Performance and energy consumption optimization method of three-valued optical computer based on serial queuing system

Abstract

The invention relates to the technical field of three-valued optical computer optimization, in particular to a method for optimizing performance and energy consumption of a three-valued optical computer based on a serial queuing system. The method comprises the following steps of: researching the influence on the response time of a system and the influence on the starting number of STOCs caused by different equipartition numbers of optical processors under different loads; according to the method, how the number of STOCs affects the system performance under various load conditions can be obtained, so that a theoretical basis is provided for optimizing the system response time, then a performance and energy consumption optimization model is obtained by researching the relationship between the system energy consumption and the number of started STOCs, and then an optimal balance point between the performance and the energy consumption is obtained under different load conditions. The method can be widely applied to the technical field of three-valued optical computer optimization.

Description

Technical Field

[0001] The invention relates to the technical field of ternary optical computer optimization, and in particular to a method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system. Background Art

[0002] Queuing theory was proposed to solve the design problem of automatic telephone systems. After decades of research and exploration, queuing systems have been widely used in the field of ternary optical computers (TOC).

[0003] The TOC model of the queuing system is adopted, and its data processing includes the following four stages: task receiving stage, data preprocessing stage, data calculation stage, and data sending stage. In the third stage of the service, the optical processor can be divided into m small optical processors to further generate m STOCs (Small TOC, STOC) for users to use.

[0004] When scheduling tasks, the system will first check whether the scheduling queue is empty. If the queue is not empty, the system will detect whether there are any idle micro-optical processors available. Once an idle processor is found, the system will execute the task scheduling process and assign tasks to these idle small optical processors on a first-come, first-served basis to achieve efficient task allocation and execution.

[0005] When the processor is divided into m STOCs, the corresponding time of the operation stage can be calculated based on the M / M / c queuing system. In the TOC service platform, the unactivated STOCs are put into sleep mode to reduce power consumption. At this time, the energy consumption on the chip is mainly generated by the STOCs in the working state. By exploring the impact of different loads and different optical processors on the system response time, as well as the impact on the number of STOCs turned on, we can find out how the number of STOCs affects the system performance under various load conditions, thus providing a theoretical basis for optimizing the system response time. By studying the relationship between the system energy consumption and the number of STOCs turned on, we can get an optimization model for performance and energy consumption, and then get the best balance between performance and energy consumption under different load conditions. Summary of the invention

[0006] The purpose of the present invention is to provide a method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system to solve the problems raised in the above background technology.

[0007] A performance and energy consumption optimization method for a ternary optical computer based on a serial queuing system is proposed. The optical processor is divided into multiple small optical processors. By controlling the number of optical processors divided and the number of small optical processors turned on, the m / m / c model in the data calculation stage is optimized to achieve multi-objective optimization of response time and energy consumption under different load conditions.

[0008] Let h be the energy consumption per STOC per unit time, and ω be the energy consumption per unit time of each task in the system. Then the objective function can be set as Formula 1:

[0009] f(c n )=hc n +ωL s

[0010] where c n It is represented as the state when the optical processor is divided into n parts and c STOCs are turned on; c and n are discrete and must be λ / cμ n <1,L S For the captain; to find the best c n , marginal analysis can be used, even if the required c n * Formula 2 must be satisfied at the same time.

[0011]

[0012] Substituting formula 1 into formula 2, we have

[0013]

[0014] Simplified:

[0015]

[0016] According to the energy consumption h required per STOC per unit time and the energy consumption ω consumed per task per unit time in the system, the optimal number of optical processors and the number of STOCs enabled are obtained to achieve the best balance between energy consumption and performance.

[0017] As a preferred embodiment; response time T = T RS +T PPS +T CS +T SS ,

[0018] T i (i=RS,PPS,CS,SS) is the time required for each stage of the model; T RS Represents the time required to receive the task, T PPS Indicates the time required for data preprocessing, T CSis the total time spent on data calculation and decoding, and T SS It represents the time when the data is sent.

[0019] As a preference; the time required for task reception

[0020] Among them, the average number of requests in the system is N RS , the arrival rate is λ, and the service rate is subject to the parameter μ rs .

[0021] As a preference; data preprocessing time

[0022] Where W is the time the system takes to provide services for this task, W q is the time the task waits in the system, N PPS is the sum of the number of tasks being processed and the number of tasks being queued in the system. The time it takes for a user request to enter the second stage follows a negative exponential distribution with parameter λ. The service rate of PPS is μ PPS .

