Scheduling a Rescueweb2.uwindsor.ca/math/hlynka/canqueue2019slide.pdf · 2019-08-21 · 0(i)...
Transcript of Scheduling a Rescueweb2.uwindsor.ca/math/hlynka/canqueue2019slide.pdf · 2019-08-21 · 0(i)...
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Scheduling a Rescue
Nian Liu1 Myron Hlynka2
1School of Mathematics and StatisticsCentral South University
Changsha, Hunan, China.
2Department of Mathematics and StatisticsUniversity of Windsor
Windsor, ON, Canada.
Canqueue 2019, Toronto
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Outline
1 Introduction
2 Additive Analysis
3 Simple Example
4 Multiplicative Analysis
5 Numerical Example
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Problem Description
In June and July of 2018, a group of thirteen persons on a soccerteam was trapped in flooded caves in Thailand. One statementthat appeared in the media stated that rescuers chose to take thestrongest boys first. This was argued to give the best chance ofsurvival. Later statements said that the weaker boys actually wereremoved first.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Problem Description
From wikipedia,“For the first part of the extraction, eighteen rescue diversconsisting of thirteen international cave divers and five Thai NavySEALs were sent into the caves to retrieve the boys, with onediver to accompany each boy on the dive out. There wereconflicting reports that the boys were rescued with the weakestfirst or strongest first. In fact, the order was which boy volunteeredfirst.”
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Problem Description
In the end, all thirteen people were rescued.
We consider different models and criteria for which removing thestronger persons first may or may not be the best strategy. Themodels to be presented do not fit precisely into standardqueueing or perishable inventory or survival analysis type models,and the objective here is different than in other settings.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Problem Description
In the end, all thirteen people were rescued.
We consider different models and criteria for which removing thestronger persons first may or may not be the best strategy. Themodels to be presented do not fit precisely into standardqueueing or perishable inventory or survival analysis type models,and the objective here is different than in other settings.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Symbols
Symbol Definitionn number of itemsi label of an item according to its
order of being processedP0(i) initial probability of success for i-th itemti the times of start of processing
for the i-th itemTi the interval between individual
processing time starts (Ti = ti+1 − ti )Xi = 1 if item i is successfully processed, 0 elseY The total number of successes Y =
∑Xi
We consider those n items to be processed one at a time, and assumeTi = T for all i .For the Thailand situation, we have n = 13.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Symbols
Symbol Definitionn number of itemsi label of an item according to its
order of being processedP0(i) initial probability of success for i-th itemti the times of start of processing
for the i-th itemTi the interval between individual
processing time starts (Ti = ti+1 − ti )Xi = 1 if item i is successfully processed, 0 elseY The total number of successes Y =
∑Xi
We consider those n items to be processed one at a time, and assumeTi = T for all i .For the Thailand situation, we have n = 13.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Measures of Success
• The expected number of successfully processed items(expected number of rescued people).• The probability that all items are successfully processed
(probability that all people are rescued).
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Analysis Outline
• Constant Probabilities P0(k)
• Variable Probabilities P1(tk )• Additive Decrease
P1(tk ) = P0(k) + f (tk )• Multiplicative Decrease
P1(tk ) = P0(k) · f (tk )
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Constant ProbabilitiesTheoremFor the generalized binomial model, both E(Y ) andP(all successes) are constant regardless of the order in whichitems occur.
Proof.Note that E(Xi) = 1pi + 0(1− pi) = pi .
E(Y ) =n∑
i=1
E(Xi) =n∑
i=1
pi
P(all successes) =n∏
i=1
pi
Since the expressions for E(Y ) and P(all successes) do notchange when we change the order in which the items areprocessed, the result follows.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Constant ProbabilitiesTheoremFor the generalized binomial model, both E(Y ) andP(all successes) are constant regardless of the order in whichitems occur.
Proof.Note that E(Xi) = 1pi + 0(1− pi) = pi .
E(Y ) =n∑
i=1
E(Xi) =n∑
i=1
pi
P(all successes) =n∏
i=1
pi
Since the expressions for E(Y ) and P(all successes) do notchange when we change the order in which the items areprocessed, the result follows.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable Probabilities
Assume that as time goes by, the probabilities change from P0 toP1, where P1(tk ) is the probability of success of processing thek -th item.
E(number of successes) =n∑
i=1
E(Xi) =n∑
k=1
P1(tk )
P(all successes) =n∏
i=1
P1(tk )
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
With constant interval T , the ordered processing start times forthe items are ~v = (0,T ,2T , . . . , (n − 1)T ) = (t1, t2, . . . , tn).
According to additive method, P1(tk ) = P0(k) + f (tk ) wheref (tk ) < 0.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
TheoremIf P0(i) + f (ti) > 0, ∀i , k, then
E(#successes) =n∑
i=1
P1(ti)
=n∑
i=1
(P0(i) + f (ti))
=n∑
i=1
P0(i) +n∑
i=1
f ((i − 1)T )
P(all successes) =n∏
i=1
P(i -th item is success)
=n∏
i=1
(P0(i) + f ((i − 1)T ))
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
According to the arithmetic/geometric mean inequality, withn∑
i=1P1(ti) fixed, the product of n terms would be larger when terms
are close(r) to each other.
That is to say, the probability that all items succeed depends onthe order of service while the expected number of successes doesnot. In this case, it is better to attend to the weaker items first.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
The former result would change if the assumption P0(i) + f (ti) > 0does not hold.
TheoremIf for some i we have P0(i) + f (ti) ≤ 0, then
E(#successes) =n∑
i=1
P1(ti) =n∑
i=1
(P0(i) + f (ti))+
P(all successes) =n∏
i=1
P1(ti) =n∏
i=1
(P0(i) + f (ti))+ = 0
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
The former result would change if the assumption P0(i) + f (ti) > 0does not hold.
