Warehouse Design: Dock Doors For Drop Trailers

Warehouse design for outbound flow and dock doors is meant for peak warehouse efficiency.

But when you’re operating, things can be different. You’ve checked the picking rates. You’ve calculated staging areas and staffing. Yet when you’re running operations, disaster strikes! You can’t get enough loads into trailers and are constantly building inventory in your shipping area. Your team is complaining that they don’t have enough doors to load. What happened?

One reason can be that the dock doors and yard throughput calculations had some faulty assumptions. Let’s explore why.

Dock Door Requirements

Conventional dock door requirement calculations (example at ChuckBerger.com ) describe dock door requirements this way.

Dock Door Calculation Formula
Useful, but you also need to know your average dock time!

This is straightforward and easy. These numbers are all readily available.

Or are they?

Peak trucks per day should come from forecasts and order profiles. The safety factor looks like a fudge factor especially since we’re already considering peak trucks forecast, but we understand wanting to be prepared for variation.

But finding out the average dock time per truck, or turn time, is trickier. How do we calculate it?

Let’s take a look.

Dock Door Turn Times

Truck turn times include the docking, loading, and shipping of the truck.

This is often estimated at an hour for Full Truck Loads (FTL). It may be less or more depending on whether the load has special handling type such as slip-sheets, dunnage, or other requirements. 

Dock door turns are the number of times a truck is shipped from a dock door in a period of time, usually expressed in turns per day or hours. So saying a dock turned three times per day is equivalent to saying that is has a turn time of 8 hours.

Dock door turns include everything that happens inside and outside of the warehouse. So it is not only how much time the truck takes to load but also how long it takes that door to “re-set” or be ready for the next load. Factors in this time include all the yard management and logistics that go on outside of a warehouse. This makes “average dock time per truck” much more complicated!

Drop Trailers vs Live Loads

Once we start looking outside of the warehouse, we can think about it like this: dock turn time should be related to how many dock doors are ready to load, in addition to the truck loading time. After all, you can’t load a trailer without a door to park it in!

First we consider whether the loads are Live Loads or Drop Loads.

Live loads are trucks that show up at a scheduled time to be loaded. Usually these take between one and three hours to arrive at the gate, check-in, get a dock door, back up, get loaded, and leave. If everyone shows up on time, that means these doors can consistently turn 8-12 times over a 24-hour period. This is straight-forward.

Drop Loads are more complex. This is where the facility keeps a pool of empty trailers onsite to load at their own schedule and wait for pickup. Drivers can arrive with an empty trailer, drop it in the yard, pick up a full trailer, and depart. This process can be very quick and allows the site operations to level-load their loading activities.

Moving the trailers into and out of doors is done by a special tractor truck which stays in the yard and moves trailers. This truck is often called the “yard jockey,” “yard dog,” or “yard hostler,” or other similar names.

Drop Trailers Require Yard Moves

Here’s the wrinkle: Turning the doors for drop-trailer loads requires the site to move loaded trailers to the yard and empty trailers to the door. Doing this for all the doors in the yard is slower than the actual loading or drop-and-hook process, so it governs the overall turns of the drop-trailer doors.

Yard Moves Mean A Queue

Imagine a building with a bunch of doors waiting to be serviced. We can think about the jockey’s work as a series of moves pulling each full trailer from a door and replacing it with an empty one. Let’s call this “resetting the door.”

As the jockey gets calls to reset each door, we see that the jockey has a queue of doors to work through. After he resets each door, he works down the list to the next door to reset. The number of doors he can do over a period of time is an arrival rate.

Now that we’ve identified a queue and an arrival rate we can use Little’s Law.

Little’s law is a relationship between queues, arrival rate, and the time of the items in the queue. It is expressed as:

L = λW

Where Queue Length (L) = Arrival Rate (λ) * Time-in-queue (W). It is a handy analytical tool for solving for unknowns in steady-state queuing systems.

We can use it to analyze our drop-trailer dock door requirements.

Longer Queues Mean More Time: Example

Let’s say your outbound pallets-per-hour pick rate is 200 pallets per hour. You are using 100% drop trailers. At 20 pallets per truckload, and with 24 loading hours per day, you need to be loading 10 trucks per hour or 240 trucks per day.

This is no problem as your loaders can load, close, and ship a truck in 60 minutes. How many doors do you need?

The intuitive answer is 10 doors, since you need 240 trucks and there are 24 hours in a day. After all, you can load each truck in an hour, hire a bunch of forklift operators, and with 10 doors you’re all set.

Now let’s analyze the problem with Little’s Law. Your yard jockey can reset doors at 20 minutes per reset (.33 hrs). This includes hooking up, pulling the trailer, parking it, finding an empty, and replacing it to the door—will only be able to service each door every 3.33 hours, or every 3 hours and 20 minutes.

