计算机网络自顶向下 第七版 第六章答案 - 图文

Problem 23

If all the 11=9+2 nodes send out data at the maximum possible rate of 100 Mbps, a total aggregate throughput of 11*100 = 1100 Mbps is possible.

Problem 24

Each departmental hub is a single collision domain that can have a maximum throughput of 100 Mbps. The links connecting the web server and the mail server has a maximum throughput of 100 Mbps. Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of 100 Mbps each, a maximum total aggregate throughput of 500 Mbps can be achieved among the 11 end systems.

Problem 25

All of the 11 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of 100 Mbps is possible among the 11 end sytems.

Problem 26

Action Link(s) packet is Explanation forwarded to B sends a Switch learns interface A, C, D, E, and F Since switch table frame to E corresponding to MAC is empty, so switch address of B does not know the interface corresponding to MAC address of E E replies with Switch learns interface B Since switch a frame to B corresponding to MAC already knows address of E interface corresponding to MAC address of B A sends a Switch learns the B Since switch frame to B interface corresponding already knows the to MAC address of A interface corresponding to MAC address of B B replies with Switch table state A Since switch a frame to A remains the same as already knows the before interface Switch Table State corresponding to MAC address of A Problem 27

a) The time required to fill L?8 bits is

L?8Lsec?msec.16128?103

b) For L?1,500, the packetization delay is

1500msec?93.75msec. 16

For L?50, the packetization delay is

50msec?3.125msec. 16

c) Store-and-forward delay ?

L?8?40 R For L?1,500, the delay is

1500?8?40sec?19.4?sec 6622?10

For L?50, store-and-forward delay ?1?sec.

d) Store-and-forward delay is small for both cases for typical link speeds. However, packetization delay for L?1500 is too large for real-time voice applications.

Problem 28

The IP addresses for those three computers (from left to right) in EE department are: 111.111.1.1, 111.111.1.2, 111.111.1.3. The subnet mask is 111.111.1/24.

The IP addresses for those three computers (from left to right) in CS department are: 111.111.2.1, 111.111.2.2, 111.111.2.3. The subnet mask is 111.111.2/24.

The router’s interface card that connects to port 1 can be configured to contain two sub-interface IP addresses: 111.111.1.0 and 111.111.2.0. The first one is for the subnet of EE department, and the second one is for the subnet of CS department. Each IP address is associated with a VLAN ID. Suppose 111.111.1.0 is associated with VLAN 11, and 111.111.2.0 is associated with VLAN 12. This means that each frame that comes from subnet 111.111.1/24 will be added an 802.1q tag with VLAN ID 11, and each frame that comes from 111.111.2/24 will be added an 802.1q tag with VLAN ID 12.

Suppose that host A in EE department with IP address 111.111.1.1 would like to send an IP datagram to host B (111.111.2.1) in CS department. Host A first encapsulates the IP datagram (destined to 111.111.2.1) into a frame with a destination MAC address equal to the MAC address of the router’s interface card that connects to port 1 of the switch. Once the router receives the frame, then it passes it up to IP layer, which decides that the IP datagram should be forwarded to subnet 111.111.2/24 via sub-interface 111.111.2.0. Then the router encapsulates the IP datagram into a frame and sends it to port 1. Note that this frame has an 802.1q tag VLAN ID 12. Once the switch receives the frame port 1, it knows that this frame is destined to VLAN with ID 12, so the switch will send the frame to Host B which is in CS department. Once Host B receives this frame, it will remove the 802.1q tag.

Problem 29

in out out label label dest interf. in out out label label dest interf. 7 A 0 7 10 A 0 12 D 0 5 8 A 1 in out out label label dest interf. 10 6 A 1 12 9 D 0 0 1 R6 0 0 0 R5 in out out label label dest interf. R4 1 R3 D 5 A 0 R2 in out out label label dest interf. 0 0 R1 in out out label label dest inter. A 8 6 A 0 6 - A 0 Problem 30

in out out label label dest interf. in out out label label dest interf. 3 12 D 0 2 4 D 1 3 D 0 in out out label label dest interf. 12 - D 0 R6 0 R5 in out out label label dest interf. 0 R4 0 1 R3 0 1 1 D 2 D 0 R2 in out out label label dest interf. 0 R1 in out out label label dest inter. A 4 1 D 0 1 12 D 1

Problem 31

(The following description is short, but contains all major key steps and key protocols involved.)

Your computer first uses DHCP to obtain an IP address. You computer first creates a special IP datagram destined to 255.255.255.255 in the DHCP server discovery step, and puts it in a Ethernet frame and broadcast it in the Ethernet. Then following the steps in the DHCP protocol, you computer is able to get an IP address with a given lease time.

A DHCP server on the Ethernet also gives your computer a list of IP addresses of first-hop routers, the subnet mask of the subnet where your computer resides, and the addresses of local DNS servers (if they exist).

Since your computer’s ARP cache is initially empty, your computer will use ARP protocol to get the MAC addresses of the first-hop router and the local DNS server.

Your computer first will get the IP address of the Web page you would like to download. If the local DNS server does not have the IP address, then your computer will use DNS protocol to find the IP address of the Web page.

Once your computer has the IP address of the Web page, then it will send out the HTTP request via the first-hop router if the Web page does not reside in a local Web server. The HTTP request message will be segmented and encapsulated into TCP packets, and then further encapsulated into IP packets, and finally encapsulated into Ethernet frames. Your computer sends the Ethernet frames destined to the first-hop router. Once the router receives the frames, it passes them up into IP layer, checks its routing table, and then sends the packets to the right interface out of all of its interfaces.

Then your IP packets will be routed through the Internet until they reach the Web server.

The server hosting the Web page will send back the Web page to your computer via HTTP response messages. Those messages will be encapsulated into TCP packets and then further into IP packets. Those IP packets follow IP routes and finally reach your

first-hop router, and then the router will forward those IP packets to your computer by encapsulating them into Ethernet frames.

Problem 32

a) Each flow evenly shares a link’s capacity with other flows traversing that link, then the 80 flows crossing the B to access-router 10 Gbps links (as well as the access router to border router links) will each only receive 10 Gbps / 80 = 125 Mbps

b) In Topology of Figure 5.31, there are four distinct paths between the first and third tier-2 switches, together providing 40 Gbps for the traffic from racks 1-4 to racks 9-12. Similarly, there are four links between second and fourth tier-2 switches, together providing 40 Gbps for the traffic from racks 5-8 to 13-16. Thus the total aggregate bandwidth is 80 Gbps, and the value per flow rate is 1 Gbps.

c) Now 20 flows will need to share each 1 Gbps bandwidth between pairs of TOR switches. So the host-to-host bit rate will be 0.5 Gbps.

Problem 33

a) Both email and video application uses the fourth rack for 0.1 percent of the time.

b) Probability that both applications need fourth rack is 0.001*0.001 = 10-6.

c) Suppose the first three racks are for video, the next rack is a shared rack for both video and email, and the next three racks are for email. Let's assume that the fourth rack has all the data and software needed for both the email and video applications. With the topology of Figure 5.31, both applications will have enough intra-bandwidth as long as both are not simultaneously using the fourth rack. From part b, both are using the fourth rack for no more than .00001 % of time, which is within the .0001% requirement.