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Valentin Fadeev

Calculus of Warehouses. Part 3

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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 lamda sub n units, where n is the number of time period concerned (day in our case). Daily gate-out rate shall be mu sub n units. Then daily stock s sub n can be found recursively:

s sub zero equals alpha

equation left hand side s sub n equals right hand side s sub n minus one plus lamda sub n minus one minus mu sub n minus one

inflow

gateout

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:

equation left hand side cap t macron equals right hand side n ary summation from i equals one to n over s sub i times open t sub i minus t sub i minus one close divided by n ary summation from i equals zero to n minus one over lamda sub i

In a continuous case it should obviously look like this:

equation left hand side cap t macron equals right hand side integral over zero under cap t open integral over zero under t open lamda of tau minus mu of tau close d tau close d t divided by integral over zero under cap t lamda of t d t

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:

equation left hand side cap t macron equals right hand side n ary summation from i equals one to n over s sub i divided by n ary summation from i equals zero to n minus one over lamda sub i

Note, that if at the start of calculations there is some initial stock s sub zero with mean storage time cap t sub zero per unit, the formula needs a small correction:

equation left hand side cap t macron equals right hand side s sub zero times cap t sub zero plus n ary summation from i equals one to n over s sub i divided by n ary summation from i equals zero to n minus one over lamda sub i

It is obvious that on the long run the effect of such correction is negligible (which is also proved by actual observations).

Dwelling time

Applying the Little's formula to the above result we obtain the expression for the required warehouse space at every moment n:

equation left hand side cap e sub n equals right hand side lamda sub n times n ary summation from i equals one to n over s sub i divided by n ary summation from i equals zero to n minus one over lamda sub i

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Valentin Fadeev

A Small proof for Little's Formula

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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:

equation left hand side cap e times t equals right hand side cap q times cap t

cap e equals cap q divided by t times cap t

But cap q divided by t equals lamda is the average daily rate of the incoming flow. Therefore:

cap e equals lamda times cap t

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

 

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Valentin Fadeev

Delta: the underrated superstar

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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:

q sub c times i of x equals cap f sub c times i times delta times open x minus x sub i close

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):

cap f sub c of x equals n ary summation over x sub i less than x over cap f sub c times i of x

Hence:

equation sequence cap f sub c of x equals n ary summation over x sub i less than x over cap f sub c times i times integral over zero under x delta times open z minus x sub i close d z equals n ary summation over x sub i less than x over cap f sub c times i times e times open x minus x sub i close

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:

q sub j of x comma a sub j comma b sub j equals q sub j of x times open e times open x minus a sub j close minus e times open x minus b sub j close close

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:

equation sequence cap f sub d equals n ary summation over x sub j less than x over integral over zero under x q sub j of z times open e times open z minus a sub j close minus e times open z minus b sub j close close d z equals

equation sequence equals n ary summation over x sub j less than x over open integral over zero under x q sub j of z times e times open z minus a sub j close d z minus integral over zero under z q sub j of z times e times open z minus b sub j close d z close equals

equation sequence equals n ary summation over x sub j less than x over open integral over a sub j under x q sub j of z times e times open z minus a sub j close d z minus integral over b sub j under x q sub j of z times e times open z minus b sub j close d z close equals

equation left hand side equals right hand side n ary summation over x sub j less than x over left square bracket open e times open z minus a sub j close times integral over a sub j under x q sub j of z d z close vertical line sub a sub j super x minus integral over a sub j under x open integral over a sub j under x q sub j of z d z close times delta times open z minus a sub j close d z postfix minus

equation left hand side negative open e times open z minus b sub j close times integral over b sub j under x q sub j of z d z close vertical line sub b sub j super x plus integral over b sub j under x open integral over b sub j under x q sub j of z d z close times delta times open z minus b sub j close d z right square bracket equals right hand side

equation left hand side equals right hand side n ary summation over x sub j less than x over open e times open x minus a sub j close times integral over a sub j under x q sub j of z d z plus e times open x minus b sub j close times integral over x under b sub j q sub j of z d z close

Here we used the fact that:

equation left hand side f of x times delta times open x minus x sub zero close equals right hand side f of x sub zero times delta times open x minus x sub zero close

