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An analytical model for designing yard layouts of a straddle carrier based container terminal

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Abstract

In designing a yard layout for a container terminal several decisions have to be made. In this paper we propose a model which provides decision support for the design of yard layouts of terminals at which straddle carrier are used. We assume that straddle carriers are used for the horizontal transport and the stacking of containers. For the proposed model we develop estimates for the expected cycle distances of straddle carriers. In this case, we distinguish between cycles to landside facilities and to the quay. Numerical results are presented for several parameter settings. For instance, we present results for a comparison of layouts where the rows in the block are orientated parallel with layouts where the rows are orientated perpendicularly to the quay.

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Acknowledgments

The authors would like to thank the anonymous referees for their useful comments on an earlier version of the paper.

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Correspondence to Jörg Wiese.

Appendices

Appendix 1: Overview of parameters and variables

The following tables summarize the used parameters (see Table 5), sets (see Table 6) and variables (see Table 7).

Table 5 Used parameters and their default values
Table 6 Used sets
Table 7 Used variables

Appendix 2: Landside-cycle estimate for perpendicular layouts

In this section we derive estimates for the expected horizontal distance of landside cycles in a perpendicular layout. Therefore, we first consider the case where b h  = 1 and b v  = 1 as in Fig. 23. As with the seaside cycles we consider the distance to single rows. In case where b h  = 1 and b v  = 1 the expected distance using the first driving strategy can be expressed by

$$ E^{(6)}_1=\left\{\begin{array}{ll} \sum\limits_{k=1}^{\frac{r}{2}} \frac{2}{r} 2k \times W=\frac{W(r + 2)}{2} & \hbox{when }r\hbox{ is even},\\ \sum\limits_{k=1}^{\frac{r-1}{2}} \frac{2}{r} 2k \times W=\frac{W(r^2-1)}{2r} & \hbox{when }r\hbox{ is odd}. \end{array}\right. $$
(46)

When r is odd we assume that the distance to the closest row is zero with a probability of \(\frac{1}{r}\). For the second driving strategy we assume for sake of brevity an additional distance similar to that in case of seaside cycle of \(\frac{AY^L}{2b_h}\) (see Sect. 4) which leads to

$$ E^{(6)}_2=\left\{\begin{array}{ll} \frac{W(r+2)}{2}+\frac{AY^L}{4b_h}&\hbox{when }r\hbox{ is even},\\ \frac{W(r^2-1)}{2r}+\frac{AY^L}{4b_h}&\hbox{when }r\hbox{ is odd}. \end{array}\right. $$
(47)
Fig. 23
figure 23

Horizontal distances for a SC landside cycle in a perpendicular layout for the first driving strategy (solid lines) and the second driving strategy (dashed lines)

Figure 24 shows the horizontal distances when b h  ≥ 1, b v  ≥ 1, and b h is even. In this case the horizontal distances are identical for both driving strategies (see Fig. 24). We consider the travel distance from the center of the TSA to the center of the blocks and back to the TSA. The travel distance to block k (with \(k=1,\ldots, \frac{b_h}{2}\)) is \(\frac{2(k-1)AY^L+AY^L}{b_h}\) and has a probability of \(\frac{2}{b_h}\). Thus

$$ E^{(7)}_o=\sum_{k=1}^{\frac{b_h}{2}} \frac{2}{b_h} \frac{2(k-1)AY^L+AY^L}{b_h}=\frac{AY^L}{2}. $$
(48)
Fig. 24
figure 24

Horizontal distances for a SC landside cycle in a perpendicular layout for the first driving strategy (solid lines) and the second driving strategy (dashed lines) with b h even

