Origin-Destination Matrix Estimation Problem in a Markov Chain Approach

Abstract

In this paper, a Markov chain origin-destination matrix estimation problem is investigated in which the average time between two incoming streams to or outgoing streams from nodes in consecutive time periods is considered as a Markov chain. Along with, a normal distribution with pre-determined parameters in each period is considered for traffic counts on links. A bi-level programming problem is introduced where in its upper level the network flow pattern in the n th period is estimated so that the probability of the estimated traffic counts is maximized, while in the lower level a traffic assignment problem with the equilibrium conditions is solved. We reduce the proposed nonlinear bi-level model to a new one level linear programming problem, where by using a trust-region method the local optimal solutions are obtained. Some numerical examples are provided to illustrate the efficiency of the proposed method.

Appendix: Convergence

Throughout this section, we provide some lemmas and theorems to assure the convergence of the proposed TRM to a local optimal solution for the bi-level O-D matrix estimation problem. It would be shown that if the stopping criteria are met, the terminated point is a local optimal solution, while if the algorithm is repeated infinitely, any accumulation point would be local optimal.

Lemma 4

The function$z^{\prime }_{0}$inProblem (14), and its gradient are Lipschitz continuous with constants p andp^′, respectively.

Proof

The variable y is bounded with regard to Inequality (22) and Inequality (23), therefore the objective function $z^{\prime }_{0}$ and its gradient are Lipschitz continuous. □

Lemma 5

The lower level objective functionz₁(f) given in Problem (24) is continuous and strictly convex with respect to thevariables$f_{rs}^{k}$.

Proof

The continuity of the function z₁(f) is derived by its definition, while the strict convexity would be concluded since the functions τ_e(.) are strictly ascending. □

Lemma 6

Let {g_i} be a sequence of real-valued continuous strictly convex functions with argument x, converging uniformly to a continuous strictly convex function g. Then arg minxg_i(x) → arg minxg(x).

Proof

Suppose that x_i = arg minxg_i(x) and x^∗ = arg minxg(x). We prove that {x_i} is a cauchy sequence. Obviously, the sequence {g_i} is cauchy. For a given 𝜖 > 0 we can choose i sufficiently large so that for m, n > i, it is resulted ||g_m − g_n||_∞ < 𝜖. To establish ||x_m − x_n||_∞ < 𝜖, it suffices to show that |g_n(x_m) − g_n(x_n)| < 𝜖, since g_is are strictly convex. Clearly $|g_{m}(x)-g_{n}(x)|<\frac {1}{2}\epsilon$ for sufficiently large m, n and all x; therefore,

$$g_{n}(x_{m})<g_{m}(x_{m})+\frac{1}{2}\epsilon<g_{m}(x_{n})+\frac{1}{2}\epsilon<g_{n}(x_{n})+\epsilon\Rightarrow g_{n}(x_{m})<g_{n}(x_{n})+\epsilon,$$

so |g_n(x_m) − g_n(x_n)| < 𝜖.

Now assume that ${x_{i}}\rightarrow \tilde {x}$ and $\tilde {x}\neq x^{*}$, then $g(x^{*})<g(\tilde {x})$ by strict convexity of g. Choosing i sufficiently large and noting continuity of g and g_i yield:

$$|g(\tilde{x})-g(x_{i})|<\frac{1}{3}\epsilon \rightarrow g(\tilde{x})<g(x_{i})+\frac{1}{3}\epsilon. $$

Furthermore, since g_i → g:

$$|g(x_{i})-g_{i}(x_{i})|<\frac{1}{3}\epsilon \rightarrow g(x_{i})<g_{i}(x_{i})+\frac{1}{3}\epsilon $$

and

$$|g(x^{*})-g_{i}(x^{*})|<\frac{1}{3}\epsilon \rightarrow g_{i}(x^{*})<g(x^{*})+\frac{1}{3}\epsilon.$$

Then

$$g(\tilde{x})<g(x_{i})+\frac{1}{3}\epsilon<g_{i}(x_{i})+\frac{2}{3}\epsilon<g_{i}(x^{*})+\frac{2}{3}\epsilon<g(x^{*})+\epsilon,$$

implies that $g(\tilde {x})<g(x^{*})+\epsilon$. This contradiction completes the proof. □

Lemma 7

Let$f^{*}(\bar {y})$and$f^{*}(\hat {y})$bethe optimal solutions to the lower level Problem (24) correspondingto$\bar {y}$and$\hat {y}$, respectively. Then for a given𝜖^′ > 0 there exists an𝜖 > 0 such that$||\bar {y}-\hat {y}||_{\infty }<\epsilon$yields$||f^{*}(\bar {y})-f^{*}(\hat {y})||_{\infty }<\epsilon ^{\prime }$.

