Underground Scheduling

8/13/2019 Underground Scheduling

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A MIP model for underground mine planning

Frederic Meunier

Work in progress with Nelson Morales
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Mine planning is a core discipline in mining. It starts with theresources and generates a production plan (a bankabledocument).Mine planning is hard, therefore it has been split into several

steps.Dening what to mine (distinguishing reserves from resources)is decoupled from determining how to do it (not onlyconceptually, but often done by different teams/consultants).
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Scheduling the tasks: a matter of prot

There is a need for a tool providing a schedule of the tasks thatmaximizes the prot.

This schedule has to be at a strategic level : time is discretized inperiods t = 1 , . . . , T .

A period (typically: about 30 days). (Typical number of periods:55).

For each period t:

the prot of performing a unit of task i: vti .the availability of resource r: Ktr .


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Notations

V: set of tasks

A: precedence constraints between tasks

T periods (t = 1 , . . . , T ) (period t takes Dt days)

R: set of resource types (for instance: tunnel boring machines,crushers, water pumps, workers...)

Ktr: set of resources of type r available during period t

1% of task i needs cri days of use of a type r resource(parallelization is allowed).
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A rst MIP formulation

variables sti: equal to 1 if task i has started by period t and 0 if not

variables eti : equal to 1 if task i has not yet ended by period t and0 if notvariables pti : percentage of task i developed during the period t
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A rst MIP formulation

max i, t vti pti

s.t. sti st+1i for all i V, t

eti et 1i for all i V, t

pti s

ti

for all

i V, tpti eti for all i V, t

1 eti s t psi for all i V, tsti , e

ti {0, 1} for all i V, t

pti 0 for all i V, t

st j 1 et 1i for all (i, j) A, t

i cripti | K

tr |D

tfor all r R, t
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This model has a drawback

The constraint

st j 1 et 1i for all (i, j) A, t = 2 , . . . , T

prohibits feasible schedule: if task i ends at the beginning of periodt , the remaining time could be used to start task j.


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A less strict constraint

st j 1 eti for all (i, j) A, t = 1 , . . . , T

may allow unfeasible schedule: task i may nish at the end of period t and task j at its beginning.


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Morales, Rocher and Rubio (2011) proposed a way to partiallysettle the problem and succeeded in allowing more feasiblesolutions. However, their model needs an exponential number of constraints and some feasible schedules are still not covered.
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The one period problem

V: set of tasks

A: precedence constraints between tasks

D days available

R: set of ressource types (for instance: tunnel boring machines,crushers, water pumps, workers...)

Kr : set of resources of type r

task i needs cri days of use of a type r ressource (parallelization isallowed).

Is it possible to achieve all tasks in D days ?
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The one period problem: necessary condition

Assume that we have a feasible schedule and denote: i the number of days devoted to task i dri the cumulated number of days a resource of type r is

assigned to task i i the number of days the resource is assigned to task i

Then we have

iV i D for all resources

dri = Kr

i for all tasks i and resource types r

i maxrR, Kr i for all tasks i

iP

i D for any path P in the digraph (V, A)

dri = c ri for all tasks i and resource types r


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The one period problem: sufficient condition

However, it is not a sufficient condition (counter-examples exist).Nevertheless if

iV

i D for all resources

dri = Kr

i for all tasks i and resource types r

i = maxrR

dri for all tasks i

iP i D for any path P in the digraph (V, A)

dri = c ri for all tasks i and resource types r

then there exists a feasible schedule.


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iP i D for any path P in the digraph (V, A) requires anexponential description.

Introducing variables yi for i V reduces the number of inequalities:

yi = 0 if i has no predecessoryi D if i has no successory j yi + i for all (i, j) A
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The one period problem: sufficient condition compactversion

PropositionIf there exist yi for all i V such that

iV cri | Kr |D for each resource type ryi = 0 if i has no predecessor

yi D if

i has no successor

y j yi + max rR cri for all (i, j) A

then there is a feasible scheduling.

This condition can be tested through linear programming.More efficiently, through dynamic programming: once

iV cri | Kr |D for each resource type r we can use a PERTapproach:put on each arc (i, j) a weight = max rR cri;

feasible schedule: iff the longest path in the graph has length D.
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Back to the multi-period problem

The constraints are added:iV ctri | Kr |Dt for each resource type r.

yi = 0 if i has no predecessoryi D if i has no successory j yi + max rR ctri for all (i, j) A
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Back to the multi-period problem: an advanced MIPformulation

We get a new model (but still with feasible solutions not covered)

max i, t vti pti

s.t. usual constraintsdening

sti , eti and sti

st j 1 eti for all (i, j) A, t

yi = 0 if i has no predecessoryi Dt if i has no successory j yi + max rR ctri for all (i, j) A

i cripti | K

tr |D

tfor all r R, t
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Open questions

For the one-period problem, is it possible to characterize feasible

instances ?Is there even a polynomial algorithm ?

A positive answer would lead to a compact MIP formulation of themulti-period problem.
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Underground Scheduling

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