CS 3733 Operating Systems, Spring 2004 Midterm Exam Comments

• 90-100: 5
• 80-89: 13
• 70-79: 17
• 60-69: 15
• 50-59: 7
• below 50: 6

• Problem 1 was like the one from the notes on Chapter 3.
• Problem 2: Part a) has three subparts and part c) has 6. Each of the 10 subparts was worth 2 points.
  scheduler dispatch means the scheduler causes a context switch.
  When the child of a process terminates, a waiting process can be moved to the ready state.
• Problem 3: Parts a), b) and c) were each 2 points. Part d) was 9 points and part e) was 5 points.
  Note that a) asks about a file descriptor table entry, not what a file descriptor table is.
  To get full credit for part d) you needed to mention the following:
       The two processes read different bytes. (3 points)
       The bytes each process reads are in alphabetical order. (2 points)
       The bytes each process reads do not have to be contiguous. (2 points)
       Each process may read 1, 2 or 3 bytes. (2 points)
  For part e) you had to say that each process reads from the beginning of the file (3 points) and that each will read 1, 2 or 3 bytes (2 points).
• Problem 4: This was just like the problem assigned in class.
• Problem 5: Note that the load average is defined as the average number of processes in the ready queue. There are a number of ways to calculate it, as in part c), but these formulas are not the definition.