Spatial Forest Planning with Integer Programming

Lecture 10 (5/4/2015)

Spatial Forest Planning

Spatial Forest Planning with

Roads

Balancing the Age-class Distribution

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Red OakRed MapleOther ZonesOther OaksOther Hrdwds.N. Hrdwds.ConifersA. Hrdwds.

Acr

es

Forest Age Class Distribution

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Red OakRed MapleOther ZonesOther OaksOther Hrdwds.N. Hrdwds.ConifersA. Hrdwds.

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Forest Age Class Distribution

Spatially-explicit Harvest Scheduling Models

• Set of management units• T planning periods• Decision: whether and when to harvest

management units– Modeled with 0-1 variables– xmt = 1 if unit m is harvested in period t, 0

otherwise

Spatially-explicit Harvest Scheduling Models (continued)

• Constraints– Logical: can only harvest a unit once, at most– Harvest volume, area and revenue flow control– Ending conditions

• Minimum average ending age• Extended rotations• Target ending inventories

– Maximum harvest area (green-up)– Others spatial concerns:

• Roads, mature patches, etc…

one constraint for each and for each t=hi,…,T

Integer Programming Model for Spatial Forest Planning

0 01

[ ]m

M T

m m m mt mtm t h

MaxZ A c x c x

0 1m

T

m mtt h

x x

0

ht

mt m mt tm M

v A x H

, 1 0l t t tb H H

, 1 0h t t tb H H

1jtj C

x C

0 01

[( ) ( ) ] 0m

M TT TT Tm m m mt mt

m t h

A Age Age x Age Age x

0,1mtx

C

subject to:

one constraint for each m=1,2,…,M

one constraint for each t=1,2,…,T

one constraint for each t=1,2,…,T-1

one constraint for each t=1,2,…,T-1

for each m=1,2,…,M and for each t=1,2,…,T

Notation

wherehm = the first period in which management unit m is old enough to be harvested,xmt = a binary variable whose value is 1 if management unit m is to be harvested

in period t for t = hm, ... T; when t = 0, the value of the binary variable is 1 if management unit m is not harvested at all during the planning horizon (i.e., xm0 represents the “do-nothing” alternative for management unit m),

M = the number of management units in the forest,T = the number of periods in the planning horizon, cmt = the net discounted net revenue per hectare plus the discounted expected

forest value at the end of the planning horizon if management unit m is harvested in period t,

Mht = the set of management units that are old enough to be harvested in period t,Am = the area of management unit m in hectares,vmt = the volume of sawtimber in m3/ha harvested from management unit m if it is

harvested in period t,Ht = the total volume of sawtimber in m3 harvested in period t,

andbl,t = a lower bound on decreases in the harvest level between periods t and t+1

(where, for example, bl,t = 1 would require non-declining harvests and bl,t = 0.9 would allow a decrease of up to 10%),

bh,t = an upper bound on increases in the harvest level between periods t and t+1 (where bh,t = 1 would allow no increase in the harvest level and bh,t = 1.1 would allow an increase of up to 10%),

C = the set of indexes corresponding to the management units in cover C, = the set of covers that arise from the problem,hi = the first period in which the youngest management unit in cover i is old

enough to be harvested, = the age of management unit m at the end of the planning horizon if it is

harvested in period t; and = the minimum average age of the forest at the end of the planning horizon.

Notation cont.

TmtAge

TAge

The Challenge of Solving Spatially-Explicit Forest Management Models

• Formulations involve many binary (0-1) decision variables– Feasible region is not convex, or even

continuous– In fact, it is a potentially immense set of

points in n-dimensional space• Solution times could increase more than

exponentially with problem size

1.) Unit Restriction Model (URM):adjacent units cannot be cut

simultaneously

2.) Area Restriction Model (ARM): adjacent units can be cut simultaneously as long as their combined area doesn’t exceed the maximum harvest opening size

Pair-wise Constraints for URM

Pair-wise adjacency constraint for AB is

1At B tx x

A, 14 B, 4 C, 10

F, 4

D, 8

E, 7

Adjacency list: AB

BC BD BE CD

AD AE

DF

What is a “better” formulation?

Maximal Clique Constraints for URM

Clique adjacency constraint for ABD is

1At B t Dtx x x

A, 14 B, 4 C, 10

F, 4

D, 8

E, 7

Maximal clique list:

BCD DF

ABD ABE

Cover Constraints for ARM

Cover constraint for FDC is:2C t D t F tx x x

A, 14 B, 4 C, 10

F, 4

D, 8

E, 7

Cover list: ABC

AE BCE BDEF CDF

AD BCD

1mtm C

x C

In general:

max 20A ha

McDill et al. 2002

GMU Constraints for ARM

A, 14 B, 4 C, 10

F, 4

D, 8

E, 7

GMU variable for BDF is: BDF tx

GMU list: A-F

BD BDE BDF BE

AB BC

CD CF

1                     , ..., jt

nt jn K

x j J and t h T

The constraintin general form:

Where is the set of GMUs that contain at least one stand in max clique j; J is the set of max cliques; and hj is the first period in which the youngest stand in clique j is old enough to be cut.

max 20A ha

McDill et al. 2002 and Goycoolea et al. 2005

jtK

The “Bucket” Formulation of ARM

• Introduce a class of “clearcuts”: , and• Introduce variables that take the

value of 1 if management unit m is assigned to clearcut i in period t.

