Robust transient analysis of multi-server queueing systems and feed-forward networks

Abstract

We propose an analytically tractable approach for studying the transient behavior of multi-server queueing systems and feed-forward networks. We model the queueing primitives via polyhedral uncertainty sets inspired by the limit laws of probability. These uncertainty sets are characterized by variability parameters that control the degree of conservatism of the model. Assuming the inter-arrival and service times belong to such uncertainty sets, we obtain closed-form expressions for the worst case transient system time in multi-server queues and feed-forward networks with deterministic routing. These analytic formulas offer rich qualitative insights on the dependence of the system times as a function of the variability parameters and the fundamental quantities in the queueing system. To approximate the average behavior, we treat the variability parameters as random variables and infer their density by using ideas from queues in heavy traffic under reflected Brownian motion. We then average the worst case values obtained with respect to the variability parameters. Our averaging approach yields approximations that match the diffusion approximations for a single queue with light-tailed primitives and allows us to extend the framework to heavy-tailed feed-forward networks. Our methodology achieves significant computational tractability and provides accurate approximations for the expected system time relative to simulated values.

Appendix: All proofs

Proof of Theorem 2

Since $\left( \nu -k+1\right) ^{1/\alpha } \le \left( \nu -k\right) ^{1/\alpha } + 1$, and given $\varGamma _m^{+}\ge 0$, we bound Eq. (23) by

$$\begin{aligned} \widehat{S}_{n}\le & {} \max _{0\le k \le \nu }\,\,\left\{ \frac{\nu -k}{\mu } + \varGamma _{m}^{+}\left( \nu -k\right) ^{1/\alpha } - \frac{m(\nu -k)}{\lambda }\right. \\&\left. +\, \varGamma _a\left[ m\left( \nu -k\right) \right] ^{1/\alpha } \right\} + \left( \frac{1}{\mu } + \varGamma _m^{+}\right) . \end{aligned}$$

By making the transformation $x=\nu -k$, where $x\in \mathbb {N}$, we can represent this problem as

$$\begin{aligned} \max _{0 \le x \le \nu , x\in \mathbb {N}} \left( \beta \cdot x^{1/\alpha } -\delta \cdot x \right) ~ \le ~ \max _{0 \le x \le \nu , x\in \mathbb {R}} \left( \beta \cdot x^{1/\alpha } -\delta \cdot x\right) , \end{aligned}$$

(82)

where $\beta = m^{1/\alpha }\varGamma _{a} + \varGamma _{m}^{+}$ and $\delta = m(1-\rho )/\lambda >0$, given $\rho <1$. If $\beta \le 0$, the function $h(x)= \beta \cdot x^{1/\alpha } -\delta \cdot x \le 0$ for all values of x, implying $\widehat{S}_{n} = 1/\mu +\varGamma _m^{+}$. For $\beta >0$, the function h is concave in x with an unconstrained maximizer

$$\begin{aligned} x^{*} = \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)} = \left( \frac{\lambda (\varGamma _m + m^{1/\alpha } \varGamma _a)}{\alpha m (1-\rho )}\right) ^{\alpha /(\alpha -1)}. \end{aligned}$$

(83)

Maximizing the function $h(\cdot )$ over the interval $[0, \nu ]$ involves a constrained one-dimensional concave maximization problem whose solution gives rise to closed-form solutions.

(a):

If $x^{*} \in [0, \nu ]$, then $x^{*}$ is the maximizer of the function h over the interval $[0, \nu ]$, leading to an expression that is independent of $\nu$:

$$\begin{aligned} \widehat{S}_{n}\le & {} \beta \left( \frac{\beta }{\alpha \delta }\right) ^{1/(\alpha -1)} - \delta \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)} + \left( \frac{1}{\mu } + \varGamma _m^{+}\right) \nonumber \\= & {} \frac{(\alpha -1)}{\alpha ^{\alpha /(\alpha -1)}}\cdot \frac{\beta ^{\alpha /(\alpha -1)}}{\delta ^{1/(\alpha -1)}}+ \left( \frac{1}{\mu } + \varGamma _m^{+}\right) . \end{aligned}$$

(84)

(b):

