Homework 2 COP The total number of paths required to reach the global state is 20 edges.
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1 Homework 2 COP 5611 Problem 1: 1.a Global state lattice 1. The total number of paths required to reach the global state is 20 edges. 2. In the global lattice each and every edge (downwards) leads to a new state. Each state is triggered due to an event. In any n dimensional lattice, though there can be multiple combinations to reach a specific global state but there can be only the same number of events occurring before leading to that state. In this problem there should be 20 events occurring. 1.b Logical clocks: 1. Logical clocks play a very important role in communication. Only the logical clocks ensure that messages are delivered to the right process at the right time. Logical clocks also ensure causal message delivery. It is essential to follow the causal message delivery principle in network communication. When causal delivery principle is not obeyed the communication channel will not reorder messages and the user will not receive the messages in order. 2. Flow control: All the flow control mechanisms (TCP protocol) use a sliding window which will extend when an acknowledgement is received. This is a form of maintaining a range of logical clocks. Taken from a TCP article The figure shows the sliding window, where each end node maintains a variable similar to logical clock and maintains a range of numbers called window. This range pointer will be moved 1
2 at every interval. The maintenance of sliding window is very crucial to enable flow control in internet. 3. Error control: A similar approach of sliding window is also used in error control. Automatic Repeat Request (ARQ) is an error control method for data transmission that uses acknowledgements and timeouts to achieve reliable transmission. The sliding window is used in different ways in different types of ARQ namely, Stop-and-wait, Go-Back-N, and Selective Repeat. 1.c Usage of real clocks in internet Real time clocks are used for the following: 1. Generating real-time packet timestamps with the help of kernel time (time in seconds calculated since Epoch). 2. Maintaining computer clock using hardware clocks. Synchronizing real-time clocks: 1. Real time clock values are always affected by a clock drift, which has to be synchronized to ensure consistency during communication between devices. The synchronizing algorithm is called NTP (Network Time Protocol). It uses UDP based message passing model and client server interaction to synchronize time. 2. Another major concern in communication is the timezone which should also be synchronized. This aspect affects Mail Transfer protocols ( s). Problem 2: Best example for supporting end-to-end argument: The best example that can precisely support the end-to-end argument is: C. Even if a chain manufacturer tests each link before assembly, he d better test the completed chain. Reason: 1. The end-to-end argument was raised by Saltzer, Reed and Clark in the early 1980 s. The argument is about creating a stable design that helps in placement of functions among modules of distributed computer system. 2
3 2. Their argument insists on the fact of end-to-end consistency and reliability in the system design. To improve the reliability it is important to place the functions that drive the system at the right places. For instance, placing the communication protocol function on both host machine and client machine improves reliability rather than having it only at one end. 3. A chain is made up minor segments which are joined together with minute links. Even if a single link breaks the whole chain will fall apart. So, to improve consistency it is important to test each link at every stage before linking all the segments to make a chain. However, to improve reliability it is always mandatory to test the final out product, i.e. the final link that connects the whole chain which makes it strong. 4. The chain manufacturer example can be related to one of the real time examples; reliable delivery of messages in FIFO queue. This example was first explained in the end-to-end argument paper by Saltzer et al. However, the FIFO example has to be modified little bit to relate to the chain manufacturer example. Chain example Segments of chain Links joining the segments Chain manufacturer checks every link of the chain at every stage It is better to check the completed chain Real time example Messages in the queue A FIFO buffer. Though the segments in the chain do not have any significance to be in a queue, each segment can be related to messages of same type. Can be related to the service rate of the FIFO buffer. Whenever a message arrives the FIFO buffer should make sure it is serviced and sent out of the buffer. Although the FIFO buffer consistently services all the messages without dropping them and sends them to the destination, an acknowledgment from the destination to the host will complete the reliable message transfer. 