Answer: 1 what do u think to help u
The correct debugged code is attached below
<h3>What is a leap year </h3>
A leap year is a year with 366 days and this year occurs once every 4 years. The debugged code line of code of the program written, is as attached below. because the line of code in the question lacks some functions and symbols
Hence we can conclude that the correct debugged code is attached below.
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Answer:
Explanation:
When you have a single copy, a large number of concurrent updates that are supposed to go to a file may result in the user obtaining incorrect information. This incorrect information obtained them leads to the file being left in an incorrect state. When you have a lot of or multiple copies, then storage waste exist and the various copies might happen not to be consistent with respect one other. In summary, what happens is that
a) Using one copy saves space, but also the change might have an effect on all the users.
b) Using multiple copies avoids eliminates the change problem, while creating its own problems, using more space.
Answer:
See explanation below
Explanation:
Previos concepts
First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".
Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".
Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".
Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"
Solution for the problem
Assuming the dataset given on the plot attached.
Part a
For this algorithm the result would be:
Job A 0-6
Job B 6-(6+3) = 6-9
Job C 9-(9+1) = 9-10
Job D 10-(10+4) = 10-14
Part b
For this algorithm the result would be:
Job A 0-6
Job C 6-(6+1) = 6-7
Job B 7-(7+3) = 7-10
Job D 10-(10+4) = 10-14
Part c
For this algorithm the result would be:
Job A 0-1 until 14
Job B 2-(2+3) = 2-5
Job C 3-(3+2) = 3-5
Job D 9-(9+5) = 9-14
Part d
For this algorithm the result would be:
Job A 0-2 , 7-9, 12-14
Job B 2-4, 9-10
Job C 4-(4+1) = 4-5
Job D 5-7, 10-12