If $x^{*} > \nu$, the function h is nondecreasing over the interval $[0,\nu ]$, with $h(\nu ) \ge h(x)$ for all $x\in [0,\nu ]$, leading to an expression that is dependent on $\nu$:

$$\begin{aligned} \widehat{S}_{n} = \beta (\nu )^{1/\alpha } - \delta (\nu ) + \left( \frac{1}{\mu } + \varGamma _m^{+}\right) . \end{aligned}$$

(85)

We obtain Eq. (24) by substituting for $\beta$ and $\delta$ expressions in parts (a) and (b). $\square$

Proof of Theorem 3

To bound the maximization problem in Eq. (32), we take a similar approach to that presented in the proof of Theorem 2 and cast the problem in the form

$$\begin{aligned}&\max _{0 \le x \le \nu -\phi , x\in \mathbb {R}} \left( \beta \cdot x^{1/\alpha } -\delta \cdot x\right) \\&\quad = \left\{ \begin{array}{ll} \beta \cdot (\nu -\phi )^{1/\alpha } - \delta \cdot (\nu -\phi ), &{}\quad \text {if}\;\nu -\phi \le \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)}{,}\\ {\displaystyle \frac{(\alpha -1)}{\alpha ^{\alpha /(\alpha -1)}}\cdot \frac{\beta ^{\alpha /(\alpha -1)}}{\delta ^{1/(\alpha -1)}} },&{} \quad \text {otherwise,} \end{array} \right. \end{aligned}$$

where $\beta =m^{1/\alpha }\varGamma _a + \varGamma _m^{+}$ and $\delta = m(1-\rho )/\lambda$. Substituting for the terms $\beta$ and $\phi$ their respective values in the above expression yields the desired result. $\square$

Proof of Theorem 5

From Eq. (50), we have that the worst case system time is given by

$$\begin{aligned} \widehat{S}_{n}= & {} \frac{J}{\mu } \\&+ \max _{0\le k_1\le \cdots \le k_{J} \le \nu }\left\{ \begin{array}{l} \left[ \varGamma _m^{(1)+}\left( k_2-k_1+1\right) ^{1/\alpha } + \cdots + \varGamma _m^{(J)+}\left( \nu -k_J+1\right) ^{1/\alpha } \right] \\ \quad +\,{\displaystyle \varGamma _a\left[ m\left( \nu -k_1\right) \right] ^{1/\alpha } - \frac{m(1-\rho )}{\lambda }\left( \nu -k_1\right) } \end{array}\right\} . \end{aligned}$$

Furthermore, since $\left( k_{j+1}-k_j+1\right) ^{1/\alpha } \le \left( k_{j+1}-k_j\right) ^{1/\alpha }+1$, for all $\hbox {j}=1,\ldots , \,{J}$, we obtain

$$\begin{aligned} \widehat{S}_{n}\le & {} \frac{J}{\mu } + \sum _{j=1}^J \varGamma _m^{(j)+}\\&+ \,\max _{0\le k_1\le \cdots \le k_{J} \le \nu } \left\{ \begin{array}{l} \left[ \varGamma _m^{(1)+}\left( k_2-k_1\right) ^{1/\alpha } + \cdots + \varGamma _m^{(J)+}\left( \nu -k_J\right) ^{1/\alpha } \right] \\ \quad +\,{\displaystyle \varGamma _a\left[ m\left( \nu -k_1\right) \right] ^{1/\alpha } - \frac{m(1-\rho )}{\lambda }\left( \nu -k_1\right) } \end{array}\right\} . \end{aligned}$$

We will isolate the problem of maximizing $\left[ \varGamma _m^{(1)+}\left( k_2-k_1\right) ^{1/\alpha } + \cdots + \varGamma _m^{(J)+}\left( \nu -k_J\right) ^{1/\alpha } \right]$ for fixed values of $k_1,\nu ,$ and make the transformations $x_1=k_2-k_1, \ldots , x_J=\nu -k_J$, where $x_{j}\in \mathbb {N}$, for all $j=1, \ldots , J$. With these transformations, the optimization problem simplifies to