[0023] As a preference; the response time of the operation phase

[0024] L c is the average number of tasks in the system, and λ is the request arrival rate in this stage.

[0025] As a preference; response time during data transmission phase:

[0026] where μ ss is the average service rate in this stage, and λ is the request arrival rate in this stage. BRIEF DESCRIPTION OF THE DRAWINGS

[0027] Figure 1 This is a state transition diagram of the first stage of the embodiment;

[0028] Figure 2 This is a state transition diagram of the M / M / n queuing model in the data calculation phase of the embodiment;

[0029] Figure 3 is a response time diagram obtained by the embodiment under low load conditions;

[0030] Figure 4 is a response time diagram obtained by the embodiment under medium and low load conditions;

[0031] Figure 5 is a response time diagram obtained under medium load conditions in the embodiment;

[0032] Figure 6is a response time diagram obtained by the embodiment under medium and high load conditions;

[0033] Figure 7 is a response time diagram obtained by the embodiment under high load conditions;

[0034] Figure 8 An energy consumption diagram is drawn according to the relationship between the system energy consumption and the evenly divided quantity and the open quantity in the embodiment. DETAILED DESCRIPTION

[0035] The technical solutions in the embodiments of the present invention are described clearly and completely below. Obviously, the described embodiments are only part of the embodiments of the present invention, not all of them. Based on the embodiments of the present invention, all other embodiments obtained by ordinary technicians in this field without creative work are within the scope of protection of the present invention.

[0036] Example

[0037] The TOC service platform used in this experiment is extended from SD16, with a total of 1152 data bits. This platform can be divided into multiple small processors and generate multiple STOCs. Although the TOC processor has the advantage of low energy consumption, considering the continuous increase in the number of TOC processor bits in recent years, TOC has the potential to develop into a cloud server, data center or supercomputer, and it will face the challenge of massive data processing tasks. In this case, the energy consumption of TOC is bound to gradually increase and will bring a huge economic burden.

[0038] In order to further explore the possibility of improving performance while reducing energy consumption, it is planned to evenly divide the optical processor of this platform and generate n independent STOCs. In order to reduce energy consumption during the computing process while ensuring performance, this study uses dynamic voltage and frequency scaling technology (Dynamic Voltage and Frequency Scaling, DVFS), through which the operating frequency and voltage of the chip can be adjusted in real time to adapt to different workloads. Dynamically adjusting voltage and frequency can effectively reduce the energy consumption of the chip while providing the required performance.

[0039] In the TOC service platform, the unactivated STOC is put into sleep mode to reduce power consumption. At this time, the STOC in the working state except the awake state will generate some energy consumption, so we focus on this part of energy consumption.

[0040] Assume that the energy consumption generated by TOC during operation is E, E work 、E sleep are the power consumption when STOC is working and sleeping respectively, then E=cE work +(nc)Esleep ; Where c is the number of STOCs in working state. When SD16 is working, the power consumption of the backlight is about 150mA×5V=750mA. The TOC service platform is expanded from 6 SD16s, and the power consumption of the backlight is about 4.5W.

[0041] This embodiment uses the system response time (Time, T) as the performance indicator, where T is defined as the time required for the system to receive the calculation result after the user sends a request. The calculation formula of T is:

[0042] T=T RS +T PPS +T CS +T SS

[0043] T i (i=RS,PPS,CS,SS) is Figure 2 The time required for each stage of the model. Among them, T RS Represents the queue receiving time, T PPS Indicates the time required for data preprocessing, T CS is the total time spent on data calculation and decoding, and T SS It represents the time when the data is sent.

[0044] Task receiving phase T RS The calculation model is as follows:

[0045] Assume that the requests submitted by users are random and each request is independent of each other and will not affect each other. In this scenario, the M / M / 1 queue model can be used for modeling. At the same time, we assume that the user's requests occur at an arrival rate λ, that is, the arrival time follows a negative exponential distribution with parameter λ. Assume that the service rate of the receiving server RS ​​follows parameter μ rs According to the M / M / 1 queue model, the RS service in the first phase is modeled and the Continuous-Time Markov Chain (CTMC) state transition is plotted as follows: Figure 1 shown.

