andreaulicna/42_netpractice

This project is a System Administration related exercise.

42_NetPractise

To start

tar -xf [file_name]
chmod +rwx [file_name]
[browser] [file_name]

Levels

Level 1

Goal 1

Starting state:

• host A and host B are on the same network
• subnet mask is 255.255.255.0 which translates to 11111111.11111111.11111111.00000000 in binary and therefore means that:
  • the first 3 bytes represent the network address
  • the 4th byte represents the host address

Solution:

• the address to change is just that of the host as we are on the same network
• given that it is the 4th byte that represents the host address, the solution is any address between 104.94.23.0 and 104.94.23.255 except:
  • 104.94.23.0: the first number, "0" represents the network
  • 104.94.23.255: the last number, "255", is reserved for the broadcast address (used to send information to all the systems on a network)
  • 104.94.23.12: already used by host B

Goal 2

Starting state:

• host D and host C are on the same network
• subnet mask is 255.255.0.0 which translates to 11111111.11111111.00000000.00000000 in binary and therefore means that:
  • the first 2 bytes represent the network address
  • the last 2 bytes represent the host address

Solution:

• the address to change is just that of the host as we are on the same network
• given that it is the last 2 byte that represents the host address, the solution is any address between 211.191.0.0 and 211.191.255.255 except:
  • 211.191.0.0: the first number, "0" represents the network
  • 211.191.255.255: the last number, "255", is reserved for the broadcast address (used to send information to all the systems on a network)
  • 211.191.81.75: already used by host C

Level 2

Goal 1

Starting state:

• host A and host B are on the same network
• subnet mask is 255.255.255.224 (given for host A and will be copied to host B as a part of the solution) which translates to 11111111.11111111.11111111.11100000 in binary and therefore means that:
  • the first 27 bits represent the network address
  • only the last 5 bits represent the host address

Solution:

• the subnet mask should be the same for both hosts and since we can change only that of host B, we copy the value of host A there
• given that it s the last 5 bits that represent the host address, the solution is any address between 11000000.10101000.00100011.11000000 and 11000000.10101000.00100011.11011111in binary and therefore between 192.168.35.192 and 192.168.35.223 in decimal, except:
  • 192.168.35.192: the first number, "0" represents the network
  • 192.168.35.223: the last number, "255", is reserved for the broadcast address (used to send information to all the systems on a network)
  • 192.168.35.222: already used by host B

Goal 2

Starting state:

• host D and host C are on the same network
• host C's subnet mask is 255.255.255.252 which translates to 11111111.11111111.11111111.11111100 in binary and therefore means that:
  • the first 30 bits (as specific in host's D subnet mask) represent the network address
  • the last 2 bits represent the host address

Solution:

• given that it is the last 2 bits that represent the host address and we are not given at least one to derive the solution from, the addresses just need to obey this criteria:
  • the first 30 bits (the network address) must be the same for client D and client c
  • they cannot be the same for client D and client C
  • the last 2 bits (that represent the host address) cannot be both 1 nor 0 as the former is the broadcast address and the later is the network address
  • they cannot be among the reserved IP addresses
  • neither address can be 127.0.0.1 as this is a 'special', loopback, address
    • packets sent to this address never reach the network but are looped through the network interface card only
    • it can be used for diagnostic purposes to verify that the internal path through the TCP/IP protocols is working

Level 3

Starting state

• hosts A, B and C are all on the same network, connected via a switch
• host C's subnet mask is 255.255.255.128 which translates to 11111111.11111111.11111111.10000000 in binary and therefore means that:
  • the first 25 bits represent the network address
  • the last 7 bits represent the host address

Solution

• since host C has a subnet mask set and all the hosts are on the same network, their subnet masks should be the same - either 255.255.255.128 or /25
• given that it is the last 7 bits that represent the host address, the solution for host A and host B addresses can be any address between .00000000 and .01111111 in binary which translates to an address between 104.198.152.0 and 104.198.152.127, except:
  • 104.198.152.0: represents the network address
  • 104.198.152.127: represents the broadcast address
  • 104.198.152.125: already taken by host A

Level 4

Starting state

• host A and B are on the same network, connected via a switch
• the switch also connects route R with them
  • a route connects multiple networks together, doing so via interefaces for each
• we need to take care only of the communication happening between the router Interface R1 and those of hosts (Interface A1 and Interface B1), so within one network

Solution

• interface R1, A1 nor B1 has a mask setup, and so the choice is on us
  • for simplicity, a mask of increments of 8 can be used as it won't require binary calculation to find the range of suitable addresses
  • mask of 255.255.0.0 or /16 can, therefore, do
• interface A1 has an IP address specified and so the first 2 bytes of addresses for Interface B1 and R1 should follow it
  • therefore, their addresses can be any between 72.96.0.0 and 72.96.255.255 expect:
    • 72.96.0.0: represents the network address
    • 72.96.255.255: represents the broadcast address
    • 72.96.116.132: already taken by interface A1

