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Figure 17. Remainder queue size of CM; for 8000 tasks. Figure 16. Remainder queue size of CM; for 7000 tasks.

Figure 17 Remainder queue size of CM; for 8000 tasks. Figure 16. Remainder queue size of CM; for 7000 tasks.

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Figure 1. The infrastructure of cloud computing. Therefore, cloud computing is the most important trend currently. The service providers can supply large computing resources, development platforms, and services, as shown in Figure 1. In various fields, the type of cloud computing can be sorted into three categories according to its applications:
Figure 1. The infrastructure of cloud computing. Therefore, cloud computing is the most important trend currently. The service providers can supply large computing resources, development platforms, and services, as shown in Figure 1. In various fields, the type of cloud computing can be sorted into three categories according to its applications:
Figure 2. The example of hierarchical virtualization topology.
Figure 2. The example of hierarchical virtualization topology.
Figure 3. Amazon Elastic Compute Cloud.
Figure 3. Amazon Elastic Compute Cloud.
Figure 4. Reference Queue based Cloud Service Architecture (RQCSA).
Figure 4. Reference Queue based Cloud Service Architecture (RQCSA).
Table 1. The FSQSM parameters and definitions. In Figure 6, the following four rounds have demonstrated the task execution and scheduling process caused by new coming tasks. The first two rounds describe the distribution mechanism of the FSQSM and show the example of RQ searching when the CM is fully loaded. The third and fourth rounds illustrate how to select the RQ by rotation when the RQs have the same remaining amount. Moreover, this example not only assumes that the execution time of each task is randomly generated and marked on the task but also defines the unit time as the CSC execution time in each round. In Round 1, the CM3 queue is fully loaded. The CSC1 and CSC3 are still executing the previous tasks and needs one-unit time. Now, a new coming task enters a fully loaded CM3, so the RQ mechanism is started and searches for available nodes. In the case of Round 1, the task will be assigned to CSC) and perform services
Table 1. The FSQSM parameters and definitions. In Figure 6, the following four rounds have demonstrated the task execution and scheduling process caused by new coming tasks. The first two rounds describe the distribution mechanism of the FSQSM and show the example of RQ searching when the CM is fully loaded. The third and fourth rounds illustrate how to select the RQ by rotation when the RQs have the same remaining amount. Moreover, this example not only assumes that the execution time of each task is randomly generated and marked on the task but also defines the unit time as the CSC execution time in each round. In Round 1, the CM3 queue is fully loaded. The CSC1 and CSC3 are still executing the previous tasks and needs one-unit time. Now, a new coming task enters a fully loaded CM3, so the RQ mechanism is started and searches for available nodes. In the case of Round 1, the task will be assigned to CSC) and perform services
Figure 5. The Fitness Service Queue Selection Mechanism (FSQSM) flow chart.
Figure 5. The Fitness Service Queue Selection Mechanism (FSQSM) flow chart.
there are three new coming tasks are sent to the CSC2. Therefore, by checking the RQ, the two RQ with the same remaining amount will be assigned for the tasks based on the rotation defined by the algorithm. As a result, the task which requires a five-unit time to be finished will be sent to the CM). In Round 4, the CSC2 already finished the prior task, and the CI distributes a new coming task to the CMB. In this case, because the CSC1, CSC2, and CSC3 are all in idle, the CM;, CM), and CM3 will distribute tasks to the CSC1, CSC2, and CSC3, respectively.
there are three new coming tasks are sent to the CSC2. Therefore, by checking the RQ, the two RQ with the same remaining amount will be assigned for the tasks based on the rotation defined by the algorithm. As a result, the task which requires a five-unit time to be finished will be sent to the CM). In Round 4, the CSC2 already finished the prior task, and the CI distributes a new coming task to the CMB. In this case, because the CSC1, CSC2, and CSC3 are all in idle, the CM;, CM), and CM3 will distribute tasks to the CSC1, CSC2, and CSC3, respectively.
Table 2. The specifications of the experimental host and analysis tools. In the experimental environment, there will be 16 CMs to manage all the service nodes in the cloud. And the queue size of each CM will be 1024 slots. Next, three different task amounts will be assigned: Low (<3000 tasks), Medium (4000-7000 tasks), and High (>8000 tasks), and the experiment will perform 1000 simulations on each amount level, respectively. In the trail environment, all execution time is randomly assigned and lost tasks will be distributed again. waiting time, and makespan, will be executed. Moreover, this experiment even uses the Mean Average Deviation (MAD) to calculate the distributed task numbers and verify if the FSQSM distributes tasks to the CM evenly to avoid the overload of a single CM in which the overall efficiency will be lower [24-26]. The experimental host and analysis tools are listed in Table 2.
Table 2. The specifications of the experimental host and analysis tools. In the experimental environment, there will be 16 CMs to manage all the service nodes in the cloud. And the queue size of each CM will be 1024 slots. Next, three different task amounts will be assigned: Low (<3000 tasks), Medium (4000-7000 tasks), and High (>8000 tasks), and the experiment will perform 1000 simulations on each amount level, respectively. In the trail environment, all execution time is randomly assigned and lost tasks will be distributed again. waiting time, and makespan, will be executed. Moreover, this experiment even uses the Mean Average Deviation (MAD) to calculate the distributed task numbers and verify if the FSQSM distributes tasks to the CM evenly to avoid the overload of a single CM in which the overall efficiency will be lower [24-26]. The experimental host and analysis tools are listed in Table 2.
Figure 7. Queue waiting time, Cluster Manager (CM) = 16, queue size = 1024. The first discussion will be on the Random and Round-Robin algorithms. When the task numbers gradually increase starting from 1000, the queue waiting time of those two algorithms is much longer than the FSQSM. The main reason for the long waiting time is that those two algorithms do not consider the remaining amount of CM queue, which frequent task reassignments and reduce the performance of the entire system. Moreover, the tasks are distributed by randomly selecting the CMs
Figure 7. Queue waiting time, Cluster Manager (CM) = 16, queue size = 1024. The first discussion will be on the Random and Round-Robin algorithms. When the task numbers gradually increase starting from 1000, the queue waiting time of those two algorithms is much longer than the FSQSM. The main reason for the long waiting time is that those two algorithms do not consider the remaining amount of CM queue, which frequent task reassignments and reduce the performance of the entire system. Moreover, the tasks are distributed by randomly selecting the CMs
PRR EN RE Ey, CREE NAAEEE DA, eae Niee MMAR RR EE CERNE MEE Ge However, when testing the makespan, the experiment is performed based on the number of tasks between 2000 to 10,000 in order to obtain a more accurate analysis of the pros and cons of the three algorithms, as shown in Figure 8. The Random algorithm has a longer makespan because its random selecting method may select an RQ with no remaining amount when distributing tasks, in which the tasks may be missing and lead to a task re-distribution. Under the Round-Robin method, although it selects an RQ by rotation, it still does not take the RQ’s current remaining amount into account. Therefore, the makespan of the Round-Robin is still longer than the FSQSM. The proposed FSQSM considers the remaining amount of all RQs and selects the RQ with the most amount to distribute tasks which makes the makespan much shorter. The goal of testing the number of tasks is to verify the number of executed/completed services of the proposed FSQSM, Random and Round-Robin in a specific period. As the simulating time increases, the Random and Round-Robin serves fewer tasks than the FSQSM, and the reason for this is that the first two algorithms do not take the current remaining amount of RQ into account when distributing tasks and the system will be slowed down because the selected RQ may be fully loaded and causes task re-distributions. On the other hand, the FSQSM takes the remaining amount of RQ to avoid the drawback. There are some inconsistent results that occur in Figure 9, and the reason for this is that the completion time of each task is randomly defined. That means that if the FSQSM happens to serve tasks with shorter completion times, it will be able to handle more tasks, otherwise, it handles fewer tasks.