[0046] At this time, the busyness of the task receiving server can be expressed as:

[0047]

[0048] Assume that the probability of task arrival and the service rate of the receiving server remain stable and they both follow an exponential distribution. When the average service rate is higher than the average arrival rate, it indicates that the system has enough time to process requests. In this case, the system will be in a stable state. In other words, when λ / μ RS<1, the system will reach a stable state, and this process conforms to the Markov queue model. According to the state transition diagram of the M / M / 1 queue model, in a stable state, for any state n, "inflow" = "outflow", and the following equation can be listed:

[0049]

[0050] Where p0 is the probability that there is one task in the system, p2 is the probability that there are two tasks in the system, and the probability of the mth state is p m (m=0,1,2,…), we get Accumulate the probabilities of all possible situations in the task receiving system, and we get p0+p1+p2+…+p n =1, that is Assume that the average number of requests in the system is N RS , According to Little's formula, the average time to receive a task is:

[0051]

[0052] Data preprocessing stageT PPS The calculation model is as follows:

[0053] The time when the user request enters the second stage follows a negative exponential distribution with parameter λ. During this period, the main task of PPS is to convert the operation number into MSD number. The service rate of PPS is μ PPS , its response time T pps The modeling can refer to the first stage. In this queuing process, the average length of the tasks in the preprocessing queue is T pps , and T pps =W+W q , where W is the time the system takes to provide services for this task, W q N is the time the task waits in the system. PPS is the sum of the number of tasks being processed and the number of tasks being queued in the system, and

[0054]

[0055] Data calculation phase response time T cs The calculation model is as follows

[0056] When the processor is evenly divided into m STOCs, the modeling of the third stage can be carried out with reference to the M / M / c queuing system. Assume that each STOC is completely consistent in specifications, has the same software and hardware configuration, and has the same probability of being assigned tasks by the scheduler, such as Figure 2 Shown is the state transition diagram of the M / M / n queuing model.

[0057] The service intensity of the entire OP is μ, and the service intensity of STOC is When the arrival rate of computing requests in this stage is also λ, the average service rate of the system is The busyness of STOC is When c <1, the system is in a stable state. Then the probability p0 that there is no task in the system can be expressed as:

[0058]

[0059] The probability that there are m tasks in the system is P m , p m (m=0,1,2,…) can be expressed as:

[0060]

[0061] At this time, the queue length of the queue system is L q , The captain is L c , The task takes a total of T from the start to the end of the calculation. cs time, then the average number of tasks in the system is the captain L c for:

[0062]

[0063] The response time of the operation phase

[0064] Data sending phase T SS The calculation model is as follows

[0065] In the final stage of the service, the results calculated by the upper level are gathered here, packaged by the data sending server and sent to the user terminal. During this period, this process can be regarded as a simple M / M / 1 queuing system. The average service rate of SS (Sending Server) is μ ss According to the task receiving stage and the modeling of the preprocessing stage, the time consumed in the data sending stage can be obtained as:

[0066]

[0067] Lian Li T RS , T PPS , T CS , T SS The response time T can be obtained.

[0068] The impact of STOC average number and open number on response time

[0069] This paper explores the impact of different averaging numbers on system response time under different load conditions, with a special focus on the role of the number of STOCs turned on. By analyzing the relationship between these variables, this study hopes to reveal how the number of STOCs affects system performance under various load conditions, thereby providing a theoretical basis for optimizing system response time.

[0070] In order to analyze the system response time under different load conditions, this study set the task arrival rates under five conditions: low load, medium-low load, medium load, medium-high load, and high load, respectively, according to 20%, 40%, 60%, 80%, and 90% of the system service capacity. Under these settings, the response time of the system under different load conditions is measured and obtained to evaluate the performance of the system under different load levels.

[0071] Table 1 System response time at low load

[0072]

[0073] Table 2 System response time at low load

[0074]

[0075] When the load is in Table 3, the system response time

[0076]

[0077] Table 4 System response time at medium and high loads

[0078]

[0079] Table 5 System response time at high load

[0080]

[0081] like Figure 3-Figure 7 Based on the response time obtained under different load conditions in the experiment, a graph is drawn showing the relationship between the response time and the average number of shares and the number of open connections.

[0082] From the diagram, we can intuitively see that under the same equal distribution conditions, as the number of STOCs c increases, the system response time gradually decreases, but the trend of reduction gradually becomes flat. In the M / M / c queuing model, as the number of servers c increases, the system response time usually decreases accordingly. More servers can process a greater number of tasks in parallel, thereby reducing the waiting time of tasks in the queue. Therefore, increasing c can usually significantly improve the overall efficiency of the system, which is manifested as a shortened response time. However, this improvement does not grow linearly. When c reaches a certain critical point, the improvement in system performance will gradually decrease because the waiting time is close to its minimum value at this time.

