Personal Blogs

I am posting it in reply to a question I came across in one social network. There is no original material here. I am just trying to reproduce the argument from one university lecture, as far as I can remember it.

The task is to calculate optimal stock level at a warehouse. Stock is replenished to a certain value (which is to be determined) immediately at fixed equally spaced moments in time. Amount of stock units cnosumed during each period is random.

Let be the stock level which is maintained. Assume that stock is replenished over a fixed period of time (take it to be unitary to simplyfy calculations). Assume that units are used during each time period with probability .

Let be unitary storage cost, is a “penalty” cost
assigned when stock is theoretically below 0, or below some
minimal admissible level (lost orders, emergency orders, etc).

Assume also that the stock is replenished immediately.

Consider two possible situations.

1) , i.e there is a positive remainder in the warehouse at
the end of the period. Storage cost is proportional to the area
under the stock graph.

Expectancy of this value is:

Therefore the average cost is:

2) . In this case the cost of both types is incurred. Again using the geometrical approach we find that cost elements are proportional to the areas of the triangles. The same argument as above gives the answer:

Thus the total cost is given by the following expression:



Marginal analysis is used to determine the optimal stock level. Optimal value of minimizing satisfies the relation:

(here my memory is abit vague, so I am giving it without proof and without guarantee)

Practically this means calculating for a series of the values of s until the above double inequality holds.

Little's formula is one of the cornerstones of the queueing theory, because it applies to a wide class of systems. It states that the number of customers in the system is the product of the average arrival rate and the mean time a customer spends in the system. Like all fundamental things it allows different proofs and interprerations. Some of them are really convincing, involving Kolmogorov-Chapman equations and related apparatus.

I will present a rather "mechanical" approach which gives the result in just three lines, but requires some preparatory discussion. We shall consider a warehouse where unitized (boxes, pallets, containers) cargo is stored. Cargo unit is a "customer" in the queue,storage time is service time.

Warehouse operator charges for (amount of cargo)X(storage time). It is reasonable to call this magnitude "warehouse work": A, (box)X(day). In a similar way "transport work" is introduced as (cargo)X(distance). Consider a limit case of a warehouse X having 1 storage place, say for 1 box, and another warehouse Y capable of storing 2 boxes at the same time.Then amount of work of 2 box-days can be performed by X in 2 days, while Y can do the same in 1 day. It is therefore natural to develop analogues with mechanics and consider capacity E as "power" of the warehouse meaning the amount of work that can be performed in a unit of time.

Now we move on to the proof. Suppose Q units of cargo are brought to the warehouse every t days. Warehouse has capacity of E units and mean dwelling T time is known based on observations. What is the condition for the system to work in a stable way, i.e. to cope with the incoming flow?

Incoming Q units will "create" warehouse work of QT box-days. The warehouse must "process" them in t days (because then the next lot is coming) having at least capacity E, that is it needs to perform work of Et box-days. The balance equation (or "law of conservation", if you like it) then looks like this:

But is the average daily rate of the incoming flow. Therefore:

Which is the required result. In this interpretation it gives the minimum required capacity to put the cargo flow through the system.

This is one of the small problems I formulated for myself during my studies, just for the sake of curiosity. Calculations are pretty simple and straightforward, but the results look nice, also with some unexpected turn in the middle.

In maritime transport one has to deal with the so-called bulk cargoes (iron ore, coal, fertilizers etc) which are often stored in the ports in large piles. I once considered determining some quantitative characteristics that are useful for designing the warehouse to store such type of cargo.

Let’s approximate the pile of cargo with the geometrical body of height H having a rectangle L₂ X B at its base and 4 facets inclined towards the centre at equal slope, thus forming the upper horizontal edge of length L_1:

Its volume can be calculated as follows:

Where

So that

Now let us determine the amount of energy required to form the pile (meaning mechanical work that has to be done against the forces of gravity). Assume that density of the cargo is γ kg/m³. To lift a small layer to height H-x requires:

Thus, total work equals:

No we can ask the following question: if volume is a given value, what should be length and breadth so that the pile occupies minimal warehouse area? To simplify calculations and cancel out some parameters let us introduce the value called “angle of natural slope”. This is the term from soil mechanics which means the angle that the substance makes against horizontal surface. It depends on density, viscosity and inter-particle friction and can be found in special tables.

Let:

Assuming that V=const, obtain expression for L and insert it in the formula for the area S=LB:

Now, assuming again that V=const, let us determine the values of L, B again which minimize the amount of energy required to form the pile. Again, substituting for L, but now in the expression for A we obtain:

MathJax failure: TeX parse error: Extra open brace or missing close brace

Remarkably, we get the same values which were obtained to minimize the base area, although the approach is now quite different.

To complete the picture we can use the derived results to determine maximum allowed volume for a single pile if unitary pressure on the ground is limited by some given value q kg/m²: