How to correctly locate the IP address?

Question

Computer networks

Good evening.
Recently, a friend was asked a problem, and I never found the right solution to it.
The bottom line is that you need to correctly allocate an IP address so that the network is functioning.
Given
the number of computers in subnets
A 425
B 32
C 341
D 95
E 353
F 86
G 253
H 308
address 183.118.32.0/20
And the scheme
To be honest, I did not even find the right approach to the solution.
It is clear from the condition that the VLSM addressing scheme must be used.
The essence of the question from the source.
design, create and verify the functional model of data network based on the TCP/IP architecture. Detiled information about IP address range and amount of PCs in each network will be sent individually by an e-mail
Allocate effectively IP addresses using VLSM addressing scheme focusing on saving IP address range. Do not forget to address all point-to-point connections between routers.
Tasks
Design IP address plan. Use VLSM while saving IP addresses.
Create network model in PacketTracer. Configure optional devices. The networks must be accesible to eachother. Use static routing.
Verify your model using ICMP tool (ping).
Make documentation. Briefly, clearly and concisely describe your solution and important parts of the configuration of routers, switches and computers. Include overview table with IP addresses used.

Answer 1

/20 is not an address, but a network with 4096 potential addresses (-2 for network and broadcast). Let's start allocating addresses for subnets, after sorting them in descending order by the number of computers. We will choose subnets so that their size is 4 more than the number of computers (address of the network itself, broadcast, addresses for the switch and router):

183.118.32.0/20 = 183.118.32.0-183.118.47.255
A | 425 | 512 | /23 | 183.118.32.0/23 = 183.118.32.0-183.118.33.255
E | 353 | 512 | /23 | 183.118.34.0/23 = 183.118.34.0-183.118.35.255
C | 341 | 512 | /23 | 183.118.36.0/23 = 183.118.36.0-183.118.37.255
H | 308 | 512 | /23 | 183.118.38.0/23 = 183.118.38.0-183.118.39.255
G | 253 | 512 | /23 | 183.118.40.0/23 = 183.118.40.0-183.118.41.255
D |  95 | 128 | /25 | 183.118.42.0/25 = 183.118.42.0-183.118.42.127
F |  86 | 128 | /25 | 183.118.42.128/25 = 183.118.42.128-183.118.42.255
B |  32 |  64 | /26 | 183.118.43.0/26 = 183.118.43.0-183.118.43.63

To connect routers, you can add another /28 subnet or 8 /30 subnets. Routers with switches are connected by trunks and do not require separate addresses for communication with each other.

Answer 2

What is thrash anyway? :-) The condition of the problem is clearly incomplete. Either you didn't fully state the condition, or you omitted most of the context. What needs to be done? What does "host an IP address" mean?