[0083] In addition, there are several places in Figure (a), (b), and (c) where the response time is abnormally higher than the others. At this time, the system service rate is slightly higher than the task arrival rate. In the M / M / c queuing model, when the task arrival rate is close to the service rate, the system load is close to saturation. In this case, the server utilization rate is very high, and almost all servers are continuously processing tasks, resulting in the accumulation of tasks in the queue and the rapid increase of the queue length. New tasks need to wait longer before they can be processed, and the average waiting time of the system will also be significantly extended. Therefore, when the arrival rate is close to the service rate, the performance of the system will deteriorate significantly and the response time will be greatly extended.

[0084] In addition, there is an interesting phenomenon in the figure, that is, when the task arrival rate and the total system service rate are fixed, the more the number of equal shares is, the longer the system response time is. In this regard, a m / m / c queuing system is specially set up for experimental verification. Take the task arrival rate λ=20,μ=60,μ=cμ c ,

[0085] Table 6 Verification test

[0086]

[0087] When the number of service stations c increases, the chance of each task waiting for service decreases, so the average waiting time L q This is because more service desks mean more tasks can be processed at the same time, reducing the time tasks wait in the queue. However, due to the increase in the number of service desks, the overall number of busy service desks Increase. s is the total number of tasks in the system, equal to the number of tasks in the queue L q and the number of task stations being serviced, L busy Although L q As c increases, it decreases, but L busy The increase is even greater. Therefore, the total number of tasks in the system is L s It will increase as c increases.

[0088] The impact of STOC average number and open number on system energy consumption

[0089] like Figure 8 As shown, according to the relationship between system energy consumption and the average number and the number of open numbers, E = cE work +(nc)E sleep , draw an energy consumption diagram.

[0090] It can be clearly observed from the figure that the more TOCs are divided, the more detailed the control of power consumption is. When the system load is about 10%, the server divided into two STOCs must keep one STOC turned on. At this time, the power consumption is significantly higher than when it is divided into 10 STOCs, 1 is turned on, and 9 STOCs are dormant. When the system load is about 60%, the server divided into 2 STOCs needs to keep 2 STOCs turned on, while the server divided into 6 STOCs only needs to turn on 5 and put 1 to sleep. At this time, the power consumption is less than when it is divided into 2 STOCs. Therefore, the more equal divisions, the more conducive to fine control of power consumption.

[0091] Multi-objective optimization model for performance and energy consumption

[0092] When studying the relationship between the performance and energy consumption of the TOC service system, the phenomenon observed is that as the number of STOCs opened c increases, the response time will decrease accordingly, showing a performance enhancement phenomenon. This shows that as the number of processors in the system increases, the execution time of the task is relatively shortened, and the overall performance of the system is improved. At the same time, the total energy consumption will increase with the increase of c. As the number of processors increases, the energy consumption of the system also increases accordingly, resulting in an increase in total energy consumption.

[0093] Obviously, there is a trade-off between performance and energy consumption. Increasing the number of processors can improve the system's parallel processing capabilities and execution efficiency, thereby shortening the task completion time, but this will also lead to more energy consumption. Therefore, when designing and optimizing the system, it is necessary to comprehensively consider the trade-off between performance and energy consumption and find the optimal number of processors to meet specific performance requirements and energy constraints.

[0094] Considering response time and system energy consumption as different objective functions, improving the value of one objective function usually has a negative impact on the other objective function, so it is necessary to make trade-offs and decisions between multiple objectives. For example, reducing the total execution time by increasing the number of processors may lead to an increase in the total energy consumption of the system; while reducing energy consumption may extend the execution time of the task. Therefore, finding the optimal number of processors requires comprehensive consideration of these factors to achieve the best balance between performance and energy consumption.

[0095] When optimizing the TOC service model, for simplicity, we consider optimizing the third-stage m / m / c model. By controlling the number of processors shared and enabled, we can achieve multi-objective optimization of adaptation time and energy consumption.

[0096] Let h be the energy consumption per STOC per unit time, ω be the energy consumption per unit time of each task in the system, then the objective function can be set as

[0097] f(c n )=hc n +ωL s

[0098] where c n It is represented as the state when the optical processor is divided into n parts and c STOCs are turned on. c and n are discrete and must be λ / cμ n <1; To find the optimal c n , marginal analysis can be used, even if the required c n * Must satisfy both

[0099]

[0100] Substituting the objective function into the above formula, we have

[0101]

[0102] Simplified:

[0103]

[0104] Calculate the m / m / c model in the third stage and find the optimal c n * , which shortens the response time while maintaining the system energy consumption at a low level.

[0105] Table 7. Captains with different numbers of STOC enabled when the average number of points is 10 under low load.