$$\begin{aligned}&\max _{0\le k_1 \le \nu , k_1\in \mathbb {N}} \left( m^{1/\alpha }\varGamma _a \left( \nu -k_1\right) ^{1/\alpha } - \frac{m(1-\rho )}{\lambda }\left( \nu -k_1\right) \right. \nonumber \\&\quad \left. + \,\left\{ \begin{array}{ll} \max &{} ~~ {\displaystyle \left[ \varGamma _m^{(1)+}x_1^{1/\alpha } + \cdots + \varGamma _m^{(J)+}x_J^{1/\alpha }\right] }\\ \text {s.t.} &{} ~~ x_1+\cdots +x_J = \nu -k_1 \\ &{} ~~x_{j}\in \mathbb {N}, \forall j=2,\ldots ,J \end{array} \right\} \right) . \end{aligned}$$

(86)

The optimal solution to the inner optimization problem satisfies

$$\begin{aligned} \varGamma _m^{(1)+}(x_1^{*})^{1/(\alpha -1)} = \varGamma _m^{(2)+}(x_2^{*})^{1/(\alpha -1)} = \cdots = \varGamma _m^{(J)+}(x_J^{*})^{1/(\alpha -1)}, \end{aligned}$$

by the first-order optimality conditions. Using the additional condition that $\sum _{j=1}^J x_j^{*} = {\nu -k_1},$ the optimal solution can be found analytically as

$$\begin{aligned} x_i^{*} = \frac{(\varGamma _m^{(i)+})^{\alpha /(\alpha -1)}}{ \displaystyle \sum \nolimits _{j=1}^J (\varGamma _m^{(j)+})^{\alpha /(\alpha -1)}}\cdot (\nu -k_1), \quad \forall i = 1,2,\ldots ,J, \end{aligned}$$

leading to an optimal value of

$$\begin{aligned} \varGamma _m^{(1)+}(x_1^{*})^{1/\alpha } + \cdots + \varGamma _m^{(J)+}(x_1^{*})^{1/\alpha } = {(\nu -k_1)^{1/\alpha }\cdot }\left( \sum _{j=1}^J (\varGamma _m^{(j)+})^{\alpha /(\alpha -1)}\right) ^{(\alpha -1)/\alpha }.\nonumber \\ \end{aligned}$$

(87)

Substituting the optimal solution of the inner problem in Eq. (86), the performance analysis reduces to solving the following one-dimensional optimization problem:

$$\begin{aligned} \max _{0\le k_1 \le \nu }~\left\{ \left( m^{1/\alpha }\varGamma _a + \left[ \sum _{j=1}^J (\varGamma _m^{(j)+})^{\alpha /(\alpha -1)}\right] ^{(\alpha -1)/\alpha }\right) \cdot \left( \nu -k_1\right) ^{1/\alpha } -\, \frac{m(1-\rho )}{\lambda }\left( \nu -k_1\right) \right\} ,\nonumber \\ \end{aligned}$$

(88)

which can be cast in the form of the optimization problem in Eq. (82), with

$$\begin{aligned} \beta = m^{1/\alpha }\varGamma _a + \left( \sum _{j=1}^J (\varGamma _m^{(j)+})^{\alpha /(\alpha -1)}\right) ^{(\alpha -1)/\alpha } ~~ \text {and} ~~ \delta = \frac{m(1-\rho )}{\lambda }. \end{aligned}$$

Referring to the proof of Theorem 2, the solution to Eq. (88) is

$$\begin{aligned} {\displaystyle \max _{0\le x \le \nu }~~\beta \cdot x^{1/\alpha } - \delta \cdot x} = \left\{ \begin{array}{ll} \beta \cdot \nu ^{1/\alpha } - \delta \cdot \nu , &{} \quad \text {if}\; \nu \le \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)},\\ {\displaystyle \frac{(\alpha -1)}{\alpha ^{\alpha /(\alpha -1)}}\cdot \frac{\beta ^{\alpha /(\alpha -1)}}{\delta ^{1/(\alpha -1)}} }, &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

We obtain the desired result by substituting for $\beta$ and $\delta$ their respective values. $\square$

Proof of Theorem 6

We maximize both terms in Eq. (54) separately as follows:

(a):

