Consider a video-on-demand system which uses a Content Delivery Network (CDN) to deliver HD (High Definition) streams at 1.9 Mbit/s (which includes video, audio and overheads). The service has 1200000 customers, of which 11% are expected to be simultaneously viewing at peak time. The central (or origin) server will hold 17000 movies of average length 91 minutes. The edge servers, nearer the customers, will also hold additional copies of the most popular 13% of movies, and these will satisfy 55% of all service requests. (The most popular 13% of titles are held in BOTH the central and edge servers.) What storage capacity must each edge server have? Give your answer in Gigabytes OR Gibibytes (either answer accepted). Do not type the units in the answer box; just type the number. The NEXT question (i.e. next numbered question below) provides space for you to show your working. Answer:

Consider a video-on-demand system which uses a Content Delivery Network (CDN) to deliver HD (High Definition) streams at 1.9 Mbit/s (which includes video, audio and overheads). The service has 1200000 customers, of which 11% are expected to be simultaneously viewing at peak time. The central (or origin) server will hold 17000 movies of average length 91 minutes. The edge servers, nearer the customers, will also hold additional copies of the most popular 13% of movies, and these will satisfy 55% of all service requests. (The most popular 13% of titles are held in BOTH the central and edge servers.) What storage capacity must each edge server have? Give your answer in Gigabytes OR Gibibytes (either answer accepted). Do not type the units in the answer box; just type the number. The NEXT question (i.e. next numbered question below) provides space for you to show your working. Answer:

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Suppose nodes A and B are on the same 10 Mbps broadcast channel, and the propagation delay between the two nodes is 245 bit-times. Suppose A and B send Ethernet frames at the same time. Suppose the transmission time of the entire frame is 295 bit-times. So, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. A node chooses the value of K at random from {0,1,2,...2n−1} where n is the number of collisions experienced on the channel – note n is set to 1 in this case). For Ethernet, the actual amount of time a node waits is K*512 bit times (i.e., K times the amount of time needed to send 512 bits into the Ethernet). Suppose A chooses a K value of 0 and B chooses a K value of 1. Assuming no other nodes are active, can the retransmissions from A and B collide? Justify your answer by showing all the intermediate steps of your calculations.
Suppose we have a web page which contains 10 objects including: a html main file, 8 jpg image, and a .mp3 background music. For simplicity, assume each of the 10 objects has a size of 4 Kbits and that each object can be completely transferred in one TCP packet. Furthermore, assume the round-trip time between the server and client is RTT= 250 ms, and the download rate is 1 Mbps. What will be the time it takes to complete transferring the web page running in the non-persistent mode? Total delay = ms
In a network, two systems are connected and the distance between the system is 5000 km. The propagation speed of the medium is 2 x 107 m/s. The bandwidth of the network is 500 MBps. If the average packet size is 108 bits and the network is used to its full capacity. The minimum size of the sequence number field is n bits. The processing delay at nodes is 0 ms. What is the value of n in case of Go-Back-N sliding window protocol?
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Imagine you have a network that is trying to predict a categorical response with three levels (categories) and uses a softmax activation function on the final layer. If there is just one node in the L-1 layer and that node has an activation value of 5 for a particular observation, use the parameters given below to calculate the weighted input that goes into the 1st node in the final layer. The value is: e : (Wf = -0.2; W = 0.4; W = 0.1, 6f = 0.3, b = -0.2, b = 0.8 O A.-1 O B. 0.3 OC-.0.7 O D. 1.8
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Consider Figure 2.12, for which there is an institutional network connected to the Internet. Suppose that the average object size is 1,000,000 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 three seconds on average (see Section 2.2.5). 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. For the average access delay, use ∆/(1 - ∆b), where ∆ is the average time required to send an object over the access link and b is the arrival rate of objects to the access link. Find the total average response time. I get ∆=1000000/15000000= 1/15, then ∆/(1 - ∆b)= (1/15)/(1-(1/15)*16)=-1, so the finally answer is : 3+(-1)=2s or 3+0=3s?
Consider Figure 2.12, for which there is an institutional network connected to the Internet. Suppose that the average object size is 1,000,000 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 three seconds on average (see Section 2.2.5). 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. For the average access delay, use ∆/(1 - ∆b), where ∆ is the average time required to send an object over the access link and b is the arrival rate of objects to the access link. Find the total average response time. I get ∆=1000000/15000000= 1/15, then ∆/(1 - ∆b)= (1/15)/(1-(1/15)*16)=-1, so the finally answer is : 3+(-1)=2s or 3+0=3s?
With persistent communication, a receiver generally has its own local buffer where messages can be stored when the receiver is not executing. To create such a buffer, we may need to specify its size. Give an argument why this is preferable, as well as one against specification of the size.
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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 three seconds on average (see Section 2.2.5). 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. For the average access delay, use Δ/(1 – Δ ), where Δ is the average time required to send an object over the access link and is the arrival rate of objects to the access link. a. Find the total average response time.b. Now suppose a cache installed in the institutional LAN, Suppose the miss rate is 0.4. Find the total response time.
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