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The
samples herein are intended to reflect the type of questions
presented on the CCIE Qualification Test. In no way does this sample
test reflect the diversity of the true test and Cisco assumes no
responsibility for the performance of any person taking the
CCIE
Qualification
Test subsequent to passing this sample.
Network Fundamentals:
1. Which
layer<s> of the OSI Reference Model provide<s> for
internetwork connectivity?
a. Data
Link
b. Physical
c. Session
d.
Network
e. Presentation
2. The DSAP
and SSAP of an IEEE 802.2 header both are set to 'AA' hex to
indicate:
a. There is
routing information included in the packet.
b. There is a
SNAP header to follow.
c. The packet suffered a
collision.
e. The Ethertype is Localtalk.
f. The
address was recognized and the frame accepted.
Bridge/Router Technology:
3. In a
bridged, IEEE 802.5/Token Ring environment, which combination of
bits indicate to the source that a packet has been forwarded across
a bridging device?
a. A is 0, C is
0 and there is a RIF.
b. A is 1, C is 0 and RII is
0.
c. A is 0, C is 1 and there is no RIF.
d. A is
1, C is 1 and RII is 1.
e. The question is not applicable
in a Token Ring environment.
4. Certain
protocols such as SNA, Netbios and LAT are not routable and
therefore must be bridged or encapsulated
because:
a. Native IBM
protocols do not conform to the OSI RM stack.
b. SNA and
Netbios cannot be routed but LAT can.
c. They would be too
CPU intensive.
d. They only run in "batch mode" processes
where routing is not an issue.
e. These protocols have no
explicit network address.
5. What is
<are> the purpose<s> of the Spanning Tree
Algorithm?
a. To keep
routing updates from being transmitted onto the same port on which
they were received.
b. To prevent a "loop free" logical
tree topology.
c. To discover a "loop free" topology and
provide, as possible, a path between every pair of
LAN's.
d. To provide a path between every LAN
segment.
e. To help ensure that messages will arrive over
all possible paths.
6. Some
disadvantages of bridging versus routing are: (more than one
answer)
a. Bridges
cannot provide any form of flow control.
b. Bridges cannot
provide congestive feedback information to end nodes.
c.
Bridges offer no form of filtering.
d. Bridges are more
difficult to administer/maintain than routers.
e. Bridges
create more overhead traffic than routers.
7. Split
Horizon is primarily designed to: (more than one
answer)
a. Prevent
routing loops between adjacent routers.
b. Ensure that
information about a route is always sent back over the path on which
it arrived.
c. Functions to provide extra algorithm
stability to "Hold Downs".
d. Replace the "Poison Reverse
Updates" algorithm.
e. Ensure that hop counts are not
exceeded.
Internet Protocols:
8. What is
the maximum number of hosts that can be assigned to a class "C",
non-subnetted network?
a.
1024
b. 65025
c. 254
d. 16
e.
48
9. The main
difference between OSPF and RIP are:
a. Both are
IGP's, but the RIP convergence time is much shorter than that of
OSPF.
b. RIP is a distance vector protocol whereas OSPF is
a link state protocol.
c. There is no difference between
the two, other than RIP is used for IP and OSPF is used for
OSI.
d. OSPF allows for a lower hop count than
RIP.
e. RIP works better in large internetworks than
OSPF.
10. Given the
IP address of 193.243.12.43 and a subnet mask of 255.255.255.128,
what is the subnet address?
a.
194.243.12.32
b. 193.243.0.0
c.
194.243.12.43
d. 193.243.12.128
e. None of the
above.
11. Which of
the following is a function of ICMP: (more than one
answer)
a. Report TCP,
time to exist exceeded..
b. Redirect UDP
messages.
c. Transport SNMP Gets
d. Discover
subnet masks.
e. Report routing failures.
