# ANALYSIS AND TESTING OF DIFFERENT TIME QUANTUM FOR ROUND ROBIN ALGORITHM

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2 4. Assign TQ to processes P j TQ, j++ 5. If(j < amount of processes) then go to step 4 6. Compute the remaining Burst time of the every processes and jump to step 3 7. If new process is arrived : Update ready queue and attend step 3 8. Compute Average Waiting Time, Average turnaround 9. End III. METHOD FOR TIME QUANTUM Average turnaround time = ( )/5 =162.8 Average waiting time = ( )/5 =102 D. Time quantum = Σ Burst time i / N In my example i get the time quantum =61,30 Let say five processes P 1, P 2, P 3, P 4 and P 5 in the ready queue and its arrival time is zero. Suppose burst time of the processes are 30,42,50,85,97 respectively. Now I am going to describe different time quantum for round robin algorithm. A. I select Time quantum (TQ) randomly then according to algorithm time quantum will vary: choose the first process from the ready queue and give the C.P.U. to it according time quantum (assume TQ= 30). If left over C.P.U burst time of the running process is fewer than 30 time quantum then assign C.P.U again to the same running process for leftover C.P.U burst time. After termination, disconnect it from the ready queue and repeat this step. Let Time quantum =30 Average turnaround time = ( )/5 =161.6 Average waiting time = ( )/5=100.8 E. If (meanvalue > medianvalue) Time quantum = ceil (sqrt((meanvalue * max burst time) +( medianvalue * min burst time))) Else If (medianvalue > meanvalue) Time quantum = ceil (sqrt((medianvalue * max burst time) +( meanvalue * min burst time))) Else Time quantum = meanvalue According to above formula, time quantum =86,11 Average turnaround time=( )/5 =147.4 Average waiting time=( )/5=86.4 F. Time Quantum = minimum burst time: So my Time quantum = 30, 12,8,35,12 B. Time quantum = (median + determinant factor) /2 Median = 1/2(X (total number of processes/2) + X(1+total number of processes/2)) if total number of process is even X(total number of processes+1)/2 if total number of processes is odd Determinant factor = (maximum burst_time + minimum burst_time)/2 Average turnaround time = ( )/5 =191.8 Average waiting time= ( )/5 =131 G. Time Quantum = (AVG+MAXBT) According to above formula, Time quantum =79, 15,3 In my example i get the time quantum=57, Average turnaround time = ( )/5 =158.4 Average waiting time=( )/5=97.6 C. Time Quantum =MAXIMUM_BURST MINIMUM_BURST So my time quantum = 67,12,12 Average turnaround time=( )/5 =162.8 Average waiting time=( ) / 5 =102 H. Time Quantum= Arithmetic Mean and Harmonic mean Compute time quantum : If (processes are not homogeneous and some processes are smaller than the others) then Time quantum = Harmonic_Mean of burst_times 41

3 else If (processes are not homogeneous and some processes are larger than the others) then Time quantum = Arithmetic_Mean of Burst_ times Time quantum = Arithmetic_mean: Hence my TQ = 61,30 K. Time Quantum = (Mid + Max)/2 Mid = (Minimum burst time + Maximum burst time)/2 According to above formula, time quantum TQ: Time quantum =80,12, 5 Average turnaround time = ( )/5 = Average waiting time = ( )/ 5 =100 I. Time Quantum = Harmonic_mean Hence TQ=51,40,6 Average turnaround time= ( )/5 =157.2 Average waiting time=( )/5 =96.4 Now Suppose another process P 6 whose arrival time is 0 and burst time is 1 also has come According to algorithm because processes are not homogeneous and one process has burst_time 1 which is smaller than others So my time quantum will be same as Harmonic_Mean of burst_times. TQ=6(harmonic mean), 55(arithmetic mean), 30(arithmetic mean), 6(arithmetic mean) Average turnaround time=( )/6 =147.6 Average waiting time=( )/6 =96.8 J. Time Quantum = mean + Standard Deviation Standard Deviation = σ2= {(1/n) Σ (xi - x) 2 } 1/2 According to above formula standard deviation=0.44,0 and TQ=61,30 Average turnaround time = ( )/5 = Average waiting time = ( )/5 = 100 Average turnaround time=( )/5 =163 Average waiting time=( )/5=102.2 L. TQ (Time Quantum) = Eavg / E TQ E (AVGTime Quantum of Even number of Processes )= total burst_time of even number of process / total even process TQ O ( AVG Time Quantum of Odd number of Processes) = total burst_time of odd process / total odd process if (TQ E >=TQ O ) Set total_tq=tq E Else Set total_tq=tq O According to my formula time quantum TQ=63,34 Average turnaround time=( )/ 5 =159.6 Average waiting time=( )/5 =98.8 M. Time Quantum = Square root of (mean*highest Burst): So my time quantum: TQ=77, 17,3 Average turnaround time= ( )/ 5 =162.4 Average waiting time=( )/5=102 N. Time quantum = MAVG / Count Medium 1 = (Lowest burst time + Highest burst time)/2 Set total_medium 2 =0, COUNTER =0 for i=medium 1 to total_process { Total_medium 2 = total_medium 2 +burst_time i COUNTER++ } TQn=total_medium 2 /COUNTER According to formula my TQ (time quantum) =91,6 42

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