Step (1c): Procedure when the $k$th packet $b_k$ of $B$ is taken from $S$.

For each $j=j_k,j_k+1,\ldots,k   (j_1=1)$ do the following computations.

1. Compute $f(k,j)$
   When we move graph $A$ and $B$ along the Y-axis so that $a_0$ and $b_{j-1}$ are at the same level 0 on the Y-axis, the two graphs (named as $A'$ and $B(j)$) are repositioned as follows:

   $$A':\quad \langle (t(a_1),a_1-a_0), (t(a_2),a_2-a_0),\ldots,(t(a_n),a_n-a_0) \rangle$$

   $$B(j):\quad \langle (u(b_1),b_1-b_{j-1}),\ldots,(u(b_{j-1}),0),\ldots,(u(b_k),b_k-b_{j-1}),\ldots \rangle\enspace .$$

   
   
   $f(k,j)$ is the index of $A'$ at which $a_{f(k,j)}-a_0$ is the lowest position above $b_k-b_{j-1}$, and is computed using $f(k-1,j)$ as the starting position by the following equation.

   $$f(k,j) = \left\{ \begin{array}{l} \min\{i \vert a_i-a_0 > b_k - b... ...-1}\}) n+1 \quad \mbox{ if } a_n-a_0 \leq b_k - b_{j-1} \end{array} \right.$$ (3)

   
   
   $f(k,j)=n+1$ is a special case indicating that the height of graph $B(j)$ above 0 exceeds that of $A'$ so that no more packet $b_i (i>k)$ of $B$ is needed for computing $M(i,j)$ for $j$.

2. Compute $g(k,j),l(k,j),$ and $M(k,j)$
   We then compute $M(k,j)$, the area surrounded by graph $A'$ and $B(j)$ in the range $[0, b_k- b_{j-1}]$ on the Y-axis. $g(k,j)$ is the maximum difference and $l(k,j)$ is the minimum difference on the X-axis between $A'$ and $B(j)$ in the range $[0, b_k- b_{j-1}]$ on the Y-axis. These values are computed using the values at $k-1$ by the following equations.

   $$\begin{split}M(k,j) &= M(k-1,j) &+ \left\{ \begin{array}{ll... ...box{ if } f(k-1,j)=f(k,j)=n+1 \end{array} \right. \end{split}$$ (4)

   $$\begin{split}L(k,j) &= v(b_k,a_{f(k-1,j)}) \times \left((a_{f... ...times \left((b_k-b_{j-1})-(a_{f(k,j)-1}-a_0)\right) \end{split}$$ (5)

   $$\begin{split}g(k,j)&=\max\{g(k-1,j), \max\{v(b_k,a_i) \vert... ...n\{v(b_k,a_i) \vert f(k-1,j)\leq i\leq f(k,j)\}\} \end{split}$$ (6)

   
   
   If $f(k,j)=n+1$, the last term of (6) is not added, and $v(b_k,a_{n+1})$ is not counted in (7) either.

3. Compute $\hat{M}(k,j)$
   If $f(k,j)=n+1$, the range on the Y-axis of $B(j)$ covers the range on the Y-axis of $A'$ ( $[0,d] \subset [b_0-b_{j-1}, b_k-b_{j-1}]$). Then $M'(k,j)$, the area surrounded by two graphs when we move $B(j)$ as close as possible to $A'$ along the X-axis, can be computed by the following equation.
   $$M'(k,j) = \min\{ \left\vert M(k,j) - g(k,j)\times d \right\vert, \left\vert M(k,j) - l(k,j) \times d \right\vert \}$$

   We can compute $\hat{M}(k,j) = \min\{M'(k,i) \vert i\leq j\}$ by the following equation if either $\hat{M}(k,j-1)$ or $M'(k,j)$ is defined.

   $$\hat{M}(k,j) = \min\{\hat{M}(k,j-1), M'(k,j)\}$$ (7)

   
   

After all the computations for $j=1,2,\ldots,k$ are done, $M(k)=\hat{M}(k,k)$ is the area surrounded by graph $A$ and $B(k)$ in the vertical range of $A$ when $B(k)$ moves horizontally and vertically without crossing $A$ so that $B(k)$ is as close as possible to $A$, where $B(k)$ is a sub array of $B$:

$$B(k):\quad \langle (u(b_1),b_1), \ldots, (u(b_k), b_k) \rangle$$

If $M(k)$ is not defined, the deviation for $B(k)$ from $A$ cannot be defined. We can delete objects (such as $b_j$) associated with $j=j_k-1,j_k+1,\ldots,j_{k+1}-2$ allocated in memory except for the ones at the minimum. ($j_k$ is defined by $j_{k+1} = \max\{j \vert j\geq j_k, f(k,j)=n+1\}$    or $j_{k+1}=1$    if $f(k,1)\leq n$.)

When we have finished processing all the packets in $B$ and suppose the number of packets in $B$ is $m$, the deviation for $B$ from $A$ is obtained by $M(m)/d$.