By Assumption 1, and applying similar arguments to those presented in the proof of Theorem 5, the first term in Eq. (54) is bounded by

$$\begin{aligned}&\max _{\begin{array}{c} 0\le k_1 \le \phi ,\\ k_1\in \mathbb {N} \end{array}} \left( {\frac{\nu -k_1}{\mu }} + \left\{ \begin{array}{ll} \max &{} {\displaystyle \left[ \varGamma _m^{(1)+}x_1^{1/\alpha } + \cdots + \varGamma _m^{(J)+}x_J^{1/\alpha }\right] }\\ \text {s.t.} &{} x_1+\cdots +x_J = \nu -k_1 \\ &{} x_{j}\in \mathbb {N}, \forall j=2,\ldots ,J \end{array} \right\} \right) \nonumber \\&\quad \qquad +\,\frac{J}{\mu } + \sum _{j=1}^{J} \varGamma _m^{(j)+} - \frac{n-n_0}{\lambda } + \gamma _a(n-n_0)^{1/\alpha } . \end{aligned}$$

(89)

The optimal objective function of the inner optimization problem in Eq. (89) is given by Eq. (87). Hence, the bound on the first term in Eq. (54) becomes

$$\begin{aligned}&\max _{\begin{array}{c} 0\le k_1\le \phi \end{array}} \left( {\displaystyle \frac{\nu - k_1}{\mu } + \varGamma _m \cdot \left( \nu - k_1\right) ^{1/\alpha } }\right) + \frac{J}{\mu } \\&\qquad + \sum _{j=1}^{J} \varGamma _m^{(j)+} - \frac{n-n_0}{\lambda } + \gamma _a(n-n_0)^{1/\alpha }, \end{aligned}$$

where $\varGamma _m$ is defined in Eq. (51). Since $\varGamma _m\ge 0$, the term $x/\mu +\varGamma _m x^{1/\alpha }$ is increasing in x, yielding

$$\begin{aligned} \max _{\begin{array}{c} 0\le k_1\le \phi \end{array}} \left( {\displaystyle \frac{\nu - k_1}{\mu } + \varGamma _m \cdot \left( \nu - k_1\right) ^{1/\alpha } }\right) = \frac{\nu }{\mu } + \varGamma _m \cdot \nu ^{1/\alpha }. \end{aligned}$$

(b):

To bound the second term in Eq. (54), we take a similar approach to that presented in the proof of Theorem 5 and cast the problem in the form

$$\begin{aligned}&\max _{0 \le x \le \nu -\phi , x\in \mathbb {R}} \left( \beta \cdot x^{1/\alpha } -\delta \cdot x\right) \\&\quad =\left\{ \begin{array}{ll} \beta \cdot (\nu -\phi )^{1/\alpha } - \delta \cdot (\nu -\gamma ), &{} \quad \text {if}\;\nu -\phi \le \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)},\\ {\displaystyle \frac{(\alpha -1)}{\alpha ^{\alpha /(\alpha -1)}}\cdot \frac{\beta ^{\alpha /(\alpha -1)}}{\delta ^{1/(\alpha -1)}} },&{}\quad \text {otherwise}. \end{array} \right\} \end{aligned}$$

Substituting $\beta = m^{1/\alpha }\varGamma _a + \varGamma _m$ and $\delta = {m(1-\rho )}/{\lambda }$ yields the desired result.

$\square$

Proof of Theorem 7

The desired result is obtained by maximizing the system time for each path $P\in \mathcal {P}_{\ell }$. In order to apply the bounds on the system times from Assumption 1 to the quantity in Eq. (67), we need to account for the number of jobs that pass through node $a_{j}$ between the arrivals of job $k_{a_{j}}$, which belongs to $\mathcal {E}_{a_{j-1}a_{j}} \subseteq \mathcal {L}_{a_{j}}$, and job $k_{a_{j+1}}$, which belongs to $\mathcal {E}_{a_{j}a_{j+1}} \subseteq \mathcal {L}_{a_{j}}$. Mathematically, we let $\varDelta _{a_{j}}$ denote this number, i.e.,

$$\begin{aligned} \varDelta _{a_{j}} = \bigg | \left\{ k: k_a{_{j}} \le k \le k_{a_{j+1}}, k\in \mathcal {L}_{a_{j}}\right\} \bigg |. \end{aligned}$$

(90)

