COP Operating Systems Design Principles (Spring 2012)

Transcription

1 Problem 1: COP Operating Systems Design Principles (Spring 2012) Homework a. Number of paths to reach the global state : Number of paths = Number of ways to reach the global state (sum of the lattice positions)/number of paths that can be reached with the individual states Number of paths = ( )! / (6! * 2! * 3! * 4! * 5!) = (20)! / (6! * 2! * 3! * 4! * 5!) = paths 1.1.b. Number of events on the longest and on the shortest path: In a global lattice only the same number of events can lead to a particular state which is equal to the sum of the events. Number of events on the longest and the shortest path = = 20 events 1.2 Example of application of Logical clocks for communication protocols: Best example of application of logical clocks for communication protocols is Transmission Control Protocol (TCP). Flow control: TCP takes care of flow control by maintaining sliding windows. Here logical clocks are used to help the process measure its own progress. The sender and receiver both maintain their logical clocks to ensure the advancement in the window so as to send/receive the next segments. Error control: TCP takes care of error control with the help of retransmissions and timeout. The sender starts its timer after it sends a segment. It waits for the receiver to send an acknowledgement until its timer expires. If no acknowledgement is received then the packet is lost somewhere in the network. After that the sender retransmits the lost segment. Here the sender and the receiver maintain their own logical clocks. 1.3 Real clocks used to implement critical mechanism for communication in the internet: In the internet there is no global clock. Each device has its own internal clock and its own notion of time. These clocks can easily drift values, accumulating significant error over a

2 A. period of time. This creates serious synchronization problems. For most applications it is important to know the time at which the event happened or the relative ordering of the events to ensure communication between devices. It is essential to have clock synchronization between the devices. This is taken care by Network Time Protocol (NTP). Here the real clocks help in synchronization and also adhere to the physical time. Problem 2: The best example of an end to end argument is C. Even if a chain manufacturer tests each link before assembly, he d better test the completed chain. The end-to-end principle states that application-specific functions ought to reside in the end hosts of a network rather than in intermediary nodes provided they can be implemented "completely and correctly" in the end hosts. According to the definition, even if the individual links are checked and tested the completed chain has to be tested to ensure if the links are integrated properly together. This ensures end to end connectivity and improves the reliability of the system. Problem 3: Given: Data rate of the network = bytes/second Packet size = 1000 bytes Propagation time = 500 microseconds 3. a. Maximum throughput for Send-and-wait: Number of packets per second in the network = /1000 = 1000 packets/second Time taken by one packet to flow in the network = 1/1000 = 1 millisecond = 1000 microseconds Time taken by the packet to reach the receiver = Time taken by one packet to flow in the network + Propagation time = =1500 microseconds In Send-and-wait the next packet is sent only after receiving acknowledgement. So the total time for sending a packet and receiving acknowledgement = = 2000 microseconds. There is one packet flow for every 2000 microsecond in Send-and-wait.

3 Maximum throughput = 1/2000 microseconds = bytes/second 3. b. Maximum throughput for Flow-control: In Flow-control there is a packet flow every 1000 microseconds since the sender does not wait for the acknowledgement to send the next packet. The sender sends until the window is open. Maximum throughput = 1/1000 microseconds = bytes/second 3. c. Maximum throughput for Blast: Here the maximum throughput will be the rate at which the sender sends the packet since only one acknowledgement is received for a blast of packets it can be ignored. Throughput for blast is same as Flow-control. Here the sender does not even care about the window. Maximum throughput = 1/1000 microseconds = bytes/second 3. d. Product to be chosen for maximum reliable operation: C. Flow-control, since Blast will be unreliable and Send-and-wait is slower. Flow-control will be the best choice for the following reasons. Since the receiver has a byes buffer it cannot handle Blast. In Blast the sender keeps sending the packets without taking into account the receiver s capacity. In this case there may be loss of packets if the number of packets reaches more than that of the receiver s capacity. Sent-and-wait is slow because the sender waits for the acknowledgement of the previous packet to send the next packet. So, Flow-control is the best choice for this scenario. Problem 4: Given: Maximum packet size = 480 bytes Round-trip time = 100 milliseconds 4. a. Maximum data rate: Since lock-step end-to-end protocol is used, the maximum data rate without loss of packets = 480/100 milliseconds = 4800 bytes/second 4. b. Data rate after installing resend timer:

