Answers to Chapter 9 Review Questions

Answers to Chapter 9 Review Questions

1:

What are the three layers of the ATM stack? What does each do?

A:

Refer back to Figure 9-3 in Chapter 9. The three layers are as follows:

• AAL Slice & dice

• ATM Build headers

• Physical Ship cells

2:

What is the difference between ATM and SONET?

A:

ATM is a cloud technology that deals with cells. SONET is a point-to-point technology that is used to carry cells.

3:

What is the difference between a Catalyst with two LANE modules and a two-port ATM switch?

A:

As an edge device, the Catalyst only switches frames between ports; the LS1010, as an ATM switch, only switches cells. See Figure 9-8 in Chapter 9.

4:

What is the difference between a VPI, a VCI, and an NSAP? When is each used?

A:

VPI and VCI values are two parts of the address placed in the header of every cell (both PVCs and SVCs). NSAPs are only used to build the SVC. After the SVC is built, the VPI/VCI values are used to switch cells along the VC.

5:

Assume you attached an ATM network analyzer to an ATM cloud consisting of one LS1010 ATM switch and two Catalysts with LANE modules. What types of cells could you capture to observe VPI and VCI values? What type of cells could you capture to observe NSAP addresses?

A:

All cells contain VPI/VCI values, but NSAP addresses can only be observed in cells that carry signaling messages (such as a UNI 4.0 call SETUP message).

6:

What are the three sections of an NSAP address? What does each part represent?

A:

The following list outlines the three sections of an NSAP address and what each represents.

• Prefix What ATM switch?

• ESI What devices on the ATM switch?

• Selector Byte What software component in the end station?

7:

How do Catalysts automatically generate ESI and selector byte values for use with LANE?

A:

The following list shows how Catalysts automatically generate ESI and selector byte values for use with LANE.

• LEC = MAC . **

• LES = MAC + 1 . **

• BUS = MAC + 2 . **

• LECS = MAC + 3 . 00

8:

What is the five-step initialization process used by LANE Clients to join an ELAN?

A:

The five-step initialization process used by LANE Clients to join an ELAN is as follows:

1. Client contacts LECS (Bouncer).

2. Client contacts LES (Bartender).

3. LES contacts Client.

4. Client contacts BUS (Gossip).

5. BUS contacts Client.

9:

What are the names of the six types of circuits used by LANE? What type of traffic does each carry?

A:

The following list outlines the names of the six types of circuits used by LANE and what type of traffic each carries.

• Configuration Direct Requests to join ELAN and NSAP of LES

• Control Direct LE_ARPs

• Control Distribute LE_ARPs that need to be flooded to all Proxy Clients

• Multicast Send Broadcast, multicast, and unknown unicast traffic that needs to be flooded to all Clients

• Multicast Forward Broadcast, multicast, and unknown unicast traffic that is being flooded

• Data Direct End-user data from LEC to LEC

10:

What is the difference between an IP ARP and an LE_ARP?

A:

IP ARPs are used to request the MAC address associated with an IP address.

LE_ARPs are used to request the NSAP address associated with a MAC address.

11:

In a network that needs to trunk two VLANs between two Catalysts, how many LECs are required? How many LECSs? How many LESs? How many BUSs?

A:

• LECs = 4 One per Client per ELAN

• LECSs = 1 One per LANE network

• LESs = 2 One per ELAN

• BUSs = 2 One per ELAN

12:

If the network in Question 11 grows to ten Catalysts and ten VLANs, how many LECs, LECSs, LESs and BUSs are required? Assume that every Catalyst has ports assigned to every VLAN.

A:

• LECs = 100

• LECSs = 1

• LESs = 10

• BUSs = 10

13:

Trace the data path in Figure 9-26 from an Ethernet-attached node in VLAN 1 on Cat-A to an Ethernet-attached node in VLAN 2 on Cat-B. Why is this inefficient?

A:

The traffic travels through the ELAN1 to the router where it is routed to ELAN2. It then travels across ELAN2 to reach the node on Cat-B. This is inefficient because: 1) the router might not be as fast as the ATM network and 2) the router was required to reassemble the cells back into a complete packet before the routing decision could be made. After the packet is routed, it has to again be segmented into cells.