Solving the Knapsack Problem Sonia Cae ri Optimisation ...recherche.enac.fr/~cafieri/slides/knapsack-1x2.pdf · Sonia Cae ri (ENAC) Knapsack Problem 2011-2012 5 / 50 Greedy algorithm

Optimisation Discrete

Solving the Knapsack Problem

Sonia Cafieri

ENAC

[email protected]

Sonia Cafieri (ENAC) Knapsack Problem 2011-2012 1 / 50

Outline

1 Introduction

2 Heuristics for the Knapsack Problem

3 Solving the Continuous Knapsack Problem

4 Exact solution of the KPBranch and Bound algorithms for IPBranch and Bound for the KPBranch and Bound for IP : Remarks


Knapsack problem (KP)

max

n∑i=1

xiui

s.t.n∑

i=1

xivi ≤ V

xi ∈ {0, 1} ∀i

V = volume of the knapsack

vi = volume of objecti

ui = value of objecti

xi = 1 if object i is selected, 0 otherwise


Solving the KP

There is no polynomial algorithm known that solves the problem to optimality.There is also no proof that such an algorithm does not exist.

How to solve the Knapsack problem?

Apply an algorithm with exponential running time to find anoptimal solution

Enumerate all possible solutions in a smart way to find anoptimal solution:branch-and-bound or dynamical programming

Apply a polynomial algorithm that constructs agood but possibly non-optimalsolution: heuristic approach.


Heuristics for KP : Greedy algorithm

Greedy : algorithm that makes the choice that looks the best at the moment.

Idea :

Sort by most rewarding items with respect to a given criterionand take as many of the top most valuable items as will fit.

Heuristic 1 : sort ui i ∈ {1, . . . , n} by non-increasing order

Heuristic 2 : sort vi i ∈ {1, . . . , n} by increasing order

Heuristic 3 : sort ui/vi i ∈ {1, . . . , n} by non-increasing order

⇒ the most efficient greedy


Greedy algorithm for KP

Let i, j be indexes such thatj = 1, . . . , n correspond to items sorted with respect to a givencriterion

W← 0

for j = 1, . . . , n do

W← vi j + W

if (W≤ V) then

xi j ← 1

else

xi j ← 0

W←W− vi j

endif

endfor


0-1 Knapsack and Continuous Knapsack

0-1 Knapsack problem :

you can only choose to take all or nothing of a particular item(0-1)

Continuous (Fractional) Knapsack :

you can take any fractions of the items to put in the knapsack

⇒ continuous relaxationof 0-1 Knapsack

For both versions :

Optimal Substructure property:

an optimal solution contains within it optimal solutions tosubproblems.

If we remove itemi from an optimal solution, then the remaining items must be optimal forthe subproblem where the bag volume isV − vi and the available items do not include itemi.Otherwise, we would be able to construct an even better solution than the optimal one.


Solving the CKP: Greedy (n.3)

Idea :

Sort by most rewarding items with respect to the ratio of value to size, and take as muchas possible of the most valuable items as will fit.

Steps :

1 compute ratios (densities)ui/vi

2 sort them by non-increasing order

3 take as much as possible of the densest item not already in thebag

Complexity :O(n logn) for sorting, O(n) for greedy choices


Solving the CKP: Greedy (n.3)Sort values/volumes by non-increasing order.Let i, j be indexes such thatui j/vi j with j = 1 is the maximum ratio:

ui1

vi1≥ ui2

vi2≥ · · · ≥ uin

vin

W← 0

for i ≤ n do

if (W + vi j ≤ V) then

take itemi :xi j ← 1W←W + vi j

else

take a fraction of itemi :

xih ←V −W

vihxi j ← 0 ∀j > h

endif

endforSonia Cafieri (ENAC) Knapsack Problem 2011-2012 9 / 50

Solving the Continuous Knapsack Problem

TheoremThe Greedy algorithm (n.3) is optimal for the Continuous Knapsack Problem (CKP)(continuous relaxation of 0-1 Knapsack)

Proof. 1

Note that the algorithm fills the knapsack completely.It may take a fraction of an item, which can only be the last selected item.

