| A1: | Answer: D. The default subnet mask for Class C is 255.255.255.0. A is incorrect, as it is the default mask for Class A. B is incorrect, as it is the default mask for Class B. C is incorrect, as this is not a valid mask. Subnet masks must be consecutive ones. E is not correct, as it is the subnet mask of all ones. | | A2: | Answer: A. Transmission Control Protocol (TCP) provides reliable end-to-end connectivity, along with error-detection capabilities. B is incorrect, as UDP does not provide any reliability. C is incorrect, as IP does not provide reliability. D is incorrect, as ICMP is used to help troubleshoot TCP/IP connections. | | A3: | Answer: C. This particular subnet mask allows for up to 62 hosts per network. A is incorrect, as it allows for 65,534 hosts. B is incorrect, as it allows up to 254 hosts. D is incorrect, as it is not a valid subnet mask. | | A4: | Answer: D. The next valid IP host address in the 172.16.12.32 network is 172.16.12.34. A is incorrect, as it is on a different subnetwork with a /28 mask. B is incorrect, as it is the network address for that particular subnetwork. C is incorrect, as it is the same IP address as the router interface. E and F are incorrect as well, as they are on different subnets than the router interface. Choice G is incorrect because it represents the broadcast address for the 172.16.12.32 subnetwork and cannot be used as a host address. | | A5: | Answer: C. The network address for the IP address is 172.16.208.0 255.255.240.0. A, B, and D are incorrect, as they are different subnetworks. | | A6: | Answer: D. This subnet mask gives you two additional subnets using VLSM, with up to 14 hosts per subnetwork. A is incorrect, as it is a higher subnet mask than your original /26, which is actually called supernetting. B is incorrect because it is your original subnet mask. C is incorrect, as it does not give you enough subnets. E is incorrect, as it gives you enough subnets (six), but you would have only six hosts. | | A7: | Answer: A. You have actually not done any subnetting with a /24 mask; it is the default Class C subnet mask of 255.255.255.0. This gives you one network with 254 possible hosts. B, C, and D are incorrect because you have not borrowed any bits to create subnetworks. | | A8: | Answers: A, C, and D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host port (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 1111110. B is incorrect, as 112 in decimal is 1110000 in binary. This is not a valid host address in this network. All its host bits are zero. E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero. | | A9: | Answer: C. The network address would be 201.100.5.64 with a subnet mask of 255.255.255.240. A is incorrect, as it is the network address of subnet zero. B is incorrect, as it is the network address of another subnetwork. D is incorrect, as it is a valid host address on the 201.100.5.64 subnetwork. E is incorrect, as it is a broadcast address on another subnetwork. F is incorrect, as it is a valid host address on another subnetwork. | | A10: | Answer: A. TCP (Transmission Control Protocol) utilizes a three-way handshake to establish a session. B is incorrect, as IP (Internet Protocol) is used at Layer 3 of the OSI Model to provide logical addressing. C is incorrect, as UDP is not reliable and does not use a three-way handshake to establish a connection. D is incorrect, as IPX is not reliable and does not use a handshake method to provide a connection. E is incorrect, as frame relay does not provide any acknowledgment or error correction. | | A11: | Answer: A. TCP provides reliable, connection-oriented sessions to ensure that data is transferred correctly. B is incorrect, as IP (Internet Protocol) is used at Layer 3 of the OSI Model to provide logical addressing. C is incorrect, as IPX is not reliable and does not use a handshake method to provide a connection. D is incorrect, as frame relay does not provide any acknowledgment or error correction, and thus is unreliable and connectionless. | | A12: | Answer: C. UDP (User Datagram Protocol) relies on Layer 7 (Application) to handle error detection, correction, and control. Because of the low overhead, it is a very fast protocol. A is incorrect, as TCP contains significant overhead to handle the reliable delivery of data. B is incorrect, as IP (Internet Protocol) is used at Layer 3 of the OSI Model to provide logical addressing. D is incorrect, as SPX, Novell's proprietary connection-based protocol, provides reliable delivery and thus more latency. E is incorrect, as AppleTalk is a Layer 3 protocol that deals with logical addressing. | | A13: | Answer: B. Applications and operating systems have the capability to adjust the window size of TCP. The window size is the amount of data that can be sent before TCP requires an acknowledgment, or ACK. A is incorrect because adjusting the Maximum Transmission Unit merely adjusts how big the packets are or how much data they can contain. C is incorrect, as it does not exist