All practice questions

Question 201Question →

A network technician is configuring a branch office router connected to headquarters via two WAN links. The primary connection uses a static route targeting the headquarters network $10.2.0.0/16$ , configured with an administrative distance of $120$ . A secondary backup connection runs Open Shortest Path First (OSPF) to reach the same network prefix. During testing, traffic unexpectedly routes through the secondary OSPF link even while the primary link remains operational. Which configuration change will ensure that traffic prioritizes the primary static route over the backup OSPF route?

Reduce the administrative distance of the static route to a value lower than

110

.

Decrease the interface metric cost on the secondary OSPF link to

1

.

Reconfigure the primary static route using NAT Overload (PAT) on the outgoing WAN interface.

Change the subnet mask of the static route to

10.2.0.0/8

to broaden prefix matching.

Show answer & explanation

Answer: Reduce the administrative distance of the static route to a value lower than $110$ .

Answer

Reduce the administrative distance of the static route to a value lower than 110.

When a router receives multiple routes to the exact same destination prefix from different sources, it evaluates Administrative Distance (AD). Standard OSPF has an AD of 110. Because the static route was manually set to an AD of 120, the router preferred OSPF. Lowering the static route's AD below 110 (for example, to 1) restores its priority over the OSPF route.

Step-by-Step Solution

1

Identify the routing source selection process.

The router compares Administrative Distance (AD) first when choosing between routes to the exact same prefix learned from different routing protocols or static configurations.

Lower AD values indicate higher preference and trustworthiness.

2

Compare the AD values of the two configured paths.

The static route currently has an AD of 120, while the standard OSPF route has a default AD of 110.

Because 110 is lower than 120, the router selects the OSPF path.

3

Determine the necessary configuration change.

Lowering the static route's AD below 110 (such as setting it to the default static AD of 1) forces the router to install the static route into the routing table.

This guarantees the primary static connection is preferred over the backup dynamic route.

Key Concept

Administrative Distance vs Protocol Metrics in Route Selection

Estimated Time:1m 30s

Question 202Question →

A network engineer is manually assigning static IPv4 addresses to industrial control gateways within an automated manufacturing segment using the address block $172.20.64.0/22$ . Which of the following IPv4 addresses are valid, usable host addresses that can be assigned to devices on this subnet? (Select TWO.)

Select all that apply

172.20.64.1

172.20.67.254

172.20.64.0

172.20.67.255

172.20.68.1

Show answer & explanation

Answer: $172.20.64.1$ ; $172.20.67.254$

Answer

The addresses

172.20.64.1

and

172.20.67.254

are valid, usable host IP addresses within the

172.20.64.0/22

subnet.

The subnet

172.20.64.0/22

covers the IP range

172.20.64.0

through

172.20.67.255

. Subtracting the network address (

172.20.64.0

) and the broadcast address (

172.20.67.255

) yields a usable host IP address range of

172.20.64.1

to

172.20.67.254

. Both

172.20.64.1

and

172.20.67.254

fall inside this range.

Step-by-Step Solution

1

Determine the subnet mask and block size for a

/22

prefix length.

A

/22

prefix corresponds to a subnet mask of

255.255.252.0

. The block size in the third octet is

256 - 252 = 4

.

Calculating the block size identifies the boundary increments between adjacent networks.

2

Identify the network ID and broadcast address for the

172.20.64.0/22

network.

The network ID is

172.20.64.0

. The next subnet begins at

172.20.68.0

, making the broadcast address

172.20.67.255

.

Network addresses and broadcast addresses are reserved and cannot be assigned as usable host IPs.

3

Determine the valid host address range.

The usable host range spans from

172.20.64.1

(first host) to

172.20.67.254

(last host).

Usable host addresses consist of all IP addresses strictly between the network ID and the broadcast address.

4

Evaluate the choices against the usable host range.

172.20.64.1

and

172.20.67.254

lie within the usable range, whereas

172.20.64.0

is the network ID,

172.20.67.255

is the broadcast address, and

172.20.68.1

belongs to another network.

Only IP addresses inside the calculated host range are valid host static assignments.

Key Concept

Usable IPv4 Host Range Calculation for Subnets Larger than

/24

Question 203Question →

A network administrator is assigning static IP addresses to new database servers within an isolated VLAN allocated the block $192.168.50.128/28$ . Which of the following represent valid, usable host IP addresses that can be assigned to these servers? (Select TWO.)

Select all that apply

192.168.50.128

192.168.50.130

192.168.50.140

192.168.50.143

Show answer & explanation

Answer: $192.168.50.130$ ; $192.168.50.140$

Answer

The addresses

192.168.50.130

and

192.168.50.140

are valid usable host IP addresses within the

192.168.50.128/28

subnetwork.

For the network block

192.168.50.128/28

, the block size is

16

(

2^{32-28}

). The network address is

192.168.50.128

and the broadcast address is

192.168.50.143

. Subtracting these two reserved addresses leaves the usable range of

192.168.50.129

to

192.168.50.142

. Both

192.168.50.130

and

192.168.50.140

lie within this range.