Proof

Recall that the variables x_e can be considered as functions of the variables $f_{rs}^{k}$. Obviously, the feasible region for x_e in the lower level problem would be invariant if $||\bar {y}-\hat {y}||_{\infty }<\epsilon$. Furthermore, the continuity of the function arg minxg(x) for continuous strictly convex function g, proved by Lemma 6, with Lemma 5 complete the proof. □

Lemma 8

Let$\bar {z}_{1}(f)$be the second orderapproximation of the functionz₁(f) at the point$(\bar {y},\bar {f})$suchthat$\bar {f}=f^{*}(\bar {y})=\arg \min _{f}z_{1}(f), s.t. \bar {y},f\in S$. If

$$f^{*}(\hat{y})=\arg\min_{f}z_{1}(f), s.t. \hat{y},f\in S\qquad\text{,}\qquad \bar{f}(\hat{y})=\arg\min_{f}\bar{z}_{1}(f), s.t. \hat{y},f\in S,$$

whereSis the feasible region of the lower level problem, then$||\bar {f}(\hat {y})-f^{*}(\hat {y})||_{\infty }$is closed to zero, for vectors$\hat {y}$with$||\hat {y}-\bar {y}||_{\infty }<\epsilon$where𝜖is sufficiently small.

Proof

Constraints (17) can be rewritten as Af = y in matrix notation. If $f^{*}(\hat {y})$ is the optimal solution to $\min z_{1}(f); \hat {y},f\in S$, then it satisfies the optimality conditions below:

$$f^{*}_{k}= 0 \Rightarrow \{ \triangledown z_{1}|_{f^{*}}+\alpha A\}_{k}\geq 0 $$

$$f^{*}_{k}>0 \Rightarrow \{ \triangledown z_{1}|_{f^{*}}+\alpha A\}_{k}= 0,$$

where α is the dual multiplier vector corresponding to the constraint Af = y. Since $\triangledown z_{1}|_{f^{*}}=\triangledown \bar {z}_{1}|_{{f^{*}}}+o||f^{*}(\hat {y})-\bar {f}||_{\infty }^{2}$ and $||f^{*}(\hat {y})-\bar {f}||_{\infty } \rightarrow 0$ by Lemma 7, so the above relationships become:

$$f^{*}_{k}= 0 \Rightarrow \{ \triangledown \bar{z}_{1}|_{f^{*}}+\alpha A\}_{k}\geq o||f^{*}(\hat{y})-\bar{f}||_{\infty}^{2}$$

$$f^{*}_{k}>0 \Rightarrow \{ \triangledown \bar{z}_{1}|_{f^{*}}+\alpha A\}_{k}= o||f^{*}(\hat{y})-\bar{f}||_{\infty}^{2}.$$

Therefore, the optimality conditions of $\min \bar {z}_{1}(f); \hat {y},f \in S$ also hold for $f^{*}(\hat {y})$ with an error of order $o||f^{*}(\hat {y})-\bar {f}||_{\infty }^{2}$ where $||f^{*}(\hat {y})-\bar {f}||_{\infty } \rightarrow 0$. Due to strict convexity of the original and approximated lower level objective functions, the optimal solution to the function $\bar {z}_{1}(f); \hat {y},f\in S,$is arbitrary close to $f^{*}(\hat {y})$. □

Theorem 1

(Convergence of the algorithm in finitely many iterations) If the trust-region algorithm Section 4.1is terminated in finitely many iterations, then the terminated point is the localoptimal solution to the original BP Problem (14).