L = λW

Little's Law Example

W = 3.33 hrs or 3 hrs and 20 minutes

With a turn time of 3.33 hours, and because your loaders can load a full truck before the jockey arrives for the next reset, you can now do the math.

With 10 doors and each door is turning every 3.33 hours, you’re getting 7.2 turns per day. 7.2 * 10 doors = 72 loads per day. Because you require 240 trucks per day to ship, you can see right away that using 10 doors and one yard jockey is a losing strategy.

What do we need to do to get back on track?

Two Levers To Pull: Yard Jockeys or Doors

We can do two things to fix this: add doors or add yard jockeys.

If we keep adding doors with one jockey, we find that the jockey can still only get through 72 resets per day. It doesn’t matter how many more doors you add. You’ll eventually fill up all of them and be bottlenecked at 72 loads per day, unless you enlist some over-the-road drivers to do moves or swaps for you!

What happens if you add jockeys instead?

With two jockeys, the 10 doors will turn every 1.67 hours, which means that doors will turn 14.4x per day, which means that there can be up to 144 (10 * 14.4) loads per day.

With three jockeys, you can turn doors every 1.1 hours or 66 minutes. This means 21.6 turns or 216 loads per day.

You can calculate the number of resets with the number of jockeys multiplied by the working hours and their reset rate per hour.

At this point, though, the constraint moves to the loaders who need to be ready to load a trailer as soon as it’s in the door and available. At 20 minutes for the trailer plus the hour for loading, the minimum turn time at the door is 1.33 hours or 18 turns per day, and 180 trailers per day.

So we can’t hit the 240 loads per day with 10 doors no matter how many jockeys we have.

We find that because the number of resets is a discontinuous function of the minimum turn time (which we consider as fixed), the number of jockeys, and the number of doors—let’s add yard jockeys and docks.

Results of adding Jockeys and Docks

We start adding dock doors and jockeys and find a surprising result: Adding more doors adds incremental load capacity only so long as jockeys are available to keep door turn times to the minimum turn time.

Using our example of 20 minutes to reset a door and 60 minutes (one hour) to load a trailer, and assuming we have enough labor and outbound load staging areas to avoid bottlenecks in the warehouse, we find that the number of outbound loads is driven by the number of jockeys, not the number of outbound docks.

Jockey and Door scenario table

In this example, for scenario 2, the warehouse can put 144 loads out the door with a minimum of 8 doors. As long as only two jockeys are available, that 144 loads max does not change.  The warehouse may have many more than 8 doors:

Loads shipped are fixed after jockey capacity is utilized

What will change is the turn times, and wait times, per door. We see the hours-per-reset climb as dock doors climb. But overall outbound throughput does not change after the maximum supportable doors number is covered by jockeys.

How loads shipped vary by jockeys used

Simplifying this further, we conclude the number of jockeys for max turns at the door is equal to the minimum dock time divided by jockey door reset time. Using our numbers, this is 1.33 hours min dock time per door (80 minutes) divided by 20 minutes per reset. This has each jockey completing a reset of each door every 1.33 hours—just enough time for the warehouse team to load the truck and the jockey to complete the reset, and letting us know the ratio is four doors per jockey for maximum door turns.

Calculating optimal doors per jockey

Implications

We’ve established that the number of loads is a function not only of the number of doors available but also of the minimum loading time and the number of jockeys. This means we can run the analysis like this, with the assumptions we used above as an example: 

We optimize the warehouse throughput with 13.3 doors (round to 14), which is the same result as in the equation at the start of the post, less the “safety factor.” 

And now we also know that the assumption is based on three yard jockeys doing door resets at 20 minutes each, regardless of the number of doors. This is critical information for determining the true capacity for shipping outbound loads.

Any of these figures can be buffered for “real-world” efficiency loss or variability. Faster loading means more jockeys are needed, or maybe the facility doesn’t want to run three yard jockeys for 24 hours continuously.

The next step would be calculating the number of live-load doors required and adding it to the drop-load door total. 

This is a sound method for finding the minimal footprint or optimal setup for a warehouse drop program, identifying the resources needed to run it, and identifying where extra capacity might be needed.

Conclusion

For sites with drop trailer programs, queuing logic is important for calculating output capacities. It takes some of the guesswork out of calculating a requirement for dock doors or whether a certain outbound plan is realistic. It also gives options for increasing output that might not be apparent to the team on the warehouse floor. It also paints a picture of the constraint that yard jockeys and yard moves place on the outbound operation.

A lot goes into planning the right number of dock doors. A good baseline shipping forecast by-mode-of-transportation is fundamental. Logistics plans for number of drop vs live loads will drive dock turn planning and yard capacity. Warehouse operations judgments of real-world conditions like average loading times, efficiencies, and buffer times are important. All modes of transportation plus receiving, returns, waste disposal, recycling, and other requirements can increase requirements.

Doing the homework on all the constraints is critical to a good design. This approach will help your teams determine what to plan and how to plan it.

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