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Valentin Fadeev

Calculus of Warehouses. Part 2

Wednesday, 20 Jan 2010, 20:20
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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

Tags: optimization, warehouse
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Valentin Fadeev

Calculus of Warehouses

Wednesday, 20 Jan 2010, 19:31
Visible to anyone in the world
Edited by Valentin Fadeev, Saturday, 19 June 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

Tags: optimization, warehouse
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