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Engineering Computer Science Operations Research : Applications and Algorithms

a. To find the number of times a job visits the CPU, disk1, and disk2 on average. Given: N u m b e r o f c l i e n t s = 8 P r o b a b i l i t y o f C P U , P C P U = 7 / 16 P r o b a b i l i t y o f d i s k 1 , P d i s k 1 = 8 / 16 P r o b a b i l i t y o f d i s k 1 , P d i s k 1 = 1 / 16 Explanation: For CPU: To calculate the average number of times the job visits CPU, the formula is given below: A v e r a g e C P U = n u m b e r o f c l i e n t s * P r o b a b i l i t y o f C P U = 8 * ( 7 / 16 ) = 3.5 ≈ 4 Therefore, the average number of times the job visits the CPU is 4. For Disk1: To calculate the average number of times the job visits disk1, the formula is given below: A v e r a g e d i s k 1 = n u m b e r o f c l i e n t s * Pr o b a b i l i t y o f d i s k 1 = 8 * ( 8 / 16 ) = 4 Therefore, the average number of times the job visits disk1 is 4. For Disk2: To calculate the average number of times the job visits disk2, the formula is given below: A v e r a g e d i s k 2 = n u m b e r o f c l i e n t s * Pr o b a b i l i t y o f d i s k 2 = 8 * ( 1 / 16 ) = 0.5 ≈ 1 Therefore, the average number of times the job visits the disk2 is 1. b. To find the amount of time a job spends in the CPU, disk1, disk2, and in the system. Given: N u m b e r o f c l i e n t s = 8 C P U t i m e = 0.01 sec o n d s D i s k 1 t i m e = 0.02 sec o n d s D i s k 2 t i m e = 0.03 sec o n d s Explanation: For CPU: To calculate the amount of time a job spends in the CPU, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * C P U t i m e = 8 * 0.01 = 0.08 sec o n d s Therefore, the average amount of time a job spends on the CPU is 0.08 seconds. For Disk1: To calculate the amount of time a job spends in the disk1, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * D i s k 1 t i m e = 8 * 0.02 = 0.16 sec o n d s Therefore, the average amount of time a job spends in disk1 is 0.16 seconds. For Disk2: To calculate the amount of time a job spends in the disk2, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * D i s k 2 t i m e = 8 * 0.03 = 0.24 sec o n d s Therefore, the average amount of time a job spends in disk1 is 0.16 seconds. For the complete system: To calculate the amount of time a job spends in the complete system, the formula is given below: A v e r a g e t i m e = C P U t i m e + D i s k 1 t i m e + D i s k 2 t i m e = 0.08 + 0.16 + 0.24 = 0.48 sec o n d s Therefore, the average amount of time a job spends in the system is 0.48 seconds.

a. To find the number of times a job visits the CPU, disk1, and disk2 on average. Given: N u m b e r o f c l i e n t s = 8 P r o b a b i l i t y o f C P U , P C P U = 7 / 16 P r o b a b i l i t y o f d i s k 1 , P d i s k 1 = 8 / 16 P r o b a b i l i t y o f d i s k 1 , P d i s k 1 = 1 / 16 Explanation: For CPU: To calculate the average number of times the job visits CPU, the formula is given below: A v e r a g e C P U = n u m b e r o f c l i e n t s * P r o b a b i l i t y o f C P U = 8 * ( 7 / 16 ) = 3.5 ≈ 4 Therefore, the average number of times the job visits the CPU is 4. For Disk1: To calculate the average number of times the job visits disk1, the formula is given below: A v e r a g e d i s k 1 = n u m b e r o f c l i e n t s * Pr o b a b i l i t y o f d i s k 1 = 8 * ( 8 / 16 ) = 4 Therefore, the average number of times the job visits disk1 is 4. For Disk2: To calculate the average number of times the job visits disk2, the formula is given below: A v e r a g e d i s k 2 = n u m b e r o f c l i e n t s * Pr o b a b i l i t y o f d i s k 2 = 8 * ( 1 / 16 ) = 0.5 ≈ 1 Therefore, the average number of times the job visits the disk2 is 1. b. To find the amount of time a job spends in the CPU, disk1, disk2, and in the system. Given: N u m b e r o f c l i e n t s = 8 C P U t i m e = 0.01 sec o n d s D i s k 1 t i m e = 0.02 sec o n d s D i s k 2 t i m e = 0.03 sec o n d s Explanation: For CPU: To calculate the amount of time a job spends in the CPU, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * C P U t i m e = 8 * 0.01 = 0.08 sec o n d s Therefore, the average amount of time a job spends on the CPU is 0.08 seconds. For Disk1: To calculate the amount of time a job spends in the disk1, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * D i s k 1 t i m e = 8 * 0.02 = 0.16 sec o n d s Therefore, the average amount of time a job spends in disk1 is 0.16 seconds. For Disk2: To calculate the amount of time a job spends in the disk2, the formula is given below: A v e r a g e t i m e = n u m b e r o f c l i e n t s * D i s k 2 t i m e = 8 * 0.03 = 0.24 sec o n d s Therefore, the average amount of time a job spends in disk1 is 0.16 seconds. For the complete system: To calculate the amount of time a job spends in the complete system, the formula is given below: A v e r a g e t i m e = C P U t i m e + D i s k 1 t i m e + D i s k 2 t i m e = 0.08 + 0.16 + 0.24 = 0.48 sec o n d s Therefore, the average amount of time a job spends in the system is 0.48 seconds.

Solution Summary: The author explains how to calculate the average number of times a job visits the CPU, disk1, and disk2.

Operations Research : Applications and Algorithms

Operations Research : Applications and Algorithms

4th Edition

ISBN: 9780534380588

Author: Wayne L. Winston

Publisher: Brooks Cole

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Operations Research : Applications and Algorithms

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