Step-by-Step Solution

1

Determine the block size from the CIDR prefix length

A

/28

prefix leaves

32 - 28 = 4

host bits, yielding a block size of

2^4 = 16

.

Calculating the block size identifies subnet boundary increments.

2

Identify the network ID and broadcast address

The network address is

192.168.50.128

. Adding the block size minus 1 (

128 + 16 - 1

) gives the broadcast address of

192.168.50.143

.

The first and last addresses in a CIDR block are reserved for network identification and subnet broadcasting.

3

Calculate the usable host address range

The usable host range is from

192.168.50.129

(first usable) to

192.168.50.142

(last usable).

Usable host addresses exclude the network ID and broadcast address.

4

Evaluate the candidate IP addresses against the usable range

192.168.50.130

and

192.168.50.140

fall within

[192.168.50.129, 192.168.50.142]

.

192.168.50.128

is the network ID and

192.168.50.143

is the broadcast address.

Only IP addresses strictly inside the usable bounds may be assigned to network hosts.

Key Concept

IPv4 Subnet Boundaries and Usable Host Allocation

Question 204Question →

An enterprise campus utilizes a Layer 3 core switch handling routing for multiple virtual LANs. Host devices on VLAN 50 ( $10.50.0.0/24$ ) fail to acquire dynamic IP address assignments upon boot, resulting in self-assigned APIPA addresses ( $169.254.x.x$ ). Packet analysis on the core switch reveals that incoming broadcast $DHCPDISCOVER$ messages arrive at the VLAN 50 interface, but the switch fails to encapsulate and forward these requests as unicast traffic to the remote DHCP server located at $172.16.100.25$ . Which configuration issue on the core switch is preventing clients on VLAN 50 from receiving DHCP leases?

The ip helper-address statement is omitted from the Switch Virtual Interface (SVI) for VLAN 50.

The ip helper-address statement was configured on the physical uplink port connecting to the DHCP server rather than the client gateway SVI.

The core switch is improperly attempting to establish a TCP three-way handshake with the DHCP server prior to relaying lease requests.

The relay agent is forwarding client discovery packets using target UDP port 68 instead of target UDP port 67.

Show answer & explanation

Answer: The ip helper-address statement is omitted from the Switch Virtual Interface (SVI) for VLAN 50.

Answer

The ip helper-address statement is omitted from the Switch Virtual Interface (SVI) for VLAN 50.

For a Layer 3 switch or router to act as a DHCP relay agent, the relay directive (such as `ip helper-address`) must be applied to the specific Layer 3 ingress interface (SVI or physical interface) that receives the client's broadcast traffic. When clients on VLAN 50 broadcast a DHCPDISCOVER message, the core switch checks SVI 50 for a helper address. If missing, the broadcast is discarded at the Layer 3 boundary. When properly configured on SVI 50, the switch converts the broadcast into a unicast UDP packet (destined for port 67 on the DHCP server) and populates the `giaddr` field with SVI 50's IP address.

Step-by-Step Solution

1

Analyze client frame traffic characteristics

Clients on VLAN 50 broadcast DHCPDISCOVER frames to destination 255.255.255.255, bounded by the Layer 2 broadcast domain.

Routers and Layer 3 switches drop broadcast packets by default unless explicitly configured to intercept and relay them.

2

Identify the ingress interface for DHCP relay placement

The core switch must intercept broadcasts at the gateway interface serving the client broadcast domain (VLAN 50 SVI).

The relay agent records the interface IP address in the giaddr (gateway IP address) field of the DHCP packet header to inform the DHCP server which address scope to allocate from.

3

Verify missing interface directive

Without the `ip helper-address 172.16.100.25` directive on SVI 50, the switch silently discards client DHCP broadcasts.

The missing command directly explains why no outbound unicast packets were generated toward the DHCP server.

Key Concept

DHCP Relay Agent Placement and Gateway SVI Configuration

Question 205Question →

A network administrator is designing an IPv4 subnetting scheme for a branch office using the $192.168.10.0/24$ network address block. Each departmental VLAN requires support for at least $28$ usable host IP addresses. Which subnet mask will satisfy this host requirement while maximizing the total number of subnets that can be created?

255.255.255.224

255.255.255.240

255.255.255.192

255.255.255.248

Show answer & explanation

Answer: $255.255.255.224$

Answer

The subnet mask

255.255.255.224

satisfies the requirement.

To support at least

28

hosts,

5

bits are required for the host portion because

2^5 - 2 = 30

usable IP addresses. A

/27

prefix (

255.255.255.224

) leaves

5

host bits, yielding

8

subnets from a

/24

block, which maximizes the subnet count while satisfying the host constraint.

Step-by-Step Solution

1

Calculate the minimum number of host bits (

h

) required using the formula

2^h - 2 \ge \text{hosts}

.

2^5 - 2 = 30 \ge 28

, so

h = 5

host bits are needed.

Two IP addresses per subnet are reserved for the network ID and the broadcast address.

2

Determine the network prefix length from the total number of IPv4 bits (

32

).

Prefix length

= 32 - 5 = /27

.

Subtracting host bits from

32

leaves the number of network bits.

3

Convert the CIDR prefix

/27

into dotted-decimal subnet mask format.

