MATERIAL-HANDLING SYSTEMS:INDUSTRIAL TRUCKS

INDUSTRIAL TRUCKS

There is a wide range of industrial trucks available. Going from simple to sophisticated and from wide-aisle to narrow-aisle trucks, a possible list would be as follows: (1) hand truck / cart, (2) pallet jack, (3) walkie stacker, (4) pallet / platform truck, (5) counterbalanced lift truck, (6) straddle truck,

(7) reach truck, (8) sideloader, (9) turret truck, (10) storage / retrieval truck, and (11) order picker truck, among others. Examples of 2, 3, 5, 6, 8, 9, and 11 are shown in Figures 1, 2, 3, 4, 5, 6, and

7, respectively. The simplest truck with a mast (which allows the load to be raised and lowered) is a walkie stacker. The most common truck found in industry, however, is the counterbalanced lift truck, which still requires wide aisles due to its turning radius, and the fact that the truck must face the rack to store or retrieve the load.

Due to its shorter size, the straddle truck has a smaller turning radius than the counterbalanced lift truck; however, outriggers (‘‘long feet’’) are added to compensate for load weight. Also, the straddle truck still has to face the rack to store / retrieve a load, which results in wide aisles. A reach truck is very similar to a straddle truck except that it is has relatively shorter outriggers and it is equipped with a pantograph mechanism that allows the truck to insert / remove a load into / from the rack without moving the truck back and forth; the truck still has to face the rack, however.

To further reduce aisle width, the sideloader truck was introduced. This truck faces only one side of the aisle; aisle width is reduced by eliminating any turns inside the aisle. It is also especially suited for handling long stock such as lumber or bar stock. Its primary drawback, of course, is the fact that in order to serve the opposite side of the aisle, the truck must leave the aisle, turn around, and reenter facing the opposite side. Serving both sides of the aisle at the same time, while requiring no turns inside the aisle, became possible with the turret truck, which resembles a counterbalanced lift truck but is able to store / retrieve loads without facing the rack. This is accomplished by giving the turret truck a swing mast or swing forks: the body of the truck remains stationary (facing forward in the aisle) while the load is turned to face the rack and the operator remains at ground level. As rack heights increased to better utilize land and cubic space, it became necessary to raise the operator along with the load, which led to the storage / retrieval truck.

Another truck that raises the operator with the load is the order picker truck. However, unlike the above trucks which are designed primarily for unit load in and unit load out, an order picker truck allows the operator to start with an empty pallet (or magazine) and stack various containers or parts on it as he / she picks the appropriate parts / products from unit loads stored in the rack.

A summary of basic features of various industrial trucks, including cost, is presented in Table 1, adapted from (Tompkins et al. 1996). The reader should keep in mind that truck prices and features change over time; nevertheless, the information provided in Table 1 is useful for comparison purposes.

An industrial truck is basically used for moving loads in and out of storage (i.e., they perform a storage / retrieval function) and / or moving loads from one point to other point(s). Counterbalanced lift trucks, and similar low-mast trucks, are also used for loading / unloading trailers at shipping / receiving docks. In the first case, a fundamental question that arises is the throughput capacity of a truck; that is, how many storages and retrievals per hour can a particular truck perform? In the second case, a similar question posed in a slightly different fashion would be: given the throughput require- ment (the number of loads that must be moved per time unit), how many trucks do we need? Both questions may appear simple on the surface, but they are not straightforward to answer.

Consider first the storage / retrieval case. The throughput capacity of the truck depends on the rack size, the storage policy, the travel speed and type of truck, and the load pick-up / deposit times. Assuming a unit load in, unit load out operation, a truck will handle a load either on a single command (SC) or dual command (DC) basis. With SC operation, the truck starts at the input / output (I / O) point, which is typically located at the lower left-hand corner of the rack, travels to the appropriate rack opening, deposits (or picks up) the load, and returns to the I / O point. With SC storage, the truck picks up the load from the I / O point and returns empty to the I / O point. With SC retrieval, the truck travels empty from the I / O point to the rack, picks up the load, and brings it to the I / O point, where it is deposited.

