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NEW QUESTION: 1
The implementations group has been using the test bed to do a 'proof-of-concept' that requires both Client 1 and Client 2 to access the WEB Server at 209.65.200.241.
After several changes to the network addressing, routing scheme, DHCP services, NTP services, layer 2 connectivity, FHRP services, and device security, a trouble ticket has been opened indicating that Client 1 cannot ping the 209.65.200.241 address.
Use the supported commands to isolated the cause of this fault and answer the following questions.
On which device is the fault condition located?
A. DSW2
B. ASW2
C. ASW1
D. R4
E. R3
F. R1
G. R2
H. DSW1
Answer: C
Explanation:
port security needs is configured on ASW1.
=================================================
Topic 13, Ticket 8 : Redistribution of EIGRP to OSPF
Topology Overview (Actual Troubleshooting lab design is for below network design)
*
Client Should have IP 10.2.1.3
*
EIGRP 100 is running between switch DSW1 & DSW2
*
OSPF (Process ID 1) is running between R1, R2, R3, R4
*
Network of OSPF is redistributed in EIGRP
*
BGP 65001 is configured on R1 with Webserver cloud AS 65002
*
HSRP is running between DSW1 & DSW2 Switches
The company has created the test bed shown in the layer 2 and layer 3 topology exhibits.
This network consists of four routers, two layer 3 switches and two layer 2 switches.
In the IPv4 layer 3 topology, R1, R2, R3, and R4 are running OSPF with an OSPF process number 1.
DSW1, DSW2 and R4 are running EIGRP with an AS of 10. Redistribution is enabled where necessary.
R1 is running a BGP AS with a number of 65001. This AS has an eBGP connection to AS
65002 in the ISP's network. Because the company's address space is in the private range.
R1 is also providing NAT translations between the inside (10.1.0.0/16 & 10.2.0.0/16) networks and outside (209.65.0.0/24) network.
ASW1 and ASW2 are layer 2 switches.
NTP is enabled on all devices with 209.65.200.226 serving as the master clock source.
The client workstations receive their IP address and default gateway via R4's DHCP server.
The default gateway address of 10.2.1.254 is the IP address of HSRP group 10 which is running on DSW1 and DSW2.
In the IPv6 layer 3 topology R1, R2, and R3 are running OSPFv3 with an OSPF process number 6.
DSW1, DSW2 and R4 are running RIPng process name RIP_ZONE.
The two IPv6 routing domains, OSPF 6 and RIPng are connected via GRE tunnel running over the underlying IPv4 OSPF domain. Redistrution is enabled where necessary.
Recently the implementation group has been using the test bed to do a 'proof-of-concept' on several implementations. This involved changing the configuration on one or more of the devices. You will be presented with a series of trouble tickets related to issues introduced during these configurations.
Note: Although trouble tickets have many similar fault indications, each ticket has its own issue and solution.
Each ticket has 3 sub questions that need to be answered & topology remains same.
Question-1 Fault is found on which device,
Question-2 Fault condition is related to,
Question-3 What exact problem is seen & what needs to be done for solution
= ====================================================================
= =========


Client is unable to ping IP 209.65.200.241
Solution
Steps need to follow as below:-
*
When we check on client 1 & Client 2 desktop we are not receiving DHCP address from R4 ipconfig ----- Client will be receiving IP address 10.2.1.3
*
IP 10.2.1.3 will be able to ping from R4 , but cannot ping from R3, R2, R1
*
This clearly shows problem at R4 since EIGRP is between DSW1, DSW2 & R4 and OSPF protocol is running between R4, R3, R2, R1 so routes from R4 are not propagated to R3, R2, R1
*
Since R4 is able to ping 10.2.1.3 it means that routes are received in EIGRP & same needs to be advertised in OSPF to ping from R3, R2, R1.
*
Need to check the routes are being advertised properly or not in OSPF & EIGRP vice-versa.


*
From above snap shot it clearly indicates that redistribution done in EIGRP is having problem & by default all routes are denied from ospf to EIGRP... so need to change route-map name.
*
Change required: On R4, in the redistribution of EIGRP routing protocol, we need to change name of route-map to resolve the issue. It references route-map OSPF_to_EIGRP but the actual route map is called OSPF->EIGRP.
-------------------------------------------------------------------------------------------------------------------------
-----

NEW QUESTION: 2
Which three capabilities are provided by MLD snooping? (Choose three.)
A. flooding control packets to the egress VLAN
B. a 5-minute aging timer
C. a 60-second aging timer
D. user-configured ports age out automatically
E. dynamic port learning
F. IPv6 multicast router discovery
Answer: B,E,F
Explanation:
Like IGMP snooping, MLD snooping performs multicast router discovery, with these characteristics:
Ports configured by a user never age out.
Dynamic port learning results from MLDv1 snooping queries and IPv6 PIMv2 packets.
If there are multiple routers on the same Layer 2 interface, MLD snooping tracks a single multicast router on the port (the router that most recently sent a router control packet).
Dynamic multicast router port aging is based on a default timer of 5 minutes; the multicast router is deleted from the router port list if no control packet is received on the port for 5 minutes.
IPv6 multicast router discovery only takes place when MLD snooping is enabled on the switch.
Reference: https://www.cisco.com/c/en/us/td/docs/switches/lan/catalyst3750/software/release/122_55_se/configuration/guide/scg3750/swv6mld.pdf

NEW QUESTION: 3
Wendy tutors math students after school every day for five days. Each day, she tutors twice as many
students as she tutored the previous day. If she tutors t students the first day, what is the average
(arithmetic mean) number of students she tutors each day over the course of the week?

A. Option E
B. Option D
C. Option A
D. Option B
E. Option C
Answer: A
Explanation:
Explanation/Reference:
If Wendy tutors t students the first day, then she tutors 2t students the second day, 4t students the third
day, 8t students the fourth day, and 16t students the fifth day. The average number of students tutored
each day over the course of the week is equal to the sum of the tutored students divided by the number of
days: