be organized in a hierarchy. Shown below are some different levels of hierarchy .

As one moves up the hierarchy, the devices are expensive but fast and as we move down the hierarchy, the cost per bit of storage decreases but access times increases (devices are slow). There usually exists a tradeoff in the amount of storage required at each level in the hierarchy. Of the various storage devices in the hierarchy, those above the disks are volatile and the rest are non-volatile.

          Thus the design of a computer system balances or makes a tradeoff of all the above factors namely speed, cost and volatility. It provides as much expensive volatile memory as necessary and as much inexpensive non-volatile memory as possible.

REGRESSION ANALYSIS CODE

#include<stdio.h>

#include<math.h>

void main()          

{

 char choice;

 void Regression();

 void Estimation();

 void Evaluation();

 void display();

 clrscr();

 do

 {

   printf("\nselect your choice\n");

   printf("0.display kloc and effort values\n");

   printf("1.Regression Analysisl\n");

   printf("2.Estimation Model\n");

   printf("3.Evaluation of Model\n");

   printf("4.Exit This Code\n");

   choice=getche();

   switch(choice)

   {

     case '0':display();

                 break;

     case '1':Regression();

                 break;

     case '2':Estimation();

                 break;

     case '3':Evaluation();

                 break;

     default:exit(1);

   }

   printf("\nsome other operation(Y/N)\n");

   choice=getche();

  }while((choice=='Y')||(choice=='y'));

}

void Regression()

{

 char choice;

 void linear();

 void hyperbola();

 void power();

 void display();

 clrscr();

 do

 {

   printf("select your choice\n");

   printf("0.display kloc and effort values\n");

   printf("1.Linear Regression\n");

   printf("2.Power Regression\n");

   printf("3.Inverse Regression\n");

   printf("4.Exit this Regression\n");

   choice=getche();

   switch(choice)

   {

     case '0':display();

                 break;

     case '1':linear();

                 break;

             case '2':power();

                 break;

     case '3':hyperbola();

                 break;   

     default:main();

   }

   printf("\nsome other operation(Y/N)\n");

   choice=getche();

  }while((choice=='Y')||(choice=='y'));

}

void linear()

{

  int i,n=13;

  float a,b,ab[13],bc=0,cd=0,de,ef=0,fg[13],gh=0,hi,kl;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  for(i=0;i<=12;i++)

  {

   ab[i]=kloc[i]*kloc[i];

   bc=bc+ab[i];

   cd=cd+effort[i];

   ef=ef+kloc[i];

   fg[i]=kloc[i]*effort[i];

   gh=gh+fg[i];

  }

  kl=ef*ef;

  de=bc*cd-ef*gh;

  hi=n*bc-kl;

  a=de/hi;

  b=(n*gh-ef*cd)/hi;

  printf("\na value is:%.3f",a);

  printf("\tb value is:%.3f",b);

}

void hyperbola()

{

  int i,n=13;

  float a,b,ab[13],bc=0,cd=0,de,ef=0,fg[13],gh=0,hi,kl;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  for(i=0;i<=12;i++)

  {

   ",a);

  printf("\tb value is:%.3f",b);

}

void power()

{

  int i,n=13;

  double a,b,aa,ab[13],kloc[13],effort[13],bc=0,cd=0,de,ef=0,fg[13],gh=0,hi,kl;

  double klocc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  double effortt[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  for(i=0;i<=12;i++)

  {

  kloc[i]=log(klocc[i]);

  effort[i]=log(effortt[i]);

  }

  for(i=0;i<=12;i++)

  {

   ab[i]=kloc[i]*kloc[i];

   bc=bc+ab[i];

  b=(n*gh-ef*cd)/hi;

  aa=pow(2.7182,a);

  printf("\na value is:%.3lf",aa);

  printf("\tb value is:%.3lf",b);

}

void display()

{

  int i;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  printf("\nSNO\tKLOC\tActual Effort\n");

  for(i=0;i<=12;i++)

  {

  printf("%d\t%.2f\t%.2f\n",i,kloc[i],effort[i]);

  }

}

float Basic()

{

  int i;

  float c,beffort[13],error[13],mm=0;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  printf("\nBASIC EFFORT \tERROR %\n\n");

  for(i=0;i<=12;i++)

  { +1,beffort[i],error[i]);

 }

 return mm;

}

float Halsted()

{

  int i;

  float a,b,c,beffort[13],error[13],mm=0;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  printf("\nBASIC EFFORT \tERROR %\n\n");

  for(i=0;i<=12;i++)

  {

   c=pow(kloc[i],1.50);

}

 return mm;

}

float Mittal()

{

  int i;

  float a,b,c,d,e,f,g,h,beffort[13],error[13],mm=0;

  float kloc[13]={39,40.5,50,128.6,161.4,164.8,200,214.4,253.6,254.2,289,449.9,450};

  float effort[13]={72,82.5,84,230.7,157,246.9,130.3,86.9,287,258.7,116,336.3,1107.31};

  printf("\nBASIC EFFORT \tERROR %\n\n");

  for(i=0;i<=12;i++)

  {

   c=pow(0.7*kloc[i],0.665);

   d=pow(kloc[i],0.665);

   e=pow(1.3*kloc[i],0.665);

   beffort[i]=(1*6*c+4*6*d+1*6*e)/6;

   error[i]=abs(((beffort[i]-effort[i])/effort[i])*100);

   mm=mm+error[i];

   printf("%d:%.3f\t%.3f\n",i+1,beffort[i],error[i]);

 }

 return mm;

}