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KIM [24]
3 years ago
11

Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running

times of 6, 4, 1, 3, and 7 minutes. Their ( externally determined ) priorities are 3, 5, 2, 1, and 4, respectively, with 5 being the highest priority. For each of the following scheduling algorithms, determine the mean process waiting time. Ignore process switching overhead.
a) Round Robin ( assume quantum = 1 )
b) Priority scheduling
c) First-come first-served
d) Shortest job first
For a), assume that the system is multitasking, and that each job gets its fair share of the CPU. For b) through d) assume that only one job at a time runs, until it finishes. All jobs are completely CPU bound.
Engineering
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1   2   3   4   5   6   7   8   9   10   11   12   13   14   15   16   17   18   19   20   21   22   23   24   25   26   27   28   29    30

A  B  C   D  E  A  B   C* D  E    A    B    D   E    A   B   D*   E     A   B     E     A    B* E   A     E  A    E*  A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6       7-14        15-24      25-26        27-30

   

 B           E             A             C            D

     AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10      11-16      17-18      19-22      23-30

 

   A            B              C            D              E

   

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2        3-6         7-12         13-20      21-30

C           D            B               E                A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

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3 years ago
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Explanation:

Series:

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Solving the above series:

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So, the series is:  9, 94, 916, 964, 9256, 91024, 94096, .................

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3 years ago
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8 0
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A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N
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Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

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6 0
3 years ago
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4 0
3 years ago
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