/27 = 255.255.255.224

.

The fourth octet contains three binary

1

s (

11100000_2 = 224

).

Key Concept

Calculating required host bits and dotted-decimal subnet mask to maximize subnets while satisfying host capacity constraints

Question 206Question →

During the operation of IEEE 802.1D Spanning Tree Protocol (STP), a switch port transitions through multiple operational states. Which of the following activities occur specifically while a port is in the Learning state? (Select TWO.)

Select all that apply

It populates the switch MAC address table using source MAC addresses from incoming data frames.

It transmits and processes Bridge Protocol Data Units (BPDUs) received from neighboring switches.

It forwards user payload data frames between connected network interfaces.

It discards all BPDUs to prevent premature topology calculations.

It reconfigures the interface 802.1Q trunk native VLAN assignment.

Show answer & explanation

Answer: It populates the switch MAC address table using source MAC addresses from incoming data frames.; It transmits and processes Bridge Protocol Data Units (BPDUs) received from neighboring switches.

Answer

In the IEEE 802.1D Learning state, the switch port processes BPDUs and populates its MAC address table, but it does not yet forward user data frames.

During the standard IEEE 802.1D Learning state, a switch port actively processes and sends BPDUs while learning source MAC addresses from received frames to populate the MAC address table. However, to prevent switching loops during convergence, it does not forward user payload frames.

Step-by-Step Solution

1

Identify the primary functions of the 802.1D Learning port state.

The Learning state is designed to prepare the switch for forwarding traffic by learning MAC addresses without actually forwarding frames.

Populating the MAC table prior to forwarding prevents excessive unicast flooding upon entering the Forwarding state.

2

Evaluate control plane vs. data plane behavior during the Learning state.

Control plane BPDUs are sent and received, MAC table entries are updated from incoming frames, but user data traffic is not forwarded.

Restricting data frame forwarding during convergence prevents temporary switching loops while topology details stabilize.

Key Concept

802.1D Spanning Tree Protocol Port State Operations

Question 207Question →

An enterprise network technician is documenting baseline security parameters and service configurations across network endpoints. Match each network protocol on the left with its corresponding default transport protocol and port specification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

SNMP matches UDP port 161; LDAP matches TCP port 389; POP3 matches TCP port 110; SFTP matches TCP port 22.

Each listed service maps directly to its official IANA transport protocol and default port assignment: SNMP polling traffic communicates over UDP port 161; unencrypted LDAP directory services communicate over TCP port 389; unencrypted POP3 mail retrieval uses TCP port 110; and SFTP uses SSH encapsulation over TCP port 22.

Step-by-Step Solution

1

Determine transport protocol and default port for SNMP polling.

SNMP manager-to-agent queries rely on low-overhead UDP port 161.

Standard status queries operate over UDP port 161, distinct from SNMP traps which use UDP port 162.

2

Determine transport protocol and default port for LDAP.

Standard LDAP operates over connection-oriented TCP port 389.

Unencrypted directory authentication lookups use TCP 389, whereas encrypted LDAPS uses TCP 636.

3

Determine transport protocol and default port for POP3.

Standard POP3 operates over TCP port 110.

Legacy post office mail download services bind to TCP 110, whereas secure POP3S binds to TCP 995.

4

Determine transport protocol and default port for SFTP.

SFTP operates over TCP port 22.

SFTP leverages the underlying SSH protocol subsystem running on TCP port 22.

Key Concept

Standard Network Protocol and Port Assignments

Question 208Question →

A network architect is subnetting the IP address block $10.112.0.0/19$ to provision dedicated VLANs for expanding branch offices. Each VLAN requires support for at least $127$ usable host IP addresses while conserving as much address space as possible. Subnets are allocated sequentially starting from the beginning of the block. What is the broadcast address of the $5^{\text{th}}$ subnet allocated?

10.112.4.255

10.112.2.127

10.112.3.255

10.112.5.255

Show answer & explanation

Answer: $10.112.4.255$

Answer

The broadcast address of the 5th allocated subnet is

10.112.4.255

.

To support

127

usable hosts, the subnet must provide at least

127

host IP addresses. Using the formula

2^h - 2

,

h=7

host bits yields

128 - 2 = 126

usable hosts, which does not meet the requirement. Therefore,

h=8

host bits must be used (

2^8 - 2 = 254

usable hosts), corresponding to a

/24

prefix with a block size of

256

IP addresses per subnet. Counting sequentially from

10.112.0.0

, the 5th subnet is

10.112.4.0/24

, which has a broadcast address of

10.112.4.255

.

Step-by-Step Solution

1

Determine the required number of host bits (

h

).

h = 8

host bits.

The usable host formula is

2^h - 2 \ge \text{required hosts}

. For

127

hosts,

h=7

gives

2^7 - 2 = 126

usable hosts, which is insufficient. Selecting

h=8

gives

2^8 - 2 = 254

usable hosts.

2

Calculate the subnet prefix and block size.

Prefix is

/24

(

32 - 8 = 24

), and block size is

2^8 = 256

addresses per subnet.

Each

/24

subnet increments the third octet by

1

.

3

List the sequential

/24

subnets starting from

10.112.0.0/19

.

