Exercise #1 

Question #1: 

q1: A CPU-intensive job would use the full time quantum. There are two such
    jobs: A and D. But A "skips a beat", i.e., it is not scheduled between two
    of D's CPU bursts. The only possible reason is that A is doing an I/O burst.
    Therefore D is the job we're looking for.

q2: C, because it has the shortest CPU burst and is I/O-bound

q3: The CPU utilization of job B is as follows:
	BBBBBB  BBB   BBB  BBBBBB

	At a minimum, this could be a first CPU burst of 9ms, 
        followed by an I/O-burst, followed by a 3ms CPU burst, followed
        by an I/O burst, followed by the beginning of a long CPU burst. So at a mininum
        2 I/O-burts could be initiazed in the sequence. At a maximum
        that would be 4 I/O-burst. 


Question #2: 

q1

IIoIIoIIoCCCCCo


q2

At time 2ms, the first I/O-bound job starts its I/O burst.
That I/O bursts lasts until time t1 = 2ms + 4ms = 6ms, at which point the job is back in the Ready Queue. 
That job will not run until time t2 = 3*(2+0.1)+5+0.1 = 11.4ms
Therefore, it was Ready for t2-t1 = 5.4ms
Same reasoning for all other I/O-bound jobs, just shifted by 5.4ms.
Final Answer: 5.4ms


q3

Let x be the desired overhead. Given the pattern in Question #1, the
fraction of the time spent doing context switching is:

   4x / (4x + 11)

So we simply need to solve for this fraction to be equal to 0.01. 

5x = 0.01 * 4x + 0.01 * 11

This gives us: x = 0.01 * 11 / (5 - 0.01 * 4) ~ .0222ms



------------------------------------

Exercise #2: 

Question #1:

q1 
    A: mid-priority queue
    B: high-priority queue
    C: mid-priority queue
    D: mid-priority queue
    E: low-priority queue

q2

    Here is the execution sequence. 

    AAAABBCCCCDDDDEEEEBB 
    01234567890123456789

    At time 4+2+4+4+4 =18 all jobs have run once. All of them but B have been demoted because 
    they have used up their entire time quantum.
    Therefore B starts its second time quantum at time 18. 


Question #2:

q1      

    AAA BBBB AAA B AAA B...


q2

    Yes. In the long run A runs for 3 ms and then job B for 1ms. Therefore
    job B runs for 1/4 of the time in the long run, which is more than 20%.

q3

AAABBBBAAABAAABAAABBBBAAABAAABAAABBBBAAAB
01234567890123456789012345678901234567890
                ^               ^ 
                |               | 
               boost          boost 

or with spaces:

AAA BBBB AAA B AAA B AAA BBBB AAA B AAA B AAA BBBB AAA B
012 3456 789 0 123 4 567 8901 234 5 678 9 012 3456 789 0
                      ^                     ^ 
                      |                     | 
                     boost                boost