To avoid empty travel to / from the I / O point, a truck can perform a DC cycle, provided that a storage request and a retrieval request are present at the same time. To perform a DC cycle, the truck starts at the I / O point, picks up the load to be stored, travels to an empty rack opening to deposit the load, then travels directly to the rack opening where the load to be retrieved is located, picks up that load, travels to the I / O point and deposits the load. Hence, two loads are handled on each DC cycle and empty travel occurs only from the storage point to the retrieval point.

The throughput capacity of a truck (operating within an aisle) depends on the expected time it takes to perform a SC and a DC cycle; it also depends on what fraction of time the truck performs a SC or DC cycle. Suppose the storage rack is H ft high and L ft long. Further suppose the truck’s lift speed (or vertical speed) is v fpm and its travel speed (or horizontal speed) is h fpm. If the truck is capable of concurrent (i.e., simultaneous) travel in the horizontal and vertical directions, Bozer and White (1984) have shown that the expected SC travel time, say E(SC), and the expected DC travel time, say E(DC), can be computed from the following equations, assuming that randomized storage is used (i.e., a load is equally likely to be stored in or retrieved from any location in the rack) and the I / O point is located at the lower left-hand corner of the rack:

open location closest in time to the I / O point.) If the average rack utilization is high, the above expected travel times are reasonably accurate (not considering acceleration / deceleration of the truck). If the average rack utilization falls below 90% or so, however, the above expected travel times are likely to overestimate the actual expected travel times since the COL rule tends to favor rack openings close to the I / O point, while randomized storage always picks each opening in the rack with equal probability.

Concurrent travel is possible whenever the lift motor (vertical motor) is separate from the travel motor (horizontal motor). While concurrent travel is often limited when the forks are raised (due to safety considerations), most trucks follow a combination of concurrent and sequential travel while operating. A conservative approach would be to assume 100% sequential travel; that is, the truck first moves horizontally down the aisle, once it stops, then the forks are raised (for vertical travel).

With sequential travel, E(SC) and E(DC) can be computed from the following equations assuming randomized storage:

Often the load pick-up / deposit time is constant and P = D.

As an example, consider a rack 20 ft tall (H = 20) and 100 ft long (L = 100). The I / O point is assumed to be located at the lower left-hand corner of the rack. Suppose the truck’s lift speed is 80 fpm (v = 80) and travel speed is 250 fpm (h = 250). The load pick-up or deposit time is equal to 0.15 min (P = D = 0.15). Using Eqs. (3) and (4), we obtain T = 0.40 and b = 0.625. That is, the rack’s longer side is 0.40 min long and its shape factor is equal to 0.625. Assuming randomized storage and concurrent travel, we obtain E(SC) = 0.4521 min and E(DC) = 0.6082 min from Eqs.

(1) and (2), respectively. Hence, for the expected cycle times we obtain TSC = 0.7521 min and TDC = 1.2082 min. Although the expected DC cycle time is longer than the expected SC cycle time, two loads (one storage and one retrieval) are handled on each DC cycle.

The throughput capacity of the truck (operating in one aisle) depends on the fraction of SC vs.

DC cycles it performs. If an operation is defined as a storage or a retrieval, the throughput capacity can be measured in operations performed per hour. Suppose 50% of the operations in the above example are storages and 50% are retrievals. (In the long-term, of course, the two percentages would be equal; however, during certain shifts or segments of the day, the truck may perform more operations of one type.) Further suppose that 30% of the storages are performed on a DC basis (i.e., 30% of the storages are matched with a retrieval and the rest are performed on a SC basis).

If the throughput capacity is denoted by z operations / hr, we have (2 X 0.15z)(1.2082 --: 2) + (0.35z + 0.35z)(0.7521) = 60(0.90) = 54 min / hr, assuming a 90% truck utilization. Note that 30% of 50% yields 0.15z, which is the fraction of storage operations performed per hour on a DC basis;

each storage operation takes 1.2082 --: 2 = 0.6041 min when performed as part of a DC cycle. Also, each such storage operation is matched with a retrieval operation, which explains the multiplication of 0.15z by two. Likewise, 70% of 50% yields 0.35z, which is the fraction of storage operations performed per hour on a SC basis. The same holds true for SC retrievals. Solving the above equation for z, we obtain z = 76.30 operations / hr as the throughput capacity of the truck at 90% utilization.

A smaller value for truck utilization may be used to allow for downtime or maintenance, battery recharge or replacement, time for the truck to change aisles, and so on.

For comparison purposes, the reader may verify that, with sequential travel, assuming no other changes, the throughput capacity of the truck drops to 61.02 operations / hr. That is, in this particular example, the truck performs approximately 25% more operations per hour with concurrent travel.

The number of trucks needed depends on the overall throughput requirement of the system. For example, if the system needs to perform 220 operations / hr, 220 / 76.30 = 2.88, or 3 trucks, would be required if concurrent travel is assumed. With sequential travel, 3.61, or 4 trucks, would be required. An additional allowance may be required for congestion if the trucks serve a small number of aisles and interfere with each other. Also, we assumed that all the loads are delivered to (and removed from) the I / O point (located at the end of each aisle) via another handling system. Additional truck travel time required can be easily computed if a single, central I / O point is used to serve all the aisles.

Consider next the case where trucks are used for moving loads from one point to other point(s). Given the throughput requirement (i.e., the number of loads that must be moved per time unit), and the layout / configuration of the points served by the trucks, we would like to know how many trucks are needed. It is assumed that one load is moved on each trip.

To determine the number of trucks needed, in addition to the throughput requirement and the location of the points, we need to specify a ‘‘dispatching policy,’’ which basically determines which truck moves which load and when. There are a number of centralized and decentralized dispatching rules one can adopt. Centralized rules require a central dispatcher (or computer) to keep track of each load in the system waiting to be moved; such loads are also known as move requests. They also require a means of communication between the dispatcher and each device so that the appropriate device can be dispatched to the appropriate load. Decentralized rules tend to be less efficient but simpler and less expensive. With decentralized rules, each device generally follows a set of fixed instructions (or preprogrammed instructions) to decide which load to move next.

Specifying a dispatching rule is important because the dispatching rule often affects how much empty travel (or ‘‘deadheading’’) each device performs in serving the move requests. Depending on the flow patterns and the location of the points served by the devices, each device may have to perform a nontrivial amount of empty travel. The model we present below assumes first-come-first- served (FCFS) dispatching, which is a simple, centralized dispatching rule used in industry. Under FCFS dispatching, when a device becomes available (i.e., it delivers a load and becomes empty), it is assigned to the oldest move request in the system, regardless of the location of the oldest move request relative to the location of the device. If there are no move requests in the system when the device becomes available, then it idles at its last delivery point and waits for the next move request to arrive. When a move request arrives, if there is an idle device, it is assigned to the move request; if there are two or more idle devices, one of them is randomly picked and assigned to the move request. If there are no idle devices when a move request arrives, then it must wait until it becomes the oldest move request in the system and a device becomes available.

If the device becomes available at point i (or station i) and the oldest move request is located at point j (or station j), the device travels empty from station i to station j. Hence, when a device is assigned to a move request (i.e., when a device is serving a move request), the service time includes the empty travel time needed for the device to reach the move request. Obviously, there are several possible improvements to the FCFS dispatching rule. For example, when a device becomes available at station i, instead of the oldest move request, it could be assigned to the move request closest to station i. Such a rule, known as the shortest-travel-time-first (STTF) rule, is also used in industry, and it tends to reduce empty device travel relative to FCFS dispatching. However, STTF dispatching does not lend itself well to analytical modeling and therefore its use often requires simulation. In contrast, FCFS dispatching can be modeled analytically, and the results obtained with it would at least serve as a benchmark.

Assuming FCFS dispatching, suppose the flow data are given as a from-to chart, where ƒij rep- resents the number of loads that must be moved per hour from station i to station j. (Recall that one load is moved on each trip.) Let t e (t ƒ) be the empty (loaded) device travel time in minutes from station i to station j. Assuming that it takes P mins to pick-up and D mins to deposit the load, for simplicity we will assume that t ƒ = t e + (P + D). (The model we show below works with any user-defined t e and t ƒ values.)

