# 3.1 Review question: synchronization  (Page 4/4)

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A           15                4

B            9                7

C            3                3

D            6                1

E           12                6

a. round robin with a time quantum of 1 minute

1 2 3 4 5   Elapsed Time

============================

A B C D E   5

A B C D E   10

A B C D E   15

A B     D E   19

A B     D E   23

A B     D E   27

A B        E   30

A B        E   33

A B        E   36

A           E   38

A           E   40

A           E   42

A               43

A               44

A               45

Process Turnaround Time

========================

A        45

B        35

C        13

D        26

E        42

Average Turnaround Time = 32.2 minutes

b. priority scheduling

Process Priority Turnaround Time

==================================

B        7         9

E        6        21

A        4        36

C        3        39

D        1        45

Average Turnaround Time = 30 minutes

c. FCFS (run in order 15, 9, 3, 6, and 12)

Process Turnaround Time

========================

A        15

B        24

C        27

D        33

E        45

Average Turnaround Time = 28.8 minutes

d. shortest job first

Process Running Time Turnaround Time

======================================

C            3          3

D            6          9

B            9          18

E           12          30

A           15          45

Average Turnaround Time = 21 minutes

## Consider a set of three periodic tasks with the execution profiles as follows:

 Process Arrival Time Execution Time Ending Deadline A(1) 0 10 20 A(2) 20 10 40 . . . . . . . . . . . . B(1) 0 10 50 B(2) 50 10 100 . . . . . . . . . . . . C(1) 0 15 50 C(2) 50 15 100 . . . . . . . . . . . .

Develop scheduling diagrams for this set of tasks.

## Suppose that a microcontroller used in an automobile has four priority levels for interrupts. the units interrupting the controller include anImpact sensor subsystem, which needs attention with 0.1 ms, along with four other subsystems as follows:

 Subsystem Interrupt Rate Max service time (ms) Fuel/ignition 500/sec 1 Engine temperature 1/sec 100 Dashboard display 800/sec 0.2 Air conditioner 1/sec 100

Discuss how priorities could be assigned to guarantee the response time of 0.1 ms for the impact sensor and to handle each of theother critical units before the next interrupt is produced.

The impact sensor must be given the highest priority because its required response time is less than the service time for each of theother interrupts types; no other subsystem can share this highest priority.

The air conditioner can be given the lowest priority level because its function is not safety-critical.

We have two priority levels left for three subsystems.

Because the engine temperature monitor has a maximum service time of 100 ms it must have a lower priority than the display andfuel/ignition subsystems, which need several hundreds of service periods per second.

Finally, the display and fuel/ignition subsystems can share the same priority level since each can be served after the other withoutexceeding the time available before the next interrupt.

For example, if the fuel/ignition subsystem service starts right before the display interrupt, a total of 1.2 ms will elapse beforethe service routines of both interrupts are completed, whereas there is 1/800 sec - 1.25 ms available before the next dashboard display interruptarrives.

Note that we have ignored the service time for impact sensor subsystem interrupts that have the highest priority because they occurvery infrequently, and when they do, all else becomes unimportant.

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