Subnet 1:

10.112.0.0/24

Subnet 2:

10.112.1.0/24

Subnet 3:

10.112.2.0/24

Subnet 4:

10.112.3.0/24

Subnet 5:

10.112.4.0/24

Subnets are assigned continuously starting from the network ID

10.112.0.0

.

4

Identify the broadcast address for the

5^{\text{th}}

subnet (

10.112.4.0/24

).

Broadcast address is

10.112.4.255

.

The broadcast address is the last IP address in the subnet block immediately preceding the next subnet (

10.112.5.0

).

Key Concept

IPv4 Subnet Mask Calculation and Boundary Identification

Question 209Question →

A workstation on a remote VLAN requires an IP address assignment from a central DHCP server located on a different subnet across a Layer 3 router. What is the correct sequence of packet delivery steps during the discovery and offering stage of this relayed DHCP request?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The sequence begins when the client broadcasts a DHCPDISCOVER packet on its local subnet. The DHCP relay agent receives this broadcast, inserts its own interface address as the gateway IP (giaddr), and forwards the packet as a unicast message to the central DHCP server. The DHCP server matches the giaddr to the appropriate IP scope and returns a unicast DHCPOFFER packet to the relay agent. Finally, the relay agent forwards the DHCPOFFER packet to the local subnet for the client host.

In a cross-subnet DHCP environment, the host initiates the process by broadcasting a DHCPDISCOVER packet locally. The relay agent on the router interface receives this broadcast, populates the giaddr header field with its interface IP, and forwards the packet as a unicast message to the central DHCP server. The server reads the giaddr, allocates an IP from the corresponding scope, and returns a unicast DHCPOFFER packet to the relay agent. The relay agent then forwards the DHCPOFFER message onto the host's subnet.

Step-by-Step Solution

1

Identify the initial client request broadcast.

The unconfigured host initiates communication by broadcasting a DHCPDISCOVER message within its local broadcast domain.

Clients lacking IP configurations rely on local broadcast packets to reach DHCP listening services.

2

Determine how the relay agent processes the broadcast request.

The configured router interface intercepts the broadcast, inserts its interface IP into the giaddr field, and sends a unicast DHCPDISCOVER packet to the remote server IP.

Routers block standard broadcast packets by default; the relay agent encapsulation enables cross-subnet transmission.

3

Trace the server response mechanism back to the relay agent.

The DHCP server identifies the subnet pool via the giaddr header and responds with a unicast DHCPOFFER packet addressed to the relay agent.

The server cannot reply directly to the unconfigured client's unrouted IP, so it targets the reachable relay agent interface.

4

Trace the final delivery to the client.

The relay agent forwards the DHCPOFFER packet onto the client's local subnet segment.

This completes the delivery of offered lease details to the requesting workstation host.

Key Concept

DHCP Relay Agent Packet Processing and Relay Sequence

Question 210Question →

A network security engineer is investigating a DNS operational failure between an internal primary name server and a DMZ-hosted secondary name server following a firewall policy hardening exercise. The secondary server fails to perform full zone transfers (AXFR), and external resolvers fail to receive DNSSEC-signed responses that exceed 512 bytes. The engineer executes the following diagnostic command from the secondary server:

text
$ dig @172.16.10.10 example.com AXFR +dnssec
;; Truncation detected; requesting TCP fallback...
;; Connection failed: connection timed out
;; Communications error to 172.16.10.10#53: timed out

Which protocol behavior explains these combined failures, and what resolution must be implemented on the perimeter firewall?

Zone transfers (AXFR) and truncated DNS responses (TC flag set) require TCP port 53 transport; the firewall must be updated to allow inbound and outbound TCP port 53 traffic between the name servers.

DNS zone transfers and large payload responses operate exclusively over UDP port 53; the primary server must be reconfigured to split records into multiple CNAME alias entries to avoid packet loss.

Fallback from UDP to TCP occurs only during administrative diagnostic queries; the firewall should maintain a block on TCP port 53 and instead rely on UDP port 53 with larger MTU sizing.

DNSSEC enforcement and zone synchronization require secure shell transport; the firewall policy must be modified to permit traffic targeted to SSH on TCP port 22.

Show answer & explanation

Answer: Zone transfers (AXFR) and truncated DNS responses (TC flag set) require TCP port 53 transport; the firewall must be updated to allow inbound and outbound TCP port 53 traffic between the name servers.

Answer

Zone transfers (AXFR) and truncated DNS responses (TC flag set) require TCP port 53 transport; the firewall must be updated to allow inbound and outbound TCP port 53 traffic between the name servers.

DNS relies on both UDP and TCP on port 53. While standard lookup queries utilize UDP for speed and low overhead, zone transfers (AXFR) and responses that exceed maximum UDP payload sizes (such as large DNSSEC key responses) trigger the Truncation (TC) flag in the DNS header. When the TC flag is set, the client or secondary server MUST re-query over TCP port 53. If a firewall blocks TCP port 53, both zone transfers and large DNSSEC query resolutions will fail.

Step-by-Step Solution

1

Analyze the diagnostic command output.

The `dig` output shows `Truncation detected; requesting TCP fallback...` followed by a timeout on port 53.

