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balu736 [363]
2 years ago
12

Suppose that the following processes arrive for execution at the times indicated. Each process will run for the amount of time l

isted. In answering the questions, use non-preemptive scheduling, and base all decisions on the information you have at the time the decision must be made.
Process Arrival Time Burst Time
P1 0.0 8
P2 0.4 4
P3 1.0 1
a) What is the average turnaround time for these processes with the FCFS scheduling algorithm?

b) What is the average turnaround time for these processes with the SJF scheduling algorithm?
Computers and Technology
1 answer:
Len [333]2 years ago
5 0

Answer:

a) 10.53

b) 9.53

Explanation:

a) Average Turnaround Time: ( (8-0)+(12-0.4)+(13-1.0) ) / 3 = 10.53

b) Average Turnaround Time: ( (8-0)+(13-0.4)+(9-1.0) ) / 3 = 9.53

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Gloria needs to show in spreadsheet form the number of students in her class who speak different languages. Which type of graph
vaieri [72.5K]
I think the most suitable chart to be used is a pie chart. A pie chart is used to show percentages and the best way to show relative sizes of data in your set. It is a good way to show which language is spoken more in the classroom, which is least spoken and at a glance. I hope my answer helps.
4 0
2 years ago
Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

5 0
2 years ago
There are no breakpoints in the "Access Customer Account" subpage however there is an error. What will happen if you choose to s
blsea [12.9K]

Answer:

Move to the breakpoint at "Get Customer Details" stage.

Explanation:

This question is incomplete, we need a diagram and four options to resolve this questions, I attached the diagram and the options.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the breakpoint at "Get Customer Details" stage.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the "Recover1" stage.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the stage containing the error on the "Access Customer Account" page.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the "Exception1" stage.

In this case, we are going to run an extra from subpage testing in <u>Process studio</u>. Process studio is a tool where we can test these processes.

The process always it's going to work all stages in the "Access Customer Account" we're going to receive an interruption when the error is activated, moving the focus to the Get Customer Details stage.

There is an error in the stage Access Customer Account, if we run the processes with the Go button, the focus should move into the Recover1 stage, and should show the error.

But in this particular example, we must use the STEP OUT BUTTON, What means this?

The process it's going be executed, but won't show any error or message, because we have used the STEP OUT BUTTON.

If we use the STEP OUT BUTTON, the process should end but in this case, we have a breakpoint in Get Customer Details stage, for that the focus and the process will end in Get Customer Details stage and not at the end or at the Recover1 stage.

8 0
2 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
2 years ago
Sam’s password is known to be formed of 3 decimal digits (0-9) in a given order. Karren and Larry are attempting to determine Sa
prohojiy [21]

Answer:

100

Explanation:

7 0
2 years ago
Read 2 more answers
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