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Advocard [28]
3 years ago
7

List the three control problems associated with competingprocesses and briefly define each.

Computers and Technology
1 answer:
hjlf3 years ago
3 0

Answer and explanation :

the three control problems associated with competing process are

  • MUTUAL EXCLUSION : We know that some resources are shareable and some are not shareable. which means only one process can access the resource at a time this type of resources are called critical resources this code can be access at only one process at a time. the other process if required to access should not be allowed
  • DEADLOCK: this hold the process without complete for example suppose there are two resources R1 and R2 and two process P1 and P2 and P1 use R1 and P2 use R2 but after some time when P1 needs R2 but R2 is not available as it is used by P2 so the all process will be on hold
  • STARVATION : when priorities are given to the process as high priorities and low priorities. And high priorities process always competing  then low priorities process have to wait for very long time this is called starvation
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When the tcp\ip translates a network layer address into a data link layer address (after not finding an entry for the network la
Pavel [41]
When the network stack translates an IP address (network address) to a MAC address (data link address) it sends a broadcast using ARP to all devices on the subnet asking who has the specified IP address.  The device with that IP address will then respond with it's MAC address to the ARP message's source.

With this in mind the answer to your question would either be broadcast or ARP request depending on what you have been learning in class.
6 0
3 years ago
Write a function called count_occurrences that takes two strings. The second string should only be one character long. The funct
iVinArrow [24]

Answer:

Here is the Python function:

def count_occurrences (string1 , string2):  # method that takes two strings as parameter and returns how many times second string occurs in first

   count = 0  #counts number of occurrence of string2 in string1

   for word in string1:  #iterates through each word of the string1

       for character in word:  #iterates through each character of each word in string1

           if character == string2:  # checks if the character in string1 is equal to the character of string2

               count = count + 1  #adds 1 to the count of string2 in string1

   return count  #returns number of times the string2 occurs in string1

#in order to check the working of the function add the following lines to the code:    

first_str = input("Enter the first string: ")  #prompts user to enter first string

second_str = input("Enter the second string (one character long): ") #prompts user to enter second string which should be one character long

occurrence = count_occurrences(first_str,second_str)  #calls count_occurrences method by passing both strings to it

print(second_str,"occurs",occurrence,"times in",first_str) #prints how many times the second_str occurs in the first_str on output screen

Explanation:

The program is well explained in the comments added to each statement of the above code. I will explain the program with an example:

Suppose

string1 = "hello world"

string2 = 'l'

So the method count_occurrences() should return how many times l occurs in hello world.

count is used to count the number of occurrences of l in hello world. It is initialized to 0.

The first loop iterates through each word of string1. The first word of string1 is "hello"

The second loop iterates through each character/letter of each word. So the first letter of "hello" word is 'h'.

The if condition if character == string2: checks if the character is equal to string2 i.e. l.

character = "h"

string2 = "l"

The are not equal so the program moves to the next character of word hello which is e. The if condition again evaluates to false because e is not equal to l. So the program moves to the next character of word hello which is l.

This time the if condition evaluates to true because

character = "l"

string2 = "l"

character == string2

l == l

So the if part is executed which has the following statement:

count = count + 1

So the count is added to 1. count was 0 previously. Now

count = 1

Next the program moves to the next character of word hello which is l. This time the if condition evaluates to true because the character l matches with string2. So count is again incremented to 1. count = 2.

Next the program moves to the next character of word hello which is o. The if condition evaluates to false as o does not match with string2 i.e. l. So the count remains 2.

Next the first loop moves to the next word of the string1 = "world". The inner (second) for loop iterates through each character of "world" from "w" to "d" and checks if any character is equal to string2. Only one character is equal to string2 in "world". So count is incremented to 1. Hence count = 3

After the loop ends the statement return count returns the number of times string2 occurs in the string1. So the output is:

l occurs 3 times in hello world  

8 0
3 years ago
A web designer is creating a web site and wants each browser to resize the text size to look the same in each device. what measu
adelina 88 [10]
The designer could use the measurements in pixels. 
4 0
3 years ago
What data unit is encapsulated inside a packet?
viva [34]

Answer: datagram

Explanation:

7 0
3 years ago
Read 2 more answers
Professor Midas drives an automobile from Newark to Reno along Interstate 80. His car’s gas tank, when full, holds enough gas
zmey [24]

Answer:

The GREEDY Algorithm

Explanation:

Based on the situation given in question, the Greedy algorithm shall give the optimal solution to professor

Suppose that the cities are at locations0 =x0< x1< . . . < x

We shall use the induction method to prove that G is the optimal solution valid for numbers less than n

We assume another solution Z which we initially consider to be optimum as well, based on that when Z fills the tank, it fills it to full level

Let us state the values in case of n intervals. Given below, we say that g1 is the first stop and z1 is also the first stop.

This can be written as ;

G=g1, g2, . . . , gk

Z=z1, z2, . . . , zk’

Here k’ <= k and k < n

Let I be an idex where for the first time gi is not equal to zi

Considering t= maxi Zi

Z′=g1, z2, z3, . . . , zk′

Now since z2, z3, . . . , zk′ should be an optimal stopping pattern for the problem otherwise we have chosen Z, with smaller gas filling (not feasible)

Using induction hypothesis we conclude thatg2, . . . , gk is an optimal stopping pattern, which is based on greedy algorithm

7 0
3 years ago
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