When a DNS response exceeds the allowable UDP buffer size or when initiating zone transfers (AXFR/IXFR), DNS transitions from UDP to TCP port 53.

2

Identify transport layer requirements for DNS Infrastructure.

Both full zone transfers (AXFR) and truncated DNSSEC queries require functional TCP port 53 communication.

UDP port 53 is used for standard small queries, but TCP port 53 is mandatory for reliable bulk data delivery (zone transfers) and fallback for oversized responses.

3

Determine the firewall remediation.

Permit TCP port 53 traffic between the primary and secondary DNS servers on the perimeter firewall.

Allowing TCP port 53 enables secondary servers to complete AXFR zone synchronization and resolves DNSSEC payload truncation fallbacks.

Key Concept

DNS Transport Layer Selection (UDP 53 vs TCP 53) and AXFR Zone Transfer Operations

Estimated Time:3m 0s

Question 211Question →

A network administrator is inspecting a router's Network Address Translation (NAT) table to verify traffic flow. Which of the following terms specifically describes the private IPv4 address assigned to a host on the internal network before any address translation takes place?

Inside Local

Inside Global

Outside Local

Outside Global

Show answer & explanation

Answer: Inside Local

Answer

Inside Local is the standard network address translation term for the un-translated private IPv4 address assigned to a device on the internal network.

Inside Local is the technical term used in networking to identify the private IP address assigned to an end device on an internal network before Network Address Translation is performed.

Step-by-Step Solution

1

Identify the location of the host and the point of observation.

The host is on the internal network ('Inside'), and the address is observed before translation occurs ('Local').

Standard NAT naming conventions combine location descriptors (Inside/Outside) with visibility scope (Local/Global).

2

Select the NAT term matching an untranslated internal host address.

Combining 'Inside' and 'Local' yields 'Inside Local'.

'Inside Local' specifically denotes the original private IP address configured on an internal endpoint.

Key Concept

NAT Terminology and Address Types

Question 212Question →

A network engineer configures an 802.1Q trunk link between two switches, `SW-CORE-01` and `SW-ACCESS-01`. Interface `GigabitEthernet0/1` on `SW-CORE-01` is configured with a native VLAN of VLAN 10, while the corresponding interface on `SW-ACCESS-01` retains the default native VLAN of VLAN 1. Which of the following statements correctly describe the operational impact of this configuration? (Select TWO.)

Select all that apply

Untagged frames transmitted from VLAN 10 on `SW-CORE-01` will be placed into VLAN 1 upon arrival at `SW-ACCESS-01`.

Control plane protocols such as CDP or LLDP will trigger warning logs indicating a Native VLAN mismatch error.

Frames explicitly tagged for VLAN 10 by `SW-CORE-01` will be automatically discarded by `SW-ACCESS-01` upon arrival.

Hosts on VLAN 10 and VLAN 1 will be able to perform Layer 3 routing between subnets without requiring a gateway device.

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 10 on `SW-CORE-01` will be placed into VLAN 1 upon arrival at `SW-ACCESS-01`.; Control plane protocols such as CDP or LLDP will trigger warning logs indicating a Native VLAN mismatch error.

Answer

Untagged frames sent from VLAN 10 on `SW-CORE-01` will be assigned to VLAN 1 on `SW-ACCESS-01`, and management protocols like CDP/LLDP will log Native VLAN mismatch warnings.

When an 802.1Q native VLAN mismatch occurs, untagged traffic transmitted by one switch is placed into the native VLAN configured on the receiving switch, leaking traffic between VLAN 10 and VLAN 1. Additionally, discovery protocols like CDP or LLDP monitor link parameters and periodically report Native VLAN mismatch errors in syslog.

Step-by-Step Solution

1

Analyze how 802.1Q handles native VLAN frame tagging

Frames belonging to the native VLAN are sent untagged across an 802.1Q trunk link.

802.1Q standard leaves native VLAN frames unencapsulated for backward compatibility with non-tagging devices.

2

Determine frame processing on the receiving switch

The receiving switch (`SW-ACCESS-01`) inspects incoming untagged frames and assigns them to its own native VLAN (VLAN 1).

Switches map incoming untagged frames directly to whatever local native VLAN ID is configured on that specific trunk interface.

3

Identify administrative and protocol indicators of mismatch

CDP and LLDP detect mismatched configuration parameters and generate log alerts.

Neighbor discovery protocols continuously exchange interface configuration details to identify misconfigurations.

Key Concept

802.1Q Native VLAN Tagging and Mismatch Consequences

Question 213Question →

A systems administrator is configuring firewall ACL rules to segregate an administrative workstation network from a management server rack. The configuration policy requires permitting remote graphical desktop administration for Windows servers and allowing network infrastructure devices to forward unencrypted system log entries to a central log collector server. Which TWO of the following port and transport protocol combinations must be allowed in the firewall rule base?

Select all that apply

TCP port 3389

UDP port 514

UDP port 3389

TCP port 514

Show answer & explanation

Answer: TCP port 3389; UDP port 514

Answer

TCP port 3389 and UDP port 514 must be permitted.

