# Deliverables 1. Assume that you have been assigned 192.168.111.129/29. 2. How many bits are borrowed to create

Discipline: Computer science

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Question

Deliverables

1. Assume that you have been assigned 192.168.111.129/29.

2. How many bits are borrowed to create the subnet field? ________________

3. What is the maximum number of subnets that can be created with this number of bits? ________________

4. How many bits can be used to create the host space? ________________

5. What is the maximum number of host addresses available per subnet? ________________

6. What is the subnet mask, in binary and decimal format? ________________

Complete the following table and calculate the subnet that this address is on, and define all the other subnets (the range of host addresses on the subnet and the directed broadcast address on the subnet).

 Subnet Number Subnet Address Range of Host Addresses Direct Broadcast Address 0 1 2 3 4 5 6 7 … Last subnet number

1. What subnet is 192.168.111.129 on?

2. A junior network administrator is trying to assign 192.168.111.127 as a static IP address for a computer on the network but is getting an error message. Why?

3. Can 192.168.111.39 be assigned as an IP address?

2) We see that the given IP address is a class C address which means that the first 3 octets belong to the network address. /29 means that the first 29 bits of the IP address is used for the network address.

Hence the number of bits borrowed to create the subnet field = 29-24 =5 bits.

3) The maximum number of subnets that can be created is 2^5= 32

4) The number of bits that can be used to create the host space = 32-29=3 bits.

5) The block size of each subnet is 2^3=8. Out of this 2 addresses cannot be used because they are reserved for subnet address and subnet broadcast address. Hence maximum number of host addresses available per subnet = 8-2=6.

6) The subnet mask in binary is 11111111.11111111.11111111.11111000

And in decimal is 255.255.255.248

 Subnet number Subnet address Range of Host Addresses Broadcast address 0 192.168.111.0 192.168.111.1 to 192.168.111.6 192.168.111.7 1 192.168.111.8 192.168.111.9 to 192.168.111.14 192.168.111.15 2 192.168.111.16 192.168.111.17 to 192.168.111.22 192.168.111.23 3 192.168.111.24 192.168.111.25 to 192.168.111.30 192.168.111.31 4 192.168.111.32 192.168.111.33 to 192.168.111.38 192.168.111.39 5 192.168.111.40 192.168.111.41 to 192.168.111.46 192.168.111.47 6 192.168.111.48 192.168.111.49 to 192.168.111.54 192.168.111.55 7 192.168.111.56 192.168.111.57 to 192.168.111.62 192.168.111.63 8 192.168.111.64 192.168.111.65 to 192.168.111.70 192.168.111.71 9 192.168.111.72 192.168.111.73 to 192.168.111.78 192.168.111.79 10 192.168.111.80 192.168.111.81 to 192.168.111.86 192.168.111.87 11 192.168.111.88 192.168.111.89 to 192.168.111.94 192.168.111.95 12 192.168.111.96 192.168.111.97 to 192.168.111.102 192.168.111.103 13 192.168.111.104 192.168.111.105 to 192.168.111.110 192.168.111.111 14 192.168.111.112 192.168.111.113 to 192.168.111.118 192.168.111.119 15 192.168.111.120 192.168.111.121 to 192.168.111.126 192.168.111.127 16 192.168.111.128 192.168.111.129 to 192.168.111.134 192.168.111.135 17 192.168.111.136 192.168.111.137 to 192.168.111.142 192.168.111.143 18 192.168.111.144 192.168.111.145 to 192.168.111.150 192.168.111.151 … 31 192.168.111.248 192.168.111.249 to 192.168.111.254 192.168.111.255

192.168.111.129 is on the subnet 16 that is 192.168.111.128.

He gets an error message because 192.168.111.127 is the broadcast address of he subnet 15 that is 192.168.111.120 and cannot be used.

No, 192.168.111.39 cannot be used as an IP address because it is the broadcast address for the subnet 192.168.111.32