Level 5

Theory

• interfaces A1 and B1 have routing tables to be completed with:
  • destination:
    • speficies the network address the host is on
    • default or 0.0.0.0/0 address is the route taken when there is no IP destination address provided, sending the packets via the next-hop address
  • next-hop:
    • refers to the closest router (it's IP address) the packets can go via

Solution

Goal 1

• interface R1 and A1 are on the same network and therefore need to have the same subnet mask of 255.255.255.128
• that subnet mask defines that the host address is represented by the last 7 bits (126 in binary is 01111110) and so is between .00000000 and 01111111 in binary, translating to between 18.143.35.0 and 18.143.35.127 in decimal, except:
  • 18.143.35.0: represents the network address
  • 18.143.35.127: represents the broadcast address
  • 18.143.35.126: already taken by interface R1

Goal 2

• interface R2 and B1 are on the same network and therefore need to have the same subnet mask of 255.255.192.0
• that subnet mask defines that the host address is represented by the last 14 bits (192 = 11000000, 254 = 11111110) and so is between .11000000.00000000 and .11111111.11111111 in binary, translating to between 153.192.192.0 and 153.192.255.255 in decimal, except:
  • 153.192.192.0: represents the network address
  • 153.192.255.255: represents the broadcast address
  • 153.192.192.254: already taken by the interface R2

Goal 3

• since there is only one route that both, A1 and B1, interfaces can send packets via saying default or 0.0.0.0/0 is enough
• the next-hop address needs to equal the IP address of the closest route

Level 6

Theory

• internet behaves like a router
• yet, if an interface is directly or indirectly connected to the internet, its address cannot have the following reserved private IP ranges:

192.168.0.0 - 192.168.255.255 (65,536 IP addresses)
172.16.0.0 - 172.31.255.255   (1,048,576 IP addresses)
10.0.0.0 - 10.255.255.255     (16,777,216 IP addresses)

Solution

• match host A's subnet mask to that of the router
• get host address of interface R1: the subnet mask defines that the host address is defined by the last 7 bits (128 = 10000000, 227 = 11100011) and so is between .10000000 and .11111111 in binary translating to 75.221.134.128 and 75.221.134.255 in decimal, except:
  • 75.221.134.128: represents the network address
  • 75.221.134.255: represents the broadcast address
  • 75.221.134.227: already taken by host A
• match next-hop of host A to the router's IP address
• reset the destination of router to default
• set the internet's destination to the network address of host A (calculated in the first step), including the subnet mask in / slash notation (/25)

Level 7

Theory

• different networks have different ranges of IP addresses
• an overlap in IP address range would imply that the interfaces are on the same network

Solution

• the IP addresses of interface R11 and R12 are set and hence this is where we should start - by setting the subnet mask for them
• since the subnet mask defines which part of the IP represents the network and host address, it needs to be set so that the range of addresses for the network the interface R11 is in doesn't overlap with the range of the addresses for the network the interface R12 is in (and then also the 3rd network present)
• possible masks:
  • mask /25 isn't enough as it creates only two ranges:
    • .0 to .127
    • .128 to .255
  • mask /26 (or higher) is more suitable as it creates 4 ranges:
    • .0 to .63
    • .64 to .127
    • .128 to .191
    • .192 to .255
• lastly, the next-hop fields need to be setup in this way:
  • each route needs to have the other one's address
  • each host needs to have the address of the router it is connected to

Level 8

Theory

• overlapping IP address as in Level 7

Solution

• set fields that are certain:
  • internet next-hop is equal to the interface R12's IP address
  • interface R13's IP address is equal to the set next-hop in the router R2 routing table
  • subnet mask of interface R23 is equal to the subnet mask of interface D1
• determine the networks range based on the internet's destination:
  • the range is between 159.23.26.0 and 159.23.26.63 (0 is 00000000 in binary and by changing the last 6 bits, we get 00111111 which is 63 in decimal)
• now we need to split this range into at least 3 parts since there are 3 networks to set IP addresses that cannot overlap for:
  • router R1 and R2
    • we need only 2 IP addresses and so the subnet mask can be set to /30
    • therefore the range (62 = 00111110) is between .00111100 and .00111111 in binary translating to between 159.23.26.60 and 159.23.26.63 in decimal (except the border addresses)
  • host D and router R2
    • the subnet mask is set to 255.255.255.240 (/28), and so the last 4 bits will be changing in the IP address, splitting the IP address into 4 parts
    • to make things easy, we can set the last byte of the IP address to 0 to determine the lowest IP range
    • and so the range will be between .00000000 and .00001111 in binary, translating to between 159.23.26.0 and 159.23.26.15 in decimal (except the border addresses)
  • host C and touter R2
    • here we again need just 2 IP addresses, so the subnet mask can be set to /30
    • given the lowest range is used for the network of host D and router R2, we can pick up the next available one
    • that is the range between .00010000 and .00010011 in binary, translating to between 159.23.26.16 and 159.23.26.19 in decimal (except the border addresses)
  • the last thing is to set the addresses in the routing tables:
    • host C:
      • destination: default
      • next-hop: interface R22's IP address
    • host D:
      • destination: default
      • next-hop: interface R23's IP address
    • router R2:
      • destination: default
    • internet I:
      • next-hop: interface R12's IP address (already set as a certain field)
    • router R1:
      • the destination and next-hop for the internet are already entered, so these are for router R2:
        • destination: network address that all the 3 networks in this level are on (which is defined as the destination for the internet I too)
        • next-hop: interface R21's IP address