The scenario calls for two distinct services: Windows graphical remote desktop management and unencrypted event log collection. Remote Desktop Protocol (RDP) establishes connection-oriented management sessions on TCP port 3389. Standard Syslog event logging transmits unencrypted log entries over UDP port 514. Therefore, allowing TCP port 3389 and UDP port 514 satisfies both operational requirements.

Step-by-Step Solution

1

Identify the required network service protocols from the administrative scenario requirements.

Graphical desktop administration for Windows servers requires Remote Desktop Protocol (RDP). Unencrypted system log entry forwarding requires standard Syslog.

Mapping enterprise management functions to their standard network service protocols is necessary for firewall rule design.

2

Determine the transport layer protocol and default destination port for Remote Desktop Protocol (RDP).

RDP uses TCP port 3389.

RDP requires a reliable connection-oriented transport protocol (TCP) to manage interactive desktop display updates and keyboard/mouse inputs without packet loss.

3

Determine the transport layer protocol and default destination port for standard unencrypted Syslog.

Standard Syslog uses UDP port 514.

Native Syslog is designed as a low-overhead, connectionless log-reporting protocol that transmits event notifications over UDP.

Key Concept

Standard Port Numbers and Transport Layer Protocol Pairings

Estimated Time:1m 30s

Question 214Question →

A network architect is designing a Variable Length Subnet Mask (VLSM) IP addressing scheme starting from the base network block $172.16.64.0/19$ . The subnets must be allocated contiguously without wasting address space in the following order:

1. Subnet A (Server Farm): Requires support for at least $1,000$ usable host IP addresses.
2. Subnet B (Corporate Clients): Requires support for at least $450$ usable host IP addresses.
3. Subnet C (VoIP Infrastructure): Requires support for at least $100$ usable host IP addresses.

Assuming each subnet is assigned the most efficient mask possible to satisfy host requirements, what is the broadcast IP address of Subnet C?

172.16.70.127

172.16.70.255

172.16.74.127

172.16.70.128

Show answer & explanation

Answer: 172.16.70.127

Answer

172.16.70.127

The option specifying 172.16.70.127 accurately calculates the minimum host bit requirements for each sequential VLSM allocation (10 bits for Subnet A, 9 bits for Subnet B, and 7 bits for Subnet C), placing Subnet C at 172.16.70.0/25 with a broadcast address of 172.16.70.127.

Step-by-Step Solution

1

Calculate mask and range for Subnet A (1,000 hosts)

Network: 172.16.64.0/22, Range: 172.16.64.0 to 172.16.67.255

For 1,000 hosts, formula 2^h - 2 >= 1,000 requires h = 10 bits (2^10 - 2 = 1,022 usable hosts). Prefix length is 32 - 10 = /22 (block size of 4 in the 3rd octet).

2

Calculate mask and range for Subnet B (450 hosts)

Network: 172.16.68.0/23, Range: 172.16.68.0 to 172.16.69.255

Subnet B begins immediately after Subnet A at 172.16.68.0. For 450 hosts, 2^h - 2 >= 450 requires h = 9 bits (2^9 - 2 = 510 usable hosts). Prefix length is 32 - 9 = /23 (block size of 2 in the 3rd octet).

3

Calculate mask, range, and broadcast for Subnet C (100 hosts)

Network: 172.16.70.0/25, Broadcast: 172.16.70.127

Subnet C begins immediately after Subnet B at 172.16.70.0. For 100 hosts, 2^h - 2 >= 100 requires h = 7 bits (2^7 - 2 = 126 usable hosts). Prefix length is 32 - 7 = /25 (block size of 128 in the 4th octet). The range is 172.16.70.0 to 172.16.70.127, making 172.16.70.127 the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) Host Capacity Allocation

Question 215Question →

A network administrator is assigned the IPv4 block $172.31.96.0/22$ for a data center expansion. The administrator must allocate the first subnet from this block to support a web server farm requiring at least $110$ usable host IP addresses. Which of the following correctly identifies the subnet ID and broadcast address for this web server subnet using the most space-efficient CIDR prefix?

Subnet ID:

172.31.96.0/25

, Broadcast Address:

172.31.96.127

Subnet ID:

172.31.96.0/25

, Broadcast Address:

172.31.96.128

Subnet ID:

172.31.96.0/26

, Broadcast Address:

172.31.96.63

Subnet ID:

172.31.96.0/24

, Broadcast Address:

172.31.96.255

Show answer & explanation

Answer: Subnet ID: $172.31.96.0/25$ , Broadcast Address: $172.31.96.127$

Answer

Subnet ID:

172.31.96.0/25

, Broadcast Address:

172.31.96.127

To host at least 110 devices, 7 host bits are required because

2^7 - 2 = 126

usable hosts. A prefix length of

/25

(

32 - 7

) creates subnets of size 128. Starting from

172.31.96.0

, the first

/25

subnet covers addresses

172.31.96.0

through

172.31.96.127

. Thus, the Subnet ID is

172.31.96.0/25

and the Broadcast Address is

172.31.96.127

.

Step-by-Step Solution

1

Determine the required number of host bits (

h

).

