Consider a source S and a destination D using the Selective Repeat Protocol (SRP) with a window size of N=4N=4. Each packet is 10001000 bits.
The delays are as follows:
- Each packet transmitted by S has a transmission time of 10 ms10ms.
- The end-to-end delay from the time that S starts transmitting a packet until D completely receives that packet is 100 ms100ms.
- Each ack transmitted by D has a transmission time of 1 ms1ms.
- The end-to-end delay from the time that D starts transmitting an ack until S completely receives that ack is 100 ms100ms.
- The timeout value is 300 ms300ms.
Here is what happens to each packet:
- Packet #0 is received by D, and the corresponding ACK is received by S.
- The first transmission of packet #1 is dropped by a router in between S and D.
- The second transmission of packet #1 is received by D, and the corresponding ACK is received by S.
- The first transmission of packet #2 is dropped by a router in between S and D.
- The second transmission of packet #2 is received by D, and the corresponding ACK is received by S.
- Packet #3 is received by D, and the corresponding ACK is received by S.
- Packet #4 is received by D, and the corresponding ACK is received by S.
- Packet #5 is received by D, and the corresponding ACK is received by S.
(a) At time t = 150 mst=150ms, what is S’s window?
(b) At what time does the retransmission of packet #1 start?
(c) At what time does the application layer at D receive packet #2?
(d) What is the throughput during the time from t = 0 mst=0ms until t = 500 mst=500ms?
Write your answers for parts (a)-(d) (as well as any diagram or equations use to obtain your answers) on paper, and upload the image here:
- Transient Overloads: One drawback of RM algorithm is that task priorities are defined by their periods. Sometimes, we must change the task priorities to ensure that all critical tasks get completed. With average execution times, all the critical and non-critical tasks may be RM-schedulable. But, if we consider worst-case execution time we wish to schedule so that all critical tasks meet the deadline and the non-critical task may miss the dead line. The solution is to boost the priority of some critical tasks higher than the non-critical tasks. This is done by altering the “Period” of some critical tasks so that their priorities increase. For example: If we reduce the period of a task by k and have k number of tasks each small sub task will have period divided by K, and execution time divided by k. When the period is smaller, the tasks automatically get higher priority. We can also lengthen the period of non-critical tasks and make it larger than the largest critical task and increase the non-critical task’s execution time also.Carry out a period transformation for this task set to ensure that all the critical tasks will meet the dead line even with worst case execution timing.
1.Draw for your modified table of tasks, timing diagram to show schedulability for average execution time and another diagram to show for worst-case execution time. Identify if any tasks miss the dead line.
- Give explanation of your period transformations and solution.
- What is CPU Utilization for average and worst-case execution times.
- Considering Context switch time to be 1 ms, redraw the execution time line for this system for both average and worst case timing.
- What is the system time-loading (CPU utilization) factor with the context switching included
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