• Objective function:

Constantino et al. 2008

itmx

00

1

[ ]m

M Ti it

m m m mt mm i t h

MaxZ a c x c x

The “Bucket” Formulation of ARM (cont.)

• Constraints:

0,

1m

Titm

t t h i

x

for m = 1, 2, ..., M

max1

Mit

m mm

a x A

for i and t = hm, . . . T

it itm Qx w for ,   ,  Q m Q i m and t = hm, . . . T

1itQ

i

w

for Q and t = hm, . . . T

The “Bucket” Formulation of ARM (cont.)

• Constraints:

, 

0ht

itmt m m t

m M i

v a x H

for t = 1, 2, . . . T

00

1

[( ) ( ) ] 0m

M TT TT i T itm m m mt m

i m t h

a Age Age x Age Age x

, 1

, 1

0

0

l t t t

h t t t

b H H

b H H

for t = 1, 2, . . . T-1

The “Bucket” Formulation of ARM (cont.)

• takes the value of one whenever a unit in maximal clique is assigned to clearcut i in period t;

• is a maximal clique in the set of maximal cliques.

itQw

Q

Q

Strengthening and Lifting Covers

13 14 43 50 3x x x x

24 38 1x x

18 31 40 2x x x

24 38 3

24 38 49

11

x x xx x x

18 31 40 6 15

18 31 40 8

22

x x x x xx x x x

13 14 43 50 32 3x x x x x

max 48A ha

Formalization

maxjj Ca A

1jj Cx C

max\ jj C la A

.l C

where C is a set of management units (nodes) that forma connected sub-graph of the underlying adjacency graph,and for which holds, but

for any

The minimal cover constraint has the general form of:

maximum harvest area

area of unit j

Strengthening the Minimal Covers

{ {0,1} : 1, C }.nii C

P x x C

( )N A

s( )( , ) max{ : and x 1}jj N s Cs C x x P

denote the set of all possible minimal covers thatarise from a certain forest planning problem (oradjacency graph).

Notation: Let

For every set of management units A, let the set of all management units adjacent, but not belonging, to A.

Define:

represent

Define:

Number of units in the forest

Proposition: Consider a minimal cover C ( ).s N C

* ( ) ( , ) 1 .N s C s C Define: *: Then, for all

1j sj C

x x C

is valid for P.

and

Strengthening the Minimal CoversCont.

s( )( , ) max{ : and x 1}jj N s Cs C x x P

Strengthening the Minimal CoversCont.

1C

\ ( ) ( )j s j j

j C j C N s j N s C

x x x x

( )

\ ( ) *jj N s C

C N s x

\ ( ) ( , ) *C N s s C

.x P 0sx

1sx

Proof: Consider If , then the inequality holds by the

If , then:

definition of minimal cover C.

* ( ) ( , ) 1N s C s C

How strong are these inequalities?

max 3A ha

332211ss

441ha1ha 1ha1ha1ha1ha

2ha2ha

1ha1ha

1 2 3 4 3x x x x

1 2 3 4 2 3sx x x x x 1 2 3 4 1 3sx x x x x

* ( ) ( , ) 1N s C s C

Sequential Lifting Algorithm

\{ }C

NO

YES

YES

Pick a cover, C from set .

Generate the set of management units, S that are adjacent to at least 2 units of C.

Pick a unit, s from set S.

* 1s ?

Determine *s using Proposition 1.

?

STOP

NO

YES

Generate the set of minimal covers, using the Path

Algorithm

For Cs Q , set * s s Solve ,A CP .

, 1?A CP C

NO

NO 0? i

Solve ,A CP .

, 1?A CP C YES NO

NO

YES

1. 12. 3. 1

i i

jt ij j

* ?C C

s ss Q s Q

YES

? Ci Q NO

YES

0? i

YES

\{ }

0

for an ?s

kk Q s

Cs Q

NO

* ?s s

YES

NO

S ?

NO

\{ }S S s

{ }C CQ Q s CQ ?

YES

1CQ

NO

YES

NO

Build E(C) YES

i=j=1

1. 2. 13. 1

j

i i

i t

j j

NO

i =i+1

? Ci Q NO

YES

YES

[1] [1] [2]

[3]

[4]

[5] [7]

[6]

[8]

[9]

[23]

[10]

[11]

[12]

CQ .

Build E(C)

Build E(C)

Build E(C)

[13] [14]

[15]

[16] [17]

[18] [19]

[20]

[21]

[22]

[24]

Strengthening and lifting

Compatibility testing

Open Questions:

1) Can one define a rule for multiple lifting?

2) Can one find even stronger inequalities?

3) Is there a better formulation?

A “Cutting Plane” Algorithm for the Cover Formulation

21 37 47 2X X X

8 18 40 2X X X