Therefore, for example:

equation sequence open integral over a sub j under x q sub j of z d z close times delta times open x minus a sub j close equals open integral over a sub j under a sub j q sub j of z d z close times delta times open x minus a sub j close equals zero

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:

cap m sub c times i of x equals cap f sub c times i times e times open x minus x sub i close times open x minus x sub i close

equation sequence cap m sub c times i times open x plus d times x close equals cap f sub c times i times e times open x plus d times x minus x sub i close times open x plus d times x minus x sub i close equals cap f sub c times i times e times open x minus x sub i close times open x minus x sub i plus d times x close

equation left hand side d times cap m sub c times i equals right hand side cap f sub c times i times e times open x minus x sub i close times d times x

equation sequence cap m sub c times i equals cap f sub c times i times integral over zero under x e times open z minus x sub i close d z equals cap f sub c times i times x minus x sub i plus absolute value of x minus x sub i divided by two

Hence, total moment is:

equation left hand side cap m sub c equals right hand side n ary summation over x sub i less than x over cap f sub i times c times x minus x sub i plus absolute value of x minus x sub i divided by two

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:

cap f sub d times j of x equals integral over a sub j under x q sub j of z d z

equation sequence cap m sub d times j equals integral over a sub j under x q sub j of z d z times open x minus integral over a sub j under x z times q sub j of z d z divided by integral over a sub j under x q sub j of z d z close equals x times integral over a sub j under x q sub j of z d z minus integral over a sub j under x z times q sub j of z d z

Differentiating both parts by x we obtain:

equation sequence d times cap m sub j times d divided by d times x equals integral over a sub j under x q sub j of z d z plus x times q sub j of x minus x times q sub j of x equals integral over a sub j under x q sub j of z d z equals cap f sub j times d of x

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:

cap f of x equals d times cap m divided by d times x

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):

equation sequence mu of x equals cap f times delta times open x plus h close minus cap f times delta of x equals cap f times h times delta times open x plus h close minus delta of x divided by h equals cap m times delta times open x plus h close minus delta of x divided by h

Or taking the limit as h tends to 0:

mu of x equals cap m times delta super prime of x

This does not imply that expression for shear force will contain terms like cap m times delta of x . 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:

cap m equals n ary summation over x sub k less than x over cap m sub k times e times open x minus x sub k close

Finally we can write the expressions for shear force and bending moment int the most general case:

cap q of x equals n ary summation over x sub i less than x over cap f sub i times c times e times open x minus x sub i close plus n ary summation over x sub j less than x over open e times open x minus a sub j close times integral over a sub j under x q sub j of z d z plus e times open x minus b sub j close times integral over x under b sub j q sub j of z d z close

cap m of x equals sum with, 3 , summands n ary summation over x sub k less than x over cap m sub k times e times open x minus x sub k close plus n ary summation over x sub i less than x over cap f sub i times c times x minus x sub i plus absolute value of x minus x sub i divided by two plus n ary summation over x sub j less than x over integral over zero under x open e times open t minus a sub j close times integral over a sub j under t q sub j of z d z plus e times open t minus b sub j close times integral over t under b sub j q sub j of z d z close d t

In an important particular case where qj(x)=qj=const these expresions reduce to:

cap q of x equals n ary summation over x sub i less than x over cap f sub i times c times e times open x minus x sub i close plus n ary summation over x sub j less than x over q sub j times open open x minus a sub j close times e times open x minus a sub j close plus open x minus b sub j close times e times open x minus b sub j close close

cap m of x equals sum with, 3 , summands n ary summation over x sub k less than x over cap m sub k times e times open x minus x sub k close plus n ary summation over x sub i less than x over cap f sub i times c times x minus x sub i plus absolute value of x minus x sub i divided by two plus n ary summation over x sub j less than x over q sub j divided by two times left square bracket left parenthesis x minus a sub j times right parenthesis squared times e times open x minus a sub j close plus open x minus b sub j times right parenthesis squared times e times open x minus b sub j close close

Permalink 1 comment (latest comment by Ralph Joseph Prescott, Sunday, 14 Mar 2010, 18:46)
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Valentin Fadeev

Calculus of Warehouses. Part 2

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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:d7609c6675f972b63794bef3b2665403.jpg

