12.4 Balanced Systems

The MVA iteration starts once the customer distribution among the devices is known. That is, knowing how n 1 customers are distributed among the devices, (i.e., the graphics/327fig02.gif(n 1)'s), the performance measures when there are n customers in the system follow directly, as seen from the MVA algorithm given in Table 12.5.

Now consider a balanced system. A system is considered to be balanced if a typical customer places the same average load (i.e., demand D) on each of the devices. This implies that all devices are equally utilized. A balanced system is not one where all devices are the same speed, only that the faster devices are either visited more often or the demand per visit to them is higher. A balanced system implies that there is no single bottleneck in the system, where improvements to that one device (i.e., the bottleneck device) would have a greater positive impact on overall performance than improvements made to any other device. In some sense, balanced systems are the ideal, with no particular device or resource single-handedly limiting performance. Balanced systems are important to consider, since they provide an upper bound on performance, a gold standard toward which to aspire.

For example, reconsider the database server example. From Tables 12.1 and 12.4 (and indeed for any number of customers in the network), the slow disk has the highest utilization. It is the system bottleneck and the system is not balanced. Because the slow disk is over-utilized compared to the other devices, one way to improve performance would be to move some of the files from the slow disk to the fast disk. This has the effect of reducing the load (and utilization) of the slow disk and increasing the load (and utilization) of the fast disk. (Another way to achieve balance between the two disks would be to replace the slower disk by another fast disk. This would further improve performance, but would be more expensive than simply moving files between the disks.)

By moving disk files so that the fast disk is visited twice as often as the slow disk, the overall system becomes balanced. Now consider this balanced system with ten customers in the system, as shown in Figure 12.4. Notice now that all device demands (i.e., the D's) are equal. The normal way to solve this system would be to run ten iterations of MVA, one for every customer level. However, because the system is balanced, only one MVA step is required.

Figure 12.4. Balanced database server example (10 customers).


To see this, recall that the only thing necessary (i.e., the iteration basis) for finding the performance measures for ten customers is for MVA to know the average number of customers at each device when there are only nine customers in the system (i.e., ñi(9) for each device i). Since the system is balanced, the nine customers are equally distributed among the devices with three customers being at each of the three devices. Knowing that ñi(9) = 3 for each i, from the MVA algorithm in Table 12.5, it follows quickly that the average residence time for each device i (i.e., graphics/326fig01.gif) is:


The overall system response time (i.e., graphics/326fig02.gif) is:


The overall system throughput (i.e., graphics/325fig02.gif) is:


The throughput for each device i (i.e., Xi(n) = Vi x X0(n)) is:


The utilization for each device i (i.e., Ui(n) = Si x Xi(n)) is:


And, finally, the average number of customers at each device i (i.e., graphics/327fig01.gif) is:


In balanced systems, all the device demands (i.e., the Di's) are equivalent. Let this common device demand be D. Therefore, finding the overall system response time can be simplified to:


and overall system throughput is simply:


As a verification in the balanced database server example, where n = 10, D = 10, and K = 3, the overall system response time is R0(10) = 10(3 + 10 1) = 120 seconds and the overall system throughput is X0(10) = 10/120 = 0.0833 customers per second. Thus, performance metrics for balanced systems, which follow directly from the MVA formulae, are extremely easy to compute. Overall response time requires one addition, one subtraction, and one multiplication. Overall system throughput requires one extra division.

Performance by Design. Computer Capacity Planning by Example
Performance by Design: Computer Capacity Planning By Example
ISBN: 0130906735
EAN: 2147483647
Year: 2003
Pages: 166

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