In order to implement the classic DoS flood attack, the attacker must generate a sufficiently large volume of packets to exceed the capacity of the link to the target organization. Consider an attack using ICMP echo request (ping) packets that are 500 bytes in size (ignoring framing overhead). How many of these packets per second must the attacker send to flood a target organization using a 0.5-Mbps link

Answers

Answer 1

Answer:

131 packets

Explanation:

Given

W (link ) =0.5 M bits,  and Ping packets ( P ) = 500bytes

To flood a target organization that has a bandwidth of W bits using a packet Mbits, you will require a W / P packet

P ( required ) = 500  * 8 = 4000 bits

therefore the W /P packet is calculated as

= 0.5 M bits / 4000 bits

given that :  1 M bits =  1,048,576 bits

0.5 M bits / 4000 bits=  131.07  

hence approximately 131 packets/ sec is required to flood the target organization


Related Questions

In a multimedia presentation, when might voice-over be a better choice than placing the text on the screen?
when you have very little to say
when your audience is bored
when you have a lot to say
when you don’t want to write

Answers

Um, it is quite the battle between C and D, but knowing school they probably think D is lazy, so my final answer is C. Sorry if I'm wrong <3

In a multimedia presentation, the option that might be the voice-over  when you have a lot to say.

What is multimedia presentation?

A multimedia presentation is known to be a presentation where a person is said to often stand by themselves to present information via slides, video, etc.

In a multimedia presentation, the option that might be in the case of voice-over is when you have a lot to say as it is better to put it in a video format.

Learn more about multimedia from

https://brainly.com/question/24138353

#SPJ2

Suppose Alice downloads a buggy browser that implements TLS incorrectly. The TLS specification says that, during the handshake, the browser should send a random 256-bit number RB. Instead of picking RB randomly the browser always sends all zeros. Describe an attack that is possible against this buggy browser and how to update the browser so that this attack is no longer feasible.

Answers

Solution :

It is given that Alice downloads the buggy browser which implements a TLS incorrectly. The specification of a TLS states that during a handshake, the browser sends a 256 bit number of RB randomly.

So in this case, a man-n-the-middle attack is possible. It can compromise the confidentiality of Alice. Updating the browser by visiting the website and checking its latest version of the browser or installing some other browser which has a more trust in the market for its security features.

You work for a small company that exports artisan chocolate. Although you measure your products in kilograms, you often get orders in both pounds and ounces. You have decided that rather than have to look up conversions all the time, you could use Python code to take inputs to make conversions between the different units of measurement.
You will write three blocks of code. The first will convert kilograms to pounds and ounces. The second will convert pounds to kilograms and ounces. The third will convert ounces to kilograms and pounds.
The conversions are as follows:
1 kilogram = 35.274 ounces
1 kilogram = 2.20462 pounds
1 pound = 0.453592 kilograms
1 pound = 16 ounces
1 ounce = 0.0283 kilograms
1 ounce = 0.0625 pounds
For the purposes of this activity the template for a function has been provided. You have not yet covered functions in the course, but they are a way of reusing code. Like a Python script, a function can have zero or more parameters. In the code window you will see three functions defined as def convert_kg(value):, def convert_pounds(value):, and def convert_ounces(value):. Each function will have a block showing you where to place your code.

Answers

Answer:

Answered below.

Explanation:

def convert_kg(value){

ounces = value * 35.274

pounds = value * 2.20462

print("Kilograms to ounces: $ounces")

print("Kilograms to pounds: $pounds)

}

def convert_pounds(value){

kg = value * 0.453592

ounce = value * 16

print("pounds to kg; $kg")

print ("pounds to ounce; $ounce")

}

def convert_ounces(value){

kg = value * 0.0283

pounds = value * 0.0625

print ("ounces to kg; $kg")

print ("ounces to pounds; $pounds")

}

A slide contains three text boxes and three images that correspond to the text boxes. Which option can you use to display a text box with its
image, one after the other, in a timed sequence?
A. transition
В. table
C. animation
D. slide master
E. template

Answers

The answer is A. Transition.

Answer:

animation

Explanation:

# change amount as an integer input, and output the change using the fewest coins, one coin type per line. The coin types are Dollars, Quarters, Dimes, Nickels, and Pennies. Use singular and plural coin names as appropriate, like 1 Penny vs. 2 Pennies
my code produces no output and i cant find why?


coin_change =int(input())

def coin_change(cents):
if cents <= 0:
print( 'Zero cents.')
else:
quarter = cents // 25
dime = (cents % 25) //10
nickle = cents % 25 % 10 // 5
penny = cents % 5



print (coin_change )
# produces no output

Answers

Answer:

Explanation:

The Python code provided was not producing any output because you were never printing out the coin variables that you created. The following code adds the needed print statements using the right singular or plural coin name as needed.

cents = int(input())

 

def coin_change(cents):

   if cents <= 0:

       print('Zero cents.')

   else:

       quarter = cents // 25

       dime = (cents % 25) // 10

       nickle = cents % 25 % 10 // 5

       penny = cents % 5

   if quarter == 0 or quarter > 1:

       print(str(quarter) + " quarters")

   else:

       print(str(quarter) + " quarter")

   if dime == 0 or dime > 1:

       print(str(dime) + " dimes")

   else:

       print(str(dime) + " dime")

   if nickle == 0 or nickle > 1:

       print(str(nickle) + " nickels")

   else:

       print(str(nickle) + " nickel")

   if penny == 0 or penny > 1:

       print(str(penny) + " pennies")

   else:

       print(str(penny) + " penny")

coin_change(cents)

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