Proof

Assume that the stopping criteria is met at iteration k, i.e. either ${\Delta }_{t}^{\mu + 1}<{\Delta }_{t}^{min}=\epsilon$ or $\bar {z}_{0}(y^{\mu },f^{\mu })-\bar {z}_{0}(y^{m},f^{m})<\epsilon$, where 𝜖 > 0 is small enough. If the stopping criteria is met by ${\Delta }_{t}^{\mu + 1}<\epsilon$, we would show that (y^m, f^∗(y^m)) is a local optimal solution to the original BP Problem (14). By contradiction suppose that (y^m, f^∗(y^m)) is not local optimal, so there exists some feasible point $(\hat {y},f^{*}(\hat {y}))$ in the interval $||y-\bar {y}||_{\infty }<{\Delta }_{t}^{\mu + 1}<\epsilon$ satisfying in the feasibility constraints of the upper level problem, so that $z_{0}^{\prime }(\hat {y},f^{*}(\hat {y}))<z_{0}^{\prime }(y^{m},f^{*}(y^{m}))$, where f^∗(y) is the optimal solution of the lower level problem (24), corresponding to the upper level variable y. Let $\hat {f}=\bar {f}(\hat {y})$ be the optimal solution for $\bar {z}_{1}(f); \hat {y},f\in S$ where $\bar {z}_{1}$ is the approximated lower level objective function at $\bar {f}=f^{*}(\bar {y})$. Then, using Lemma 7 and Lemma 8 together with triangular inequality, $||\hat {f}-\bar {f}||_{\infty }\leq ||\hat {f}- f^{*}(\hat {y})||_{\infty }+||f^{*}(\hat {y})-\bar {f}||_{\infty }$ yield $||\hat {f}-\bar {f}||_{\infty }\rightarrow 0$, provided that 𝜖 is sufficiently small. Now, applying the Lipschitz continuity conditions of function $z^{\prime }_{0}$ and its gradient results in the following inequalities:

$$\begin{array}{@{}rcl@{}} \bar{z}_{0}(\hat{y},\hat{f})& \leq& z_{0}^{\prime}(\hat{y},\hat{f})+\frac{1}{2}p^{\prime}(||\hat{y}-\bar{y}||^{2}+||\hat{f}-\bar{f}||^{2}) \\ &\leq& z_{0}^{\prime}(\hat{y},f^{*}(\hat{y}))+\frac{1}{2}p^{\prime}(||\hat{y}-\bar{y}||^{2}+||\hat{f}-\bar{f}||^{2})+p||\hat{f}-f^{*}(\hat{y})||\\&<&z_{0}^{\prime}(y^{m},f^{*}(y^{m}))+\frac{1}{2}p^{\prime}(||\hat{y}-\bar{y}||^{2}+||\hat{f}-\bar{f}||^{2})+p||\hat{f}-f^{*}(\hat{y})||\\&\leq& z_{0}^{\prime}(y^{m},f^{m})+\frac{1}{2}p^{\prime}(||\hat{y}-\bar{y}||^{2}+||\hat{f}-\bar{f}||^{2})+p||\hat{f}-f^{*}(\hat{y})||+p||f^{m}-f^{*}(y^{m})||\\&\leq& \bar{z}_{0}(y^{m},f^{m})+\frac{1}{2}p^{\prime}(||\hat{y}-\bar{y}||^{2}+||\hat{f}-\bar{f}||^{2})+p||\hat{f}-f^{*}(\hat{y})||+p||f^{m}-f^{*}(y^{m})||\\&&+\frac{1}{2}p^{\prime}(||y^{m}-\bar{y}||^{2}+||f^{m}-\bar{f}||^{2}). \end{array} $$

Therefore, $\bar {z}_{0}(\hat {y},\hat {f})<\bar {z}_{0}(y^{m},f^{m})+\epsilon ^{\prime }$ where 𝜖^′→ 0, which contradicts the optimality of the point (y^m, f^m) to Problem (21) in the interval $||y-\bar {y}||_{\infty } < {\Delta }_{t}^{\mu + 1}<\epsilon$.

Now assume that the stopping criteria is met by $\bar {z}_{0}(y^{\mu },f^{\mu })-\bar {z}_{0}(y^{m},f^{m})<\epsilon$ where 𝜖 is small enough; therefore, (y^μ, f^μ) is a local optimal solution to the Problem (21). By contradiction suppose that (y^μ, f^μ) is not local optimal for the original BP problem (14); therefore, for each 𝜖^′ > 0 there exists a point $(\hat {y},f^{*}(\hat {y}))$ so that $||\hat {y}-y^{\mu }||<\epsilon ^{\prime }$ and $z^{\prime }_{0}(\hat {y},f^{*}(\hat {y}))<z^{\prime }_{0}(y^{\mu },f^{\mu })$, where $\bar {f}(y)$ and f^∗(y) are the optimal solutions of the approximated and original lower level problems corresponding to y, respectively. Similar to the first case, the following inequalities hold:

