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Q&A for June 15, 2001
This week, we continue our TCP/IP exploration with two questions about basic IP addressing concepts and a question that tests whether you know the functions that common TCP/IP services perform.
Morris Lewis
June 13, 2001
4 Min Read
Welcome to Certifiable, your exam prep headquarters. Here you'll find questions about some of the tricky areas that are fair game for the certification exams. Following the questions, you'll find the correct answers and explanatory text. We change the questions weekly.
Questions (June 15, 2001)
Answers (June 15, 2001)
This week, we continue our TCP/IP exploration with two questions about basic IP addressing concepts and a question that tests whether you know the functions that common TCP/IP services perform. As is often the case, answering correctly isn't really the goal. Instead, you should try explain to yourself or your study partner WHY the answers are correct. Memorizing facts about TCP/IP might help you pass exams, but successful network administrators are successful because they learned why TCP/IP works the way it does. Behind these seemingly simple questions is a wealth of information about why IP addresses must be in a certain format, how routers work, and how computers find each other on the network. Knowing the theory behind the facts will help you make intelligent guesses when you face a new problem—as well as help you prepare for the exams.
Questions (June 15, 2001)
Question 1
In the IP address 129.200.1.254, what is the network class and default subnet mask?
Class A, 255.255.255.0
Class C, 255.255.255.0
Class A, 255.0.0.0
Class B, 255.255.0.0
Question 2
Assuming a default subnet mask, what is the network ID for a host with an IP address of 192.200.254.1?
192.200.254.0
192.200.0.0
192.0.0.0
255.255.255.0
255.255.255.255
Question 3
Scenario
The company network is routed and consists of 20 UNIX Machines, 30 Windows 2000 servers, 50 machines running Win2K Professional, and 1000 machines running Windows 98. All machines must communicate with one another. Routers connect the 14 subnets.
Required Objective
To have a Win2K server act as a router between two subnets and have the route table updated automatically.
Optional Objectives
To have the Microsoft machines initiate sessions with the servers using NetBIOS names.
To have the UNIX machines and the Win2K machines be able to communicate using the FTP command. This includes the ability for a Microsoft or UNIX machine to initiate the FTP session.
Proposed Solution
Enable the routing by placing multiple network adapters in the Win2K server and configure each with a different IP address. The IP address for card 1 will be 131.107.2.25 and the address for card 2 will be 131.107.3.26. These addresses are appropriate for the subnets to which the cards are cabled. The subnet mask is 255.255.255.0.
Activate and configure RRAS.
Enable and configure Routing Information Protocol version 2 (RIP2) for the Internet Protocol in RRAS.
Manually configure appropriate IP addresses and masks on all the clients.
Configure two of the Win2K servers to be WINS servers.
Add the addresses of the WINS servers to the Microsoft client configurations.
Add DNS to one of the Win2K servers.
Create a zone that will include A records for all the UNIX machines.
Enable WINS lookup for the zone.
Add the address of the WINS server.
Configure all the Microsoft and UNIX machines as DNS clients and provide the DNS server as their primary server.
The proposed solution
Achieves the required objective but doesn't achieve either of the optional objectives.
Achieves the required objective and one of the optional objectives.
Achieves the required objective and both of the optional objectives.
Doesn't achieve the required objective.
Answers (June 15, 2001)
Answer to Question 1
The correct answer is D—Class B, 255.255.0.0. The leftmost bits of the first octet are 10, which indicates that the network address is a Class B address. Although the predominant practice now is to use classless addresses, Windows 2000 chooses 255.255.0.0 as the default subnet when you enter this address into the "Internet Protocol (TCP/IP) Properties" dialog box.Answer to Question 2
The correct answer is A—192.200.254.0. An address with a 192 value in the first octet is a Class C address because it ranges between 192 and 223 (i.e., the leftmost bits in the first octet are 11). Because Class C networks have 24-bit subnets (i.e., 255.255.255.0), the network portion is 192.200.254.0. The notation for classless subnetting is 192.200.254.0/24.Answer to Question 3
The correct answer is C—Achieves the required objective and both of the optional objectives. You've taken all the steps to enable dynamic routing, and the interface addresses are compatible with the assigned mask. You've properly configured a Windows 2000 server as a router whose route table will be updated automatically. By installing WINS and configuring the Microsoft clients to use WINS, you've met the first optional objective. You can communicate with the Win2K machines using the NetBIOS or machine names because WINS will resolve and return the IP addresses.You've accomplished the second optional objective by installing a DNS server. The Microsoft machines can resolve the host name to the IP address of the UNIX machines because the records are part of the zone file. The Win2K machines can initiate an FTP session using the host name. The UNIX machines can resolve the Microsoft machine names because the DNS zone is configured to query the WINS server for the IP address of the Microsoft machines.
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