in the TCP/IP protocol stack. D is incorrect, as the Frame Check Sequence (FCS) is used at the Data-link layer of the OSI Model to check for corrupted frames. | | A14: | Answer: B. When a host receiving information misses a segment from the TCP window, it sends an ACK message that shows the next segment number it is ready to receive. If this segment number is the same as a segment that was sent, the sending host knows to resend the information. A is incorrect, as the TCP segment number is included in the ACK sent. C and D are not valid acknowledgments and are thus incorrect. | | A15: | Answer: C. A Class C network can support 254 hosts. A is incorrect, as a Class A network supports more than 16 million hosts. B is incorrect, as a Class B network supports more than 65 thousand hosts. D is incorrect, as a Class D network is for use by multicast applications. E is incorrect, as Class E addresses are experimental and used only by the government and research organizations. | | A16: | Answers: A, C, and D. These addresses are all part of the private ranges of 10.x.x.x and 172.16.x.x 172.31.x.x. B is incorrect, as it does not fall into any of the RFC 1918-specified private ranges. Instead, it is a public IP address that is routable on the Internet. E is incorrect, as it is not a part of the private IP address range. It falls outside the 192.168.x.x range and thus is a public IP address. | | A17: | Answer: C. The subnet mask 255.255.255.224 gives you six usable subnetworks, assuming you are not using ip subnet-zero, with up to 30 available hosts per subnet. This subnet mask meets the criteria you have been given and even allows a bit of future growth. A is incorrect, as it is the default Class C subnet mask. It provides one network with 254 hosts. B is incorrect, as it gives only two subnetworks, assuming you are not using ip subnet-zero. D is incorrect, as it gives enough subnetworks but not enough hosts for the Customer Support or Engineering departments. E is incorrect, as it gives enough subnetworks but not enough hosts for most of the departments. | | A18: | Answer: A. User Datagram Protocol (UDP) relies on the applications to provide error correction or detection. B is incorrect, as TCP is reliable and provides connection-oriented sessions. C is incorrect, as SNMP is a management protocol used by devices to communicate information to a management system such as CiscoWorks or HP's OpenView. D is incorrect, as ICMP is used by such troubleshooting utilities as ping and traceroute. | | A19: | Answers: B and C. TCP numbers the sequence of the segments sent out to ensure that all data arrives safely. The receiving side must send back acknowledgments that all segments were received. If it sends back negative acknowledgments, TCP resends the data. A is incorrect, as Address Resolution Protocol (ARP) or Inverse ARP provide MAC address resolution. D is incorrect, as ping is a utility that checks TCP/IP connectivity. E is incorrect, as routing updates are used by routing protocols such as RIP or OSPF to update information stored on another router's routing table. | | A20: | Answer: A. By using windowing, you can increase the amount of data sent before an ACK is sent back by the receiving side. This increases throughput. B is incorrect, as increasing window size increases, not decreases, throughput. C is incorrect, as increasing window size decreases latency. D is incorrect, as the protocol still requires an ACK to be sent by the receiving side. This does not decrease reliability in the slightest. | | A21: | Answer: F. The 255.255.255.248 subnet mask used on a Class C IP network produces 30 usable subnets. A is incorrect; it does not create any subnets, as it is the default Class C subnet mask. B is incorrect, as it creates only 2 subnets that is, if the IOS recognizes IP subnet zero. C is incorrect, as 255.255.255.192 creates only 2 usable subnets. D is incorrect, as 255.255.255.224 creates only 6 usable subnets. E is incorrect, as 255.255.255.240 creates only 14 usable subnets. | | A22: | Answer: A. Address Resolution Protocol (ARP) resolves IP addresses to MAC addresses. B is incorrect, as Reverse Address Resolution Protocol (RARP) resolves MAC addresses to IP addresses, thus the "reverse." C is incorrect, as Serial Line Address Resolution Protocol (SLARP) automatically assigns IP addresses to serial interfaces if AutoInstall is being used and HDLC is the protocol in use on that interface. D is incorrect, as DHCP automatically assigns IP addresses to interfaces. | | A23: | Answer: D. The IP address of a Layer 2 switch is strictly for management purposes and doesn't affect Layer 3 connectivity. A, B, and C are incorrect, as they are valid reasons why a user cannot ping a remote host. E is incorrect because a remote host is in a different subnet by definition. | | A24: | Answer: D. According to the IETF, the Class D IP address range is used for multicast group addresses. A is incorrect, as Class A addresses are used by large corporations and governments for unicast purposes. B is incorrect, as Class B addresses are used by large-to-medium corporations and service providers for unicast purposes. C is incorrect, as service providers, users, and small businesses use Class C addresses for unicast purposes. Unicast is based on one-to-one communication, whereas multicast is used for one-to-many communication. | | A25: | Answer: D. Prefix notation, or / notation, shows the number of subnet mask bits that are turned to one, signifying a network portion of the address. A is incorrect, as it does not show the number of hosts on a subnetwork. B is incorrect, as it does not show the number of subnetworks in use. C is incorrect because, even though you can figure out whether it is a Class A, B, or C address based on the prefixes of /8, /16, or /24, it does not necessarily show the class of the address. | | A26: | Answer: D. The custom subnet mask of 255.255.255.224 gives you 30 hosts. Even though this number exceeds the requested 16, if you borrow an additional bit, you have only 14 available hosts. A, B, and C do not give you any subnets at all, and thus are incorrect. E is incorrect, as it gives you only 14 available hosts. | | A27: | Answer: A. Class A networks have all been assigned to governments and extremely large corporations. B is incorrect, as small-to-medium size companies would receive only a Class C network. C and D are incorrect, as a small office/home office (SOHO) or individual would not receive a network only a few IP addresses from a Class C network. | | A28: | Answers: B and D. The address 192.168.5.98/27 has a decimal subnet mask of 255.255.255.224 and uses an increment of 32. This ends up making the network range 192.168.5.96 192.168.5.127. Answer A is incorrect because this is the broadcast address of the previous subnet. Answer C is incorrect because it is the network address of the following subnet. | | A29: | Answer: D. The IP address 172.31.45.34 is part of the RFC 1918-defined private IP address range for a Class B network. A, B, C, and E are incorrect, as they are all public IP addresses that can be routed on the Internet. | | A30: | Answer: B. The default Class B subnet mask is 255.255.0.0, or /16 in prefix notation. A is incorrect, as it shows a broadcast. C is incorrect, as it is a custom subnet mask, more than likely used for a Class A network. D is incorrect, as it is a special mask typically used for default routes. | | A31: | Answer: C. It is not a valid host address; 192.168.5.95/27 is a directed broadcast address for the 192.168.5.64 network. A is incorrect, as you can certainly assign Class C addresses to any type of interface. B is incorrect, as the /27 mask is the 255.255.255.224 subnet mask, which is perfectly valid. D is incorrect because it is a private IP address. E is incorrect, as the fact that it is a private IP address will not cause it to be refused by an interface. | | A32: | Answer: B. This gives you 126 subnets, assuming you are not using ip subnet-zero, with 510 hosts per subnet. A is incorrect, as it does not provide you with enough hosts to meet the criteria. C is incorrect, as it is an invalid subnet mask. Subnet masks must have contiguous ones. D is incorrect, as it is the default Class B subnet mask, and as such, you would not have any subnets. | | A33: | Answer: B. The address 192.168.2.37 is a valid host address on the 192.168.2.32 network, which has a directed broadcast address of 192.168.2.39. A, C, and D are incorrect, as 192.168.2.37 does not fall on any of those subnetworks with the given subnet mask. | | A34: | Answer: A. The number 11011010 = 218 in decimal. B is incorrect, as 11011011 = 219. C is incorrect, as 11011100 = 220. D is incorrect, as 11011101 = 221. | | A35: | Answer: C. The number 01011010 = 90 in decimal. A is incorrect, as 01001011 = 75. B is incorrect, as 01010011 = 83. D is incorrect, as 01100001 = 97. | | A36: | Answer: B. The number 11010110 = 214 in decimal. A is incorrect, as 11000110 = 198. C is incorrect, as 11111100 = 252. D is incorrect, as 11111111 = 255. | | A37: | Answer: A. The number 10110110 = 182 in decimal. B is incorrect, as 11000000 = 192. C is incorrect, as 11001010 = 202. D is incorrect, as 11010100 = 212. | | A38: | Answers: B, C, and D. They are valid host addresses when using a 255.255.255.224 subnet mask against the address. A is incorrect, as it is a broadcast address on the 16.234.118.32 network. E is incorrect, as it is a broadcast address on the 210.45.116.128 network. F is incorrect, as it is the network address for 237.63.12.192. | | A39: | Answer: E. The 255.255.255.248, or /28 subnet mask, gives you 16 subnetworks and 14 usable hosts, using the 2^n formula for subnets and (2^n 2) formula for usable hosts, where n equals the number of bits borrowed for the subnet, or number of bits left for the hosts. You are able to use the first and last subnetworks because of the ip subnet-zero command on your router. A is incorrect, as it would be a /30 mask. B is incorrect, as it would be a /27 