Level 9

Goal 1

• set the subnet mask based on the specified mask of interface R11 for all the systems in the network
• keeping in the original IP address of interface R11, we have a range between 192.168.172.0 and 192.168.172.128 (except border addresses) for this network to use
• for the routing tables of both, host B and A, we set destination to default and next-hop to the IP address of switch S

Goal 2

• host D - gluon to router:
  • set the subnet mask of interface D1 to the same one the interface R23 has
  • the host D routing table specifies the IP address of interface R23
  • based on that, we can determine the range for this network that is (45 = 00101101) 28.168.0.0 to 28.168.63.255 (except the border addresses)
• host C - cation to router:
  • we can keep in the subnet mask as well as the original IP addresses as we're free to choose them
• routing tables:
  • set the destination of host D to default

Goal 3

• one of the destination in the internet I's rounting table needs to be the network address of the network host A - meson is in
  • that should be 192.168.172.0/25, however, it is an address that belongs to a range of reserved private IP ranges that cannot be used when an interface is directly or indirectly connected to the internet
  • we, therefore, need to change it to an address that is not in those ranges, in all these places:
    • each system in the network -> host B, host A and switch S
    • routing table of each host as the next-hop
    • and only then add it to the routing table of internet I

Goal 4 (and Goal 5 as a side effect)

• setup connection between router R1 and R2 by:
  • setting the subnet mask of interface R21 to interface R13
  • keeping the original IP address of the interface R13 and adjusting that of interface R21 to match the conditions for the IP addresses range given by the subnet mask
  • set a line in routing tables for router R2 and R1 to:
    • destination: default
    • next-hop: the IP address of the other router
• setting one line of router R1's routing table as follows:
  • destination: network address of host D - gluon, which should end with .0.0
  • next-hop: IP address of interface R21

Goal 6

• the routing table of internet I needs to contain a line that has the destination set to the network address of the network host C - cation is in:
  • that should be 10.0.0.0/24, but it is an address that belongs to a range of reserved private IP ranges that cannot be used when an interface is directly or indirectly connected to the internet
  • it, therefore, needs to be changed to an adress that is not those ranges, in all these places:
    • each system in the network -> host C and interface R22
    • routing table of host C
    • and only then add the network address to the routing table of internet I

Last touches

• the routing table of router R1 need to connect:
  • host C with the internet and therefore needs to have the host C's network address set as destination to a next-hop of interface R21's IP address
  • host A and host D and therefore needs to have the host D's network address set as destination to a next-hop of interface R21's IP address

Level 10

R11-H21-H11 network

• set the subnet mask from interface R11 to all other ones
• calculate the IP range for the network:
  • 128 = 10000000 and so the last 7 bits can be changed
  • that gives us the range (1 = 00000001) of 128.1.26.0 to 128.1.26.127 with mask /25
• choose an address from that range for interface H21

R23-H41 network

• set the subnet mask from interface H41 to interface R23 too
• calculate the IP range for the network:
  • 192 = 11000000 so the last 6 bits can be changed
  • that gives us the range (131 = 10000011) of 128.1.26.128 to 128.1.26.191 with mask /26
• instead of just choosing an address from this range, use the address already specified in the host H4's rounting table for the interface R23's IP address

R13-R21 network

• set the subnet mask (equal to /30) from interface R21 to interface R13 too
• calculate the IP range for the network:
  • 252 = 11111100 so the last 2 bits can be changed
  • that gives us the range (253 = 11111101) of 128.1.26.252 to 128.1.26.255 with mask /30

R22-H31 network

• let's summarize the IP address ranges used so far to determine a one that can be used for this network:

128.1.26.0 - 128.1.26.127 with mask /25		# R11-H21-H11
128.1.26.128 - 128.1.26.191 with mask /26	# R23-H41 
128.1.26.252 - 128.1.26.255 with mask /30	# R13-H21

• therefore, there is a range of .192 to .251 that can be used for this network
• since we need only 2 IP addresses we can use mask /30 with some value from within that range
• if it the value of 200 (200 = 11001000), the range would be .200 to .203
• now we can set the next-hop in the routing table of host H3 to the IP address of interface H31
• and also put the network address of this network, which is 128.1.26.200, to the routing table or router R1, including the mask /30

Connection to the internet

• the destination of the internet I's routing table need to cover the IP address ranges of all 4 networks as there is at least one system in each network that needs to be connected to the internet
• those ranges are:

128.1.26.0 - 128.1.26.127 with mask /25		# R11-H21-H11
128.1.26.128 - 128.1.26.191 with mask /26	# R23-H41 
128.1.26.252 - 128.1.26.255 with mask /30	# R13-H21
128.1.26.200 - 128.1.26.203 with mask /30	# R22-H31

• therefore the destination can be set 128.1.26.0/24 which would cover them all