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server; and (3) the transfer capacity of the spare server is all used in rebuild. The rst assumption excludes manual and hardware replacement schemes. The second assumption excludes the use of back-up devices to reload data into the spare server. The third assumption guarantees the transfer capacity available at the spare server. Note that the rebuild of unavailable units is performed either at the spare server, the remaining (N S 1) servers, or partially done at both. Therefore, we have the next lemma. Lemma 14.1. For any unavailable data unit rebuilt and stored into the spare server, we need at least (N S 1) transmissions from the remaining active servers. Proof. If the unavailable data unit is rebuilt by the spare server, then the (N S 1) data/redundant units of the same parity group will need to be sent to the spare server, resulting in (N S 1) transmissions. If the unavailable data unit is rebuilt by one of the remaining servers, then (N S 2) data/redundant units from servers other than the rebuild server and the spare server will need to be sent to the rebuild server, resulting in (N S 2) transmissions. After that, the rebuild server will send the rebuilt data unit to the spare server for storage, incurring one more transmission. Together the whole process thus generates (N S 1) transmissions. Each server contributes its idle transfer capacity to the rebuild process either for transmission, reception, or both. Let {0 1} be the proportion of transfer capacity each server has used for transmission in the rebuild process. The sum of the transmission capacities of the remaining servers is then equal to SS (1 )(N S 1) (14.12)

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Note that the transmitted data will have to be received. Therefore, the total reception capacities must be at least as large as the total transmission capacities, i.e., (1 )SS (1 )(N S 1) + SS SS (1 )(N S 1) (14.13)

ARRAY f[n] | -- declare a vector of sample values f[i] = ff[i]/n irule 0 | -- this DYNAMIC segment handles only sampled data t = 0 | TMAX = 1 a = (select the range a = X2 X1 of the X-sweep) b = (select the starting value b = X1 of X) NN = (select the number of estimated values) h = (select the Parzen-window width) alpha = 1/(2 * h^2) | beta = 1/(h * n * sqrt(2 * PI)) drun PARZEN

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where on the L.H.S. the rst term is the reception capacities of the remaining servers and the second term is the reception capacity of the spare server. Now Lemma 14.1 shows that the rebuild rate is proportional to the sum of transmission rates by the remaining servers. Therefore, to maximize the rebuild rate we need to maximize the transmission capacity of the system, subject to the constraint in equation (14.13). Noting the constraint 0 1 we can solve for by rearranging equation (14.13) to give 1 + (1 )(N S 1) , for (1 1/(N S 1)) 2(1 )(N S 1) = (14.14) 1, otherwise Next we substitute equation (14.14) into equation (14.12) and invoke Lemma 14.1 to obtain the maximum achievable data rebuild rate: SS [1 + (1 )(N S 1)] for (1 1/(N S 1)) 2(N S 1) (14.15) Rmax = S (1 ) otherwise S

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at the end of our experiment-protocol script, say of Figure 4-3 or 4-4 These script lines set parameter values and then execute an extra DYNAMIC program segment labeled PARZEN, which will produce probability-density estimates F(X) for NN values of x between x = X1 and X = X2 as t increases from t = 0 to t = TMAX = 1 Note that the values selected for NN, scale, and TMAX are different from the values used for the simulation itself The extra DYNAMIC program segment computes and averages the expression (4-18) to produce NN values of F(X) between X = t0 and X = t0 + TMAX:

which is exactly the same as equation (14.11). We state the result in the following theorem: Theorem 14.2. The mixed distributed-baseline rebuild scheme achieves the optimal rebuild rate and hence requires the minimum amount of rebuild time. Proof. The result follows directly from equations (14.11) and (14.15).

The previous sections focus on deriving the rebuild rate and time. It is clear that the rebuild time increases with server utilization . To control the rebuild time, the server could limit its utilization to reserve transfer capacities for the rebuild process. First, by setting = 0 in equation (14.15) we can obtain the minimum achievable rebuild time: Tmin = U 2(N S 1)U = Rmax | =0 N S SS (14.16)

label PARZEN -X = a * t + b | -this sweeps X from X1 to X2 as t increases -compute n samples ff[i] = f[i]/n Vector f = beta * exp(- alpha * (X - x)^2)) DOT F = f * 1 | -- sum to average (note that 1/n is included in beta) dispxy X, F

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