Cisco-Specific Technology:
12. What EXEC
command would display the hardware configuration of a Cisco router:
(more than one answer)
a. list
ports
b. sh ha
c. disp hard
d. sh
all
e. sh ver
13. What EXEC
command will copy the current configuration information to
nonvolatile memory?
a. write
memory
b. write erase
c. write term
d.
write ram
e. write nvm
Network Scenarios
14. Refer to
Exhibit A below. Router R1 and R2, T0 and T1 ports are configured,
functionally, the same. The primary path between these router nodes
will be:
a. If OSPF
routing, the primary path is T0.
b. If RIP routing, the
primary path is T0.
c. If IGRP routing, the primary path is
T1.
d. This is a misconfiguration and would cause beaconing
on T0, therefore T1 is the primary.
e. These rings would
always be handled equally.
Exhibit A
Router R1#show
interface tokenring 0
1.) TokenRing 0
is up, line protocol is up
2.) Hardware is Dual Token Ring,
address is 0000.3080.5fca (bia 0000.3080.5fca)
3.) Internet
address is 1.1.0.7, subnet mask is 255.255.0.0
4.) MTU 8136
bytes, BW 4000 Kbit, DLY 630 usec, rely 255/255, load
1/255
5.) Encapsulation SNAP, loopback not set, keepalive
set (10 sec)
6.) ARP type: SNAP, ARP Timeout
4:00:00
7.) Ring speed: 16 Mbps
8.) Single ring
node, Source Route Transparent Bridge capable
9.) Source
bridging enabled, srn 300 bn 2 trn 1000 (ring group)
10.)
proxy explorers disabled, spanning explorer enabled, NetBIOS cache
disabled
11.) Group Address: 0x00000000, Functional
Address: 0x0000011A
12.) Ethernet Transit OUI:
0x0000F8
13.) Last input 0:00:00, output 0:00:00, output
hang never
14.) Last clearing of "show interface" counters
never
15.) Output queue 0/40, 0 drops; input queue 1/75, 0
drops
16.) Five minute input rate 19000 bits/sec, 32
packets/sec
17.) Five minute output rate 0 bits/sec, 0
packets/sec
18.) 4282800 packets input, 267762944 bytes, 0
no buffer
19.) Received 75020 broadcasts, 0 runts, 0
giants
20.) 0 input errors, 0 CRC, 0 frame, 0 overrun, 0
ignored, 0 abort
21.) 2710102 packets output, 178779701
bytes, 0 underruns
22.) 0 output errors, 0 collisions, 1
interface resets, 0 restarts
23.) 4
transitions
Router R1#show
interface tokenring 1
1.) TokenRing 1
is up, line protocol is up
2.) Hardware is Dual Token Ring,
address is 0000.3080.5fba (bia 0000.3080.5fba)
3.) Internet
address is 1.3.0.8, subnet mask is 255.255.0.0
4.) MTU 8136
bytes, BW 16000 Kbit, DLY 630 usec, rely 255/255, load
1/255
5.) Encapsulation SNAP, loopback not set, keepalive
set (10 sec)
6.) ARP type: SNAP, ARP Timeout
4:00:00
7.) Ring speed: 16 Mbps
8.) Single ring
node, Source Route Transparent Bridge capable
9.) Source
bridging enabled, srn 561 bn 2 trn 1000 (ring group)
10.)
proxy explorers disabled, spanning explorer enabled, NetBIOS cache
disabled
11.) Group Address: 0x00000000, Functional
Address: 0x0000011A
12.) Ethernet Transit OUI:
0x0000F8
13.) Last input 0:00:00, output 0:00:00, output
hang never
14.) Last clearing of "show interface" counters
never
15.) Output queue 0/40, 0 drops; input queue 1/75, 0
drops
16.) Five minute input rate 19000 bits/sec, 32
packets/sec
17.) Five minute output rate 0 bits/sec, 0
packets/sec
18.) 4365800 packets input, 297762942 bytes, 0
no buffer
19.) Received 6020 broadcasts, 0 runts, 0
giant
20.) 0 input errors, 0 CRC, 0 frame, 0 overrun, 0
ignored, 0 abort
21.) 3610303 packets output, 1264779801
bytes, 0 underruns
22.) 0 output errors, 0 collisions, 1
interface resets, 0 restarts
23.) 2
transitions
Answers:
1) d. The
key phrase here is "internetwork connectivity" While connectivity of
different levels are provided at each layer of the OSI model, it is
the Network layer that provides Network addresses, permitting the
routing of traffic between networks.
