Open-pit mining problem and the BZ algorithm

Transcription

1 Open-pit mining problem and the BZ algorithm Eduardo Moreno (Universidad Adolfo Ibañez) Daniel Espinoza (Universidad de Chile) Marcos Goycoolea (Universidad Adolfo Ibañez) Gonzalo Muñoz (Ph.D. student Columbia Univ.)

2 Outline The Open-Pit Mine Scheduling Problem BZ algorithm The General Algorithm Template Uncertainty in Open-pite mining BZ Algorithm & Stochastic Programming

3 Open pit mining scheduling problem Which block should be extracted, when should they be extracted and how should they be processed First integer programming model due to Thys Johnson ( 69)

4 The problem Assumptions & Notation blocks B T periods Precedences G =(B, A) By decision variables x b,t Side constraints Dx apple d Discount rate ĉ b,t = t c b

5 The problem Time-Consistency constraints: x b,t x b,t+1 Precedence (wall-slope) constraints: x a,t x b,t Capacity (knapsack) constraints: b B a b (x b,t x b,t 1 ) C t Multi-period precedence constrained knapsack

6 The problem The by formulation (PCP by ) max cx x b,t apple x b,t+1, 8 t<t 1 x a,t apple x b,t, 8(a, b) 2 A, 8t Dx apple d x b,t 2 [0, 1]

7 Computational results (LP Relaxation) We count with several REAL instances of this problem. Most of them can t be solved using CPLEX 12.1 Intel(R) Xeon 2.33 GHz, 16 GB RAM. Mine Blocks Prec. Periods CPX1Time Tinkerbell PeterPan Pinochio Mowgli Hansel MEM! Ariel MEM! SnowWhite MEM! GreEel MEM! Simba CinderellaILimit MEM! Dorothy MEM! Rapunzel MEM! Beauty MEM! Aladdin MEM! Alice MEM! BigBadWolf MEM! Cinderella MEM!

8 Ad-hoc LP Algorithms Critical Multiplier Algorithm max cx st x P i apple x j 8(i, j) 2 A ai x i apple b x i 2 [0, 1] max(c a)x st x i apple x j 8(i, j) 2 A x i 2 [0, 1] Explicit construction of optimal multipliers BZ Algorithm

9 The BZ algorithm 1. Start with µ 1 = 0, C 0 1 = B, C 0 = {C 0 1} and k = Let y k be the solution of L(P, µ k 1 ), problem where the side-constraints are penalized with multipliers µ k 1. If k>1 and y k satisfies H k 1 x = 0 stop. 3. Define C k partition of B and impose the constraints H k x = 0 consistent of x i = x j, 8i, j 2 C k h. Also the solution yk must satisfy the constraints H k x = Solve P 2 k, obtained by adding the constraints H k x = 0 to the problem P, and call the solution x k with dual variables µ k associated to the sideconstraints. If µ k = µ k 1 stop. 5. Set k k + 1 and go to step 2.

10 BZ algorithm The algorithm L(P, µ k 1 ) Max-Closure Get duals and penalize with that duals Refine partitions P 2 k Small Problem

11 BZ algorithm P 2 0 Define Solve L(P,~0) Graphically and solve it get duals µ 0 (side constraints)

12 BZ algorithm Refine Solve L(P, the partition µ 0 )

13 BZ algorithm Define Refine Pthe 2 1 partition and solve it get duals µ 1 (side constraints)

14 BZ algorithm And on...

15 BZ algorithm BZ algorithm (+ a LOT of tricks) Mine Blocks Prec. Periods CPX1Time BZ1Time Tinkerbell PeterPan Pinochio Mowgli Hansel MEM! Ariel MEM! 7.89 SnowWhite MEM! GreEel MEM! Simba CinderellaILimit MEM! Dorothy MEM! Rapunzel MEM! Beauty MEM! Aladdin MEM! Alice MEM! BigBadWolf MEM! MEM! Cinderella MEM!