TheoremIf for some i we have P0(i) + f (ti) ≤ 0, then
E(#successes) =n∑
i=1
P1(ti) =n∑
i=1
(P0(i) + f (ti))+
P(all successes) =n∏
i=1
P1(ti) =n∏
i=1
(P0(i) + f (ti))+ = 0
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesAdditive Decrease
The expected number of successes could be written as
E(#successes) =n∑
i=1
P1(ti)
=n∑
i=1
(P0(i) + f (ti))−∑
P0(i)+f (ti )<0
(P0(i) + f (ti))
=n∑
i=1
(P0(i) + f (ti)) +∑
P0(i)<−f (ti )
|P0(i) + f (ti)|
If ddt f (t) < 0, then items with bigger P0 should be processed
earlier, so that the number of terms s.t . P0(i) < −f (ti) and valuesof those terms will all increase, which would lead to a higherexpected number of successes. In this case, it is better to attendto the stronger items first, assuming that the expected number ismore important than P(all Successes).
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Simple Example
P0 = (.9, .8, .7). t = (0,1,2). f (t) = −.1 ∗ t = (0,−.1,−.2)Then P1 = P0 − f (t) = (.9− 0, .8− .1, .7− .2) = (.9, .7, .5)E(#successes) = .7. P(allsuccess) = .9 ∗ .7 ∗ .5 = .315===P0 = (.7, .8, .9). t = (0,1,2). f (t) = −.1 ∗ t = (0,−.1,−.2)Then P1 = (.7− 0, .8− .1, .9− .2) = (.7, .7, .7)E(#successes) = .7. P(allsuccess) = .7 ∗ .7 ∗ .7 = .343Choose weaker items first.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Simple Example
P0 = (.9, .1). t = (0,1). f (t) = −.2 ∗ t = (0,−.2)Then P1 = P0 − f (t) = (.9− 0, .1.− .2) = (.9,−.1)← (.9,0)E(#successes) = .45. P(allsuccess) = 0===P0 = (.1, .9). t = (0,1). f (t) = −.2 ∗ t = (0,−.2)Then P1 = (.1, .9− .2) = (.1, .7)E(#successes) = .4. P(allsuccess) = .1 ∗ .7 = .07Choose stronger items first to maximize total expected success.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesMultiplicative Decrease
According to multiplicative method,
P1(tk ) = P0(k) · f (tk )
where f (tk ) ∈ (0,1).
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Variable ProbabilitiesMultiplicative Decrease
TheoremWith multiplicative decrease, there are
E(#successes) =n∑
i=1
P1(ti) =n∑
i=1
(P0(i) · f (ti))
P(all successes) =n∏
i=1
P1(ti) =n∏
i=1
(P0(i) · f (ti))
=n∏
i=1
P0(i) ·n∏
i=1
f (ti)
CorollaryE(#successes) is maximized if higher prob. of success items areprocessed first.P(all successes) is independent of the order of service.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Numerical Example
The Thailand rescue serves as a numerical example of the modelwe have built.
Considering people who were trapped in the cave would getweaker over time, we set d
dt f (t) < 0 for both additive andmultiplicative situation.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Numerical ExampleProbabilities with additive decrease
For example, set T = 1 and choose f (t) = −0.06t for additivedecrease, generate 13 random values uniformly on (0.5,1) torepresent the probabilities P0(j) of success for person j at time 0.
As tk =∑k−1
i=1 T = k − 1,we have P1(tk ) = (P0(k)− 0.06(k − 1))+, k = 1,2, . . . ,13.
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Numerical ExampleProbabilities with additive decrease
The probabilities are presented in a matrix shown below, where thesquare at i-th row and j-th column corresponds to the j-th person’sprobability of survival if he is saved at the i-th stage. The bigger thesquare is and the darker its color, the bigger the probability.
Figure 1: Prob that customer j survives if served at stage i22 / 26
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Numerical ExampleProbabilities with additive decrease
Set P0 = (0.5337, 0.5897, 0.6306, 0.6336, 0.6916, 0.7068,0.7229, 0.7593, 0.7649, 0.8395, 0.9046, 0.9339, 0.9733), theexpected number of survivors and the probability that all thosetrapped survive are shown in table 1.
Table 1: E(#survivors) and P(all survive)
Stronger First Weaker FirstE(#survivors) 5.2612 5.0046P(all survive) 0 2.6720×10−6
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Numerical ExampleProbabilities with multiplicative decrease
Choose f (t) = 0.95t for this condition. With the same vector P0,the expected number of survivors and the probability that all thosetrapped survive are shown in table 2.
Table 2: E(#survivors) and P(all survive)
Stronger First Weaker FirstE(#survivors) 7.4918 7.0136P(all survive) 0.0003 0.0003
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
Summary Table
MaxExp MaxP(ALL)WeakFirst StrFirst WeakFirst StrFirst
Add All large p Tie Tie Win LoseDecr Some small p Lose Win Win LoseMult All p Lose Win Tie TieDecrAdd Some large p Lose Win Lose WinIncr No large p Tie Tie Lose WinMult Some large p Lose Win Lose WinIncr No large p Win Lose Tie Tie
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Scheduling aRescue
Nian Liu, MyronHlynka
Introduction
Additive AnalysisConstantProbabilities
Variable Probabilities
Additive Decrease
Simple Example
MultiplicativeAnalysis
MultiplicativeDecrease
NumericalExampleAdditive Decrease
MultiplicativeDecrease
The End
Thank you for your attention!
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