Figure 25 shows the horizontal distances in case where b h  ≥ 1, b v  ≥ 1 and b h is odd. Where b h  = 1 and b v  > 1 the estimate \(E^{(6)}_o\) has to be extended as with estimates for the seaside cycles. The distance to the lower block (the block closer to the TSA) is \(E^{(6)}_o\) with a probability of \(\frac{1}{b_v}\), and the distance to any other block is \(\frac{2AY^L}{b_h}\) with a probability of \(\frac{b_v-1}{b_v}\). Thus

$$ E^{(8)}_o=\frac{1}{b_v} E^{(6)}_o+\frac{b_v-1}{b_v} \frac{2AY^L}{b_h}. $$
(49)

Where b h  > 1 the distance to the blocks above the TSA is expressed by E (8) o with a probability of \(\frac{1}{b_h}\). The distance to the other blocks k (\(k=1,\ldots,\frac{b_h-1}{2}\)) is \(\frac{2kAY^L}{b_h}\) with a probability of \(\frac{2}{b_h}\). Hence

$$ E^{(9)}_o=\frac{1}{b_h} E^{(8)}_o+\sum_{k=1}^{\frac{b_h-1}{2}} \frac{2}{b_h} \frac{2kAY^L}{b_h}=\frac{1}{b_h} E^{(8)}_o+\frac{AY^L(b_h^2-1)}{2b_h^2}. $$
(50)

Please note that E (9) o reduces to E (8) o when b h  = 1 and to \(E^{(6)}_o\) when b v  = 1. This leads to the following expected horizontal distance for a perpendicular layout with a TSA in the middle of the yard:

$$ Dh^{pe,HM}_{o}=\left\{\begin{array}{ll} E^{(7)}_o &\hbox{ when }b_h\hbox{ is even},\\ E^{(9)}_o &\hbox{ when }b_h\hbox{ is odd}. \end{array}\right. $$
(51)
Fig. 25
figure 25

Horizontal distances for a SC landside cycle in a perpendicular layout for the first driving strategy (solid lines) and the second driving strategy (dashed lines) with b h odd

Where we assume a TSA in the left corner of the yard, as in Fig. 26, we need to change the horizontal distance estimates. We first consider the case where b h  = 1. In this case we consider each individual row of the block. We assume that the distance of a SC from the center of the TSA to the first row is zero. Consequently the distance to the second row is 2 × W for both driving strategies (see Fig. 26). The probability that a SC has to travel to a specific row is \(\frac{1}{r}\). Thus

$$ E^{(10)}_o=\sum^r_{k=1} \frac{1}{r} 2(k-1) \times W=W (r-1). $$
(52)
Fig. 26
figure 26

Horizontal distances for a SC landside cycle in a perpendicular layout for the first driving strategy (solid lines) and the second driving strategy (dashed lines) with a TSA in the lower left corner and b h  = 1

The case where b h  > 1 is shown in Fig. 27. We assume that the distance from the middle of the TSA to the middle of the first block (and back) is \(\frac{AY^L}{b_h}\). The distance to the second block is \(\frac{3AY^L}{b_h}\). The probability that a SC has to travel to a specific block is \(\frac{1}{b_h}\). Hence

$$ E^{(11)}_o=\sum^{b_h}_{k=1} \frac{1}{b_h} \frac{2(k-1)AY^L+AY^L}{b_h}=AY^L. $$
(53)

In sum, this leads to

$$ Dh^{pe,HL}_{o}=\left\{\begin{array}{ll} E^{(10)}_o &\hbox{ when }b_h=1,\\ E^{(11)}_o &\hbox{ when }b_h>1. \end{array}\right. $$
(54)
Fig. 27
figure 27

Horizontal distances for a SC landside cycle in a perpendicular layout for the first driving strategy (solid lines) and the second driving strategy (dashed lines) with a TSA in the lower left corner and b h  > 1

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Wiese, J., Suhl, L. & Kliewer, N. An analytical model for designing yard layouts of a straddle carrier based container terminal. Flex Serv Manuf J 25, 466–502 (2013). https://doi.org/10.1007/s10696-011-9132-1

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