16 General algorithm template We consider the following problem (P ) max cx s.t Ax apple b Dx apple d Easy Hard

17 General algorithm template Feasible solution z 0 k =1 µ 0 = ~0 L(P, µ k 1 )

18 General algorithm template Feasible solution z 0 k =1 µ 0 = ~0 max cx µ k 1 (Dx d) s.t L(P, µ k 1 ) Ax apple b

19 General algorithm template Feasible solution z 0 k =1 µ 0 = ~0 L(P, µ k 1 ) w k

20 General algorithm template Feasible solution z 0 k =1 µ 0 = ~0 L(P, µ k 1 ) (P 2 k ) max cx s.t P 2 k Ax apple b Dx apple d H k x = h k

21 General algorithm template Feasible solution z 0 k =1 µ 0 = ~0 L(P, µ k 1 ) Get duals and penalize with that duals P 2 k

22 General algorithm template Clearly H k x = h k defines an affine subspace, h k =0 but for simplicity let s assume so the subspace is linear

23 General algorithm template This way we can rewrite the P 2 k problem as (P 2 k ) max c X i ix i! s.t A X i ix i! apple b D X i ix i! apple d where x i is a generator of {x H i k x =0}

24 Differences? BZ Dantzig-Wolfe In the master problem we move in an affine subspace that contains the generated point We have the freedom of choosing the affine subspace In the master problem we move in the convex hull of the generated extreme points We add one column (extreme point) at a time We solve the same pricing problem in both

25 Graphically Maybe DW would need to generate... before moving out of the upper lid

26 Graphically But BZ only needs... to describe the entire upper lid

27 Case Study Selective Mine, two metallurgical process Prices, Met. Recoveries, Mining, Process and Selling costs based on the usual numbers of the industry; Asymmetrical contribution of value by process: 10% of the reserves (Process A) give 45% of the cash flow of the project. Some comparison between both process (Process A / Process B): Process Cost 12:1

28 Inventory Reserves Total Tonnes

29 Inventory Reserves Metal

30 Inventory Reserves Cash Flow

31 Mine Schedule - Tonnage

32 Mill Feed - Tonnage

33 Mine Schedule Cumulated Metal Production

34 Mine Schedule Cash Flow

35 Ultimate Pit Whittle Pit Opt Pit

36 Mining Sequence Opt Pits Whittle Nested Pits

37 An important extension x b,d,t = 1 block b is extracted in time 1..t and destination 1..d 0 otherwise

38 Marvin Extracted/processed,weigth, Fixed, Extracted/processed'weigth' Model' " " " " " " Weigth(tons), " " " " " processed" extracted" restric5on_processed" Weight'(tons)' " " " " " restric0on_processed" restric0on_extracted" processed" " " restric5on_extracted" " " extracted" 0" 0" 0" 2" 4" 6" 8" 10" 12" 14" 16" 0" 2" 4" 6" 8" 10" 12" 14" 16" Period, Period' Grade&CU& Grade&AU& 0.7" 0.8" 0.6" 0.7" 0.5" 0.6" Grade& 0.4" 0.3" 0.2" model" fixed" Grade& 0.5" 0.4" 0.3" 0.2" model" fixed" 0.1" 0.1" 0" 0" 0" 2" 4" 6" 8" 10" 12" 14" 16" 0" 2" 4" 6" 8" 10" 12" 14" 16" Period& Period&

39 Delphos Titre,Extracted/Processed,weight, Fixed, Extracted/Processed'weight' Model' " " " " " " Weight(TONN), " " processed" extracted" Weight'(TONN)' " " processed" extracted" " " " " 0" 0" 5" 10" 15" 20" 25" 30" 35" 40" 45" 0" 0" 5" 10" 15" 20" 25" 30" 35" 40" 45" Period, Period' Average&grades&in&processed&blocks& Fixed& Average&grades&in&processed&blocks&& Model& 1.6" 1.8" 1.4" 1.6" average&grade& 1.2" 1" 0.8" 0.6" 0.4" 0.2" Cu" Au" As" average&grade& 1.4" 1.2" 1" 0.8" 0.6" 0.4" 0.2" Cu" Au" As" 0" 0" 5" 10" 15" 20" 25" 30" 35" 40" 45" 0" 0" 5" 10" 15" 20" 25" 30" 35" 40" 45" Period& Period&