PRR EN RE Ey, CREE NAAEEE DA, eae Niee MMAR RR EE CERNE MEE Ge However, when testing the makespan, the experiment is performed based on the number of tasks between 2000 to 10,000 in order to obtain a more accurate analysis of the pros and cons of the three algorithms, as shown in Figure 8. The Random algorithm has a longer makespan because its random selecting method may select an RQ with no remaining amount when distributing tasks, in which the tasks may be missing and lead to a task re-distribution. Under the Round-Robin method, although it selects an RQ by rotation, it still does not take the RQ’s current remaining amount into account. Therefore, the makespan of the Round-Robin is still longer than the FSQSM. The proposed FSQSM considers the remaining amount of all RQs and selects the RQ with the most amount to distribute tasks which makes the makespan much shorter. The goal of testing the number of tasks is to verify the number of executed/completed services of the proposed FSQSM, Random and Round-Robin in a specific period. As the simulating time increases, the Random and Round-Robin serves fewer tasks than the FSQSM, and the reason for this is that the first two algorithms do not take the current remaining amount of RQ into account when distributing tasks and the system will be slowed down because the selected RQ may be fully loaded and causes task re-distributions. On the other hand, the FSQSM takes the remaining amount of RQ to avoid the drawback. There are some inconsistent results that occur in Figure 9, and the reason for this is that the completion time of each task is randomly defined. That means that if the FSQSM happens to serve tasks with shorter completion times, it will be able to handle more tasks, otherwise, it handles fewer tasks.
Figure 9. Number of tasks, CM = 16, queue size = 1024. remaining queue size of the CM; under each of the three algorithms will be calculated in low, medium, and high task-amount levels, respectively. The parameters and calculation of the MAD are shown in Table 3 and Equation (1). Subsequently, the results of the experiment are shown in Figures 10-19 and are used to calculate the MAD values of each of the three algorithms. With the MAD values, the distributions of the three algorithms can be verified.
Figure 9. Number of tasks, CM = 16, queue size = 1024. remaining queue size of the CM; under each of the three algorithms will be calculated in low, medium, and high task-amount levels, respectively. The parameters and calculation of the MAD are shown in Table 3 and Equation (1). Subsequently, the results of the experiment are shown in Figures 10-19 and are used to calculate the MAD values of each of the three algorithms. With the MAD values, the distributions of the three algorithms can be verified.
Table 3. The parameters of the Mean Average Deviation (MAD). Figure 10. Remainder queue size of CM; for 1000 tasks.
Table 3. The parameters of the Mean Average Deviation (MAD). Figure 10. Remainder queue size of CM; for 1000 tasks.
Figure 11. Remainder queue size of CM; for 2000 tasks.
Figure 11. Remainder queue size of CM; for 2000 tasks.
Figure 13. Remainder queue size of CM; for 4000 tasks. Figure 12. Remainder queue size of CM; for 3000 tasks.
Figure 13. Remainder queue size of CM; for 4000 tasks. Figure 12. Remainder queue size of CM; for 3000 tasks.
Figure 14. Remainder queue size of CM; for 5000 tasks.
Figure 14. Remainder queue size of CM; for 5000 tasks.
Figure 15. Remainder queue size of CM; for 6000 tasks.
Figure 15. Remainder queue size of CM; for 6000 tasks.
Figure 18. Remainder queue size of CM; for 9000 tasks.
Figure 18. Remainder queue size of CM; for 9000 tasks.
Figure 20. Mean Average Deviation (MAD) for all tasks
Figure 20. Mean Average Deviation (MAD) for all tasks
Figure 19. Remainder queue size of CM; for 10,000 tasks.
Figure 19. Remainder queue size of CM; for 10,000 tasks.
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