To accommodate at least

110

usable hosts, use the formula

2^h - 2 \ge 110

. For

h=6

,

2^6 - 2 = 62

(insufficient). For

h=7

,

2^7 - 2 = 126

usable hosts (sufficient).

Two addresses must be subtracted from the total block size for the network ID and broadcast address.

2

Calculate the CIDR prefix length.

Prefix length =

32 - 7 = /25

.

IPv4 addresses consist of 32 bits. Subtracting 7 host bits leaves 25 network/subnet bits.

3

Determine the block size and network range for the first subnet.

Block size =

2^7 = 128

. Starting at

172.31.96.0

, the subnet spans from

172.31.96.0

to

172.31.96.127

.

The first subnet starts at the beginning of the parent block (

172.31.96.0

).

4

Identify the Subnet ID and Broadcast Address.

Subnet ID =

172.31.96.0/25

, Broadcast Address =

172.31.96.127

. Usable host range is

172.31.96.1

through

172.31.96.126

.

The lowest address in the block is the Subnet ID, and the highest address in the block is the Broadcast Address.

Key Concept

Calculating host requirements, prefix lengths, and broadcast boundaries in IPv4 subnetting.

Question 216Question →

A network technician is assigning static IP addresses to end-user devices within a subnet configured as $192.168.1.0/26$ . Which of the following IP addresses are valid usable host addresses on this network segment? (Select TWO.)

Select all that apply

192.168.1.15

192.168.1.50

192.168.1.0

192.168.1.63

192.168.1.64

Show answer & explanation

Answer: 192.168.1.15; 192.168.1.50

Answer

The valid usable host IP addresses are 192.168.1.15 and 192.168.1.50.

For the CIDR prefix

/26

, the subnet mask is

255.255.255.192

, yielding a total block size of

64

IP addresses (

192.168.1.0

to

192.168.1.63

). Subtracting the network address (

192.168.1.0

) and the broadcast address (

192.168.1.63

) leaves a usable host range of

192.168.1.1

through

192.168.1.62

. Both 192.168.1.15 and 192.168.1.50 fall within this valid range.

Step-by-Step Solution

1

Determine the block size for the CIDR prefix /26

Block size =

2^{(32-26)} = 2^6 = 64

IP addresses

Calculating the total number of IP addresses in the subnet block.

2

Identify the network ID and broadcast address

Network ID =

192.168.1.0

, Broadcast address =

192.168.1.63

The network address is the first IP address (

192.168.1.0

) and the broadcast address is the last IP address (

192.168.1.0 + 64 - 1 = 192.168.1.63

).

3

Define the usable host address range

Usable host range =

192.168.1.1

through

192.168.1.62

Usable hosts exclude the network ID and broadcast address.

4

Evaluate choices against the usable range

192.168.1.15 and 192.168.1.50 lie inside the valid range [192.168.1.1, 192.168.1.62]

192.168.1.0 is the network ID, 192.168.1.63 is the broadcast address, and 192.168.1.64 belongs to the next subnet.

Key Concept

Subnet Host Ranges and Reserved Addresses

Question 217Question →

A network administrator is evaluating the behavior of a Layer 2 switch configured with two isolated VLANs: VLAN 10 (Sales) and VLAN 20 (Engineering). Host A, connected to a port on VLAN 10, sends a broadcast frame onto the network. Almost simultaneously, Host B, also on VLAN 10, transmits a unicast frame addressed to Host C (on VLAN 10), but Host C's MAC address is not currently listed in the switch's CAM table. Which of the following statements accurately describe how the Layer 2 switch processes these frames? (Select TWO)

Select all that apply

The switch forwards the broadcast frame from Host A out all active ports assigned to VLAN 10 except the ingress port, but suppresses transmission to VLAN 20 ports.

The switch floods the unicast frame from Host B out all active member ports of VLAN 10 except the ingress port because the destination MAC address is unknown.

The switch forwards the broadcast frame from Host A across all active ports on both VLAN 10 and VLAN 20 because broadcast traffic spans the entire physical switch by default.

The switch strips the Layer 2 frame header and re-encapsulates Host A's broadcast payload into a Layer 3 packet prior to forwarding it across ports.

The switch drops the unicast frame from Host B immediately because frames with unknown destination MAC addresses are discarded until a static entry is manually added to the CAM table.

Show answer & explanation

Answer: The switch forwards the broadcast frame from Host A out all active ports assigned to VLAN 10 except the ingress port, but suppresses transmission to VLAN 20 ports.; The switch floods the unicast frame from Host B out all active member ports of VLAN 10 except the ingress port because the destination MAC address is unknown.

Answer

The switch forwards the broadcast frame from Host A out all active member ports of VLAN 10 except the receiving port, while isolating it from VLAN 20. In addition, because Host C's MAC address is unknown in the CAM table, the switch performs unknown unicast flooding by transmitting Host B's frame out all ports belonging to VLAN 10 except the ingress port.

VLANs create separate Layer 2 broadcast domains, meaning a broadcast frame originating on VLAN 10 will be forwarded out all active ports assigned to VLAN 10 (except the receiving port) but will never cross over to ports on VLAN 20. Furthermore, when a switch receives a unicast frame whose destination MAC address is not listed in the CAM table, it performs unknown unicast flooding, transmitting the frame to all active ports within that frame's VLAN except the ingress port.

