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Bess [88]
3 years ago
7

Let A and B be two stations attempting to transmit on an Ethernet. Each has a steady queue of frames ready to send; A’s frames w

ill be numbered ????1, ????2 and so on, and B’s similarly. Let T = 51.2 ????????????c be the exponential backoff base unit. Suppose A and B simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0 × T and 1 × T, respectively. As a result, Station A transmits ????1 while Station B waits. At the end of this transmission, B will attempt to retransmit ????1 while A will attempt to transmit ????2. These first attempts will collide, but now A backs off for either 0 × T or 1 × T (with equal probability), while B backs off for time equal to one of 0 × T, 1 × T, 2 × T and 3× T (with equal probability).(a) Give the probability that A wins this second backoff race immediately after his first collision.(b) Suppose A wins this second backoff race. A transmits ????2 and when it is finished, A and B collide again as A tries to transmit ????3 and B tries once more to transmit ????1. Give the probability that Awins this third backoff race immediately after the first collision.(c) What is the probability that A wins all the ???? backoff races. (???? is a given constant)(d) Assume that there are 3 stations sharing the Ethernet. Will the chance for A to win all the backoffraces decrease or increase? Why?
Computers and Technology
1 answer:
Mariana [72]3 years ago
3 0

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete and correct question in the attachment file.  

For Point a:

For the second round,

A is selects kA(2)  either 0 or 1, so for each of them, that is \frac{1}{2}.

B selects kB(2)\  from\  (0, 1, 2, 3) for each choice with the probability of \frac{1}{4}.

If kA(2) < kB(2) wins the second rear race.

\to P[A \ wins] = P[kA(2) < kB(2)]

                     = P[kA(2) = 0] \times P[kB(2) > 0] + P[kA(2) = 1] \times P[kB(2) > 1]\\\\= \frac{1}{2} \times \frac{3}{4}  + \frac{1}{2}  \times \frac{2}{4} \\\\=\frac{3}{8}  +\frac{2}{8} \\\\= \frac{3+2}{8}\\\\= \frac{5}{8}

For Point b:

Throughout this example, kA(3) also selects to be either 0 or 1 with such a \frac{1}{2} probability. So, although B chooses kB(3) from (0, 1, 2, 3, 4, 5, 6, 7), the probabilities each are \frac{1}{8}:

\to P[A \  wins] = P[kA(3) < kB(3)]

                     = P[kA(3) = 0] \times P[kB(3) > 0] + P[kA(3) = 1] \times P[kB(3) > 1]\\\\= \frac{1}{2} \times \frac{7}{8} + \frac{1}{2} \times \frac{6}{8}\\\\= \frac{7}{16} + \frac{6}{16}\\\\= \frac{7+6}{16} \\\\= \frac{13}{16}\\\\

For point c:

Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is: P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ]

Let the k value kA(i) be A for the backoff race I select. Because A retains the breed

=(kA(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \geq P[kA(i) + 1

For point d:

Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs

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#define function

def discount(prices, isPet, nItems):

 # declare and initialize variables to 0

 i = 0

 cost = 0

 pets = 0

 item = 0

 item_Cost = 0

 #set the while loop

 while(i < nItems):

   #check the items of the variable

   if isPet[i]:

     #add and initialize in the cost

     cost += prices[i]

     #increament in the variable by 1

     pets += 1

   #otherwise

   else:

     #add and initialize in the item_Cost

     item_Cost += prices[i]

     #increament in the variable by 1

     item += 1

   #then increment in the variable by 1

   i += 1

 #check the pet is greater than equal to 1  

 #and item is greater than equal to 5

 if(pets >= 1 and item >=5):

   #then, print the message

   print("You receive discount")

   #set discount of the items

   discount = 0.2 * item_Cost

   # calculate bill after deducting discount

   bill = cost + item_Cost - discount

   #print the bill

   print("Final Bill amount is", bill)

 #otherwise

 else:

   #print the message

   print("You do not receive any discount")

   #calculate bill without deducting discount

   bill = cost + item_Cost

   #print the final bill

   print("Final Bill amount is", bill)

#set empty list type variables

price = []

pet = []

#set the infinite loop

while True:

 #get input from the user

 sprice = int(input("Enter the price (-1 to quit): "))

 #check that price is not equal to the -1

 if(sprice != -1):

   #add price in the list

   price.append(sprice)

   #ask for choice from the user

   choice = input("Is it a pet (Y / N)? ")

   #check the choice of the user is y or Y

   if(choice == 'Y' or choice == 'y'):

     # then, add in the pet

     pet.append(True)

   #otherwise

   else:

     #not add in the pet

     pet.append(False)

   print("")

 #break the loop

 else:

   break

#declare variable that store the length

items = len(price)

#call and initialize the variable

discount(price, pet, items)

<u>Output</u>:

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? y

Enter the price (-1 to quit): 69

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 41

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 52

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 74

Is it a pet (Y / N)? n

Enter the price (-1 to quit): -1

You receive discount

Final Bill amount is 361.6

Explanation:

<u>The following are the description of the program</u>.

  • Firstly, define the function 'discount()' and pass the arguments 'prices', 'isPet' and 'nItems' in its parameter.
  • Then, declare and initialize the variables 'i' to 0, 'item_Cost' to 0
  • , item to 0, 'cost' to 0 and 'pets' to 0.
  • Then, set the while loop to add the price of the items and the cost of the pets.
  • Set the if-else conditional statement to check the pet is greater than equal to 1 and the items are greater than equal to 5 then, set the discount and print the bill after deduction.
  • Otherwise, print the bill without deduction.
  • Finally, set two list type variable then, set the infinite while loop that gets input from the user and check that the variable is not equal to -1 then append the data in the list then again check the user input then append in pet otherwise not append in pet and break the loop. Then, set the variable that stores the length of the price and call the following function by passing arguments in its parameter.
8 0
3 years ago
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