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Disks and storage devices, similarly, may not be associated with a particular server. The data, and possibly the application code itself, can move between servers. The unit that performs a particular service is no longer a single computer, but rather the pair or cluster on whose computers the application may be running at a particular moment. It should not matter which computer runs the service, as long as the service runs. Consider that the cluster acts as a black box providing a particular service. Your users connect to a network address and exchange data with that address. Assuming that the speed at which the responses return is adequate, it doesn t matter to the user what is providing the service. Theoretically, the server could be a Unix computer, a Windows computer, a mainframe computer, or even a fast-typing monkey with a keyboard. As long as the service is being provided and the responses are accurate, the users should, at least in theory, be happy. There are technical reasons why a particular server is used, and why you d never mix server types, but purely from a user s perspective, it truly should not matter which type of system is doing the work.



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Assuming that the relevant time is not too large, we might approximate the conditional mean bandwidth using a Taylor-series approximation E B t jB 0 x x tr x ; 17:49

17:50





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A failover pair or cluster requires more than simply having two servers placed near each other. Let s examine the various components that are required to turn the two servers into a failover pair. (We elaborate on each of these required components in the next section, and we discuss larger and more complex combinations of servers later on.) A simple cluster is depicted in Figure 15.2. The components that are required to build a cluster are the following: Servers. You need two servers a primary server and a takeover server. (We refer to the takeover server as the secondary server in some cases, but it s the same idea.) A critical application that once ran but has now failed on the primary server migrates to the takeover server in a process called failover. The servers should be running the same operating system version, have the same patches installed, support the same binary executables, and as much as possible, be configured identically. Of course, clusters can grow larger than two nodes, but the basic premise is the same; an application moves from one node to another. Whether a cluster has two nodes or 2,000 nodes, the application still migrates in the same basic manner. Network connections. There are two different types of network connections that are required to implement failover, and a third that is strongly recommended. A pair (for redundancy) of dedicated heartbeat networks, which run between all the servers in a cluster, is a basic requirement.

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which has the advantage that no numerical inversion is required. Suppose the service capacity is c > E B 0 and we condition on B 0 > c. If r B 0 < 0, we can use Eq. (17.49) to approximate the rst time to return to c, the recovery time, by t x c =r x : 17:51

15

Suppose in addition that B is reversible; this will happen if the matrix P is reversible. Then since both the residual lifetime and the current age have distribution Fjke , E B t jB 0 E B t jB 0 . Consequently, B 0 x is a local maximum, at t 0, of E B t jB 0 . Now suppose that there are n independent sources. Then as in Duf eld and Whitt [14], it follows by use of an appropriate functional law of large numbers that, as n 3 I under regularity conditions, the stochastic paths of the B process converge to this mean path. Thus we can identify, asymptotically as n 3 I, t 0 as a hitting time for the level x. Thus, we can use Eq. (17.42) to approximate the probability of this hitting time and t in Eq. (17.51) to approximate the associated recovery time. Example 17.8.1. Consider homogeneous two-level sources, that is, j P f1; 2g, 0 b1 < b2 , p1 p2 1 with mean lifetimes m1 , m2 , and P11 P22 1 P12 1 P21 0. With n sources and B 0 x we can calculate the pj in Eq. (17.46) directly from the relation x n 1 b1 p2 b2 for x P fnb1, p n b b1 b2 ; . . . ; b1 n 1 b2 ; nb2 g. Then r x n 1 b2 b1 =m1 p2 b1 b2 =m2 p nb2 x =m1 x nb1 =m2 : 17:52

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