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Vika [28.1K]
2 years ago
6

You are an ISP. For the Address Block 195.200.0.0/16 a. If you have 320 Customers that need 128 addresses/customer - will there

be enough addresses to satisfy them? and explain why? b. Say the first block of 64 customers want 128 addresses/customer - what would the address space look like?
Computers and Technology
1 answer:
12345 [234]2 years ago
5 0

Answer:

a. The network will not satisfy the customers because the required addresses is 128 but what can be offered is 126.

b. 195.200.0.0/22

Explanation:

195.200.0.0/16

The number of bits to give 512 is 9

2^9=512

2^8=256 which is not up to our expected 320 customers that requires a network ip

Note we have to use a bit number that is equal or higher than our required number of networks.

The number of host per each subnet of the network (195.200.0.0/25) is (2^7)-2= 128-2=126

The network will not satisfy the customers because the required addresses is 128 but what can be offered is 126.

b. 64 customers requires  6 bits to be taken from the host bit to the network bit

i.e 2^6 = 64

195.200.0.0/22

The number of host per each subnet of the network (195.200.0.0/22) is (2^10)-2=1024 - 2 = 1022 hosts per subnet

This network meet the requirement " 64 customers want 128 addresses/customer "

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Give a proof for each statement.
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Answer:

  • If a group of 9 kids have won a total of 100 trophies, then at least one of the 9 kids has won at least 12 trophies.
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Explanation:

1)

Suppose that  each kid has less than 12 trophies

Total trophies = 100

Maximum trophies won by one kid = 11

total kids = 9

total number of trophies = 9 * 11 = 99 which contradicts the fact the total number of trophies are 100

2)

Suppose that  person has less than 2 cups of coffee a day

Total cups of coffee = 400

he has bought at least one cup of coffee each day

which means

total number of cups of coffee = 1* 366 = 366 which contradicts the fact the person buys at least 400 cups of coffee in a year

3)

Average of three number = (a+ b+ c)/3

suppose that there are real numbers a, b, and c such  that all three numbers are less than the average of the three numbers.

Let m be the average (a+b+c )/3 = m. Then our assumption states that (a < m) and  (b < m) and (c < m). By adding all the inequalities we get a + b + c < 3m. But m is  defined to be (a+b+c) /3  , so a + b + c = 3m.  But now we have that 3m = a + b + c < 3m. So 3m < 3m which is an obvious  contradiction. Thus our claim is true

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2 years ago
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jasenka [17]

Answer:

public class Main

{

public static void main(String[] args) {

    int userNum = 40;

    while(userNum > 1){

        userNum /= 2;

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    }

}

}

Explanation:

*The code is in Java.

Initialize the userNum

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Basically, this loop will iterate until userNum becomes 1. It will keep dividing the userNum by 2 and print this value.

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