5. Inference: In spite of enforcing the destination to send acknowledgment, the FIFO buffer might not service a message if continuous test on FIFO buffer service is not employed. Similarly, even if the consistent service rate is ensured the message should be acknowledged to satisfy the host and improve reliability. Thus, the chain manufacturer should better test the final product in addition to the inspection on every link. 6. The other examples namely, example about web hackers, per-packet checksums and network function in application layer do not support the end-to-end argument. In addition, just pushing the network function into application layer will not ensure integrity in the communication between devices. There is also opposing an argument which suppresses the placing of error correction function in application layer which will increase the cost of computation*. The per-packet checksums might seem like supporting the end-to-end argument, but in reality per parity checksums augmented by parity improves redundancy and data failure tolerance, but will not improve the end-to-end 3
4 reliability. The pipe example just shows how data is transferred from write end to read end of the pipe in UNIX but cannot be used to support end-to-end argument. *Referring to A critical review of End-to-End arguments by Tim Moors. Problem 3: 3.a Maximum throughput for Send-and-wait: Given: Network data rate: 1,000,000 bytes per second Packet size: 1000 bytes Propagation time: 500 microseconds Total capacity of network: 1,000,000/1000 bytes = 1000 packets per second. Send-and-wait sends a packet and waits for acknowledgment to send the next one: Each byte takes 1 micro second to travel in the network. Thus, 1000 bytes takes 1000 microseconds. Total time to reach receiver = (propagation time) = 1500 microseconds Total time for next packet to start = (ACK) = 2000 microseconds. Throughput = 1/2000 microseconds = 500,000 bytes/second. 3.b Maximum throughput for Flow-control: Given: Network data rate: 1,000,000 bytes per second Packet size: 1000 bytes Propagation time: 500 microseconds Total capacity of network: 1,000,000/1000 bytes = 1000 packets per second. Send-and-wait sends a packet and waits for acknowledgment to send the next one: Each byte takes 1 micro second to travel in the network. Thus, 1000 bytes takes 1000 microseconds. 4
5 Packets continuously get acknowledged and will be pushed into the network without delay. Throughput = 1/1000 microseconds = 1,000,000 bytes/second. 3.c Maximum throughput for Blast: Given: Network data rate: 1,000,000 bytes per second Packet size: 1000 bytes Propagation time: 500 microseconds Total capacity of network: 1,000,000/1000 bytes = 1000 packets per second. Send-and-wait sends a packet and waits for acknowledgment to send the next one: Each byte takes 1 micro second to travel in the network. Thus, 1000 bytes takes 1000 microseconds. Similar to the flow-control the packets will be continuously pushed except that it will not get any acknowledgment. This fact will not change the throughput but can reduce network overhead due to acknowledgements at the cost of reliability. Throughput = 1/1000 microseconds = 1,000,000 bytes/second. 3.d Which protocol to choose: It is better to choose C. Flow control. Having a finite buffer, the most optimistic protocol to use here is flow-control. This has practically reliable sliding window with maximum throughput. Problem 4. 4.a Calculate the maximum data rate possible: Given: Max packet size = 480 bytes RTT = 100 milliseconds. Data rate = 480/100 milliseconds = 4800 bytes/second. 5