By Eq. (61), the fraction of jobs passing through queue $a_j$ is $\phi _{a_j}$, yielding

$$\begin{aligned} \varDelta _{a_j} = \phi _{a_j}\cdot \left( k_{a_{j+1}} - k_{a_j} + 1\right) . \end{aligned}$$

By Assumption 1, and given that $\tilde{\varGamma }_s^{(j)}\le \tilde{\varGamma }_s^{(j)+}$, for all $j\in \mathcal {J}$, we bound the service times by

$$\begin{aligned} \max _{\mathcal {U}^{s}_{a_j}} \sum _{i=k_{a_j}}^{k_{a_{j+1}}}{X}_{i}^{(a_j)}= & {} \frac{\varDelta _{a_j}}{\mu _{a_j}} + \varGamma _s^{(a_j)+}\cdot \varDelta _{a_j}^{1/\alpha } = \frac{\phi _{a_j}\cdot \left( k_{a_{j+1}} - k_{a_j} + 1\right) }{\mu _{a_{j}}} \\&+\, \varGamma _{s}^{(a_j)+}\cdot \left[ \phi _{a_j}\cdot \left( k_{a_{j+1}} - k_{a_j} + 1\right) \right] ^{1/\alpha }. \end{aligned}$$

By applying Assumption 1, Eq. (67) becomes

(91)

where $\tilde{\mu }_{j} = \mu _{j} / \phi _j$ and $\tilde{\varGamma }_{s}^{(j)} = \varGamma _{s}^{(j)}\cdot \phi _{j}^{1/\alpha }$, for all $j\in \mathcal {J}$. We let $\tilde{\mu }_{P} =\displaystyle \min \left\{ \tilde{\mu }_{a_j}, a_j\in P\right\}$, $\rho _P = \lambda /\tilde{\mu }_{P}$. By making the change of variable $x_{a_j} = k_{a_{j+1}} - k_{a_j}$, for all $a_j\in P$, we bound the maximization problem in Eq. (91) by

$$\begin{aligned}&\max _{1\le k_{a_1} \le n } \left( \varGamma _a\left( n-k_{a_1}\right) ^{1/\alpha } - \frac{1-\rho _P}{\lambda }\left( n-k_{a_{1}}\right) \right. \nonumber \\&\quad \left. +\, \left\{ \begin{array}{ll} \text {max} &{} ~ \left[ \tilde{\varGamma }_{s}^{(a_1)+} \cdot x_{a_1}^{1/\alpha } + \cdots + \tilde{\varGamma }_{s}^{(\ell )+} \cdot x_{a_J}^{1/\alpha } \right] \\ \text {s.t.} &{} ~x_{a_1} + \cdots + x_{\ell } = n-k_{a_1} \end{array} \right\} \right) . \end{aligned}$$

(92)

The optimal objective function for the inner optimization problem is given in Eq. (87). The performance analysis reduces to solving the following one-dimensional optimization problem:

(93)

which can be cast in the form of the optimization problem in Eq. (82), with

$$\begin{aligned} \beta = \varGamma _a + \left( \, \sum _{j\in P} \left( \tilde{\varGamma }_s^{(j)+}\right) ^{\alpha /(\alpha -1)}\right) ^{(\alpha -1)/\alpha } ~~ \text {and} ~~ \delta = \frac{1-\rho _P}{\lambda }. \end{aligned}$$

Referring to the proof of Theorem 2, the solution to Eq. (93) is

$$\begin{aligned} {\displaystyle \max _{0\le x \le n}~~\beta \cdot n^{1/\alpha } - \delta \cdot n} = \left\{ \begin{array}{ll} \beta \cdot n^{1/\alpha } - \delta \cdot n, &{} \quad \text {if}\; n \le \left( \frac{\beta }{\alpha \delta }\right) ^{\alpha /(\alpha -1)},\\ \\ {\displaystyle \frac{(\alpha -1)}{\alpha ^{\alpha /(\alpha -1)}}\cdot \frac{\beta ^{\alpha /(\alpha -1)}}{\delta ^{1/(\alpha -1)}} }, &{}\quad \text {otherwise}. \end{array} \right. \end{aligned}$$

We obtain the desired result by substituting for $\beta$ and $\delta$ their respective values. $\square$

Proof of Proposition 2

We prove the result using the technique of mathematical induction.