4 Resend timer = 1000 milliseconds 1% of the packets are getting lost. If there are 100 blocks sent there is a flow of 200 packets in the network (considering the RTT). Loss of 1% packets implies loss for 2 blocks. Average time taken for 100 blocks = 98 * 100 milliseconds + 2 * 1100 milliseconds (additional resend timer) = 9.8 seconds seconds = 12 seconds Time taken for 1 block =.12 seconds = 120 milliseconds Data rate = 480/120 milliseconds = 4000 bytes/second 4. c. Data rate with satellite delay: 50% of the roundtrips now take extra time 100 milliseconds. Average time take for 100 blocks = 50 * 100 milliseconds + 50 * 200 milliseconds (satellite delay) = = 15 seconds Time taken for 1 block =.15 seconds = 150 milliseconds Data rate with satellite delay = 480/150 milliseconds = 3200 bytes/second 4. d. Data rate on Tuesdays: The 50% of the roundtrips that undergo satellite delay are now resent. But again there are chances for this resent packet to undergo satellite delay and so on. The problem here is recursive. The equation can be, Time taken for 1 block = 0.5 * 100 milliseconds (without delay) 0.5 * (150 + time taken for 1 block) (with delay and recursion) T = 0.5 * * (150 + T) 2T 100 = T T = 250 milliseconds Date rate on Tuesdays= 480/250 milliseconds =1920 bytes/second 4. e. Usage of data rate of the slowest link in the network: The roundtrip time given in the problem already includes the data rate of the slowest link. The data rate of the slowest link has been used in the problem.

5 Problem 5: The automatic dialer did not get activated because Ben would have handled the outputs from the fail-fast detectors wrongly. If the fail-fast detectors notify that it has failed then its inputs should not be taken for the majority voter as it is invalid. There are two ways the question can be perceived if the question is what mistake Ben made then the answer would be B. He should have used a voter that ignores failed inputs from fail-fast sources. In this case the non-active inputs will be taken into consideration as the voter cannot identify if it is from a failed source. If the question is what Ben could have done to avoid the mistake then the answer would be D. He should have done both A and B. (Where A. He should have used fail-fast smoke detectors and C. He should have used a voter that ignores non-active inputs.). In this case the detector would have notified the voter that it failed and the voter would not have taken this inactive input. Problem 6: Given: Mean time to failure of plane A = 6000 hours Mean time to failure of plane B = 5000 hours Time of flight with plane A = 6 hours Time of flight with plane B = 5 hours Failures occur according to a memoryless random process Since plane s failure process is memoryless, The conditional failure probability for plane A =1/6000 The conditional failure probability for plane B =1/5000 The probability of failure of plane A during its flight = 6 * 1/6000 = 1/1000 The probability of failure of plane B during its flight = 5 * 1/5000 = 1/1000 Both the planes have the same probability of failure. The passenger can choose either of the flights to minimize the chance of plane failure during the flight.

6 Problem 7: 7. a. Advantage of RAID 5 over RAID 4: B. Read performance in the absence of errors is enhanced and C. Write performance in the absence of errors is enhanced. Read in RAID 5 is faster than read in RAID 4 because of distributed parity. Multiple writes are possible in RAID 5 i.e. parallel reads and writes are possible. In RAID 4 there is a single disk for parity so there is always a bottleneck at the parity disk. Whereas in RAID 5 parities are written in different disks so simultaneous writes are possible. 7. b. Workload for which RAID 4 has better write performance than RAID 5: RAID 4 has better write performance than RAID 5 for a workload that consists of single and small write operation (no simultaneous writes). The parity disk in RAID 4 becomes a bottleneck only when there are multiple writes. Performance is better as there is an additional computation of the disk number for writing the party in the case of RAID 5 but in RAID 4 it can directly be written on disk N. 7. c. Number of disks required between RAID 1 and RAID 5: Minimum number of disks required for RAID 1 is 2 and for RAID 5 is 3 (Only with 3 disks EX-OR and computation of parity is possible). RAID 1 uses mirroring where the same data is written in both the disks. There is no parity used. 7. d. Better performance for a workload consisting of small reads and small writes: RAID 1 has better performance for a workload consisting of small reads and small writes. This is because there is no additional parity overload in RAID 1. And also RAID 1 offers excellent read speed and a write-speed that is comparable to that of a single disk.