First Come, First Served (FCFS) - PowerPoint PPT Presentation

• The simplest algorithm
• When a process becomes ready, it enters the FIFO queue.
• When the currently-running process ceases to execute, the oldest process in the queue is selected for running.
• Non-preemptive

FIFO Ready Queue

CPU

dispatched

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P1

P2

P3

P4

P5

• ex 1 Process Arrival Time Process Exec Time P1 0 24 msec P2 1 3 msec P3 2 3 msec
• Calculate the average waiting time, turnaround time and throughput.
• ex 2 Process Arrival Time Process Exec Time P2 0 3 msec P3 1 3 msec P1 2 24 msec
• Calculate the average waiting time, turnaround time and throughput.

• Process Arrival T Exec TimeP1 0 24 msecP2 1 3 msecP3 2 3 msec

• Process Arrival T Exec TimeP2 0 3 msecP3 1 3 msecP1 2 24 msec

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0

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P1

P2

P2

P3

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P1

Waiting time = (0 + 23 + 25) / 3 = 16 sec

Turnaround t = (24 + 26 + 28) / 3 = 26 sec

Throughput = 3/30 = 0.1

Waiting time = (0 + 2 + 4) / 3 = 2 sec

Turnaround t = (3 + 5 + 28) / 3 = 14.3 sec

Throughput = 3/30 = 0.1

• The ratio of turnaround time to process exec time.
• A relative delay experienced by a process
• The longer process exec time, the greater absolute amount of tolerable delay.
• ex 1 Process Arrival Time Process Exec Time P1 0 24 msec P2 1 3 msec P3 2 3 msec
• ex 2 Process Arrival Time Process Exec Time P2 0 3 msec P3 1 3 msec P1 2 24 msec
• Calculate the turnaround time and the normalized turnaround time for each process.

• Process Arrival T Exec TimeP1 0 24 msecP2 1 3 msecP3 2 3 msec

• Process Arrival T Exec TimeP2 0 3 msecP3 1 3 msecP1 2 24 msec

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P1

P2

P2

P3

P3

P1

Waiting time = (0 + 23 + 25) / 3 = 16 sec

Turnaround t = (24 + 26 + 28) / 3 = 26 sec

NT = 24/24 + 26/3 + 28/3 = 1 + 8.6 + 9.3

= 18.9 sec

Throughput = 3/30 = 0.1

Waiting time = (0 + 2 + 4) / 3 = 2 sec

Turnaround t = (3 + 5 + 28) / 3 = 12 sec

NT = 3/3 + 5/3 + 28/24 = 1 + 1.6 + 1.1

= 3.7 sec

Throughput = 3/30 = 0.1

• This problem occurs whenever a short-lived process arrives right after a long-lived process.
• When the variance of process exec time is high, FCFS penalizes short processes.
• FCFS is not a fair scheduling algorithm; it does not treat long-lived and short-lived processes equally.

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T NT

3-0 = 3 3/3=1…

9-2=7 7/6=1…

13-4=7 7/4=1…

18-6=12 12/5=2…

20-8=12 12/2=6

Avg T Avg NT

8.60 2.56

P1

P2

P3

P4

P5

• Clock-based preemption of process (time slicing)
• When a clock interrupt occurs, the currently-running process is placed in the end of the Ready queue, and the next ready process is executed.
• A straightforward way to reduce the penalty that short-lived processes suffer with FCFS.
• A fairer algorithm than FCFS

Ready Queue

FIFO and Circular Queue

CPU

dispatched

timeout

• How to determine the length of time quantum (time slice)?
• If it’s too short,
• Processing overhead becomes high to handle clock interrupts and perform context switches.
• (the time to handle a clock interrupt and perform a context switch) / (time quantum) should be small enough.
• If it’s too long,
• RR degenerates to FCFS.

• Process Arrival Time Process Exec Time1 0 3 msec2 2 6 msec3 4 4 msec4 6 5 msec5 8 2 msec

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RRtq=1

P3

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P2

RRtq=1

P3

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P5

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T NT

4-0 = 4 4/3=1…

18-2=16 16/6=2…

17-4=13 13/4=3…

20-6=14 14/5=2…

15-8=7 7/2=3.5

Avg T Avg NT

10.8 2.71

P1

P2

RRtq=1

P3

P4

P5

0

5

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T NT

4-0 = 4 4/3=1…

18-2=16 16/6=2…

17-4=13 13/4=3…

20-6=14 14/5=2…

15-8=7 7/2=3.5

Avg T Avg NT

10.8 2.71

P1

P2

RRtq=1

P3

P4

P5

17 context switches

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T NT

3-0 = 3 3/3=1…

9-2=7 7/6=1…

13-4=7 7/4=1…

18-6=12 12/5=2…

20-8=12 12/2=6

Avg T Avg NT

8.60 2.56

P1

P2

FCFS

P3

P4

P5

• Process Arrival Time Process Exec Time1 0 3 msec2 2 6 msec3 4 4 msec4 6 5 msec5 8 2 msec

Avg process exec

time = 4 msec

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P1

P2

RRtq=4

P3

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P5

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RRtq=4

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0

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T NT

4-0 = 4 4/3=1…

18-2=16 16/6=2…

17-4=13 13/4=3…

20-6=14 14/5=2…

15-8=7 7/2=3…

Avg T Avg NT

10.8 2.71

P1

P2

RRtq=1

P3

P4

P5

17 context switches

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T NT

3-0 = 3 3/3=1

17-2=15 15/6=2…

11-4=7 7/4=1…

20-6=14 14/5=2…

19-8=11 11/2=5.5

Avg T Avg NT

10 2.71

P1

P2

RRtq=4

P3

P4

P5

6 context switches

• Process Arrival Time Process Exec Time1 0 3 msec2 2 6 msec3 4 4 msec4 6 5 msec5 8 2 msec

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RRtq=6

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RRtq=6

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RRtq=6

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FCFS

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• Should be slightly longer than typical process execution time
• Should be shorter than the longest process exec time.
• Generally 10 to 100 msec

• RR treats long-lived and short-lived processes equally.
• However, it does not treat CPU-bound and I/O-bound processes equally.
• CPU-bound process
• Mainly performs computational work and occasionally uses I/O devises
• I/O-bound process
• spends more time using I/O devises than using the CPU.

• An I/O-bound process uses for a short period and then is blocked for an I/O operation. It returns to the Ready queue when the I/O operation is completed.
• A CPU-bound process uses a complete time quantum and returns to the Ready queue.

Ready Queue

CPU

exit

dispatched

admitted

interrupted

CPU-bound processes tend touse more CPU time, whichleads to poor performancein I/O-bound processes.

I/O 1completed

I/O 1 Blocked Queue

I/O type 1 wait

I/O N completed

I/O N Blocked Queue

I/O type N wait

• Read Paper #1 (see the course web site), and summarize how the proposed scheduling algorithm addresses this fairness issue.
• Use a queuing diagram in your explanation.