Consider a storage disk with $$4$$ platters (numbered as $$0, 1, 2$$ and $$3$$), $$200$$ cylinders (numbered as $$0, 1,$$ … , $$199$$), and $$256$$ sectors per track (numbered as $$0, 1, … , 255$$). The following $$6$$ disk requests of the form [sector number, cylinder number, platter number] are received by the disk controller at the same time: $$[120, 72, 2] , [180, 134, 1] , [60, 20, 0] , [212, 86, 3] , [56, 116, 2] , [118, 16, 1]$$

Currently the head is positioned at sector number $$100$$ of cylinder $$80,$$ and is moving towards higher cylinder numbers. The average power dissipation in moving the head over $$100$$ cylinders is $$20$$ milliwatts and for reversing the direction of the head movement once is $$15$$ milliwatts. Power dissipation associated with rotational latency and switching of head between different platters is negligible.

The total power consumption in milliwatts to satisfy all of the above disk requests using the Shortest Seek Time First disk scheduling algorithm is _______.

Consider a process executing on an operating system that uses demand paging. The average time for a memory access in the system is $$M$$ units if the corresponding memory page is available in memory, and $$D$$ units if the memory access causes a page fault. It has been experimentally measured that the average time taken for a memory access in the process is $$X$$ units.

Which one of the following is the correct expression for the page fault rate experienced by the process?

Consider the following C program.

#include< stdio.h >
struct Ournode{
 char x,y,z;
};
int main(){
 struct Ournode p = {'1', '0', 'a'+2};
 struct Ournode *q = &p;
 printf ("%c, %c", *((char*)q+1), *((char*)q+2));
 return 0;
}

The output of this program is:

Consider the following C program:

#include< stdio.h >
void fun1(char *s1, char *s2){
  char *tmp;
  tmp = s1;
  s1 = s2;
  s2 = tmp;
}
void fun2(char **s1, char **s2){
  char *tmp;
  tmp = *s1;
  *s1 = *s2;
  *s2 = tmp;
}
int main(){
  char *str1 = "Hi", *str2 = "Bye";
  fun1(str1, str2); printf("%s %s ", str1, str2);
  fun2(&str1, &str2); printf("%s %s", str1, str2);
  return 0;
}

The output of the program above is