1. 4330/6310 FIRST ASSIGNMENT Spring 2015 Jehan-François Pâris jfparis@uh.edu

2. The model We have One quad-core CPU One disk One input device Three queues  CPU queue "ready queue"  Disk queue  Device queue RQRQ DQDQ Disk CC CC IQIQ Input

3. AN EXAMPLE

4. P0 begins at t = 0 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 RQRQ DQDQ Disk CC CC IQIQ Input NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40

5. P0 gets core until t = 0 + 200 = 200 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 RQRQ DQDQ Disk CC CC IQIQ Input NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40

6. What's next? RQRQ DQDQ Disk CC CC IQIQ Input NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40

7. P1 arrives at t = 100 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

8. P1 gets core until t = 100 + 30 = 130 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

9. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

10. P1 waits for input until t = 130 +800 = 930 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

11. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

12. P0 waits for input device NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

13. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

14. P1 gets core until t = 930 + 40 = 970 P0 waits for input until t = 930 + 900 = 1830 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

15. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

16. P1 terminates at t = 970 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

17. Your program will display (I) Process 1 terminates at t = 970 Core 0 is IDLE Core 1 is IDLE Core 2 is IDLE Core 3 is IDLE Disk is IDLE Average number of busy cores: Ready queue contains: -- Disk queue contains: --

18. Your program will display (II) Process IDStart timeCPU timeStatus 00200WAITING 110070TERMINATED

19. How to compute CPU utilization Keep track of total time for all CPU requests: 200 + 30 + 40 = 270 ms Divide by elapsed time:  270/970 = 0.278 (rounded) Since there are four cores, the maximum CPU utilization is 4.0

20. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

21. P0 gets core until t = 1830 +10 = 1840 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

22. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

23. P0 gets disk until t = 1840 + 10 = 1850 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

24. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

25. P0 gets CPU until t = 1850 + 30 =1880 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

26. What's next? NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

27. P0 terminates at t = 1880 NEW 0 START 0 CPU 200 INPUT 900 CPU 10 I/O 10 CPU 30 NEW 1 START 100 CPU 30 INPUT 800 CPU 40 RQRQ DQDQ Disk CC CC IQIQ Input

28. Your program will display (I) Process 0 terminates at t = 1880 Core 0 is IDLE Core 1 is IDLE Core 2 is IDLE Core 3 is IDLE Disk is IDLE Average number of busy cores: 0.174 Ready queue contains: -- Disk queue contains: --

29. Your program will display (II) Process IDStart timeCPU timeStatus 00240TERMINATED

30. How to compute CPU utilization Keep track of total time for all CPU requests: 200 + 10 + 30 + 30 + 40 = 310 ms Divide by elapsed time:  310/1880 = 0.165 (rounded) Since there are four cores, the maximum CPU utilization is 4.0

31. HANDLING ZERO-DELAY DISK ACCESSES

32. Zero-delay disk accesses Represent disk requests that can be satisfied without actually accessing the disk:  Data can be read from the I/O buffer  Data are written to the I/O buffer A process performing a zero-delay I/O request will:  Skip the disk  Start without further delay its next processing step

33. An example (I) … CPU 5 I/O 0 CPU 20 … RQRQ DQDQ Disk CC CC IQIQ Input Assume next step for red process is I/O 0

34. An example (II) … CPU 5 I/O 0 CPU 20 … RQRQ DQDQ Disk CC CC IQIQ Input Red process will immediately request a core (and get one because one or more cores were idle)

35. ENGINEERING THE SIMULATION

36. Simulating time Absolutely nothing happens to our model between two successive "events" Events are  Arrival of a new process  Start of a computing step  Completion of a computing step We associate an event routine with each event

37. Arrival event routine Process first request of process  It will always be a CPU request

38. CPU request routine current time is clock request time is crt if a core is free : mark core busy until clock + crt add crt to corebusytimes else : enter process in ready queue

39. CPU request completion routine if ready queue is empty : mark core idle else: pick first process P' in ready queue crt' is request time for P' mark core busy until clock + crt' add crt' to corebusytimes proceed with next request for completing process

40. Disk request routine current time is clock request time is drt if drt == 0: proceed with next process request if disk is free : mark disk busy until clock + drt add drt to diskbusytimes else : enter process request in disk queue

41. Disk request completion routine if disk queue is empty : mark disk idle else : pick first process request P' in disk queue drt' is request time for P' mark disk busy until clock + drt' add drt' to diskbusytimes proceed with next request for completing process

42. Input request routine current time is clock request time is irt if input device is free : mark input device busy until clock + irt else : enter process request in device queue

43. Input request completion routine if input queue is empty : mark input device idle else : pick first process request P' in input queue irt' is request time for P' mark disk busy until clock + irt' proceed with next request for completing process

44. The simulation scheduler 1.Find next event by looking at: CPU request completion times Disk request completion time time Input request completion time Arrival time of next process 2.Set current time to event time 3.Process event routine 4.Repeat until all processes are done

45. Organizing our program (I) Most steps of simulation involve scheduling future completion events Associate with each completion event an event notice  Time of event  Device  Process sequence number

46. Organizing our program (II) Do the same with process starts  Time of event  Process start  Process sequence number

47. Organizing our program (III) Process all event notices in chronological order Release disk 247 Release core 250 New process 245 New process 270 New process 310 First notice to be processed

48. Organizing our program (IV) Keep the event list sorted (priority queue) Release disk 247 Release core 250 New process 245 New process 270 New process 310 First notice to be processed is head of the list

49. Organizing our program (V) Overall organization of main program schedule first event # will be a process start while event list is not empty : process next event in list print simulation results

50. Organizing our event list Priority queue Two kinds of entries  Computational steps completion times: Created and inserted "on the fly"  Process arrivals: Created during input phase Already sorted

51. An implementation My main data structures would be:  Data table  Process table  Device table

52. The data table Stores the input data Line indices are used in process table OperationParameter NEW 5 CPU 10 INPUT 0 CPU 20 NEW 20 CPU 50 … …

53. The process table (I) Start Time First Line Last Line Current Line 503 varies 204…… …………

54. The process table (I) Start Time First Line Last Line Current Line 503 varies 204…… …………

55. The process table (II) One line per process  Line index is process sequence number! First column has start time of process First line, last line and current line respectively identify first line, last line and current line of the process in the input table Last column is for processes waiting for end of an INPUT step

56. The device table (I) ProcessCompletion time P015 -- -- First column is not needed: use row index to specify device

57. The device table (II) One line per device (core, disk and Input) Contains  A busy/free flag  A process sequence number if device is busy  A completion time Zero (or a very large value) if device is free

58. Finding the next event (I) If you use a priority list for your events, you should start by "seeding" your priority list with the arrival times of all processes After that, you add successive events to the list each time you can allocate a resource to a process. Your simulation will end once the list is empty (and you cannot add anything to it).

59. Finding the next event (II) If you do not use a priority list for your events, you can find the next event to process by searching the lowest value in  The process start times in the process table  Completion times of INPUT steps  The completion times in the device table

60. A full list implementation INT… 0 INT… 5 RUN 10 RUN 20 … … … Very elegant but harder to debug

61. Reading your input You must use I/O redirection  assign1 < input_file Advantages  Very flexible  Programmer writes her code as if it was read from standard input No need to mess with fopen(), argc and argcv