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Multiprogramming P = efisiensi waktu tunggu proses  P= I/O wait ……(%) CPU IDLE  P n  P n = CPU Idle pada saat terdapat n process di memory  CPU Idle.

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Presentasi berjudul: "Multiprogramming P = efisiensi waktu tunggu proses  P= I/O wait ……(%) CPU IDLE  P n  P n = CPU Idle pada saat terdapat n process di memory  CPU Idle."— Transcript presentasi:

1 Multiprogramming P = efisiensi waktu tunggu proses  P= I/O wait ……(%) CPU IDLE  P n  P n = CPU Idle pada saat terdapat n process di memory  CPU Idle = P n  n = jumlah process(banyak program)

2 Multiprogramming CPU Busy : tingkat kesibukan prosesor  CPU Busy = 1 - cpu idle = 1 - P n CPU / PROCESS  Cpu busy / jml job TABEL CPU UTILIZATION  cpu idle, cpu busy, cpu/process CONTOH SOAL!!

3 Contoh soal Job (n)Arrival TimeCPU Time 110.005 210.071 310.153 410.172 Jika I/O Wait 80% maka buatlah : Table CPU Utilization ! Grafik proses ! Diketahui data proses sbb:

4 Table CPU Utilization Job1234 CPU Idle ( P n ) CPU Busy ( 1- P n ) CPU/Process ( cpu busy/job )

5 Table CPU Utilization Job1234 CPU Idle ( P n )0.80.640.510.41 CPU Busy ( 1- P n )0.200.360.490.59 CPU/Process ( cpu busy/job ) 0.200.180.160.15

6 Grafik proses 1 2 3 4 00 Job Time

7 JobArrival TimeCPU Time 110.005 210.071 310.153 410.172

8 grafik proses 1 2 3 4 00 07 12.5151727.929.332.5 0.360.511.4 0.261.740.36 1 1.74 0.261.74 0.64 Job Time     10.00  1 job 0.2. 7 = 1.4 10.07  2 job 0.18. x1 = 1 x1 = 5.5 10.12.5  1 job 0.2. 2.5 = 0.5 10.15  2 job 0.18. 2 = 0.36 10.17  3 job 0.16. x2 = 1.74 x2 = 10.9 10.27.9  2 job 0.18. x3 = 0.26 x3 = 1.4 10. 29.3  1 job 0.2. x4 = 0.64 x4 = 3.2 x1 x2x3 x4


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