The greedy solution is such that there exists ah such that

1 = xi1 = xi2 = · · · = xih > xih+1 > xih+2 = · · · = xin = 0.

We compare this solution with another (feasible) solutionyi1, . . . , yin.

Let k be the smallest index such thatyik < 1, andlet ℓ be the smallest index such thatk < ℓ and yiℓ > 0.(such anℓ exists, otherwise they solution would be equal to thex solution).



Proof. (cont.)

We increaseyik and decreaseyiℓ , while keeping all other values equal.Let ǫ = min{vik(1− yik), viℓyiℓ}.We increaseyik by

ǫ

vikand decreaseyiℓ by

ǫ

viℓ, obtaining a new solution.

This new solution is no worse than they solution.Moreover, eitheryik has become 1, oryiℓ has become 0.

Repeating the same argument we find the solutionx of the greedy.



TheoremThe Greedy algorithm (n.3) is optimal for the Continuous Knapsack Problem (CKP)(continuous relaxation of 0-1 Knapsack)

Proof. 2

Let O = {o1, o2, . . . , on} be the optimal solution andG = {g1, g2, . . . , gn} be thegreedy solution(where the items are ordered with respect to the greedy choice).We show that there exist an optimal solution such that it alsotakes the greedy choicein each iteration.

First, we show that an optimal solution selects (a fraction or 1 unit of) itemg1.



Proof. 2 (cont.)

Two cases:1 G takes 1 unit ofg1.

If O also takes 1 unit ofg1, done.If O doesn’t take 1 unit ofg1, then we take away volumevg1 from O and put 1unit of g1 to it ⇒ new solutionO′.Of courseO′ has the same volume constraintV asO.Moreover,g1 has maximum ratiou/v.⇒ O′ is as good asO ⇒ g1 ∈ O′ andO′ is an optimal solution.

2 G takes a fractionf of g1 (the volume added toO is f × vg1).If O also takes a fractionf of g1, done.If O takes less thanf of g1 (cannot take more becauseo optimal), then we takeawayf × vg1 from O and putf of g1 to it ⇒ new solutionO′.As before,O′ is as good asO (otherwise contradiction).



Proof. 2 (cont.)

Second, we observe thatafter each greedy choice is made, we have a problem of the sameform asthe original one.

The problem exhibits the optimal substructure property⇒ G≡ O.



(Another) Alternative Proof:

The simplex algorithm for solving the CKP takes exactly the same steps as the greedy.

Let ih be such that :

∀ j ∈ {1, . . . , h} xi j ← 1

∀ j ∈ {h + 2, . . . , n} xi j ← 0

xih+1 ← V −∑hj=1 vi j

vih+1

that is,ih is such that:h∑

j=1

vi j ≤ V andh+1∑j=1

vi j > V

Let us apply the simplex algorithm with the dictionary representation.



(Another) Alternative Proof: (cont.)

xn+1 = 1− xi1

xn+2 = 1− xi2

...x2n = 1− xin

x2n+1 = V −∑nj=1 vi j xi j

z =∑n

j=1 ui j xi j




pivot: xi1 ↔ xn+1

xi1 = 1− xn+1

xn+2 = 1− xi2

...x2n = 1− xin

x2n+1 = V −∑nj=2 vi j xi j − vi1(1− xn+1) = (V − vi1)−

∑nj=2 vi j xi j + vi1xn+1

z = ui1 +∑n

j=2 ui j xi j − ui1xn+1

We continue with following dictionaries.




