Edited by Valentin Fadeev, Saturday, 27 Mar 2010, 20:42
Further to the discussion of the cargo traffic through a port started in the previous post, i want to get together some thoughts and results obtained so far. What is missing here is only a leap into probabilistic formulas (i believe Poisson is just inevitable), but i'm still lacking confidence to assemble the whole thing in a working model.
Assume, we have cargo coming in at a daily rate of units, where n is the number of time period concerned (day in our case). Daily gate-out rate shall be units. Then daily stock can be found recursively:
Mean storage time can be calculated as the total amount of warehouse work (see the previous post) divided by the total amount of cargo put through:
In a continuous case it should obviously look like this:
I took this definition form some old textbook and just made it complicated. However, it seems very natural and works perfectly in practice. It looks like formula for the center of gravity of the daily stok graph, but with changed order of summation (integration)
If calculations are made on a daily basis, the formula can be simplified as folows:
Note, that if at the start of calculations there is some initial stock with mean storage time per unit, the formula needs a small correction:
It is obvious that on the long run the effect of such correction is negligible (which is also proved by actual observations).
Applying the Little's formula to the above result we obtain the expression for the required warehouse space at every moment n:
Edited by Valentin Fadeev, Tuesday, 4 Oct 2011, 22:29
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.
Edited by Valentin Fadeev, Sunday, 16 Jan 2011, 23:22
Delta-function introduced by one of the founders of quantum electrodynamics P.A.M. Dirac belongs to the very abstract concepts of the function theory. An important fact is that it arises, as the result of differentiation of discontinuous functions. This links delta to many practical problems which would not allow for a rigorous solution without it.
I encountered the problem which inspired me to write this article during the course on mechanics in the university. The textbook proof of the relation between bending moment and shear force did not make use of delta function. Point forces were considered as mere constant terms in the sum. I tried to give it a mathematically correct treatment. This finally allowed me to develop a VB application that used simple algorithm based on the derived results to calculate shear force and bending moment. So here it is, open to critics:
Consider a stiff horizontal beam of length L. Introduce a horizontal axis X directed along it. Assume the following types of forces are applied:
1) point forces Fci at xi (i=1,2,...,n), 0<xi<L
2) distributed forces with density given by continuous functions qj(x) at intervals (aj;bj) (j=1,2,...,m) 0<aj-1<bj-1<aj<bj<L
3) moments (pairs of forces) Mk with axes of rotation at xk (k=1,2,...,p) 0<xk<L.
The task is to find the formulas for shear force and total bending moment a functions of x and establish the relation between them.
Although Fci are applied at certain points by definition, in reality the force is alwas applied to a certain finite area. In this case we can consider the density of the force being very large within this area and dropping to 0 outside it. Hence, it is covenient to define the distribution density as follows:
Shear force at point x is defined as the sum of all forces applied before that point (we are moving from the left end of the beam to the right one):
Hence:
Where e(x)=1 if x>0 and e(x)=0 otherwise (sometimes called the Heavyside function).
Now we shall find the expression for distributed forces. For qj(x) may be defined of the whole axis, we have to "cut away" the unnecessary branches and leave only what stays within the set intevals. Consider the following expressions:
Indeed it is easy to ascertain that the right side is equal to qj(x) within (aj;bj) and is 0 outside it. Calculating shear force demonstrates some useful properties of δ:
Here we used the fact that:
Therefore, for example:
Now we shall calculate bending moments created by all types of forces involved. Consider Fci applied at xi. Bending moment created by this force evaluated at x>xi can be determined as follows:
Hence, total moment is:
To calculate moment created by distributed forces we use the approach adopted in mechanics. Replace the distributed force to the left of the point of summation with a point force applied at the center of gravity of the figure enclosed by the graph of qj(x) and lines x=aj, x=bj. If aj<x<bj:
Differentiating both parts by x we obtain:
In fact, we could as well include point forces in the above derivation considering total density q(x)=qc(x)+qd(x). We can therefore derive the general relation between shear force and bending moment:
Now we need to calculate the contribution made by moments of pairs of forces. Each of these moments can be considered as a pair of equal large forces F applied at a small distance h of each other and oppositely directed. They will make the following contribution to the expression for total density q(x):
Or taking the limit as h tends to 0:
This does not imply that expression for shear force will contain terms like . This is due to the fact that shear force consists of vertical components of all forces and for pairs of forces those components will cancel out each other.
Therefore total bending moment of the pairs of forces is expressed as:
Finally we can write the expressions for shear force and bending moment int the most general case:
In an important particular case where qj(x)=qj=const these expresions reduce to:
Edited by Valentin Fadeev, Sunday, 14 Mar 2010, 14:43
Another surprising appearance of a fine method in a clumsy problem is offered when calculating optimal parameters for a pile of unitary cargo (bales, cases, crates etc).
Say we have bales of cotton stowed in several layers. According to safety regulations the pile must have a pyramid-like structure with "stairs" after certain amount of layers. Suppose we have 1 layer in the upper tier x bales long and y bales wide:
The next tier has 3 layers and is 2 bales longer (x+2) and 1 bale wider (y+1). What should be the values of x and y so that area occupied by the pile is minimized? Assume we have a fixed and sufficiently large total amount of bales, say 60.
If we behave negligently enough to give x and y continuous values (which is not so bad when total cargo count is high), the problem yields a nice solution using.. Lagrange multipliers:
Really, if one can use a scalpel as a screwdriver, that's what I am doing, but like in the previous case, it just looks nice
Edited by Valentin Fadeev, Saturday, 19 Jun 2010, 09:46
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 L1:
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:
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