What have we discussed?
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1. To understand the parts of one whole (i.e. a unit) we represent a unit by a block
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block divided into 10 equal parts means each part is 10 (one-tenth) of a unit.
be written as 0.1 in decimal notation. The dot represents the decimal point
comes between the units place and the tenths place.
2. Every fraction with denominator 10 can be written in decimal notation and vice-
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3. One block divided into 100 equal parts means each part is (one-hundredt
a unit. It can be written as 0.01 in decimal notation.
4. Every fraction with denominator 100 can be written in decimal notation
vice-versa.
5. In the place value table, as we go from left to the right, the multiplying factor beca
10 of the previous factor.
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Answers
Answer:
Out of all the available processes, CPU is assigned to the process having largest burst time.
In case of a tie, it is broken by FCFS Scheduling.
LJF Scheduling can be used in both preemptive and non-preemptive mode.
Preemptive mode of Longest Job First is called as Longest Remaining Time First (LRTF).
Advantages-
No process can complete until the longest job also reaches its completion.
All the processes approximately finishes at the same time.
Disadvantages-
The waiting time is high.
Processes with smaller burst time may starve for CPU.
PRACTICE PROBLEMS BASED ON LJF SCHEDULING-
Problem-01:
Consider the set of 5 processes whose arrival time and burst time are given below-
Process Id Arrival time Burst time
P1 0 3
P2 1 2
P3 2 4
P4 3 5
P5 4 6
If the CPU scheduling policy is LJF non-preemptive, calculate the average waiting time and average turn around time.
Solution-
Gantt Chart-
Now, we know-
Turn Around time = Exit time – Arrival time
Waiting time = Turn Around time – Burst time
Also read- Various Times of Process
Process Id Exit time Turn Around time Waiting time
P1 3 3 – 0 = 3 3 – 3 = 0
P2 20 20 – 1 = 19 19 – 2 = 17
P3 18 18 – 2 = 16 16 – 4 = 12
P4 8 8 – 3 = 5 5 – 5 = 0
P5 14 14 – 4 = 10 10 – 6 = 4
Now,
Average Turn Around time = (3 + 19 + 16 + 5 + 10) / 5 = 53 / 5 = 10.6 unit
Average waiting time = (0 + 17 + 12 + 0 + 4) / 5 = 33 / 5 = 6.6 unit
Problem-02:
Consider the set of 4 processes whose arrival time and burst time are given below-
Process Id Arrival time Burst time
P1 1 2
P2 2 4
P3 3 6
P4 4 8
If the CPU scheduling policy is LJF preemptive, calculate the average waiting time and average turn around time.
Solution-
Gantt Chart-
Now, we know-
Turn Around time = Exit time – Arrival time
Waiting time = Turn Around time – Burst time
Process Id Exit time Turn Around time Waiting time
P1 18 18 – 1 = 17 17 – 2 = 15
P2 19 19 – 2 = 17 17 – 4 = 13
P3 20 20 – 3 = 17 17 – 6 = 11
P4 21 21 – 4 = 17 17 – 8 = 9
Now,
Average Turn Around time = (17 + 17 + 17 + 17) / 4 = 68 / 4 = 17 unit
Average waiting time = (15 + 13 + 11 + 9) / 4 = 48 / 4 = 12 unit
Problem-03:
Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF, ties are broken by giving priority to the process with the lowest process id. The average turn around time is-
13 unit
14 unit
15 unit
16 unit
Solution-
We have the set of 3 processes whose arrival time and burst time are given below-
Process Id Arrival time Burst time
P1 0 2
P2 0 4
P3 0 8
Gantt Chart-
Now, we know-
Turn Around time = Exit time – Arrival time
Process Id Exit time Turn Around time
P1 12 12 – 0 = 12
P2 13 13 – 0 = 13
P3 24 14 – 0 = 14
Now,
Average Turn Around time = (12 + 13 + 14) / 3 = 39 / 3 = 13 unit
Thus, Option (A) is correct.
To gain better understanding about LJF Scheduling,
Answer:
.
Step-by-step explanation: