The effects of picker-oriented operational factors on hand-off delay in a bucket brigade order picking system

Abstract

During collaboration between neighboring pickers in a bucket brigade order picking, the downstream picker frequently experiences productivity loss in the form of a hand-off delay while waiting for a new tote from the upstream picker. This paper proposes an analytical model to quantify the hand-off delay of downstream pickers under non-deterministic pick times and non-instantaneous walk times of the upstream pickers. Numerical analyses show the effects of the magnitude of pick time, the variation of pick time, and the forward walk time on the hand-off delay, and simulation models show the effects of the number of pickers and their skill differences. We conclude that stable pick times are especially important to reduce hand-off delays and that slowest-to-fastest picker assignments offset hand-off delays with the blocking delays.

Appendix

1.1 The interdependency in the hand-off location and the hand-off delay

Let X _i be the ith hand-off. Then, we evaluate the interdependency in the hand-off location and the hand-off delay. We conduct a simulation to observe the hand-off location and the hand-off delay between ith hand-off and i + 1st hand-off. Figure 13 shows the scatter plot of ith hand-off and the i + 1st hand-off. Figure 13a indicates the correlation between the hand-off locations, which is about − 0.47 correlation value. The correlation coefficient is relatively high as shown in the scatter plot comparison. Vice versa, in the scatter plot of the hand-off delay between ith hand-off and i + 1st hand-off, we look for a clear linear trend, which is close to a relative randomness. The correlation coefficient appears smaller than 0.1.

Fig. 13

Scatter plot of X _i and X_i+1 in n = 50, p = 0.1, m = 10, and unit pick time. a Hand-off location b hand-off delay

The autocorrelation analysis in Fig. 14 generalizes the result above. In Fig. 14a, the autocorrelation analysis of the hand-off location shows relatively high autocorrelation values and patterns over the increase in lags. However, in Fig. 14b, the autocorrelation analysis of the hand-off delay shows smaller values and it appears that there is little likelihood of dependence between delays.

Fig. 14

Autocorrelation plot over lags in n = 50, p = 0.1, and unit pick time. a Hand-off location b hand-off delay

An experiment over larger p and other pick time distribution in Figs. 15 and 16 confirms the observations. We consider an exponential pick time situation with higher pick probability. The similar results repeat over the hand-off locations and the hand-off delay. Consistently, the scatter plots and autocorrelation plots show a relatively high dependency in the hand-off location and a relatively high randomness in the hand-off delay.

Fig. 15

Scatter plot of X _i and X_i+1 in n = 50, p = 0.5, m = 1, and exp(1) pick time a hand-off location b hand-off delay

Fig. 16

Autocorrelation plot over lags in n = 50, p = 0.1, m = 1, and Exp(1.0) pick time. a Hand-off location b hand-off delay

In conclusion, the dependency in the hand-off location does not affect the hand-off delay under the stochastic pick time and the finite walk time. Intuitively, hand-off locations “remember” their last hand-offs while the arrival times of downstream pickers do not repeat the previous hand-off times in the stochastic picking.

1.2 Detailed calculation of the expected waiting time (E[W])

Excess-time model The expected waiting time (E[W]) is E[PT²]/2E[PT] when pickers return randomly, the walk speed is infinite, and there is no blocking delay.

Proof

The renewal process is derived based on the definition in Ross (2010). It follows that there is a renewal process {N(t), t ≥ 0} having inter-arrival times PT _n , n ≥ 1, and assuming a hand-off when a downstream picker meets an upstream picker. We assume the hand-off to be a reward. We define W(t) to equal the time from t until the next stop, a renewal, and, as such, it represents the remaining (residual or excess) life of the pick at time t. Over the renewal process {PT(t), t ≥ 0}, we measure the time of W(t) as the excess time Y(t) = A_n+1(t) – t.

Then, according to the renewal reward process (Ross 2010), the time to completion at time t is the time from t until the next renewal and becomes the remaining time to completion at time t. We assume the new waiting time t ₀ be a reward during a renewal cycle. We write the average value of the excess as,

$$ \begin{aligned} E\left[ W \right] & \equiv \mathop {\lim }\limits_{s \to \infty } \frac{{\int_{0}^{s} {W(t){\text{d}}t} }}{s} = \frac{{E\left[ {{\text{ Reward}}\;{\text{during}}\;{\text{a}}\;{\text{renewal}}\;{\text{cycle }}} \right]}}{{E\left[ {{\text{ Time}}\;{\text{of}}\;{\text{a}}\;{\text{renewal}}\;{\text{cycle}}} \right]}} \\ & = \frac{{E\left[ {{\text{ Hand - off delay}}\;{\text{during}}\;{\text{a}}\;{\text{stop }}} \right]}}{{E\left[ {{\text{ Pick}}\;{\text{time}}\;{\text{of}}\;{\text{a}}\;{\text{stop}}} \right]}} \\ \end{aligned} $$

In addition, over the given cycle length PT, we obtain the following expected value,

$$ \begin{aligned} E\left[ {{\text{Hand - off}}\;{\text{delay}}\;{\text{during}}\;{\text{a}}\;{\text{stop}}} \right] & = E\left[ {\int_{0}^{PT} {W(t) \cdot {\text{d}}t} } \right] \\ & = E\left[ {\int_{0}^{PT} {\left( {PT - t} \right) \cdot {\text{d}}t} } \right] \, \\ \end{aligned} $$

The expectation for the overall PT ranges from 0 to infinite and PT has a distribution F,

$$ \begin{aligned} E\left[ {\int_{0}^{PT} {\left( {PT - t} \right) \cdot {\text{d}}t} } \right] & = E\left[ {\int_{0}^{PT} {\left( {PT - t} \right) \cdot {\text{d}}t} } \right] = E\left[ {\left. {\left( {PTt - \frac{{t^{2} }}{2}} \right)} \right|_{0}^{PT} } \right] = E\left[ {\frac{{PT^{2} }}{2}} \right] \\ & = \frac{{E\left[ {PT^{2} } \right]}}{2} \\ \end{aligned} $$

where $\, E\left[ {\int_{0}^{X} {\left( {X - t} \right) \cdot {\text{d}}t} } \right] \, = \, \int_{0}^{\infty } {\int_{0}^{X} {\left( {X - t} \right) \cdot {\text{d}}t} \cdot {\text{d}}F\left( X \right)}$.

The time of a pick cycle is an inter-arrival time between the picks having the distribution function F. Then, $E\left[ {{\text{Pick}}\;{\text{time}}\;{\text{of}}\;{\text{a}}\;{\text{stop}}} \right] = E\left[ {PT} \right]$.

Thus, we express the expected waiting time (E[W]) as,

$$ E[W] = \frac{{E\left[ {{\text{Hand - off}}\;{\text{delay}}\;{\text{during}}\;{\text{a}}\;{\text{stop}}} \right]}}{{E\left[ {{\text{Pick}}\;{\text{time}}\;{\text{of}}\;{\text{a}}\;{\text{stop}}} \right]}} = \frac{{E\left[ {PT^{2} } \right]}}{{2 \cdot E\left[ {PT} \right]}} $$

□.

1.3 Proof of Theorem 1

When a downstream picker returns randomly and meets an upstream picker who is picking, the expected waiting time of the downstream picker (E[HD _p ]) is

$$ E[HD_{p} ] = \frac{{E\left[ {PT^{2} } \right]}}{{2 \cdot E\left[ {PT} \right]}} = \frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1) $$

Proof

The expected delay time per hand-off occurrence (HD _p ) is derived when an upstream picker is picking by replacing W with HD _p in Fig. 5. First, we estimate E[PT]. Given these pick and walk decision patterns, the expected pick time as a stop (PT) is equal to the expected number of picks at a pick face · unit pick time (τ) = the expected number of picks at a pick face. Formulate E[PT] as:

$$ \begin{aligned} E[PT] & = pq + 2p^{2} q + 3p^{3} q + \cdots + (m - 1)p^{m - 1} q + mp^{m} \\ & = p(1 - p) + 2p^{2} (1 - p) + 3p^{3} (1 - p) + \cdots + (m - 1)p^{m - 1} (1 - p) + mp^{m} \\ & = p + ( - p^{2} + 2p^{2} ) + ( - 2p^{3} + 3p^{3} ) + \cdots + ( - (m - 1)p^{m} + mp^{m} ) \\ & = p + p^{2} + p^{3} + \cdots + p^{m - 1} + p^{m} \\ & = \sum\limits_{i = 1}^{m} {p^{i} } = \frac{{p(1 - p^{m} )}}{(1 - p)}. \\ \end{aligned} $$

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Second, E[PT²] becomes

$$ \begin{aligned} E[PT^{2} ] & = \sum\limits_{i = 1}^{m} {i^{2} p^{i} q} \\ & = pq + 2^{2} p^{2} q + 3^{2} p^{3} q + \cdots + (m - 1)^{2} p^{m - 1} q + m^{2} p^{m} \\ & = p(1 - p) + 2^{2} p^{2} (1 - p) + 3^{2} p^{3} (1 - p) + \cdots + (m - 1)^{2} p^{m - 1} (1 - p) + m^{2} p^{m} \\ & = (p - p^{2} ) + (2^{2} p^{2} - 2^{2} p^{3} ) + \cdots + ((m - 1)^{2} p^{m - 1} - (m - 1)^{2} p^{m} ) + mp^{m} \\ & = p + ( - p^{2} + 2^{2} p^{2} ) + ( - 2^{2} p^{3} + 3^{2} p^{3} ) + \cdots + ( - (m - 1)^{2} p^{m} ) + m^{2} p^{m} ) \\ & = \sum\limits_{i = 1}^{m} {(2i - 1)p^{i} } = 2\sum\limits_{i = 1}^{m} {i \cdot p^{i} } - \sum\limits_{i = 1}^{m} {p^{i} } \\ & = \frac{2}{1 - p}\left[ {\frac{{p(1 - p^{m} )}}{(1 - p)} - m \cdot p^{m + 1} } \right] - \frac{{p(1 - p^{m} )}}{(1 - p)} \\ & = \frac{{p(1 - p^{m} )(p + 1)}}{{(1 - p)^{2} }} - \frac{{2m \cdot p^{m + 1} }}{(1 - p)}. \\ \end{aligned} $$

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Finally, we obtain the expected hand-off delay during picking(HD _p ) as

$$ \begin{aligned} E[HD_{p} ] & = \frac{{E\left[ {PT^{2} } \right]}}{{2 \cdot E\left[ {PT} \right]}}\quad \quad \quad \quad \quad {\text{from }}\left( {12} \right)\;{\text{and}}\;\left( {13} \right) \\ & = \frac{{\frac{{p(1 - p^{m} )(p + 1)}}{{(1 - p)^{2} }} - \frac{{2m \cdot p^{m + 1} }}{(1 - p)}}}{{2\frac{{p(1 - p^{m} )}}{(1 - p)}}} \\ & = \frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}} \\ \end{aligned} $$

□.

1.4 Calculation of Eq. (8)

E[HD _w ] becomes

$$ \begin{aligned} E[HD_{w} ] & = p\left( {E[PT] + \frac{{E[FT^{2} ]}}{2E[FT]}} \right) + (1 - p)\frac{{E[FT^{2} ]}}{2E[FT]} \\ & = p\left( {\sum\limits_{i = 1}^{m} {p^{i} } + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)\quad \quad \quad \quad {\text{from}}\;\left( {12} \right),E\left[ {FT} \right] \, = ft,\;{\text{and}}\;E\left[ {FT^{2} } \right] \, = ft^{2} \\ & = p\left( {\frac{{p(1 - p^{m} )}}{(1 - p)} + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right) \\ \end{aligned} $$

1.5 Calculation of Eq. (9)

The expected delay time per hand-off occurrence and the expected delay time per order are derived when there are instantaneous walk times, a large number of picks, and no picker blocking. Obtain the expected hand-off delay as

$$ \begin{aligned} E[HD] & = E\left[ {HD_{p} } \right]\frac{E[PT]}{E[PT] + E[FT]} + E\left[ {HD_{w} } \right]\frac{E[FT]}{E[PT] + E[FT]} \\ &= \left[ {\frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}} \right]\frac{{\frac{{p(1 - p^{m} )}}{1 - p}}}{{\frac{{p(1 - p^{m} )}}{1 - p} + ft}} + \left[ {\frac{{p^{2} (1 - p^{m} ) + ft(1 - p)}}{2(1 - p)}} \right]\frac{ft}{{\frac{{p(1 - p^{m} )}}{1 - p} + ft}} \\ & = \left[ {\frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}} \right]\frac{{\frac{{p(1 - p^{m} )}}{1 - p}}}{{\frac{{p(1 - p^{m} )}}{1 - p} + ft}} + \left[ {\frac{{p^{2} (1 - p^{m} ) + ft(1 - p)}}{2(1 - p)}} \right]\frac{ft}{{\frac{{p(1 - p^{m} )}}{1 - p} + ft}} \\ & = \left[ {\frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}} \right]\frac{{\frac{{p(1 - p^{m} )}}{1 - p}}}{{\frac{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}}{1 - p}}} + \left[ {\frac{{p(1 - p^{m} )}}{(1 - p)}p + \frac{ft}{2}} \right]\frac{ft}{{\frac{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}}{1 - p}}} \\ & = \left[ {\frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}} \right]\frac{{p(1 - p^{m} )}}{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}} + \left[ {\frac{{p^{2} (1 - p^{m} ) + ft(1 - p)}}{2(1 - p)}} \right]\frac{{ft\left( {1 - p} \right)}}{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}} \\ & = \left[ {\frac{p + 1}{2(1 - p)} - \frac{{m \cdot p^{m} }}{{(1 - p^{m} )}}} \right]\frac{{p(1 - p^{m} )}}{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}} + \frac{ft}{2}\frac{{p^{2} (1 - p^{m} ) + ft(1 - p)}}{{p(1 - p^{m} ) + ft\left( {1 - p} \right)}} \\ \end{aligned} $$

1.6 Proof of Theorem 3

From Eq. (9), we define the lower bound and upper bound of the HD. The lower bound inequality holds by simplifying $E[PT^{2} ] = \sum\nolimits_{i = 1}^{m} {i^{2} \cdot p^{i} q} + m^{2} p^{m + 1} \ge m^{2} p^{m + 1}$ as follows:

$$ \left[ {\frac{{m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}} \right]\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {\sum\limits_{i = 1}^{m} {p^{i} } + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{E[FT]}{E[PT] + E[FT]} \le E[HD] $$

The following limiting property shows that the lower bound is bounded by (m + ft)/2.

$$ \begin{aligned} &\mathop {\lim }\limits_{p \to 1} \left[ {\frac{{m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}\frac{{\sum\nolimits_{i = 1}^{m} {p^{i} } }}{{\sum\nolimits_{i = 1}^{m} {p^{i} } + ft}} + \left[ {p\left( {\sum\nolimits_{i = 1}^{m} {p^{i} } + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{ft}{{\sum\nolimits_{i = 1}^{m} {p^{i} } + ft}}} \right] \\ & \mathop { = \lim }\limits_{p \to 1} \left[ {\frac{{m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}\frac{m}{m + ft} + \left[ {\left( {m + \frac{ft}{2}} \right)} \right]\frac{ft}{m + ft}} \right] \\ & = \left[ {\frac{{m^{2} }}{2m}} \right]\frac{m}{m + ft} + \left( {m + \frac{ft}{2}} \right)\frac{ft}{m + ft} \\ & = \left[ {\frac{m}{2}} \right]\frac{m}{m + ft} + \left( {\frac{2m + ft}{2}} \right)\frac{ft}{m + ft} \\ & = \frac{{m^{2} + 2m \cdot ft + ft^{2} }}{{2\left( {m + ft} \right)}} = \frac{{\left( {m + ft} \right)^{2} }}{{2\left( {m + ft} \right)}} = \frac{m + ft}{2} \\ \end{aligned} $$

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Similarly, the upper bound inequality holds by replacing $E[PT^{2} ] = \sum\nolimits_{i = 1}^{m} {i^{2} \cdot p^{i} q} + m^{2} p^{m + 1} \le q\sum\nolimits_{i = 1}^{m} {i^{2} } + m^{2} p^{m + 1}$ as follows:

$$ E[HD] \le \left[ {\frac{{q\sum\nolimits_{i = 1}^{m} {i^{2} } + m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}} \right]\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {\sum\limits_{i = 1}^{m} {p^{i} } + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{E[FT]}{E[PT] + E[FT]} $$

The upper bound also is bounded by (m + ft)/2 (see “Proof of Theorem 3” section in Appendix for details):

$$ \begin{aligned} &\mathop {\lim }\limits_{p \to 1} \left[ {\frac{{q\sum\nolimits_{i = 1}^{m} {i^{2} } + m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}\frac{pt}{pt + ft} + \left[ {p\left( {\sum\nolimits_{i = 1}^{m} {p^{i} } + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{ft}{pt + ft}} \right] \\ & = \mathop {\lim }\limits_{p \to 1} \left[ {\frac{{q\frac{m(m + 1)(2m + 1)}{6} + m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}\frac{pt}{pt + ft} + \left[ {p\left( {m + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{ft}{{\sum\nolimits_{i = 1}^{m} {p^{i} } + ft}}} \right] \\ & = \mathop {\lim }\limits_{p \to 1} \left[ {\frac{{(1 - p)\frac{m(m + 1)(2m + 1)}{6} + m^{2} \cdot p^{m + 1} }}{{2\sum\nolimits_{i = 1}^{m} {p^{i} } }}\frac{pt}{pt + ft} + \left[ {p\left( {m + \frac{ft}{2}} \right) + (1 - p)\left( {\frac{ft}{2}} \right)} \right]\frac{ft}{{\sum\nolimits_{i = 1}^{m} {p^{i} } + ft}}} \right] \\ & = \left[ {\frac{{m^{2} }}{2m}} \right]\frac{m}{m + ft} + \left( {m + \frac{ft}{2}} \right)\frac{ft}{m + ft} = \frac{m + ft}{2} \\ \end{aligned} $$

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As both the lower bound (Eq. (14) and the upper bound (Eq. (15)) converge to the same value (m + ft)/2, E[HD] results in (m + ft)/2:

$$ \begin{aligned} & \frac{m + ft}{2} \le \mathop {\lim }\limits_{p \to 1} E[HD] \le \frac{m + ft}{2}, \\ & \mathop {\lim }\limits_{p \to 1} E[HD] = \frac{m + ft}{2} \\ \end{aligned} $$

□.

1.7 General distributions and calculation steps of Eq. (11)

$$ \begin{aligned} E[HD] & = \frac{{E[PT^{2} ]}}{2E[PT]}\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {E[PT] + \frac{{E[FT^{2} ]}}{2E[FT]}} \right) + (1 - p)\frac{{E[FT^{2} ]}}{2E[FT]}} \right]\frac{E[FT]}{E[PT] + E[FT]} \\ & = \frac{{E[PT^{2} ]}}{2E[PT]}\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {E[PT] + \frac{E[FT]}{2}} \right) + (1 - p)\frac{E[FT]}{2}} \right]\frac{E[FT]}{E[PT] + E[FT]} \\ & = \frac{{E[PT]^{2} + Var[PT]}}{2E[PT]}\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {E[PT] + \frac{E[FT]}{2}} \right) + (1 - p)\frac{E[FT]}{2}} \right]\frac{E[FT]}{E[PT] + E[FT]} \\ & = \left[ {\frac{E[PT]}{2} + \frac{Var[PT]}{2E[PT]}} \right]\frac{E[PT]}{E[PT] + E[FT]} + \left[ {p\left( {E[PT] + \frac{E[FT]}{2}} \right) + (1 - p)\frac{E[FT]}{2}} \right]\frac{E[FT]}{E[PT] + E[FT]} \\ \end{aligned} $$

1.8 General distributions and calculation steps

If the pick duration follows the uniform distribution of Unif(a,b) where b ≥ a ≥ 0.

$$ \begin{aligned} E[HD_{\text{unif}} ] & = \left[ {\frac{{p((b - a)^{3} /6)}}{2p(a + b)/2}} \right]\frac{p}{p + ft} + \left[ {p\frac{a + b}{2} + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ & = \left[ {\frac{{(b - a)^{3} }}{6(a + b)}} \right]\frac{p}{p + ft} + \left[ {p\frac{a + b}{2} + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ \end{aligned} $$

If the pick duration follows the exponential distribution of mean = a where a > 0.

$$ \begin{aligned} E[HD_{\exp } ] & = \left[ {\frac{p(2a)}{2pa}} \right]\frac{p}{p + ft} + \left[ {pa + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ & = \frac{p}{p + ft} + \left[ {pa + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ \end{aligned} $$

If the pick duration follows the triangular distribution of triangular(a,b,c) where a ≥ c ≥ b ≥ 0.

$$ \begin{aligned} E[HD_{\text{tri}} ] & = \left[ {\frac{{p\left[ {\frac{{a^{2} + b^{2} + c^{2} - ab - bc - ca}}{18} + \left( {\frac{a + b + c}{3}} \right)^{2} } \right]}}{{2p\left( {\frac{a + b + c}{3}} \right)}}} \right]\frac{p}{p + ft} + \left[ {p\left( {\frac{a + b + c}{3}} \right) + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ & = \left[ {\frac{{p\left[ {\frac{{a^{2} + b^{2} + c^{2} + ab + bc + ca}}{6}} \right]}}{{2p\left( {\frac{a + b + c}{3}} \right)}}} \right]\frac{p}{p + ft} + \left[ {p\left( {\frac{a + b + c}{3}} \right) + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ & = \left[ {\frac{{a^{2} + b^{2} + c^{2} + ab + bc + ca}}{4(a + b + c)}} \right]\frac{p}{p + ft} + \left[ {p\left( {\frac{a + b + c}{3}} \right) + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ \end{aligned} $$

There are numerical examples of three pick time distributions: Uni = uniform [min,max] = uniform [a,b] = [0.0, 2.0]; Exp = exponential [mean] = exp[a] = [1.0]; and Tri = triangular [min, mode, max] = triangular [a, b, c] = [0.5, 1.0, 1.5], where the time unit represents the time spent to retrieve an item.

$$ \begin{aligned} E[HD_{unif} ] & = \left[ {\frac{2}{3}} \right]\frac{p}{p + ft} + \left[ {p + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ E[HD_{\exp } ] & = \frac{p}{p + ft} + \left[ {p + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ E[HD_{tri} ] & = \left[ {\frac{6.25}{12}} \right]\frac{p}{p + ft} + \left[ {p + \frac{ft}{2}} \right]\frac{ft}{p + ft} \\ \end{aligned} $$