发布时间 : 星期二 文章哈尔滨工业大学 深圳 高级计算机网络 2017 习题集更新完毕开始阅读
??ReffR(1?Pe)?(nf?nh)/t0R(1?Pe)?nf?nhR(2tproc?2tprop?tf?ta)nf?nh(1?Pe)?nf?nhR(2tproc?2tprop?nf/R?na/R)nf?nhtreactionR?nf?na78(1?Pe)(1?Pe)?(1?Pe)?2(tproc?tprop)R?nf?na(1?Pe)
(1?Pe)=(1250?25)bytes*8bits/byte0.001s*1Mbps?(1250?25)bytes*8bits/byte?When Pe=10-6,η=87.50%, when Pe=10-5,η=87.50%, when Pe=10-4,η=87.49%. 5. Describe the TCP congestion control scheme. Derive the delay modeling for TCP traffic in fixed congestion window when WS/R < RTT+S/R, assume that link bandwidth is R bps; maximum segment size is S bits, file object size is F bits, fixed window size is W segments, RTT is round trip time. Solution:
TCP must use end-to-end congestion control rather than network-assisted congestion control, since the IP layer provides no explicit feedback tothe end systems regarding network congestion.
(1) The basic idea of TCP congestion control scheme
The TCP congestion-control mechanism operating at the sender keepstrack of an additional variable, the congestion window. The congestion window,denoted cwnd, imposes a constraint on the rate at which a TCP sender can send trafficinto the network. Specifically, the amount of unacknowledged data at a sender maynot exceed the minimum of cwndand rwnd, that is:LastByteSent-LastByteAcked≤min{cwnd, rwnd}. (2) The delay model for TCP traffic
From the above condition, cwnd=W ≤ RTT*R/S + 1 and the cwnd is fixed. So, when R and S both are fixed and RTT is dynamic, the cwnd is the minimum (RTT*R/S + 1). When the network is perfect (RTT is smallest), though the network is not in congestion and the rwnd is big enough, the send rate can’t be increased. It is because the cwnd is limited the send rate. *************
Link-State (LS) Routing algorithm OSPF
In practice this is accomplishedby having each node broadcast link-state packets to all other nodes in thenetwork, with each link-state packet containing the identities and costs of itsattached links.
Theresult of the nodes’ broadcast is that all nodes have an identical and complete viewof the network. Each node can then run the LS algorithm and compute the same setofleast-cost paths as every other node. *************
Two routing protocols have been used extensivelyfor routing within an autonomous system in the Internet: the Routing InformationProtocol (RIP) and Open Shortest Path First (OSPF).
RIP is a distance-vector protocol.The maximum cost of a path is limited to 15,
thus limiting the use of RIP toautonomous systems that are fewer than 15 hops in diameter.In RIP, routing updatesare exchanged between neighbors approximately every 30 seconds using aRIP response message. The response message sent by a router or host containsa list of up to 25 destination subnets within the AS, as well as the sender’sdistance to each of those subnets. Response messages are also known as RIPadvertisements.
OSPF is a link-state protocol that usesflooding of link-state information and a Dijkstra least-cost path algorithm.With OSPF, a router broadcasts routing
information to all other routers in theautonomous system, not just to its neighboring routers. A router broadcasts linkstateinformation whenever there is a change in a link’s state (for example, a changein cost or a change in up/down status). It also broadcasts a link’s state periodically(at least once every 30 minutes), even if the link’s state has not changed.
When multiple paths to a destination have the samecost, OSPF allows multiple paths to be used.(balance the network burden)Integrated support for unicast and multicast routing.Support for hierarchy within a single routing domain.
****************************************************************************************************************************************** ? How long does it take to send a file of 640,000 bits from host A to host B over
a circuit-switched (TDM) network?
? All links are 1.536 Mbps
? Each link uses TDM with 24 slots/sec ? 500msec to establish end-to-end circuit
2. Packet-switching: store-and-forward
? Takes L/R seconds to transmit (push out) packet of L bits on to link of R bps ? Entire packet must arrive at router before it can be transmitted on next link: store and forward
? delay = 3L/R (assuming zero propagation delay Example: ? L = 7.5MbitsR = 1.5 Mbps ? delay = 15 sec
3. Packet switching versus circuit switching Packet switching allows more users to use network! ? 1 Mb/s link ? each user:
? 100 kb/s when “active” ? active 10% of time ? circuit-switching: ? 10 users
? packet switching:
? with 35 users, probability > 10 active less than .0004
Q: how did we get value 0.0004? 4.Caravan analogy
? Cars “propagate” at 100 km/hr
? Toll booth takes 12 sec to service a car (transmission time) ? car~bit; caravan ~ packet
? Q: How long until caravan is lined up before 2nd toll booth?
? Time to “push” entire caravan through toll booth onto highway = 12*10 = 120 sec
? Time for last car to propagate from 1st to 2nd toll both: 100km/(100km/hr)= 1 hr
? A: 62 minutes
? Cars now “propagate” at 1000 km/hr ? Toll booth now takes 1 min to service a car
? Q: Will cars arrive to 2nd booth before all cars serviced at 1st booth? ? Yes! After 7 min, 1st car at 2nd booth and 3 cars still at 1st booth.
? 1st bit of packet can arrive at 2nd router before packet is fully transmitted at 1st router!
? See Ethernet applet at AWL Web site
5.
6.Could the congestion problem be solved with a large buffer space? Too little memory:too much traffic will lead to buffer overflow and packet loss ? Too much memory:the queues and the delays can get so long that by the time the packets come out of the switch, most of them have already timed out and have been retransmitted by higher layers packets (or their retransmissions) have to be dropped after they have consumed precious network re-sources.