(a) Initial step As presented in [9], the system time in an m-server queue,

$$\begin{aligned} \widehat{S}_{n}\left( \mathbf {T}\right) = \widehat{S}^{\,(1)}_{n}\left( \mathbf {T}\right) = \max _{0\le k_1 \le \nu } \left( ~\max _{\mathbf {X}^{(1)} \in \mathcal {U}^{s}_{m}}\sum _{i=k_{1}}^{\nu } {X}_{r(i)}^{(1)} - \sum _{i=r\left( k_1\right) +1}^{n} T_{i} \right) , \end{aligned}$$

and therefore the result holds for $J=1$.

(b) Inductive step We now suppose that the result holds for $J-1$ queues in series, which expresses the system time across queues 2 through J as

$$\begin{aligned}&\widehat{S}_{n}^{\,(2)}\left( \mathbf {T}\right) + \cdots + \widehat{S}_{n}^{\,(J)}\left( \mathbf {T}\right) \nonumber \\&\quad = \max _{0\le k_2 \le \cdots \le k_{J} \le \nu } \left( \max _{\mathcal {U}^{s}_{2}}\sum _{i=k_2}^{k_3}X_{r(i)}^{(2)} + \cdots + ~\max _{\mathcal {U}^{s}_{J}}\sum _{i=k_J}^{n}X_{r(i)}^{(J)} ~ - \sum _{i=r\left( k_2\right) +1}^{n} T_{i}^{\,(2)} \right) ,\nonumber \\ \end{aligned}$$

(94)

where $\mathbf {T}^{\,(2)} = \left\{ T_{1}^{\,(2)}, \ldots , T_{n}^{\,(2)} \right\}$ denotes the sequence of inter-arrival times to the second queue. Note that the arrival to the second queue is simply the departure from the first queue, and therefore, denoting the inter-departure times from the first queue by $\mathbf {D}^{(1)} = \left\{ D_{1}^{(1)}, \ldots , D_{n}^{(1)} \right\}$, we have

$$\begin{aligned} \sum _{i=r\left( k_2\right) +1} T_{i}^{\,(2)} = \sum _{i=r\left( k_2\right) +1} D_{i}^{(1)} = \sum _{i=\left( k_2\right) +1}^{n}T_{i} + \widehat{S}_n^{\,(1)}\left( \mathbf {T}\right) - \widehat{S}_{r\left( k_2\right) }^{\,(1)}\left( \mathbf {T}\right) , \end{aligned}$$

(95)

where the last equality is due to the fact that no overtaking occurs at the first queue in the worst case approach. Combining Eqs. (94)–(95), we obtain

$$\begin{aligned} \widehat{S}_{n}\left( \mathbf {T}\right)= & {} \widehat{S}_{n}^{(1)}\left( \mathbf {T}\right) + \widehat{S}_{n}^{(2)}\left( \mathbf {T}\right) + \cdots + \widehat{S}_{n}^{(J)}\left( \mathbf {T}\right) \nonumber \\= & {} \max _{0\le k_2 \le \cdots \le k_{J} \le \nu } \left( \max _{\mathcal {U}^{s}_{2}}\sum _{i=k_2}^{k_3}X_{r(i)}^{(2)} + \cdots + \max _{\mathcal {U}^{s}_{J}}\sum _{i=k_J}^{n}X_{r(i)}^{(J)} \right. \nonumber \\&\left. -\,\sum _{i=r\left( k_2\right) +1}^{n}T_{i}^{\,(2)} + \widehat{S}_{r\left( k_2\right) }^{\,(1)}\left( \mathbf {T}\right) \right) . \end{aligned}$$

(96)

Since no overtaking occurs in the first queue, and given that $\lfloor r\left( k_2\right) /m \rfloor = k_2$, the system time of the $r\left( k_2\right) {\text {th}}$ job can be expressed as

$$\begin{aligned} S_{r\left( k_2\right) }^{\,(1)}\left( \mathbf {T}\right) = \max _{0\le k_1 \le k_2} \left( \max _{\mathcal {U}^{s}_{1}}\sum _{i=k_1}^{k_2} {X}_{r(i)}^{(1)} - \sum _{i=r\left( k_1\right) +1}^{r\left( k_2\right) } T_{i} \right) . \end{aligned}$$

