first question
Why is the IP address of the next hop not masked?
Because the mask is known, the ip address and mask of each port of the router must be known when setting the route, so it is known.
second question
Answer: The GARP request is always a broadcast address, not a separate mac address. Note that it is an ARP request, which is an inquiry, so it is FF-FF-FF (but if it is a PING command, it contains the ICMP response request frame The destination mac address is the mac address of its default gateway (connected router port), because the transmission range of the frame is limited and will not spread across the network , so it can only be sent to the gateway)
third question
Answer: The scope of the C frame is limited to the route (one network), and a frame can only reach its gateway at most
fourth question
Answer: For this question, you need to understand the overall process and content of PING, as well as the understanding of frames, which can only be transmitted in one network.
The first is my ip address and the ip address of the school server, which must be available.
The second is the mac address of my own host, and the mac address of the gateway of my network (the destination mac address in the ICMP response request frame is its default gateway).
In fact, the root of the problem is a ping process (indirect arp feeling) to know the role of the arp protocol. The operation of ping still needs to be placed in the mac frame, which is also based on arp.
Combination questions
first question
Answer: Looking at this ARP, it can be seen that it is not an arp request, because it is not FF-FF-FF, so it is an arp response, combined with the arp flag, 00 02 can also be found to be a response. So host 1's mac address is 30-20-55-44-77-88. So check its ip address (sender and receiver) in the data content behind 00-02, the sender’s ip is 40-40-40-02 (hexadecimal) and the receiver (pc1) is 40- 40-40-00, so it must not be a 24-bit mask, so its ip address is the network number, so it must be smaller than 24 bits, so it is 255.255.0.0.
This question is very important. You must be able to disassemble the arp message (it is a mac frame, but the data content in it is ip and mac address)
second question
Answer: This question is still based on the above analysis. It can be seen that this arp message is an arp response, so its source address is the address of the gateway, and it is a response to pc1, so when pc1 performs a ping operation at this time, The mac address requested by arp is the gateway address
third question
Answer: 10.10.10.10
understand the routing table
Direct means direct connection (port), static means static route
127.0.0.1 loopback test address InLoop0
After understanding the routing table, analyze a little bit to find the network segment where the second host is located.
fourth question
Answer: Look at the port table, you can deduce that 10.10.10.10 is the address of host 2, so 4.4.4.1 is the gateway of E5, a little derivation.
fifth question
Answer: There is no address for host 1, so you can only use the default route to go there. The network number is all 0, and the mask writes 4 0s and 1 0 are the same, so finally look at the next hop address, E2 is 30.30.30.0 Yes, so you can use 30.30.30.34 as the IP address of E1. so it can be used as next hop
Sixth question
If there is no match, it will not be forwarded and discarded. There is no forwarding to all ports in the route, and those that do not match will be discarded.
Transport layer class records:
The link layer and the network layer are still together. The network layer solves the problem of connection between various networks, how to connect and address large networks (routing), and is essentially point-to-point.
The transport layer rises one layer, end-to-end, which is the communication between two application processes of communication
1. Why is the transport layer needed
? 2. How is the tcp connection established at the transport layer? Its three-way handshake process
can be understood in conjunction with the issue of the two armies. At this time, you will know why it is three times instead of two. Both parties need to confirm , Finally, a confirmation confirmation is required to establish the connection.
Three-tier exchange quiz (all wrong, study slowly)
Three-layer switching realizes communication between different vlans.
Note that in order to make full use of three-layer switching, B cannot be selected, because choosing B will become a routing mode, and it is no longer a three-layer switching, so it must be divided into vlans. In this way, different networks can use Layer 3 switching to communicate between different vlans.
First of all, you need to know what is the next hop address. The next hop is the address of the next nearest router port, so it should be the next router, but this obviously does not have the next router, so it is not listed. Same as the previous question,
grab Live what the next hop is. Because the network number is 30, only two host numbers can be used, and one of them has been given to E1, so it is a point-to-point network. This point-to-point link mode needs to write the E1 port number, and the other big ones It can be written
that the layer-3 switch is regarded as a router, so it
is necessary to set up static routing. Layer-3 switching, so E3 must be added to a vlan to play the role of layer-3 switching. The router port is directly configured because it does not have the concept of vlan
. se=1461 needs to be confirmed, it needs to be +1, but len is 1460, so 1460 needs to be added,
this question does not know why it is not
The key to this question is the 8100 protocol field, 0x8100 is the protocol of the vlan tag tag, indicating that the mac frame contains the vlan tag tag
The most important thing is to understand the conversion between symbols and bits. There are 32 amplitudes, which means that there are 32 states, and 2 to the 5th power is 32, so a symbol needs to be represented by 5 binary bits, so it is 10000*5=50kbps
Binary Exponential Backoff Algorithm: 2 to the kth power -1
10kHZ is the bandwidth, 200kbps is the channel data transmission rate
200k=10klog2 (1+S/N) S/N=10 to the 6th
power SNR=10log10(S/N)=60db
This question is a trap. Go back N frames - GBN always has only one acceptance window, only select retransmission - SR has multiple acceptance windows, and the size of the SR acceptance window should not exceed half of the serial number range. If this question is SR should be less than 32/2=16, up to 15.
Let the data transmission rate be x, (actual sending time) t=5000/x, 0.25=t/(t+0.06) and finally calculate x=250kbps, pay attention to the unit conversion between s and ms.
rote
The most important point of this question is don’t forget about NAPT . In fact, you can’t get all the information of the private network except ip, which will be isolated. You can only get your own ip address, your own mac address and the location of the server. ip address, so what is the destination mac address? Since the scope of the frame can only be within one network segment, in fact your destination mac address is the mac address of your gateway, so you can only get these things.
This will also lead to another problem, maybe you can ping a certain ip address, but this does not mean that you can get her corresponding mac address.