With FCFS dispatching, using the results presented by Chow (1986), the number of empty trips performed per hour from station i to j, say, eij, can be shown to be given by the following equation:

where F is the total loaded trips performed per hour, that is, F = 2-i 2-j ƒij. Note that the empty trips performed per hour from station i to j is proportional to the number loaded trips / hr ending at station i and the number of loaded trips / hr originating at station j. Each time the device completes a loaded trip, it becomes empty by definition. Hence, if more loaded trips end at station i, the device becomes empty more often at station i. Likewise, if more loaded trips originate at station j, more empty devices must be dispatched to station j. Equation (9) is also presented by (Egbelu 1987); however, no dis- patching rule is specified. We note that the two terms shown in Eq. (9) can be multiplied to obtain eij only when the next load to be served by an empty device is independent of the current location of the device, which holds true under FCFS dispatching.

In Eq. (9), we allow the case where j is equal to i; that is, an empty trip may be performed from station i to station i, which implies that when a device delivers a load at station i, it is possible that the oldest move request in the system is located at station i. Depending on the exact locations of the pick-up and deposit points associated with station i, such an empty trip may or may not require a nonnegligible travel time.

Let aƒ(ae) denote the total loaded (empty) device travel required (in min / hr). It is straightforward to obtain aƒ and ae as follows:

The reader may verify that the entries in the above matrix sum up to 30 trips / hr as expected. (Recall that each loaded trip is followed by an empty trip, although the duration of an empty trip may be negligible.) Also, we note that no empty trips are performed out of station 1 because no loads are delivered into station 1. (A device becomes empty only after it has delivered a load.) Likewise, no empty devices are dispatched to stations 3 and 4 because these two stations do not send any loads, which means they never request an empty device.

Using the loaded and empty trips performed per hour, from Eq. (10) we obtain aƒ = 65 min / hr and ae = 31.64 min / hr. That is, the total (loaded plus empty) travel time requirement comes to 96.64 min / hr. Of course, there are only 60 min / hr; hence the preceding result simply implies that we are going to need more than one lift truck. Assuming t = 5 min / hr and u = 0.90, from Eq. (11) we obtain N = 1.952, that is, two lift trucks are needed.

During its available time (i.e., during the 55-minute portion of an hour), the expected state of each lift truck can be broken down as follows: (65 / 2) / 55 = 59.09% of the time traveling loaded (including load pick-up / deposit); (31.64 / 2) / 55 = 28.76% of the time traveling empty; and the re- maining portion of the time (12.15%) waiting idle for the next move request. The expected device utilization during a 55-minute period comes to 59.09% + 28.76% = 87.85%, which is less than the target utilization of 90% because two lift trucks are provided when the exact mathematical require- ment was 1.952 trucks. Note that each device travels empty almost one-third of the time that it is busy (28.76 / 87.85 = 0.3273).

One possible approach to reduce empty travel time (or deadheading) is to change the dispatching rule from FCFS to STTF. Generally speaking, STTF reduces empty device travel, and a system that meets throughput under FCFS will continue to meet throughput under STTF. However, with STTF, even if the system meets throughput and all the move requests are eventually served, some move requests placed by certain stations may have to wait longer than usual depending on the layout of the stations. This may occur especially if the stations are clustered. Note that under STTF, a device will continue to serve all the move requests in a cluster of stations, including those that arrive while it is busy, before it is dispatched to a move request placed by a station outside the cluster.

Another possible approach to reduce empty device travel is to eliminate dispatching and run each device according to a predetermined schedule. If there is little or no variance in the system, that is, if the arrival of the move requests can be predicted with reasonable certainty, then running the devices according to a schedule may be the best approach. In fact, in some systems, such as lean manufac- turing systems, since each cell or station operates to takt time, the device routes, and each move they make, can be set up ahead of time. With shrinking unit load sizes, this approach has led to what is known as a ‘‘milk run,’’ where each device, on each trip, makes a preplanned number of deliveries to a particular set of stations at specific times. In many instances, ‘‘tuggers’’ with carts attached perform the milk runs. Such handling systems are known as tractor-trailer systems, which were the forerunners to automated guided vehicle (AGV) systems discussed in Section 7.