Substituting the above expression in Eq. (96), the overall system time becomes

$$\begin{aligned} S_{n}= & {} \max _{0\le k_2 \le \cdots \le k_{J} \le \nu } \left( \max _{\mathcal {U}^{s}_{2}}\sum _{i=k_2}^{k_3} {X}_{r(i)}^{(2)} + \cdots +\max _{\mathcal {U}^{s}_{J}} \sum _{i=k_J}^{n}{X}_{r(i)}^{(J)} - \sum _{i=r\left( k_2\right) +1}^{n} T_{i} \right. \\&\left. +\, \max _{0\le k_1 \le k_2} \left( \max _{\mathcal {U}^{s}_{1}}\sum _{i=k_1}^{k_2} {X}_{r(i)}^{(1)} - \sum _{i=r\left( k_1\right) +1}^{r\left( k_2\right) } T_{i} \right) \right) . \end{aligned}$$

Rearranging the terms in the above expression proves the inductive result. This concludes the inductive step, and by mathematical induction, we have the desired result. $\square$

Proof of Proposition 3

We use the principle of mathematical induction to prove this result. Specifically, we assume that the result is true for any job $j\le n-1$ passing by some node q from the feed-forward network (disregarding where the $j{\text {th}}$ job goes next in the network after q), i.e.,

$$\begin{aligned} S_{j}\left( \mathcal {P}_q\right)= & {} \max _{P \in \mathcal {P}_{q}} \left\{ \max _{\begin{array}{c} 1\le k_{a_1}\le k_{a_2} \le \cdots \le k_q\le j\\ k_{i+1}\in \mathcal {E}_{a_ia_{i+1}} \end{array}} \left( \sum _{\begin{array}{c} i=k_{b_1}\\ i\in \mathcal {L}_{a_1} \end{array}}^{k_{a_2}}{X}_{i}^{(a_1)} +\sum _{\begin{array}{c} i=k_{a_2}\\ i\in \mathcal {L}_{a_2} \end{array}}^{k_{a_3}}{X}_{i}^{(a_2)} + \cdots \right. \right. \nonumber \\&\left. \left. +\,\sum _{\begin{array}{c} i=k_q\\ i\in \mathcal {L}_{q} \end{array}}^{j}{X}_{i}^{(q)} - \sum _{i=k_{a_1}+1}^{j}T_{i} \right) \right\} , \end{aligned}$$

(97)

where $\mathcal {P}_{q}$ denotes the set of all paths $P=\left( a_0, a_1, \ldots , q\right)$ that pass by q (disregarding the network after q). We next proceed to show that the result holds for job n exiting the network at queue $\ell$.

The system time of the $n{\text {th}}$ job at queue $\ell$ can be expressed as

$$\begin{aligned} S_{n}^{(\ell )} = \max _{\begin{array}{c} 1 \le k_\ell \le n\\ k_{\ell } \in \mathcal {L}_{\ell } \end{array}} \left( \sum _{\begin{array}{c} i=k_\ell \\ i\in \mathcal {L}_{\ell } \end{array}}^n X_i^{(\ell )} - \sum _{\begin{array}{c} i=k_\ell +1\\ i\in \mathcal {L}_{\ell } \end{array}}^n T_i^{(\ell )}\right) . \end{aligned}$$

(98)

Suppose $k_\ell \in \mathcal {E}_{q \ell }$, i.e., job $k_\ell$ enters queue $\ell$ from queue q, and without loss of generality, suppose that job n enters queue $\ell$ from queue r, i.e., $n\in \mathcal {E}_{r \ell }$. Then,

$$\begin{aligned} \sum _{\begin{array}{c} i=k_\ell +1\\ i\in \mathcal {L}_{\ell } \end{array}}^n T_i^{(\ell )} = \left( \sum _{i=1}^{n} T_{i} + S_{n}\left( \mathcal {P}_{r}\right) \right) - \left( \sum _{i=1}^{k_\ell } T_{i} + S_{k_{\ell }}\left( \mathcal {P}_{q}\right) \right) . \end{aligned}$$

