Tutorial 3 1 WEB CACHING

Transcription

1 Tutorial 3 1 WEB CACHING Consider the following figure, for which there is an institutional network connected to the Internet. Suppose that the average object size is 850,000 bits and that the average request rate from the institution s browsers to the origin servers is 16 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is 3 seconds on average. Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. or the average access delay, e Δ / (1 Δβ), where Δ is the average time required to send an object over the access link and β is the arrival rate of objects to the access link. a. ind the total average response time. b. Now suppose a cache is installed in the institutional LAN. Suppose the miss rate is 0.4. ind the total response time. c. Calculate the improvement in performance before and after web caching is ed. d. List the advantages of deploying web caching in the Internet.

2 SOLUTION a. Total average response time = average access delay + average Internet delay Average access delay = Δ / (1 Δβ) Δ = average time required to send one object over the access link = transmission time of one object = average object size / transmission data rate of access link = 850,000 / (15 * 106) = sec β = arrival rate of objects to access link = 16 requests per sec Δβ = traffic intensity of access link = ratio of the time during which access link is occupied to the time that the link is available for occupancy = (0.057) (16) = Average access delay = (0.057) / ( ) = sec Average Internet delay = 3 sec Total average response time = = sec b. Miss rate of a cache = number of misses/ number of cache accesses Hit rate of a cache = 1 miss rate = = 0.6 Since the miss rate of the cache is 0.4, this means that the access link between institution and Internet is only ed 40% of the time. As such, the traffic intensity on the access link is reduced by 40% : New traffic intensity = 40% of old traffic intensity = (0.4) (0.912) = New average access delay = (0.057) / ( ) = 0.09 sec Total average response time in case of cache miss = = 3.09 sec The response time is approximately 0 sec if the request is satisfied by the cache, which happens with probability 0.6. As such, the total average response time in case of cache hit is 0 sec Total average response time = (0.4) (response time in case of miss) + (0.6) (response time in case of hit) = (0.4) (3.09) + (0.6) (0) = sec c. With web caching, the average response time is reduced from 3.09 sec to sec. This means that ing web caching is 2.5 times faster than not ing it. d. 1) A web cache can substantially reduce the response time for a client request, particularly if the bottleneck bandwidth between the client and the origin server is much less than the bottleneck bandwidth between the client and the cache. If there is a high-speed connection between the client and the cache, as there often is, and if the cache has the requested object, then the cache will be able to deliver the object rapidly to the client. 2) Web caches can substantially reduce traffic on an institution s access link to the Internet. By reducing traffic, the institution does not have to upgrade bandwidth as quickly, thereby reducing costs.

3 DNS (DOMAIN NAME SYSTEM) TASK A: DNS LOOKUP Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so a DNS lookup is necessary to obtain the IP address. Suppose than n DNS servers are visited before your host receives the IP address from DNS; the successive visits incur an RTT of RTT1,,RTTn. urther suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let RTT0 denote the RTT between the local host and the server containing the object. a. Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the object? b. Is this DNS lookup iterative or recursive? Explain. SOLUTION a. The total amount of time to get the IP address is RTT1+RTT2+RTT3+ RTTn Once the IP address is known, RTT0 elapses to set up the TCP connection, and another RTT0 elapses to request and receive the small object (which is the base HTML file). The total response time is 2RTT0+RTT1+RTT2+..RTTn b. The DNS lookup is both recursive and iterative. More specifically, the query from the requesting host to the local DNS server is recursive becae the query asks the local DNS server to obtain the mapping on its behalf. The remaining queries, on the other hand, are iterative becae all of the replies are returned directly to the host.

4 TASK B: DNS CACHING Suppose you can access the caches in the local DNS servers of your department. Can you propose a way to roughly determine the Web servers (outside your department) that are most popular among the ers in your department? Explain. SOLUTION We can periodically take a snapshot of the DNS caches in those local DNS servers. The Web server that appears most frequently in the DNS caches is the most popular server. This is becae if more ers are interested in a particular Web server, then DNS requests for that server are more frequently sent by ers. Th, that Web server will appear in the DNS caches.

