Answer Key for TCP/IP LAN Plan CheckPoint, Due in Week Four 8. C, 255.255.240.0. 150.50.0.0 Subnet 150.50.0.0 to support 7 subnets. 150.50.0.0 is a Class B with a default subnet mask of 255.255.0.0 (11111111.11111111.00000000.00000000 in binary) To figure out how many bits we need to change from 0 to 1 we can use the formula (2^n)-2=Requirement. When you do the math you come up with the “Binary Truth Line” which is : 2-4-8-16-32-64-128-256-512-1024-2048…………. Count from left to right along the line until you pass seven. You might think 8 would be correct, but you must remember to subtract 2 for the network/wire address and the broadcast address that you cannot use. So, we must go to 16. Now, 16 is the 4 number over …show more content…
When you do the math you come up with the “Binary Truth Line” which is : 2-4-8-16-32-64-128-256-512-1024-2048…………. Count from left to right along the line until you pass 177. Since we have to remember about the 2 address that we subtract I changed the number to 177 to account for them so we make sure we have enough addresses. 128 is not enough so we must go to 256. Now, 256 is the 8th number over so we need to RESERVE 8 bits for the host portion of the address. Since we are solving for host we count those 8 bits from right to left starting at the end of the address with 0’s. In Binary it would look like this: (11111111.11111111.11111111.00000000). Everything to the right stays a 0 indicating the number of host in each subnet. Everything to the left becomes a 1. The new subnet mask is 255.255.255.0. You get this number by adding the 1’s together in each octet (128+64+32+16+8+4+2+1=255). Your networks would look something like this: 150.50.0.0 Gateway (150.50.0.1); Range for Host (150.50.0.2 – 150.50.0.254) Broadcast (150.50.0.255) 150.50.1.0 Gateway (150.50.1.1); Range for Host (150.50.1.2 – 150.50.1.254) Broadcast (150.50.1.255) 150.50.2.0 Gateway (150.50.2.1); Range for Host (150.50.2.2 – 150.50.2.254) Broadcast (150.50.2.255) The networks “increment” by 1 in the third octet. Why? If you look at the subnet mask in binary the last 1 bit is in the 1 position in the
37) DHCP ________ are configurable parameters that determine which subnets the DHCP server will serve.
Section 1Computing Usable Subnets and Hosts vLab—40 Points Total * Task 1 * Task 2 * Summary Paragraph
Sub netting: The procedure of apportioning a solitary TCP/IP system into various separate system sections called subnets.
If ans <> "Y" And ans <> "y" And ans <> "N" And ans <> "n" Then
numbers fit all of the rules that applied. I remembered that it could perfectly go into a group of
135.46.52.2. In binary this is 10000111.00101110.00110100.00000010. Again, we are looking at one of the Interfaces, so we take the 22 MSB’s, which
I think I need to understand the symbols better and how they work in Python, right now my head is thinking as I usually use them as in (1+1) = 2 , (5-2) =7 , 2*7 =14 ...but the answer is 8. how?
When setting up a network that will consist of many host computers, one of the first things that an administrator must do is to determine what class of networks that they must administer to a given business. This is the point where every administrator must know how to implement classful and classless IP addressing. A classful network is a network addressing architecture used in the internet from 1981 until the introduction of Classless Inter-Domain Routing (CIDR) in 1993. Classful IP addressing divides the addtess space on the internet into five address classes. Each class is coded in the first four bits of the address. Today
If we take the 172.16.0.0 range, that is /12. So that means that the network mask is 255.240.0.0. If we take 256-240 = 16 we get the block size which is 16. So that means that the range is from 172.16.0.0 to 172.31.255.255 because 172.32.0.0 is the next block and would be a public IP.
I found the secret formula, it was (w+L)-2 but w/l had to be reduced so it
It creates a VPC with a /16 IPv4 CIDR block. It is a network with 65,536 private
0 1 1 0 0 0 0 1 64+32+1=97, This is the ASCII code for A
and the second path is where ten is added to the value of the entrance and then the subtraction occurs. On the side you can see a series of logic gates that generate a function that will produce an output of 1 only when digit 1 of the exit is bigger than that of the entrance. This output will be sent to two different multiplexers, one for each digit. In the digit 1 diagram it will be sent to a multiplexer that will decide whether to use the subtraction where the entrance has been increased by 10 or whether to use the other where the entrance has had no alteration. If it is a 1, then it will use the first option where ten was added to the entrance. If it is a zero, then the latter will be used. This can be seen below in figure 2.
Limoncell, T. A., Hogan, C. J., & Chalup, S. R. (2007). The Practice of System and Network Administration, Second Edition. Addison-Wesley Professional.
Key tools utilized, variable length subnet masking and route summarization are explained as well. Here choosing the appropriate routing protocol is equally critical for a successful design. To implement different masks for the same major network it is necessary to have a routing protocol that supports VLSM. Such routing protocols are called classless routing protocols. They carry the mask information along with the route advertisements therefore allowing for the support of more than one mask.