40 NCL Extracted/processed'weight' Fixed' Extracted/Processed'weight' Model' " " " " Weight'(TONN)' " " " " " " processed" extracted" restric3on_processed" restric3on_extracted" Weight'(TONN)' " " " " " " restric.on_processed" restric.on_extracted" processed" extracted" 0" 0" 2" 4" 6" 8" 10" 0" 0" 2" 4" 6" 8" 10" Period' Period' Quan1ty'Su'in'processed'blocks' Mill$hours$for$processed$blocks$ " 9000" 8000" " 7000" Period'(TONN)' " " restric2on_su" model" fixed" Mill$(hours)$ 6000" 5000" 4000" 3000" restric2on_mill" model" fixed" 50000" 2000" 1000" 0" 0" 1" 2" 3" 4" 5" 6" 7" 8" 9" 10" Period' 0" 0" 1" 2" 3" 4" 5" 6" 7" 8" 9" 10" Period$

41 Uncertainty in Open-pit Mine Planning Grade Uncertainty Ore Prices x e b = ( 1 if b is extracted 0 if not x p b = ( 1 if b is processed 0 if not x p b apple xe b 8b Objective Function: (x e,x p )=(p p c p )x p c e x e ( b := grade of ore in block b)

42 Scenario approaches Multiple Blockmodels Opt. Multiple plans Extraction plan Multiple Blockmodels Opt. Extraction plan Most common in mining literature PH on ore prices Dimitrakopoulos et al. DP on grade uncertainty Risk minimization Robust programming

43 Our approach for ore prices uncertainty mean reversion process of ore prices (Ornstein Uhlenbeck process) (P ) max cx p p 2 U = {p p 9u, u 0 u apple 1,p p = p + u} A second-order conic constraint s.t Ax apple b Dx apple d Easy Hard

44 Other uses of BZ Algorithm Value-at-Risk VaR " ( ĉx) =min{t : F ĉx (t) 1 "} t Conditional Value-at-Risk (expected shortfall) CVaR " ( ĉx) =E( ĉx ĉx VaR " ( ĉx)) apple =min t t + 1 " E(( ĉx t)+ ) Convex function!

45 CVaR formulation probability of scenario i min CVaR " ( ĉx) s.t. Ax = b x 0 SAA min t + 1 " NX p i i i=1 s.t. Ax = b i c i x t 8i x, 0 scenario i CVaR " ( ĉx) =min t apple t + 1 " E(( ĉx t)+ )

46 Our idea x : A x = b min t + 1 " NX p i i Easy to solve! Upper bound i=1 s.t. c i x + t + i 0 8i NX 0 min t + 1 " p i i s.t. Ax = b i=1 x {P j } c i x + t + i 0 8i x, 0 min t + 1 " kx ˆp j ˆ j Smaller problem! N = {P j } k j=1 Lower bound ˆp j = X p i c j = 1ˆp X p i c i i2p j j i2p j j=1 s.t. Ax = b c j x + t +ˆ j 0 8j 2 {1,...,k} x, ˆ 0

47 Dual problems max y t b s.t. y t A + 1 " NX p i i c i apple 0 i=1 NX p i i = " i=1 0 apple i apple 1 8i x : A x = b Upper Lower bound N = {P j } k j=1 Lower Upper bound Additional constraints: i = i 0 8i, i 0 2 P j max s.t. max y t b 1 " NX i=1 p i i c i x NX p i i = " i=1 s.t. y t A + 1 " kx j=1 0 apple i apple 1 8i x kx j=1 ˆp j ˆj = " ˆp j ˆj c j apple 0 0 apple ˆj apple 1 8j 2 {1...K} (i) (i+1) = =1i apple i (i) =0i>i +1

48 Computational results for large N values for " =1%,10%,50%and75%. N =100, 000 N =1, 000, 000 N =10, 000, 000 Name time Iter. N time Iter. N time Iter. N adlittle afiro agg bandm beaconfd boeing bore3d brandy capri e etamacro f ganges grow grow grow israel kb lotfi perold pilot.ja

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50 Thanks...