Ask for the first half of 09 network engineer morning questions

In order to establish a complete knowledge system, it is necessary to have a deep understanding of the architecture of the network, the characteristics of the technology, the life cycle of the technology (the causes and processes of the generation, development and death of the technology), the relationship between various technologies, and the development and application direction of the technology.

It is necessary to understand the positioning and role of each knowledge point in the knowledge system, and integrate it in order to truly master the knowledge. In this way, even if you encounter knowledge that you have not been exposed to during the exam, you will have the opportunity to guess the correct answer.

A poor knowledge base often leads to candidates losing a lot of marks that they shouldn't have lost when taking the exam. For example, many candidates have learned the basics of communication and know what is a CRC code and what is a Hemming code, but they are unwilling to bother to master the calculation method of CRC code and Hemming code, and as a result, they often lose points that should not be lost. Therefore, candidates must correct their mentality and lay a solid foundation of knowledge.

Review Method:

Some test takers like to use the sea of questions tactic and do a lot of mock questions before the exam, which is questionable in its effectiveness. Before the exam, you can experience the feeling of taking the exam, grasp the difficulty of the exam, and test your strength, which is beneficial to the candidates. Occasionally doing some mock questions can also help to test your knowledge and mastery of knowledge.

But superstitious tactics can do more harm than good.

I was there a few years ago. Because I was working in this area at the time, I read the book for about three months and did some exercises. Share your experience:

1. First of all, go through the textbook, and it is recommended to study - the designated book "Network Engineer Course" for the National Computer Technology and Software Professional and Technical Qualification (Level) Examination, to buy the latest version;

2. Then read the chapters intensively and practice the questions. For the exercises, please refer to the network engineer workbook series "Network Engineer Exam Pre-Exam Sprint and Test Center Analysis", "Network Engineer Exam Sprint Guide", etc.

3. The afternoon experiment questions will test the operation commands of routers, switches and Linux systems, and after going through the textbook, you can first familiarize yourself with the basic configuration commands and refer to the question types on the exercise set.

No, you go to the library to borrow a network engineer tutorial, the question can be done a little bit Familiarize yourself with the reminder, the most important thing is the proficiency of the operating system and the use of routing switches, etc., I don't know what major you learned how to learn, anyway, I personally feel that I can't pass it once (I took the exam 2 times to pass).

The sea tactics are no longer available! Since 2013, the test questions of network workers have changed a lot.

As shown in the following figure, ports A and B are set to the same VLAN 100, ports C and D are set to the same VLAN 200, and ports E and F are set to VLAN 100 and 200 together.

Set on the R1 router with the address of vlanif100 and the address of 200.

PC1 sets the gateway to.

PC2 sets the gateway to.

Because the formula for the loss rate of the key cluster is r=d t, and d is the size of the data. The unit of sales orange is bit, in this question d = 150 (number of users) * 8 (number of transactions per user) * unit to be unified) where bright d is a minute of data volume 60s to get 8MB s

If the C disk index block is 1 KB byte and the size of each address item is 4 bytes, each disk index block can store 1024 4=256 physical addresses. There are 8 address entries in the file inode draft, of which 5 addresses are direct address indexes, which means that 0 4 is direct address indexes; The two addresses are first-level indirect address indexes, which means that the physical block number of 5 260 is stored in the physical block indicated by the first address item, and the physical block number of 261 516 is stored in the physical block indicated by the first address item. One address entry is a second-level indirect address index, and the physical block indicated by the address item stores the d addresses of 256 indirect state Huna index tables, and these 256 indirect index tables store the physical block number of 517 66052.

So choose C

Because the two network segments of and 26 are in the same LAN, in fact, the topic is also very clear. "Both the hosts of Network 1 and Network 2 are assigned IP addresses by the DHCP Server. "Why is it, because the subnet mask is 26 bits.

24 bits, excluding the FTP server address and the DHCP server address itself.

It's very simple, 32-26 = 6, 6 bits host number, (11000000) = 192, because when the host number is all zero, the IP address is used as a network address, so it can't be used, so (11000001) = 193, so the range comes out.

This is very simple, the left network 1 directly delivers the address through DHCP, and the right network sends the address through the DHCP relay, which must be two addresses.

You can look up what IP 24 and 26 mean to see why.

11111111 11111111 11111111 10000000

100 IP addresses to the 7th power of 2 = 128-2>100 Subnet interval 128

127 broadcast address.

255 is the broadcast address.

I feel that this question is a reverse test variable long subnet mask, but this question is not concretely reflected.

If we analyze the topic in the same direction, we will focus on the sentence "which has been assigned to the personnel department and the finance department".

It is from borrowing three digits, which involves a variable long subnet mask, which means that if you are divided by the subnet, there is still a demand that needs to be divided once in the subnet of the subnet, that is, in or in the division.

In the network segment is the same.

It's not interesting here, the person who wrote the question doesn't know what to think, and he didn't get the knowledge points at all.

It should be d I will give you a positive calculation example, in fact, what does the questioner want to test!!

For example, we want to plan 6 subnets.

Borrow three to divide into 8.

19 subnet spacing 32

2 to the power of 13 of host 2 in each subnet = 8190

255 broadcasts.

We found that at this time, there were 8190 subnet hosts accommodating, and there was a lot of IP waste.

There may still be demand in addition.

We can draw three subnets on it (if the question is asked, it can be said that the IP address of some other subnet has been used, and there may be an IP address requirement, so it is not necessarily borrowing 3 digits, I am equivalent to an example).

Borrow 2 bits and 21 bits at intervals of 8

For the latter port aggregation, in fact, the professional term is CIDR classless inter-domain routing.

This is mainly to reduce the size of the routing table and increase the scalability of the network.

In layman's terms, this is mainly a matter of accuracy, otherwise it doesn't make any sense.

The high position is required to have the same as the low position to have continuous bits.

It should be 29 or 28, it can't be 30

Why? Let's look in the opposite direction.

2 of 14 to the power of 16384 subnets The subnet interval is 4

In this case, a valid IP address is.

It's a broadcast address, so you can't.

You can know it under IPv4 settings on the network card, and set the IP to.

The subnet mask is set to.

There will be hints, take a look at it yourself.