2) b.
When usedas the Destination Service Access Point (DSAP) and
Source Service Access Point (SSAP), the bit pattern 0xAA is used to
indicate the presence of a SNAP, or Sub-Network Access Protocol,
Header. This is used to transfer certain data for which there is no
specific SAP value assigned within an 802.2 frame.
3) d. The
A or Address Recognised; and C, or Frame Copied; bits are contained
within the Frame Status byte in a Token Ring frame. When a frame is
copied by a source-route bridge to be put out on to another ring,
the bridge sets these bits, much as if it were actually the
destination station.
4) e.
These protocols have "flat" addressing schemes - that is, there is
no topological significance to the addresses assigned to each
station.
5) c. The
Spanning Tree Algorithm is a requirement in meshed Transparent
Bridging. As a transparently bridged frame contains no record of the
path it has taken, and contains no mechanism for aging, any bridge
loop will result in storms of packets continually circulating around
the loop. By providing a loop free path between any two points in
the network, the Spanning Tree Algorithm prevents these storms. A
form of this algorithm is used for Explorer control in Source-route
bridges.
6) a,b.
Flow Control and Congestive Feedback are functions of the
Network layer. As bridges operate at the Data Link layer, these
responsibilities fall to the end station in a bridged environment.
Most bridges do offer some filtering capabilities, at least on MAC
addresses, and they are much simpler to administer. At first glance,
it may appear that bridges create more overhead than routers, due to
the BPDU frame that is sent out every second. However, these frames
are very small compared to route advertisements sent out by routers,
which can grow to be quite large in complex networks. Also, routers
must tie Network addresses to Data Link addresses, which often
creates more overhead (Example IP, ARP)
7) a,c.
When using Split Horizon, a router does not advertise a route to the
same interface from which it learnt that route. This is one
mechanism by which the Slow Convergence Problem can be solved, and
routing loops prevented.
8) c. A
class "C" network has 8 bits available for host machines. Thus the
total number of hosts that can be addressed on such a network is 2
to the 8th power, or 256; minus the network number (x.y.z.0) and the
broadcast address (x.y.z.255), for a total of 254
hosts.
9) b. Far
and away the greatest functional difference between these IP routing
protocols is the mechanism by which they operate - i.e. distance
vector vs link state. OSPF allows for information about specific
links to be exchanged between routers - with RIP, the only available
information is the entire routing table, with only networks and
metrics being advertised periodically.
10) e.
The Subnet Address is found by setting all the host bits in an
address to 0. According to the mask there are 7 host bits, so
applying this rule yields a subnet address of 193.243.12.0. However,
closer inspection yields that there is only one subnet bit in this
class C address - therefore, this is an illegal address/mask
pair.
11) d,e.
Many different kind of routing failures can be reported via the
Destination Unreachable ICMP frame (type 3). Masks can be found
using the Address Mask Request and Reply ICMP frame (type 17, 18).
TTL (Time to Live) and Redirection exist at the Network layer (IP)
rather than the Transport layer (TCP, UDP).
12) b,e.
These commands, abbreviated forms of "show hardware" and "show
version", both produce the same output. The others would all result
in parser error messages.
13) a.
This can also be abbreviated to just "write" or "wr". Write
erase removes any configuration in NVRAM. Write Term shows you the
configuration currently being used by the box - this may not be the
same as the config in NVRAM (show config) if changes have been made
since the last "write mem" was performed. The other two will result
in parser error messages.
14) c. By
default, IGRP uses the bandwidth defined on a port as part of the
routing metric. As the defined bandwidths on these interfaces is
different (see line 4 of each display), IGRP will use interface T1
over T0, all else being equal.
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