[0106]

[0107] Under low load, when the number of equal shares is 10, each additional small optical processor will increase the power consumption by 350mW. The cost per unit time for each task to stay in the system is about 500mW. It falls within the interval (0.1341, 0.7150), so c n * =4.

[0108] Similarly, we can calculate the value of c under different loads and other equal distribution conditions.n What value of is optimal?

[0109] Table 8 Low load

[0110]

[0111] Table 9 Low and medium load

[0112]

[0113] Table 10 Load

[0114]

[0115] Table 11 Medium and high load

[0116]

[0117]

[0118] Table 12 High load

[0119]

[0120] In this embodiment, we deeply explore the M / M / c queuing model of the third stage, and optimize the number of processors enabled according to different load conditions and balancing strategies to achieve a balance between performance and energy consumption. Increasing the number of processors can significantly improve the parallel processing capability and execution efficiency of the system, thereby shortening the task completion time. However, this will also lead to an increase in energy consumption. Therefore, in the process of system design and optimization, the trade-off between performance and energy consumption must be comprehensively considered.

[0121] Through the analysis of the M / M / c queuing model, we determined how to reasonably configure the number of processors under different loads to achieve multi-objective optimization of the best response time and system energy consumption. Specifically, by experimenting with the equal distribution strategy under different load conditions, we obtained the optimal number of processors under various conditions. This process not only helps to improve the overall performance of the system, but also effectively controls energy consumption, thereby achieving the goals of high efficiency and energy saving in practical applications.

[0122] Based on the M / M / 1 and M / M / n models, this paper deeply analyzes the service response time of each stage of the ternary optical computer and the energy consumption characteristics of different components of the system. By establishing an optimization model between execution time and energy consumption, while ensuring the high performance of the system, it is committed to reducing its energy consumption and optimizing the system configuration, thereby achieving dual-objective optimization of performance and energy efficiency. Specifically, by reasonably adjusting the system resource configuration, it analyzes how to achieve the best balance between performance and energy efficiency under different load conditions.

[0123] In the research process, this paper not only carefully analyzed and calculated the various parameters of the model, but also verified the effectiveness of the optimization model through experiments. The final strategy can significantly reduce the system energy consumption under certain load conditions, while maintaining or even improving the overall performance of the system. In the future, the performance and energy consumption of TOC will be further studied based on more complex queuing models such as serial queuing models.

Claims

1. A method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system, characterized in that: The optical processor is divided into multiple small optical processors. By controlling the number of optical processors divided and the number of small optical processors turned on, the m / m / c model in the data calculation stage is optimized to achieve multi-objective optimization of response time and energy consumption under different load conditions. Let h be the energy consumption per STOC per unit time, and ω be the energy consumption per unit time of each task in the system. Then the objective function can be set as Formula 1: f(c n )=hc n +ωL s where c n It is represented as the state when the optical processor is divided into n parts and c STOCs are turned on; c and n are discrete and must be λ / cμ n <1,L S For the captain; to find the best c n , marginal analysis can be used, even if the required c n * Formula 2 must be satisfied at the same time. Substituting formula 1 into formula 2, we have Simplified:

2. The method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system according to claim 1, characterized in that: Response time T = T RS +T PPS +T CS +T SS , T i (i=RS,PPS,CS,SS) is the time required for each stage of the model; T RS Represents the time required to receive the task, T PPS Indicates the time required for data preprocessing, T CS is the total time spent on data calculation and decoding, and T SS It represents the time when the data is sent.

3. The method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system according to claim 2, characterized in that: Time required to receive the task During the task receiving phase, the average number of requests in the system is N RS , the arrival rate is λ, and the service rate is subject to the parameter μ rs .

4. The method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system according to claim 2, characterized in that: Data preprocessing time Where W is the time the system takes to provide services for this task, W q is the time the task waits in the system, N PPS is the sum of the number of tasks being processed and the number of tasks being queued in the system. The time it takes for a user request to enter the second stage follows a negative exponential distribution with parameter λ. The service rate of PPS is μ PPS .

5. The method for optimizing the performance and energy consumption of a ternary optical computer based on a serial queuing system as claimed in claim 2, characterized in that: Response time of operation stage L c is the average number of tasks in the system, and λ is the request arrival rate in this stage.

6. The method for optimizing the performance consumption of a ternary optical computer based on a serial queuing system according to claim 2, characterized in that: Response time during data sending phase: where μ ss is the average service rate in this stage, and λ is the request arrival rate in this stage.

Images