(99)

Combining Eqs. (98) and (99), we obtain

$$\begin{aligned} S_{n}^{(\ell )} + S_{n}\left( \mathcal {P}_{r}\right)= & {} \max _{\begin{array}{c} 1 \le k_\ell \le n\\ k_{\ell } \in \mathcal {L}_{\ell } \end{array}} \left( \sum _{\begin{array}{c} i=k_\ell \\ i\in \mathcal {L}_{\ell } \end{array}}^n X_i^{(\ell )} + S_{k_{\ell }}\left( \mathcal {P}_{q}\right) - \sum _{i=1}^{n} T_{i} + \sum _{i=1}^{k_\ell } T_{i} \right) \\= & {} \max _{\begin{array}{c} 1 \le k_\ell \le n\\ k_{\ell } \in \mathcal {L}_{\ell } \end{array}} \left( \sum _{\begin{array}{c} i=k_\ell \\ i\in \mathcal {L}_{\ell } \end{array}}^n X_i^{(\ell )} + S_{k_{\ell }}\left( \mathcal {P}_{q}\right) - \sum _{i=k_\ell +1}^{n} T_{i} \right) . \end{aligned}$$

By the induction hypothesis, we substitute the value of $S_{k_{\ell }}\left( \mathcal {P}_q\right)$ in the above equation and obtain

$$\begin{aligned} S_{n}^{(\ell )} + S_{n}\left( \mathcal {P}_{r}\right)= & {} S_{n}\left( \mathcal {P}_{r\ell }\right) = \max _{P \in \mathcal {P}_{q\ell }} \left\{ \max _{\begin{array}{c} 1\le k_{a_1} \le \cdots \le k_q \le k_\ell \le n\\ k_{i+1}\in \mathcal {E}_{a_ia_{i+1}} \end{array}} \left( \sum _{\begin{array}{c} i=k_{a_1}\\ i\in \mathcal {L}_{a_1} \end{array}}^{k_{a_2}} {X}_{i}^{(a_1)} + \cdots \right. \right. \\&\left. \left. +\,\sum _{\begin{array}{c} i=k_q\\ i\in \mathcal {L}_{q} \end{array}}^{k_\ell }{X}_{i}^{(q)} + \sum _{\begin{array}{c} i=k_\ell \\ i\in \mathcal {L}_{\ell } \end{array}}^n X_i^{(\ell )} - \sum _{i=k_{b_1}+1}^{n}T_{i} \right) \right\} , \end{aligned}$$

where $\mathcal {P}_{r\ell }$ and $\mathcal {P}_{q\ell }$ are the sets of paths that end at node r and q, respectively, and then feed in to node $\ell$ (disregarding what comes next in the network). Given that q and r were chosen arbitrarily, the result holds for any nodes q and r that feed into queue $\ell$, i.e., for all $q,r\in \mathcal {P}_{\ell }$. Hence,

$$\begin{aligned} S_{n}\left( \mathcal {P}_{\ell }\right)= & {} \max _{P \in \mathcal {P}_{\ell }} \left\{ \max _{\begin{array}{c} 1\le k_{a_1} \le \cdots \le k_q \le k_\ell \le n\\ k_{i+1}\in \mathcal {E}_{a_ia_{i+1}} \end{array}} \left( \sum _{\begin{array}{c} i=k_{a_1}\\ i\in \mathcal {L}_{a_1} \end{array}}^{k_{a_2}} {X}_{i}^{(a_1)} + \cdots \right. \right. \nonumber \\&\left. \left. +\,\sum _{\begin{array}{c} i=k_q\\ i\in \mathcal {L}_{q} \end{array}}^{k_\ell }{X}_{i}^{(q)} + \sum _{\begin{array}{c} i=k_\ell \\ i\in \mathcal {L}_{\ell } \end{array}}^n X_i^{(\ell )} - \sum _{i=k_{a_1}+1}^{n}T_{i} \right) \right\} . \end{aligned}$$

(100)

This concludes the inductive step and proves the result for job n. Next considering the base case of $n=1,$ it is trivial to check the validity of the inductive hypothesis. Therefore, the result follows from induction. This concludes the proof. $\square$