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s j 0 for each s > r and each 1 j . in .NET Access barcode 3 of 9 in .NET s j 0 for each s > r and each 1 j .




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s j 0 for each s > r and each 1 j . generate, create barcode 3 of 9 none for .net projects upc (6.15). Control Techniques for Complex Networks Draft copy April 22, 2007. We thus arriv e at the desired expression for W (x) in (i). The proof of part (ii) is similar to (i): Proposition 6.1.

3 gives, T (x) = max. 1 s v The ratio s , x /(os s ) is non-positive for s > r : (6.15) implies that s , x 0 and os s = . s . for such s. Hence the maximum can be restricted to s r as claimed. 6.

1.0.5 Workload and linear programs.

s, x , os s x R . + The minimal d .net vs 2010 Code 39 Extended raining time T (x) can be expressed as the value of a linear program analogous to (6.4).

Combining this observation with Proposition 6.1.4 we obtain an alternate representation of workload.

Consider the computation of the minimal time to reach one state from another. For any two states x1 , x2 X, a feasible allocation process z taking q from x1 to x2 in T seconds must satisfy q(T ) = x1 + Bz + T = x2 and T 1 z U. These constraints on z and T can be expressed as linear constraints, and we thereby obtain the LP, T (x1 , x2 ) = min s.

t. T x1 + Bz + T = x2 Cz 1T z 0. z 0, T 0.

(6.16). To construct a dual of (6.16) we express the constraints in matrix form, B C 1 z T = x1 x2 , 0. We de ne two 39 barcode for .NET dual variables, denoted R and R m . The vector is not sign + constrained since it corresponds to the equality constraint above, and is non-negative since it corresponds to the inequality constraint Cz 1T .

The dual of (6.16) takes the form T (x1 , x2 ) = max , x1 x2 s. t.

B T C T 0 T + 1T 1 0. (6.17).

The primal li near program (6.16) and its dual (6.17) have the same value, provided a bounded solution to (6.

16) exists (see Theorem 1.3.2.

) Equation (6.17) can be applied to compute W (x), which is de ned to be T (x1 , x2 ) with x1 = x, x2 = 0, and = 0. In this special case the linear program (6.

17) becomes, W (x) = max , x s. t. B T C T 0 1T 1 0.

(6.18). Control Techniques for Complex Networks Draft copy April 22, 2007. Hence, provid 3 of 9 barcode for .NET ed the origin is reachable from the initial condition x, the minimal draining time can be expressed W (x) = max , x , where the maximum is over all pairs ( , ) satisfying the constraints in (6.18).

This maximum can be taken over extreme points in the constraint region (6.18). Recall that a workload vector s de nes a face of V0 that does not pass through the origin.

Similarly, in Proposition 6.1.5 we see that ( , 0) is never an extreme point if is non-zero.

Proposition 6.1.5.

Suppose that the arrival-free model is stabilizable. Then, the nonzero extreme points of the constraint set in (6.18) are of the form ( , ) with , 1 = 1.

Proof. We rst establish that 1T = 0 or 1 if ( , ) is extremal. This follows from the form of the constraint region (6.

18): If 1T (0, 1) then (r , r ) will be feasible for r in a neighborhood of 1. Suppose now that 1T = 0. It follows that = 0 since 0, and the constraint on reduces to B T 0.

If = 0 then (r , 0) is feasible for each r > 0 when ( , 0) is feasible. This shows that ( , 0) cannot be extremal if = 0. All of these results point to a relationship between extreme points of (6.

18), and the workload vectors { s } de ned previously. The question we must answer is, If s is a workload vector, can it be extended to form a feasible solution to (6.18) Proposition 6.

1.6 provides an af rmative answer. Proposition 6.

1.6. If s is a workload vector, then there exists s R m such that + ( s , s ) is feasible for (6.

18), with s , 1 = 1. Proof. If s is a workload vector then s , v 1 for each v V0 , and this lower bound is attained at some v s .

From the de nition of V0 we have v s = B s for some s U. These observations can be expressed in a linear program, 1 = max B T s , s. t.

C 1, z 0 ,. where the ine quality s , v 1 has been replaced by the equivalent inequality B T s , = s , B 1. Letting R m denote a Lagrange multiplier for + the inequality constraint C 1, we obtain from Theorem 1.3.

2, 1 = min s. t. , 1 C T B T s 0.

. Any optimizer .net vs 2010 barcode 3/9 of this dual satis es the desired conclusions: 0, , 1 = 1, and C T + B T s 0. Hence we can take s := .

Proposition 6.1.6 is the basis of the construction of pooled resources in general network models.

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