Step-by-Step Solution

1

Analyze the broadcast domain boundaries for Host A's broadcast frame.

VLANs define logical Layer 2 broadcast domains. Broadcast traffic originating in VLAN 10 is flooded to all active ports assigned to VLAN 10 (excluding the receiving port) and blocked from crossing into VLAN 20.

Layer 2 switches do not forward broadcast traffic across different VLAN boundaries.

2

Analyze switch forwarding behavior for Host B's unicast frame with an unlearned destination MAC address.

The switch looks up Host C's destination MAC address in its CAM table. Finding no match, it performs unknown unicast flooding within VLAN 10.

Unknown unicast traffic is flooded to all ports in the source VLAN except the port of entry so the intended recipient receives the frame and can reply, allowing the switch to learn its MAC address.

Key Concept

VLAN Broadcast Domain Boundaries and Unknown Unicast Flooding Mechanics

Estimated Time:1m 30s

Question 218Question →

A network administrator needs to configure an enterprise router so that multiple internal workstations assigned private IP addresses can access the internet simultaneously using a single registered public IP address. Which address translation technology should the administrator implement?

Port Address Translation (PAT)

Static Network Address Translation (Static NAT)

Dynamic Network Address Translation (Dynamic NAT)

Layer 4 Port Forwarding

Show answer & explanation

Answer: Port Address Translation (PAT)

Answer

Port Address Translation (PAT) enables multiple internal hosts to access external networks using a single public IP address by assigning unique source port numbers to each traffic session.

Port Address Translation (PAT), or NAT Overload, modifies both the IP address and Layer 4 source port numbers of outbound packets. This allows a router to maintain a translation table mapping thousands of individual private sessions to a single public IP address.

Step-by-Step Solution

1

Analyze the technical requirement

Multiple private IP hosts need outbound internet access sharing a single public IP address.

Public IPv4 addresses are limited, requiring port multiplexing to distinguish host traffic streams.

2

Compare address translation mechanisms

Static NAT and standard Dynamic NAT use one-to-one IP mappings, whereas PAT uses a one-to-many mapping by varying TCP/UDP source port numbers.

PAT appends unique port numbers to the shared public IP address in the router's translation table.

Key Concept

Port Address Translation (PAT) / NAT Overload

Question 219Question →

A biomedical technician is deploying a network of wireless wearable vital-sign sensors that continuously transmit telemetry data to a tablet located within a two-meter radius of a patient's bed. Which of the following network types best classifies this short-range wireless architecture?

Personal Area Network (PAN)

Local Area Network (LAN)

Campus Area Network (CAN)

Metropolitan Area Network (MAN)

Show answer & explanation

Answer: Personal Area Network (PAN)

Answer

Personal Area Network (PAN) is the correct network classification for extremely short-range communications centered around an individual or personal workspace.

A Personal Area Network (PAN) is designed for extremely short-range communication (typically under 10 meters) between personal devices or localized sensors around an individual.

Step-by-Step Solution

1

Analyze the operational radius and scope described in the scenario.

The sensors transmit data over a two-meter radius around a single patient.

Geographic scope is the primary criterion for classifying network types.

2

Match the scope to standard network classification definitions.

A network limited to a few meters focused around an individual device or patient is classified as a Personal Area Network (PAN).

PANs are specifically designed for low-power, short-range inter-device communication within immediate personal space.

Key Concept

Network Types and Geographic Scopes

Question 220Question →

In a local network where multiple switches are deployed with default Spanning Tree Protocol (STP) configuration settings, all switches possess the same bridge priority of 32768. How does STP break this tie to select the root bridge?

The switch with the lowest numerical MAC address is elected as the root bridge.

The switch with the highest numerical MAC address is elected as the root bridge.

The switch with the highest numerical priority value overrides the tie and becomes the root bridge.

The switch that experiences the highest untagged traffic volume on its native VLAN is selected.

Show answer & explanation

Answer: The switch with the lowest numerical MAC address is elected as the root bridge.

Answer

The switch with the lowest numerical MAC address is elected as the root bridge.

In 802.1D Spanning Tree Protocol, the switch with the lowest Bridge ID (BID) becomes the root bridge. The BID consists of a 2-byte Priority field followed by a 6-byte MAC address. When all switches use the default bridge priority (32768), the priority comparison results in a tie. STP resolves this tie by selecting the switch with the lowest numerical MAC address.

Step-by-Step Solution

1

Analyze the Bridge ID structure used for root election.

The Bridge ID consists of a priority value concatenated with the switch's MAC address.

STP elects the bridge with the lowest overall Bridge ID as the root bridge.

2

Compare priority values across the switches.

All switches have an identical default priority of 32768.

Since the priority values are equal, priority alone cannot determine a winner.

3

Evaluate the secondary tiebreaker.

The switch with the lowest numerical MAC address wins the tiebreaker.

STP election rules designate the lowest MAC address as the decisive factor when priorities match.

Key Concept

Spanning Tree Protocol Root Bridge Election Tiebreaker

Answer

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