1. How many device addresses are available in a /30 network? How many computers

Question

1. How many device addresses are available in a /30 network? How many computers can be available in a /30 network?

2. How many device addresses are available in a 177.56.42.0 /31 network? How many computers can be available in a 177.56.42.0 /31 network?

3. Compute the result of the "bitwise-AND" operation between the IP address 192.228.17.57 and the subnet mask 255.255.255.224. Show the result in binary and in decimal form.

4. Assume that the IP address of your computer is 177.56.42.10/26, what is your-

a) Subnet mask

b) NetworkIP

c) First Host IP

d) Last host IP

e) Broadcast IP

Explanation / Answer

4)

Given 177.56.42.10/26

equivalent binary form = 10110001.00111000.00101010.00001010

b) since 26 bits form prefix and remianing 6 bits are for host

so use prefix bits to form network ip = 10110001.00111000.00101010.00 = 177.56.42.0

e) to form broadcast address keep the prefix as it is . to form the host number add the decimal values of all the binary host bits and finally append it to prefix like below :

177.56.42. ( 0+0+0+0+1+0+1+0) = 177.56.42. ( 8+2 ) = 177.56.42.10

c) for the first host, the last host bit in network IP should be 1 ,then IP is :

177.56.42. (0+0+0+0+0+0+0+1) =177.56.42.1

d) for the last host, the host is always one less than broadcast address ,then IP is :

177.56.42. (10 - 1) =177.56.42.9

1)

11111111.11111111.11111111.11111100

2 usable addresses are available. But in original zero are available because both of them are extremes of the network( i.e. network address & broadcast address).

Hence 0 address & 0 computers are available .

2)

11111111.11111111.11111111.11111110

1 usable addresses are available. But in original zero are available because both of the extremes of the network( i.e. network address & broadcast address). are accomodated.

Hence 0 address & 0 computers are available

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