# Operating Systems and Networks Sample Solution 1

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1 Spring Term 2014 Operating Systems and Networks Sample Solution 1 1 byte = 8 bits 1 kilobyte = 1024 bytes 10 3 bytes 1 Network Performance 1.1 Delays Given a 1Gbps point to point copper wire (propagation speed 200m/µs) with a length of 3km. a) What is the transmission time for a 16 MB file? t transmission = L R = 16MB 1Gbps = byte 10 9 bit s 1 = bit 10 9 bit s 1 = 0.128s b) What is the propagation delay? t propagation = d s = 3km 200m/µs = 3000m 200m/10 6 s = 15µs 1.2 Throughput What is the throughput when retrieving a file across a 1Gbps network with a round trip time of 150msec, again ignoring ACKs? Assuming that the file transfer has to be initiated by a request: R = 150msec + 1Gbps = 150msec + 32msec = 4 8Mbit = M bps 0.182s Otherwise: R = 75msec + 1Gbps = 75msec + 32msec = 4 8Mbit = M bps 0.107s 1.3 Total Time Calculate the total time required to transfer a KB file in the following cases, assuming an RTT of 100 ms, a of 1-KB data, and an initial 2 x RTT of handshaking before data is sent. The following solutions assume that the sender does not wait for a reply from the receiver after the last packet(s) have been sent, thus requiring only a wait time of 1 2RT T instead of a full RT T. However, the exercise could also be interpreted differently. In that case, the total transfer time would be increased by 1 2 RT T.

2 a) The total transfer time is the sum of the propagation delay: 1 2 RT T the transmission time: 1000 bandwidth total transfer time = 0.2s s bit bit s b) The total transfer time is the sum of the transfer time for 999 packets: 999 (RT T s bandwidth ) 1 the transfer time for the last packet: 2RT T + bandwidth To put it another way, the total transfer time is the sum of the propagation delay and the time we have to wait due to the protocol: RT T ( ) total transfer time = 0.2s s bit c) The total transfer time is the sum of bit s bandwidth ) s the propagation delay and the time we have to wait due to the protocol: 49.5 RT T bandwidth ) = 0 total transfer time = 0.2s s + 0s = 5.15s d) We need 10 RTTs to transmit the 1000 packets. The total transfer time is the sum of the propagation delay and the time we have to wait due to the protocol: 9.5 RT T bandwidth ) = 0 total transfer time = 0.2s s + 0s = 1.15s 1.4 Transmission Time How long does it take to transmit x KB over a y Mb/s link? Give your answer as a ratio of x and y. t = 8192x y or with the rule of thumb t = x y = x 125 y 2 Network Characteristics 2.1 Width of a Bit Determine the width of a bit on a 10 Gbps link. propagation is m/s m s bit s 1 = 0.023m/bit A single bit has a width of 2,3 cm on the wire. Assume a copper wire, where the speed of 2

4 (a) (b) The RTT can be determined by measuring the time where the request packet has been sent and the time when the response has been received. This measurement uses the same clock since it happens on the same machine. (c) The latency, in contrast, requires a measurement on two different machines that connect each other with the same path. Otherwise, it would require the two independent clocks in these machines to be perfectly time-synchronized. In practice, this is hard to achieve. b) Calculate the bandwidth using ping (a) Considering a connection between hosts A and B, we measure the RTT using ping. We assume that this RTT Z of a connection having bandwidth R for a given package size L consists of a transmission delay ) L R and a further delay V. Thereby, we ignore any processing delays on the machine B. ( ) L Z = 2 R + V The factor 2 comes from the fact that the ping response package has the same size as the request package. This formula is unable to determine the bandwidth having only L and Z since V is unknown. However, taking two measurements for different package sizes L 1 and Z 1 respectively L 2 and Z 2, assuming: L 1 > L 2 und Z 1 > Z 2, one gets two versions of the equation and is able to solve both equations for V. For L 1 and Z 1, this is: V = Z 1 2 L 1 (2) R Under the assumption that the further delays V do not change much for different package sizes, one can identify the two equations for V. Now solving the resulting equation for R, the formula for the bandwidth is: R = 2 L 1 L 2 Z 1 Z 2 (3) The exercise can also be solved by empirically measuring the RTT for different package sizes, drawing a diagram, and approximating the measurements through a line. Then, the bandwidth corresponds to two times the gradient of the line. (b) The ping package size can be changed on Linux using the -s switch (e.g., ping -s for sending 100 Bytes of data). On UNIX machines, the -s uses a slightly different format (ping -s 100). On Windows machines, the option to use is -l (ping -l (c) We calculate the mean value of the RTTs of the measurements for different package sizes. With the formula determined before, we calculate R. The value can vary significantly due to variations in network utilization and the relatively small package sizes. The package sizes, however, should not exceed the size limitations of and Ethernet frame (1518 Bytes, 1500 payload plus IP and ICMP headers); larger packages get fragmented which affects the measurement. A typical value for R is in the Mbps range. (1) 3.2 traceroute a) Hosts in Switzerland typically have domain names ending with.ch. Our path to contains these hosts: 1 rou-cx-1-service-inf-isg-cx-server-1.ethz.ch ( ) 2 rou-ref-hci-service-inf.ethz.ch ( ) 3 rou-fw-hci-service-inf-isg.ethz.ch ( ) 4 rou-fw-rz-fw-cla.ethz.ch ( ) 5 rou-rz-gw-fwrz-gwrz-core.ethz.ch ( ) 6 swiez2.ethz.ch ( ) 7 swiix2-10ge-3-1.switch.ch ( ) 4

5 b) One can try to find a machine in a geographically remote location or in a country with a low level of industrialization. For instance, the webserver of the University of Tokio ( is 30 hops away from ETHZ. c) The -t option (resp. -i for Windows) changes the time to live (TTL) of the packets that ping generates and the value of TTL corresponds to the maximum number of hops taken. Using this option, one can iteratively determine the maximum number of hops that a packet can traverse. d) We make use of the following observation: if the TTL expires for a ping request, then the hop on which the TTL has expired will respond and tell us. This way, we can get a response from each machine on the path to the target machine when beginning with a TTL of 1 and iteratively increment the TTL until we reach the target host. 3.3 curl a) discussed in the recitation session b) sends you a 301 Moved Permanently and tells you that the new URL is This page, in turn is very likely sending you a 302 Found with a rederict to the swiss google homepage Wireshark a) discussed in the recitation session b) We can use Wireshark to capture the traffic and inverstigate the packet payload. 5

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