mask. C is incorrect, as there would never be 32 hosts per subnet; you always have to subtract 2 from the 2^n formula because of the network ID and the directed broadcast address on each subnetwork. D is incorrect, as you would never have 16 hosts per subnet; you always have to subtract 2 from the 2^n formula because of the network ID and the directed broadcast address on each subnetwork. | | A40: | Answer: A. TCP adds overhead with acknowledgments and session handshakes. B is incorrect, as TCP tags with sequence numbers. C is incorrect, as TCP is reliable and ensures data delivery. D is incorrect, as TCP has built-in mechanisms to handle sequencing and delivery and does not require the application's assistance. | | A41: | Answer: A. The mask /30 gives you 62 subnetworks and two available hosts per subnet using the (2^n [ms] 2) formula, where n equals the number of bits borrowed for the subnet, or number of bits left for the hosts. Because the router is not using ip subnet-zero, you cannot use the first or last subnet, which is why you subtract 2 from both the subnetworks and hosts. B is incorrect, as that would be a /27 mask. C is incorrect, as there would never be 32 hosts per subnet; you always have to subtract 2 from the 2^n formula because of the network ID and the directed broadcast address on each subnetwork. D is incorrect, as you would never have 16 hosts per subnet; you always have to subtract 2 from the 2^n formula because of the network ID and the directed broadcast address on each subnetwork. E is incorrect, as the /28 subnet mask gives you 14 subnetworks and 14 possible hosts. | | A42: | Answers: A and B. They are considered network ID addresses. They have all zeros in the host portion of the IP address with the subnet mask given. C, D, and E are incorrect, as they are all directed Broadcast Addresses on subnets. | | A43: | Answers: A, C, and D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host port (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 1111110. B is incorrect, as 112 in decimal is 1110000 in binary, which is not a valid host address in this network all host bits are zero. E is incorrect, as 175 in decimal is 10101111 in binary all host bits are ones. This is the local broadcast address and cannot be used as a host address. F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, as it has all host bits of zero. | | A44: | Answer: E. The fourth byte in the IP address is 159, the binary value of which is 1011111. So, this is the broadcast address for 198.57.78.0/27 network. A is incorrect, as the binary value for 33 is 00100001. B is incorrect, as the binary value for 64 is 01000000. C and D are incorrect, as the binary value for 97 is 01100001. F is incorrect, as the binary value for 254 is 11111110. These are not broadcast addresses for the 198.57.78.0/27 network. | | A45: | Answer: C. If you use the subnet mask prefix value /28, 4 bits are left for the host portion. The total number of hosts is 16 (2 are reserved for network and broadcast in each subnetwork). The 165.100.5.68 host resides in subnetwork 165.100.5.64. Valid hosts in this network are 165.100.5.65 165.100.5.79. A is incorrect, as it is the network address for the first subnet, also known as subnet 0. B is incorrect, as it is the network address for 165.100.5.32. Valid hosts are 165.100.5.33 165.100.5.46. D is incorrect, as it is one of the valid hosts in subnetwork 165.100.5.64. E is incorrect, as it is the broadcast address. F is incorrect, as it is a valid host in subnetwork 165.100.5.0. | | A46: | Answer: C. TCP stands for Transmission Control Protocol. A, B, and D are incorrect, as they are nonexisting protocols. | | A47: | Answer: D. UDP stands for User Datagram Protocol. It is part of the TCP/IP protocol suite and operates at Layer 4 of the OSI Model. A, B, and C name nonexistent protocols or programs. | | A48: | Answer: D. UDP relies on applications to provide error correction and reliability of transmission. A is incorrect, as it describes Remote Copy Protocol (RCP). B is incorrect, as it describes Simple Mail Transfer Protocol (SMTP). C is incorrect, as it describes File Transfer Protocol (FTP). | | A49: | Answers: C, E, and F. These addresses are not private addresses defined by RFC 1918. These addresses can be routed across the public Internet. A is incorrect, as it is part of the private range of 10.x.x.x. B is incorrect, as it is part of the private range of 172.16.x.x 172.16.31.x.x. D is incorrect, as it is part of the private range of 192.168.x.x. | | A50: | Answers: A and C. A is correct because you must have a public IP address assigned to your interface to be able to communicate across the Internet. C is correct, as you must turn on the interface by using the no shutdown command. B is incorrect, as you must turn on the interface, not disable it with the shutdown command. D is incorrect, as duplex settings are not valid on serial interfaces, only on Ethernet interfaces. |
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