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:

x times y plus three times open x plus two close times open y plus one close equals 60

g of x comma y equals sum with, 3 , summands four times x times y plus three times x plus six times y

multirelation f of x comma y equals open x plus two close times open y plus one close right arrow m times i times n

g of x comma y equals zero

normal cap phi of x comma y equals f of x comma y plus lamda times g of x comma y

equation sequence normal partial differential times normal cap phi divided by normal partial differential times x equals sum with, 3 , summands y times open four times lamda plus one close plus three times lamda plus one equals zero

equation sequence normal partial differential times normal cap phi divided by normal partial differential times y equals x times open four times lamda plus one close plus two times open three times lamda plus one close equals zero

equation sequence normal partial differential times normal cap phi divided by normal partial differential times lamda equals x times open four times y plus three close plus six times open y minus nine close equals zero

x almost equals four reverse solidus full stop y almost equals two

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

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Valentin Fadeev

Calculus of Warehouses

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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:

figure

Its volume can be calculated as follows:

cap v equals integral over zero under cap h cap s of x d x

Where

equation sequence cap s of x equals a of x times b of x equals open cap l sub one plus cap l sub two minus cap l sub one divided by cap h times x close times cap b times x divided by cap h equals cap b times open cap l sub one times x divided by cap h plus open cap l sub two minus cap l sub one close close left parenthesis x divided by cap h times right parenthesis squared

So that

equation sequence cap v equals cap b times cap h times integral over zero under one left parenthesis cap l sub one times x divided by cap h plus open cap l sub two minus cap l sub one close times open x divided by cap h times right parenthesis squared close times d of x divided by cap h equals cap h times cap b divided by six times open two times cap l sub two plus cap l sub one close

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:

equation sequence d times cap a equals gamma times g times d times cap v times open cap h minus x close equals gamma times g times cap b times open cap l sub one times x times open cap h minus x close divided by cap h plus open cap l sub two minus cap l sub one close times x squared times open cap h minus x close divided by cap h squared close times d times x

Thus, total work equals:

equation sequence cap a equals gamma times g times cap b times integral over zero under cap h open cap l sub one times x times open cap h minus x close divided by cap h plus open cap l sub two minus cap l sub one close times x squared times open cap h minus x close divided by cap h squared close d x equals gamma times g times cap b times cap h squared divided by 12 times open cap l sub one plus cap l sub two close

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:

two times cap h divided by cap b equals tangent of chi

cap h equals one divided by two times cap b times tangent of chi semicolon equation sequence cap l sub one equals cap l sub two minus two times cap h divided by tangent of chi equals cap l minus cap b

cap v equals cap b squared times tangent of chi divided by 12 times open three times cap l minus cap b close

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

cap l equals four times cap v divided by cap b squared times tangent of chi plus cap b divided by three

equation sequence cap s equals cap l times cap b equals four times cap v divided by cap b times tangent of chi plus cap b squared divided by three

equation sequence cap s super prime of cap b equals negative four times cap v divided by cap b squared times tangent of chi plus two times cap b divided by three equals zero

cap b equals root of order three over three

cap l equals open four divided by root of order three over three plus two divided by three times root of order three over three close times root of order three over three

cap s equals four times open one divided by root of order three over three plus one close left parenthesis cap v divided by tangent of chi times right parenthesis super two solidus three almost equals 4.4 left parenthesis cap v divided by tangent of chi times right parenthesis super two solidus three

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:

equation sequence cap a equals gamma times g times cap b times cap h squared divided by 12 times open cap l sub one plus cap l sub two close equals gamma times g times cap h squared divided by 48 times open eight times cap v times cap b divided by tangent of chi minus cap b super four divided by three close

equation sequence cap a super prime of cap b equals gamma times g times tangent squared of chi divided by 48 times open eight times cap v divided by tangent of chi minus four times cap b cubed divided by three close equals zero

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:

cap v times gamma divided by cap s less than q

multirelation cap v less than q times cap s divided by gamma equals q divided by gamma times 4.4 left parenthesis cap v divided by tangent of chi times right parenthesis super two solidus three

multirelation cap v less than left parenthesis 4.4 times q times right parenthesis cubed divided by gamma cubed times tangent of chi almost equals 85.3 times q cubed divided by gamma cubed times tangent of chi

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