Uniprocessor Scheduling III Solution CPSC 410 Operating Systems

Uniprocessor Scheduling III (Solution) CPSC 410 Operating Systems Department of Physics, Computer Science and Engineering Christopher Newport University

Question (1) z Consider the following set of processes: Process Name Arrival Time Processing Time 1 0 2 2 1 4 3 2 8 4 3 6 Perform the analysis of a comparison of scheduling policies: FCFS, RR(q=1), RR(q=4), SPN, SRT, HRRN, Feedback(q=1), and Feedback(q=2 i).

Solution of Question 3 FCFS 00000111112 01234567890 P 1 P 2 P 3 P 4 Arrive time 0 1 2 3 1122223333444444 (2)

Solution of Question 3 - FCFS (3) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 FCFS 1 1 2 2 3 3 3 3 4 4 4 * Each square represents one time unit. The number refers to the running process First Come First Served (FCFS) Process Pi 1 2 3 4 Arrival Time Ta 0 1 2 3 Service Time Ts 2 4 8 6 Finish Time Tf 2 6 14 20 Turnaround Time Tq 2 5 12 17 Tq = Tf - Ta Normalized Tq/Ts 1. 0 1. 25 1. 5 2. 8 Turnaround Time Average 9. 00 1. 64

Solution of Question 3 Round Robin q=1 (4) 00000111112 01234567890 P 1 P 2 P 3 P 4 0 Next Ready Time 1 1 2 5 2 8 4 3 7 6 10 9 13 12 15 14 1719 16 1213243243243433

Solution of Question 3 - RR (q=1) RR q=1 (5) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 1 2 1 3 2 4 3 4 3 3 * Each square represents one time unit. The number refers to the running process Round Robin (RR q=1) Process Pi 1 2 3 Arrival Time Ta 0 1 2 Service Time Ts 2 4 8 Finish Time Tf 3 11 20 Turnaround Time Tq 3 10 18 Tq = Tf - Ta Normalized Tq/Ts 1. 5 2. 25 Turnaround Time 4 3 6 18 15 2. 5 Average 11. 5 2. 19

Solution of Question 3 Round Robin q=4 (6) 00000111112 01234567890 P 1 P 2 P 3 P 4 Next Ready Time 0 1 2 10 3 14 11222233334444333344

Solution of Question 3 - RR (q=4) RR q=4 (7) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 1 1 2 2 3 3 4 4 * Each square represents one time unit. The number refers to the running process Round Robin (RR q=4) Process Pi 1 2 3 Arrival Time Ta 0 1 2 Service Time Ts 2 4 8 Finish Time Tf 2 6 18 Turnaround Time Tq 2 5 16 Tq = Tf - Ta Normalized Tq/Ts 1. 0 1. 3 2. 0 Turnaround Time 4 3 6 20 17 2. 8 Average 10 1. 8

Solution of Question 3 SPN 00000111112 01234567890 P 1 P 2 P 3 P 4 Shortest Process Time 2 4 8 6 1122224444443333 (8)

Solution of Question 3 - SPN (9) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 SPN 1 1 2 2 4 4 4 3 3 3 3 * Each square represents one time unit. The number refers to the running process Shortest Process Next (SPN) Process Pi 1 2 3 4 Arrival Time Ta 0 1 2 3 Service Time Ts 2 4 8 6 Finish Time Tf 2 6 20 12 Turnaround Time Tq 2 5 18 9 Tq = Tf - Ta Normalized Tq/Ts 1. 0 1. 25 2. 25 1. 5 Turnaround Time Average 8. 5 1. 5

Solution of Question 3 SRT 00000111112 01234567890 P 1 P 2 P 3 P 4 Shortest Remaining Time 2 1 4 3 2 1 87 6 5 4 3 2 1 1122224444443333 (10)

Solution of Question 3 - SRT (11) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 SPN 1 1 2 2 4 4 4 3 3 3 3 * Each square represents one time unit. The number refers to the running process Shortest Remaining Time (SRT) Process Pi 1 2 3 4 Arrival Time Ta 0 1 2 3 Service Time Ts 2 4 8 6 Finish Time Tf 2 6 20 12 Turnaround Time Tq 2 5 18 9 Tq = Tf - Ta Normalized Tq/Ts 1. 0 1. 25 2. 25 1. 5 Turnaround Time Average 8. 5 1. 5

Solution of Question 3 HRRN (12) 00000111112 01234567890 P 1 P 2 P 3 P 4 P 3 RR=1. 5 P 4 RR = 1. 5 1122223333444444 P 2 RR =((2 -1) + 4)/4 = 5/4 = 1. 25; P 3 RR =((2 -2) + 8)/8 = 8/8 = 1. 0 P 3 RR =((6 -2)+8)/8 = 12/8 = 1. 5; P 4 RR =((6 -3) +6)/6 = 9/6 =1. 5

Solution of Question 3 - HRRN (13) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 HRRN 1 1 2 2 3 3 3 3 4 4 4 * Each square represents one time unit. The number refers to the running process Highest Response Ratio Next (HRRN) Process Pi 1 2 3 4 Average Arrival Time Ta 0 1 2 3 Service Time Ts 2 4 8 6 Finish Time Tf 2 6 14 20 Turnaround Time Tq 2 5 12 17 9 Tq = Tf - Ta Normalized Tq/Ts 1. 0 1. 25 1. 5 2. 83 1. 65 Turnaround Time

Solution of Question 3 FB q=1 00000111112 01234567890 PQ 1 PQ 2 PQ 3 PQ 4 P 1 0 1 PQ 5 0 1 2 3 P 2 0 1 2 3 4 5 6 P 3 1 2 3 4 5 P 4 PQ 0 0 1234234234343444 (14)

Solution of Question 3 FB q=1 (15) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 3 4 3 3 * Each square represents one time unit. The number refers to the running process Feedback (FB q=1) Process Pi 1 2 3 4 Arrival Time Ta 0 1 2 3 Service Time Ts 2 4 8 6 Finish Time Tf 5 12 20 18 Turnaround Time Tq 5 11 18 15 Tq = Tf - Ta Normalized Tq/Ts 2. 5 2. 75 2. 25 2. 5 Turnaround Time Average 12. 25 2. 5

Solution of Question 3 FB q= 2 i (16) 00000111112 01234567890 PQ 1 PQ 2 PQ 3 P 1 0 1 2 P 2 0 1 2 3 P 3 1 2 P 4 PQ 0 0 12341223344233334443

Solution of Question 3 FB q=2 i FB q= 2 i (17) 0 0 0 0 0 1 1 1 1 1 2 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 1 2 2 3 3 4 4 4 3 * Each square represents one time unit. The number refers to the running process Feedback (FB q = 2 i) Process Pi 1 2 3 Arrival Time Ta 0 1 2 Service Time Ts 2 4 8 Finish Time Tf 5 12 20 Turnaround Time Tq 5 11 18 Tq = Tf - Ta Normalized Tq/Ts 2. 5 2. 75 2. 25 Turnaround Time 4 3 6 19 16 Average 2. 67 2. 54 12. 5