6 4.b 1% packet loss and resend timer: Given: 1% packet loss For that 1% resend timer = 1000 milliseconds. Out of 100 blocks we send from source to destination there is a 1% packet loss. Considering round trip, we should consider a packet loss of 2 packets out of 100 blocks. Thus 98 blocks have 100 milliseconds RTT and 2 packets have 1100 milliseconds. Total RTT = seconds = 120 milliseconds/block. Data transfer delayed by 20 milliseconds, thus data rate = 480/120 = 4000 bytes/second. 4.c 50% taken over by satellite link Given: 50% packets get additional delay of 100 milliseconds. 50 blocks have 100 ms and 50 blocks have 200 ms. Total = 150 milliseconds/block. Data rate = 480/150 milliseconds = 3200 bytes/second. 4.d. With resend timer We are not aware of the number of retries the packet undergoes. Thus let us consider the overall time for packets to be resent be T. Thus, T = 50% of 100 ms + 50% (150ms + T) - Recursive On solving we get T = 250 milliseconds. Using that, the new data rate = 480/250 milliseconds = 1920 bytes/second. 4.e Why didn t we use slowest link data-rate? For calculating data rates we need packet sizes and round trip time. Generally the safest way to calculate round trip time is using the slowest link latency. Latency relates to the time delay in the link. This will help in calculating the round trip time typically. 6
7 Problem 5. Fire alarm: Answer: Ben should have done Both A and C. He should have used fail-fast detectors and should have used a voter that ignores non-active inputs. Reason: 1. The majority voter would have given wrong reading when out of 1 out of 7 detectors fail and the remaining create a tie situation for the voter. 2. When 4 (majority) of the detectors send false detections (no fire) Solutions: 1. Ben should have used fail-fast detectors. Since, fail-fast detectors notify when the detector fails Ben can replace the detector to avoid false alarms in spite of having a voter. 2. Consider that Ben uses fail-fast detectors, the detectors would automatically stop its normal working and notifies about the failure. Despite its failure the voter will count the detector as a participant in voting and considers its input. If 3 detectors work well and report fire and if 4 detectors stop its normal operation, due to fail-fast rule the voter will still count the 4 detectors response as no fire and will not trigger alarm. Thus it is mandatory for Ben to make the voter understand non-active inputs. When the fail-fast detector goes to non-active state after failure the voter should reduce the count of those detectors. In the example we considered, the voter would reduce the count to only 3 and will trigger the alarm. Problem 6 Flight journey Given: MTTF (Flight 1) = 6000 hours MTTF (Flight 2) = 5000 hours Journey = 6 hours Journey = 5 hours Answer: On an average the Flight 1 and Flight 2 has the probability of 1/6000 and 1/5000 to fail respectively. For their respective journey time, Flight 1 has 1/1000 and Flight 2 has 1/1000 probability to fail. Since both the Flights have the same probability to fail it is left to the passenger to choose his flight. 7
8 Note: Both are safe to travel for a short time, in reality it is obviously very unsafe to travel more than 20 hours in this flight. The flight will surely fail. Problem 7 7.a Advantage of RAID 5 over RAID 4 Answer: C. Write performance is the absence of errors is enhanced. Also B. Read performance is enhanced. It can be seen that RAID 5 writes the parity in the best calculated disk and it speeds up the process of writing. In RAID 4 the single parity disk reduces the bandwidth for writing to 1 block/second. In addition RAID 5 is known for its large-sequential fast read operations. Moreover, RAID 5 promotes parallel reads due to distributed parity. 7.b. Workload for RAID 4 RAID 4 has to involve in 4 IO penalty or workload during write operation. It requires 2 reads and 2 writes. However, even RAID 5 involves in 4 IO workload yet RAID 5 can help in flexible writes due to distributed parity. 7.c Comparison on number of disks involved for RAID 1 and 5 RAID 1 uses a technique called mirroring where the data is written twice in a pair of disk. RAID 1 requires a minimum of 2 disks whereas RAID 5 uses N disks, where N is typically 5. Though, RAID 1 can have faster reads and writes the ability to be fault tolerant is less as compared to RAID 5. The RAID 5 has a distributed parity scheme that can help in both fault-tolerance as well as parallel reads. There are few specific small-scale applications that can use RAID 1, but RAID 5 is the best and secure way to efficiently use data storage. 7.d RAID 1 vs RAID 5 for small reads and writes 1. RAID 1 (2 operations) has lesser workload and IO operations than RAID 5 (4 operations). 2. RAID 1 is also fit for small reads and writes. Most of the mission critical applications and small servers use RAID 1. 8
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