$$\begin{array}{@{}rcl@{}} \bar{z}_{0}(\hat{y},\bar{f}(\hat{y}))&\leq& {z}^{\prime}_{0}(\hat{y},\bar{f}(\hat{y}))+\frac{1}{2}p^{\prime}(||\hat{y}-y^{\mu}||^{2}+||\bar{f}(\hat{y})-f^{\mu}||^{2}) \\ &\leq& {z}^{\prime}_{0}(\hat{y},f^{*}(\hat{y}))+\frac{1}{2}p^{\prime}(||\hat{y}-y^{\mu}||^{2}+||\bar{f}(\hat{y})-f^{\mu}||^{2})+p||\bar{f}(\hat{y})-f^{*}(\hat{y})|| \\ &<& z^{\prime}_{0}(y^{\mu} ,f^{\mu})+\frac{1}{2}p^{\prime}(||\hat{y}-y^{\mu}||^{2}+||\bar{f}(\hat{y})-f^{\mu}||^{2})+p||\bar{f}(\hat{y})-f^{*}(\hat{y})||\\ &=&\bar{z}_{0}(y^{\mu} ,f^{\mu})+\frac{1}{2}p^{\prime}(||\hat{y}-y^{\mu}||^{2}+||\bar{f}(\hat{y})-f^{\mu}||^{2})+p||\bar{f}(\hat{y})-f^{*}(\hat{y})||. \end{array} $$

Therefore, $\bar {z}_{0}(\hat {y},\bar {f}(\hat {y}))\leq \bar {z}_{0}(y^{\mu },f^{\mu })+\epsilon ^{\prime \prime }$ for sufficiently small 𝜖^″ > 0 that contradicts the optimality of (y^μ, f^μ) for Problem (21). □

Theorem 2

(Convergence of the algorithm in case with infinite iterations) Let the trust-regionalgorithm Section 4.1repeated infinitely, then any accumulation point is a local optimalsolution to the original BP Problem (14).

Proof

Suppose $\{y_{\mu _{i}}\}$ is a subsequence created by the trust-region Algorithm Section 4.1, convergent to an accumulation point y^∗ while the trust-region radius is not converging to zero. Then the ratio ρ_μ is greater than or equal to η₁ infinitely ; therefore,

$$ z^{\prime}_{0}(y^{\mu_{i}},f^{\mu_{i}})-z^{\prime}_{0}(y^{\mu_{i}+ 1},f^{*}(y^{\mu_{i}+ 1})) \geq \eta_{1}(\bar{z}_{0}^{\mu_{i}}(y^{\mu_{i}},f^{\mu_{i}})- \bar{z}_{0}^{\mu_{i}}(y^{\mu_{i}+ 1},\bar{f}(y^{\mu_{i}+ 1})), $$

(25)

where $\bar {z}_{0}^{\mu _{i}}$ denotes the approximation of the upper level objective function at point $(y^{\mu _{i}},f^{\mu _{i}})$. Moreover, $f^{\mu _{i}}=f^{*}(y^{\mu _{i}})$ where f^∗(y) = arg minfz₁(f); y, f ∈ S. Also, let $f_{*}(y)=\arg \min _{f} \bar {z}_{1}(f); y,f \in S$, where $\bar {z}_{1}(f)$ is the approximated lower level objective function at the point (y^∗, f^∗(y^∗)). In addition, $\bar {f}(y^{\mu _{i}+ 1})$ denotes the optimal solution to the approximated lower level problem at the point $(y^{\mu _{i}},f^{\mu _{i}})$ corresponding to the upper level variable $y^{\mu _{i}+ 1}$. If (y^∗, f^∗(y^∗)) is not local optimal for original BP Problem (14), then for any given 𝜖 > 0 there exists a point $\tilde {y}$ so that $||\tilde {y}-y^{*}||_{\infty }<\epsilon$ and $z^{\prime }_{0}(\tilde {y},f^{*}(\tilde {y}))<z^{\prime }_{0}(y^{*},f^{*}(y^{*}))$. Then,

$$z^{\prime}_{0}(\tilde{y},f_{*}(\tilde{y}))\leq z^{\prime}_{0}(\tilde{y},f^{*}(\tilde{y}))+p||f^{*}(\tilde{y})-f_{*}(\tilde{y})|| <z^{\prime}_{0}(y^{*},f^{*}(y^{*}))+p||f^{*}(\tilde{y})-f_{*}(\tilde{y})||. $$