5 TASK C: DNS CACHING Suppose that your department has a local DNS server for all computers in the department. You are an ordinary er (i.e. not a network/system administrator). Can you determine if an external Web site was likely accessed from a computer in your department a couple of seconds ago? Explain. (Hint: Use the dig tool available on Unix and Linux hosts to explore the hierarchy of DNS servers.) SOLUTION The dig command returns the query time for finding the external Web site. If the Web site was jt accessed a couple of seconds ago, an entry for this Web site is cached in the local DNS cache, so the query time is 0 msec. Otherwise, the query time is large. 2 ILE DISTRIBUTION TASK A: CLIENT-SERVER AND P2P ILE DISTRIBUTION Consider distributing a file of = 15 Gbits to N peers, as shown in the following figure. The server has an upload rate of u s = 30 Mbps, and each peer has a download rate of di = 2 Mbps and an upload rate of u. or N = 10, 100, and 1,000 and u = 300 Kbps, 700 Kbps, and 2 Mbps, prepare a chart giving the minimum distribution time for each of the combinations of N and u for both client-server distribution and P2P distribution.

6 SOLUTION Before solving this problem, ensure that you are working with the same units throughout the entire problem. In this case, we will work with Mbits and Mbps. ( Hint : To do so We multiply the Number in Giga with 1024 not 1000 ) = 15 Gbits = 15 * 1024 Mbits u s = 30 Mbps = di = 2 Mbps or calculating the minimum distribution time for client-server distribution, we e the following formula: D CS = max { N, } As an example, let demonstrate how D CS is calculated for u = 300 Kbps and N = 10 N = d = min 2 max { N, =5120 sec. = 7680 sec. }=7680 sec. ollow these same steps for each cell of the following table: u N kbps kbps Mbps or calculating the minimum distribution time for P2P distribution, we e the following formula: D P2P = max {,, N + n i=1 ui As an example, let demonstrate how D P2P is calculated for u = 300 Kbps and N = 10 ui= u = = Mbits for all i [1, N] becae all peers have the same upload rate u =512 sec } = = 7680 sec N + n i=1 max {, = sec sec. ui, N + n i=1 ui }=7680 sec

7 u N kbps kbps Mbps TASK B: CLIENT-SERVER ILE DISTRIBUTION Refer to the figure as in task A, and consider distributing a file of bits to N peers ing a client-server architecture. Assume a fluid model where the server can simultaneoly transmit to multiple peers, transmitting to each peer at different rates, as long as the combined rate does not exceed u s. a. Suppose that d N min. Specify a distribution scheme that has a distribution time of N. b. Suppose that d N min. Specify a distribution scheme that has a distribution time of c. Conclude that the minimum distribution time is in general given by max { N, } SOLUTION a. Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of. Based on the assumption that d N N min, the rate at which the server sends the file to each client is less than each client s download rate. Th, each client can also receive at rate. Since each client receives at rate, the time for each client to receive the entire file is overall distribution time is also N. N = N N N. Since all the clients receive the file in, the N b. Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of. Based on the assumption that d N min, which can also be simplified to N, the aggregate rate at which the server sends the file to all clients is less than the server s link rate. Th, each client receives at rate. Since each client receives at rate Since all the clients receive the file in c. rom (a), DCS = N rom (b), DCS = when N, the time for each client to receive the entire file is, the overall distribution time is also. when d N min..by combining these two findings, we get that DCS = max { N, }.

8 Why max? Becae the file is always distributed at the lower present transmission rate. The lower the transmission rate, the higher the distribution time. OVERLAY AND UNDERLAY NETWORKS Consider an overlay network with N active peers, with each pair of peers having an active TCP connection. Additionally, suppose that the TCP connections pass through a total of M routers. How many nodes and edges are there in the corresponding overlay network? Number of nodes = N The edges of the overlay network are formed by the individual TCP connections, so number of edges = N(N 1) 2