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bonanova

Baggage on a conveyor belt

Question

At a busy airport a conveyor belt stretches from the runway, where all the planes land, to the baggage claim area inside the terminal. At any given time it may contain hundreds of pieces of luggage, placed there at what we may consider to be random time intervals. Each bag has two neighbors, one of which is nearer to it than the other. Each segment of the belt is bounded by two bags, which may or may not be near neighbors (to each other.)

On average, what fraction of the conveyor belt is not bounded by near-neighbors?

Example:

----- belt segment bounded by near neighbors
===== belt segment not bounded by near neighbors

... --A-----B==========C---D--------E=============F---G-H---I-- ...

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7 answers to this question

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Spoiler

 

Randomly placed is NOT sufficient info to establish the   % of unbounded belt on the in going (top side ) half of the belt . Also while implied there is no mention  of any belt length segment   Therefore the avg will b.e over 50%.  How much over 50%.  Assuming each piece of luggage as placed on belt was 2 ft wide.     2nd piece is 1 inch from 1st, and the 3rd is 2 inches away;  Then the  4th must be at least 3 inches away from the 3rd and becomes the 1st in the next segment of luggage . if this were to repeat on and on, then the % of unbound belt would be 3 inches divided by the total length of the segment.  i,e, (24 +1 + 24 + 2 + 24 +3 ) = 78;  Therefore 3 /78 =0.03845 or 3.8% for the top belt and in the scenario added to the 50% = 53.8 % This would be the lowest % using a full inch as a minimum unit.

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Belt "segment" just refers to portions of belt between adjacent bags. Top or bottom is not an issue. Consider bags are randomly placed  points on a line if you like.  The fraction of length that is not a near-neighbor segment changes with time. To imply a limit, the OP asks "on average" over time.

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One approach is to

Spoiler

simulate the bags with a random number generator.

Then it would be helpful to know that the result is

Spoiler

a rational number.

 

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Ok I tried using random numbers, got a fraction...

.8237

Bonanova, Since you said rational, I’ll guess 5/6. I’ll bet you have a simple but deep proof. I’m eager to see it! I can see that it has to be more than 2/3, because a nearest neighbor gap is smaller than either adjacent gap, hence is less than 1/3.

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On 2/19/2018 at 12:06 AM, CaptainEd said:

Ok I tried using random numbers, got a fraction...

 

  Hide contents

 

.8237

Bonanova, Since you said rational, I’ll guess 5/6. I’ll bet you have a simple but deep proof. I’m eager to see it! I can see that it has to be more than 2/3, because a nearest neighbor gap is smaller than either adjacent gap, hence is less than 1/3.

 

 

Gah! Mine was not a useful clue. The denominator (greater than 6 but not huge) is too large for a simulation to tell you the fraction. Also I think your simulation value is high. But you're thinking in the right direction. Here may be more useful clues.

Spoiler

The intervals can be classified small if both neighboring intervals are larger, large if both neighboring intervals are smaller, and medium otherwise.

Spoiler

Every interval has an equal probability of being large, medium or small. i.e. for every interval P(small) = P(medium) = P(large) = 1/3.

Spoiler

The (Poisson) probability that any interval is smaller than say a distance x can be taken, with appropriate scaling, to be 1 - e-x.

Spoiler

Part of what is needed is the expected size <large> of a large interval.

 

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Here’s my attempt at an analytical solution


Instead of a quasi infinite belt, I chose a belt of length 1, and instead of a Poisson distribution, I chose uniform distribution of segment lengths.

Consider a segment of length x in (0,1). The probability of its left neighbor being larger is (1-x), the probability of it being smaller is x. 
A segment is non-nearest if at least one of its neighbors is smaller, so the Probability of a segment being non nearest is
P(only first neighbor is smaller) + p(only second neighbor is smaller
+ p( both neighbors are smaller

Pnn(x) = 2x*(1-x)+x^2

= 2x -2x^2 + x^2 = 2x - x^2
Integrate for x =0,1
2x^2/2 - x^3/3
2/2-1/3
1-1/3 =2/3

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4 hours ago, CaptainEd said:

Here’s my attempt at an analytical solution

  Hide contents

A segment is non-nearest if at least one of its neighbors is smaller,

 

Spoiler

Consider segment D--------E in the OP

 

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