This inequality together with Lemma 8 yields $z^{\prime }_{0}(\tilde {y},f_{*}(\tilde {y}))<z^{\prime }_{0}(y^{*},f^{*}(y^{*}))+\epsilon ^{\prime }$ where 𝜖^′→ 0. Also

$$\begin{array}{@{}rcl@{}} \bar{z}_{0}(\tilde{y},f_{*}(\tilde y))&\leq& z^{\prime}_{0}(\tilde{y},f_{*}(\tilde y))+\frac{1}{2}p^{\prime}(||\tilde{y}-{y^{*}}||^{2}+||f_{*}(\tilde{y})-f^{*}(y^{*})||^{2})\\ &<&z^{\prime}_{0}(y^{*},f^{*}(y^{*}))+\frac{1}{2}p^{\prime}(||\tilde{y}-{y^{*}}||^{2}+||f_{*}(\tilde{y})-f^{*}(y^{*})||^{2}+\epsilon^{\prime} .\end{array} $$

Therefore, $\bar {z}_{0}(\tilde {y},f_{*}(\tilde {y}))<z^{\prime }_{0}(y^{*},f^{*}(y^{*}))+\epsilon ^{\prime \prime }=\bar {z}_{0}(y^{*},f^{*}(y^{*}))+\epsilon ^{\prime \prime }$, where 𝜖^″→ 0 and $\bar {z}_{0}$ is the approximated upper level objective function at the point (y^∗, f^∗(y^∗)). It yields that $\bar {z}_{0}(\tilde {y},f_{*}(\tilde {y})) \leq \bar {z}_{0}(y^{*},f^{*}(y^{*}))$. If $\bar {z}_{0}(\tilde {y},f_{*}(\tilde {y}))=\bar {z}_{0}(y^{*},f^{*}(y^{*}))$, then, since f_∗(y^∗) = f^∗(y^∗), there is no difference between $(\tilde {y},f_{*}(\tilde {y}))$ and (y^∗, f^∗(y^∗)). Therefore, the point $(\tilde {y},f_{*}(\tilde {y}))$ can also be considered as the optimal solution of Problem (21). So only the case $\bar {z}_{0}(\tilde {y},f_{*}(\tilde {y})) < \bar {z}_{0}(y^{*},f^{*}(y^{*}))$ is considered. Suppose $f_{\mu _{i}}(\tilde {y})$ is the optimal solution to the lower level problem approximated at the point $(y^{\mu _{i}},f^{\mu _{i}}=f^{*}(y^{\mu _{i}}))$, corresponding to $\tilde {y}$ as the upper level variable. Considering the continuity of function z₁ together with the continuity of function arg minxg(x) over strictly convex functions would result in $||f^{\mu _{i}}-f^{*}(y^{*})||\rightarrow 0$ and $||f_{\mu _{i}}(\tilde {y})-f_{*}(\tilde {y})||\rightarrow 0$. Furthermore,

$$\begin{array}{@{}rcl@{}} \bar{z}_{0}^{\mu_{i}} (y^{\mu_{i}},f^{\mu_{i}})-\bar{z}_{0}^{\mu_{i}}(y^{\mu_{i}+ 1},\bar{f}(y^{\mu_{i}+ 1}))&\!\geq\!& \bar{z}_{0}^{\mu_{i}}(y^{\mu_{i}},f^{\mu_{i}})-\bar{z}_{0}^{\mu_{i}}(\tilde{y},f_{\mu_{i}}(\tilde{y}))\\& \!\geq\!& \frac{1}{2} (\bar{z}_{0}(y^{*},f^{*}(y^{*}))-\bar{z}_{0}(\tilde{y},f_{*}(\tilde{y}))), \end{array} $$

(26)

where i is sufficiently large. It is resulted from Inequality (25) and Inequality (26) that the series $\sum (z^{\prime }_{0}(y^{\mu _{i}},f^{\mu _{i}})-z^{\prime }_{0}(y^{\mu _{i}+ 1},f^{*}(y^{\mu _{i}+ 1}))$ diverges where $f^{*}(y^{\mu _{i}+ 1})=f^{\mu _{i}+ 1}$. Since the TRM generates a decreasing sequence, we obtain:

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{\infty}(z^{\prime}_{0}(y^{\mu_{i}},f^{\mu_{i}})-z^{\prime}_{0}(y^{\mu_{i}+ 1},f^{\mu_{i}+ 1})) &\leq& \sum\limits_{i = 1}^{\infty}(z^{\prime}_{0}(y^{\mu_{i}},f^{\mu_{i}})-z^{\prime}_{0}(y^{\mu_{i + 1}},f^{\mu_{i + 1}}))\\ &\leq& z^{\prime}_{0}(y^{\mu_{1}},f^{\mu_{1}})-z^{\prime}_{0}(y^{*},f^{*}(y^{*}) <\infty. \end{array} $$

This contradiction proves that (y^∗, f^∗(y^∗)) is a local optimal solution to the original BP Problem (14). □