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Anonymous users2024-02-08I think it's a network, because it's a private IP network, but a public IP network. The specific division is as follows:
Hope mine can help you
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Anonymous users2024-02-07Upstairs, yes, I'm not going to say anything more, take my two points.
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Anonymous users2024-02-06Summary. 1.It can be seen from the first byte range of Class A network is 1-126, the first byte range of Class B network is 128-191, and the first byte range of Class C network is 192-223.
The network where 100 is located belongs to a Class B network. 2.The subnet mask for a standard Class B network is:
The subnet mask of the network is the subnet division of the network. 3.Since the subnet mask , 2 can be known, and 4 bits are borrowed as the subnet number.
Therefore, the number of subnets is 24-2=144Network address, 97=(01100001) 240=(11110000) See (01100000)2=96 both network address is. Broadcast address, 2, 111111111111, 2(0110111111111111) 2=both broadcast address.
Valid P address, requirements: two hosts per network segment (one set to minimum P and one set to maximum P), and use packet tracer for testing and verification.
The computer network technology course has four classic questions. 1. A unit has 510 hosts, and its network address is the network number (for example, if the student number is added 10210110128 the network segment is, please calculate the network segment.
network address, broadcast address, and section.
The maximum valid p-address and the minimum of the subnet segment of the first and second subnets.
The computer network technology course has four classic questions. 1. A unit has 510 hosts, and its network address is the network number (for example, if the student number is added 10210110128 the network segment is, please calculate the network segment.
Valid P address, requirements: two hosts per network segment (one set to minimum P and one set to maximum P), and use packet tracer for testing and verification.
network address, broadcast address, and section.
The maximum valid p-address and the minimum of the subnet segment of the first and second subnets.
The computer network technology course has four classic questions. 1. A unit has 510 hosts, and its network address is the network number (for example, if the student number is added 10210110128 the network segment is, please calculate the network segment.
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Anonymous users2024-02-05Assuming that the IP address of a computer in a network is assumed, the maximum number of machines+ that can be accommodated in the network
Dear, I'm glad to answer for you: Suppose the IP address of a computer in a network is, then the network can accommodate the largest machine + Answer: Dear, hello, this is related to the number of bits of the host bit of the IP address of the virtual address.
For IP addresses, the first 24 bits of the IP address are network bits, and the last 8 bits are the host bench burn bits, which belong to Class C IP addresses and the subnet mask is. n=8, then the number of hosts = 2 8 -2 = 254. If you have any questions, you can give me feedback in time!
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Anonymous users2024-02-04If there are no more than 15 computers in each subnet, then the host bit is 5 bits and the mask is 0000, that is.
The 4 subnets are:
Available IP address ranges from broadcast addresses.
Available IP address ranges from broadcast addresses.
Available IP address ranges from broadcast addresses.
Available IP address ranges from broadcast addresses.
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Anonymous users2024-02-03Judging as a Class C IP, according to the number of subnet hosts, it can be concluded that the n-power of 2 -2>15 (n is the host bit) needs to be greater than or equal to 5, that is, the network bit borrows up to 3 bits and is a 27-bit subnet mask.
If you borrow 3-bit network bits, you can divide 2 to the 3rd power = 8 subnets, and the increment of the subnet is 2 to the 5th power = 32 (the host bit is 5), which meets the requirements of the problem. So.
You can select any 4 subnets.
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Anonymous users2024-02-02The payload containing 1500 in the longest frame 1518b, i.e. 12000b.
The transmission delay of the longest frame sent is 1518 8b 10mbps = transmit propagation time 2km (2 10 8m s) = 10 s, the propagation delay of the received reply frame is 64 8b 10mbps = the received propagation time is also 10 s
Therefore, the effective transmission rate is 12000b ( =
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Anonymous users2024-02-01The Class C network is divided into 16 subnets, each of which has 14 available host addresses.
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Anonymous users2024-01-3118b is the Ethernet header part.
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Anonymous users2024-01-30Summary. It is now necessary to divide the 1,000 machines of a department into local area networks, where the first local area network contains 500 computers, the second local area network contains 250 computers, and the third local area network contains 250 computers. If an address block assigned to the department requires each LAN to be in different network segments, assign an appropriate address block (including the network prefix) to each LAN and write out the broadcast address, subnet mask, and host address range for each LAN.
It is now necessary to divide the 1,000 machines of a department into local area networks, where the first local area network contains 500 computers, the second local area network contains 250 computers, and the third local area network contains 250 computers. If an address block assigned to the department requires each LAN to be in different network segments, assign an appropriate address block (including the network prefix) to each LAN and write out the broadcast address, subnet mask, and host address range for each LAN.
It is now necessary to divide the 1,000 machines of a department into local area networks, where the first local area network contains 500 computers, the second local area network contains 250 computers, and the third local area network contains 250 computers. If an address block assigned to the department requires each LAN to be in different network segments, assign an appropriate address block (including the network prefix) to each LAN and write out the broadcast address, subnet mask, and host address range for each LAN. Hello dear, 1000 computers, divided into 4 subnets, 250 computers per subnet, mask use hope can help you.
What about the LAN's broadcast address, subnet mask, and host address range?
What about the LAN's broadcast address, subnet mask, and host address range? Hello dear, take the IP address and subnet mask as an example, calculate the network address, broadcast address, address range 1, convert the IP address and the subnet mask to binary numbers, the subnet mask is all 1 consecutive network address, the first 3 digits, that is, the network address in front of the thick line, and the host address in the back. 2. The network address part of the IP address and the subnet mask is calculated, and the final host address becomes 0, and the result is the network address, that is, the first IP address in the specified network segment.
Note that this address cannot be assigned to any computer. 3. If the network address part in the result of the above and operation remains unchanged, and the host address becomes 1, then the obtained is the broadcast address, which can also not be assigned to the client. A valid range of IP addresses in that network is:
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