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3 years ago

Consider the following set of processes, with the length of the CPU burst given in milliseconds:

Mathematics

1 answer:

Degger [83]3 years ago

5 0

Answer:

a) Drawing for grant charts that illustrates execution of the process is in the image attached

b) To find turn around time we use: completion time - arrival time

•Turn around time for FCFS scheduling algorithm will be:

P1 = 2-0= 2

P2= 3-0 = 3

P3 = 11-0 = 11

P4 = 15-0 = 15

P5 = 20-0 = 20

• Turn around time for SJF scheduling algorithm

P1 = 3-0= 3

P2 = 1-0 = 1

P3= 20-0 = 20

P4= 7-0 = 7

P5 = 12-0 = 12

• Turnaround time for non-preemptive algorithm

P1 = 15-0 = 15

P2 = 20-0 = 20

P3 = 8-0 = 8

P4 = 19-0 = 19

P5 = 13 - 0 = 13

• Turnaround time for RR

P1 = 2-0=2

P2= 3-0= 3

P3= 20 - 0 = 20

P4 = 19-0 = 19

P5 = 18-0 = 18

c) To find waiting time we use (turnaround time - burst time)

•Waiting time for FCFS

P1= 2-2 = 0

P2 = 3-1 = 2

P3 = 11-8 = 3

P4 = 15-4 = 11

P5 = 20-5= 15

• Waiting time for SJF

P1= 3-2 = 1

P2 = 1-1 = 0

P3 = 20-8 = 12

P4 = 7-4 = 3

P5 = 12-5 = 7

• Waiting time for non-preemptive

P1= 15-2 = 13

P2 = 20-1 = 19

P3 = 8-8 = 0

P4 = 19-4 = 15

P5 = 13 - 5 = 8

• Waiting time for RR

P1 = 2-2 = 0

P2 = 3-1 = 2

P3 = 20-8 =12

P4 = 13-4= 9

P5= 18-5=13

d) Average waiting time

•'For FCFS

= (0+2+3+11+15)/5

= 31/5

= 6.2milliseconds

•average waiting time For SJF

(1+0+12+3+7)/5

=23/5

= 4.6milliseconds

• Average waiting time For non-preemptive

(13+9+0+15+8)/5

=55/5

=11milliseconds

• average waiting time For RR

(0+2+12+9+13)/5

=7.2milliseconds

Since the SJF algorithm has 4.6milliseconds, it results in the minimum average waiting time.

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sasho [114]

Answer:

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Step-by-step explanation:

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( $\sqrt{\frac{6a+r}{5r} )^{2}$ = m^2

Then, you simplify each of the equation.

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Remove any parentheses if needed.

Solve for r.

Multiply each term by r and simplify."

Multiply both sides of the equation by 5.

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Remove parentheses.

Move 5 to the left of (m ^2) r

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Subtract 5m^2)r from both sides of the equation.

6a+r-5m^2)r=0

Subtract 6a from both sides of the equation.

r-5m^2)r=-6a

Factor r out of r-5m^2)r

r(1-5m^2)=-6a

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r = - $\frac{6a}{1-5m^2}$

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3 years ago

Find the slope of the line that passes through the points (3, 6) and (5, 3).

Temka [501]

(3,6) (5,3)
(x1,y1)(x2,y2)

The slope equation is: y2-y1
---------
x2-x1

Replace with numbers: 3-6/5-3

Minus 3 and 6 is equal to -3

Minus 5-3 is 2

So the fraction is -3/2 (no need to divide because you can't simplify it)

So the slope is -3/2

For the equation you have to replace the letters by 0's

3×0-2y=6
Times 3 and 0
0-2y=6
Now you want to get the y alone, so minus 0 from both sides
0-2y=6
-0 -0

-2y=6

Divide -2y from the 6

-2y=6
---- ---
-2 -2

y=3

The y intercept is 3 with the slope of -3/2

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