Answer:
Round trip times required = log2N
Explanation:
The round-trip times required before TCP can send N segments using a slow start is log2N. we can arrive at this by looking at the mode of operation of TCP which is at the 1st time of using a TCP it starts the congestion window as 1 then it sends an initial segment. When the acknowledgement of the initial segment arrives, TCP increases the congestion window to 2 and then sends 2 segments, When the 2 acknowledgements of the segments sent out arrives, they each increase the congestion window by one, thereby increasing the congestion window to 4 . therefore it takes log2N round trips before TCP can send N segments
Answer:
Instead of using a key or entering a code to open a door, a user can use an object, such as an ID badge, to identify themselves in order to gain access to a secure area. What term describes this type of object?
Explanation:
Answer:
The solution code is written in Python 3.
- import random
-
- count = 0
- flag = False
- guess = int(input("Input your guess (2-12): "))
-
- while(count <=3):
- dice1 = random.randint(1, 7)
- dice2 = random.randint(1, 7)
-
- if((dice1 + dice2) == guess):
- flag = True
-
- count += 1
-
-
- if(flag):
- print("User wins!")
- else:
- print("Computer wins!")
Explanation:
A Random generator is needed for this question and therefore we start by importing Python random class (Line 1)
Next, create one counter variable,<em> count</em>, to ensure there will be only three rolling of the dices (Line 3). We need another variable, <em>flag</em>, to track the status if the two dices equal to the <em>guess</em> number chosen by user (Line 4).
Next, prompt use to input a guess number (Line 5).
Within the while loop, we can use random class method <em>randint() to </em>generate random integer. The arguments 1 and 7 will give one random number ranged from 1 to 6 for <em>dice1</em> and<em> dice2</em>, respectively (Line 8 - 9).
If the total of<em> dice1 + dice2</em> equal to user <em>guess</em>, we turn the<em> flag </em>to <em>True</em>. If not, the <em>flag </em>will remain <em>False</em> after completing entire while loop.
If the <em>flag </em>turned to <em>True</em>, print the message "User Wins!" else print the message ("Computer wins!")