Dictionaryh :pivot: xih↔ xn+h

xi1 = 1− xn+1

xi2 = 1− xn+2

...xih = 1− xn+h

xn+h+1 = 1− xih+1

x2n = 1− xin

x2n+1 = (V −∑hj=1 vi j )−

∑nj=h+1 vi j xi j +

∑hj=1 vi j xn+j

z =∑h

j=1 ui j +∑n

j=h+1 ui j xi j −∑h

j=1 ui1xn+j


Solving the Continuous Knapsack Problem(Another) Alternative Proof: (cont.)

Dictionaryh + 1 :pivot: xih+1 ↔ x2n+1

xi1 = 1− xn+1

xi2 = 1− xn+2

...xih = 1− xn+h

x2n = 1− xin

xih+1 =(V −∑h

j=1 vi j )vih+1

−∑n

j=h+2 vi j xi j

vih+1

+

∑hj=1 vi j xn+j

vih+1

− x2n+1

vih+1

xn+h+1 = 1− (expression ofxih+1)

z =∑h

j=1 ui j + uih+1(expression ofxih+1) +∑n

j=h+2 ui j xi j −∑h

j=1 ui j xn+j




Dictionaryh + 1 :pivot: xih+1 ↔ x2n+1

z =h∑

j=1

ui j + uih+1

((V −∑h

j=1 vi j )vih+1

)

+n∑

j=h+2

(ui j − uih+1

vi j

vih+1

)xi j

+h∑

j=1

(vi j

vih+1

− ui j

)xn+j

−x2n+1

vih+1




We check the signs of the coefficients of the variables :

∀j = h + 2, . . . , n sign

(ui j − uih+1

vi j

vih+1

)= sign

(ui j

vi j− uih+1

vih+1

)= −1

because j > h + 1

∀j = 1, . . . , h sign

(uih+1

vih+1

vi j − ui j

)= sign

(uih+1

vih+1

− ui j

vi j

)= −1

because j < h + 1

Hence the simplex algorithm stops.

⇒ the (feasible) solution found is also the optimal solution.


Solving the KP exactly

Can we solve the KP exactly ?⇓

Branch-and-Bound (B&B)(séparation et évaluation)


Branch and Bound

Branch-and-Bound (B&B) is a general framework for solvingexactly

integer and discrete optimization problems.

Intelligentimplicit enumeration of all possible solutions

only potentially good solutions are enumerated,

solutions that cannnot lead to improve the current solutionare not explored

based on the observation that the enumeration of integer solutions has a

tree structure.


Branch and Bound - search treeSearch Tree :

Nodes: represent stages of construction of the solution

Arcs : represent choices made to build the solution

For the KP :

Nodes : represent a step in whichsome objects have been placed in the bag,some others have been left out of the bag,and others for which no decision has yetbeen taken.

Arcs : indicate the action of puttingan object in the bag or not putting it in thebag


Branch and Bound - idea

The nodes of the search tree represent subsets of all of the possible solutions.In particular, the root node representsall solutions that can be generated bygrowing the tree.

The main idea is to build step by step the search tree avoidinggrowing the whole tree

as much as possible.

Only the most promising nodes are generated and explored at any stage.

The node that is the most promising is determined by estimating a bound on the bestvalue of the objective function that can be obtained by growing that node to later stages.


Branch and Bound - main steps

Branching:

a node is selected for further growth and

two new subproblems are created

Bounding:

upper and lower bounds on the best value

attained by growing a node are estimated

Pruning:

nodes that are showed to be never either

feasible or optimal are permanently discarded



During the search, a node must be in one of the following states :

(i) has no feasible solution

(ii) is such that its associated solutionxS is feasible and(a) xS has a value better than the incumbent *

(b) xS has a value no better than the incumbent *

(iii) neither of states (i) and (ii) has yet been established.

A node in either state (i) or (ii) is said to befathomed.

A node that has not yet been fathomed is said to beactive.

A node in states (i) or (iib) cannot contain the optimum, it isprunedfrom the tree.

* incumbent= best integer solution found so far



The search continues, branching from the active nodes untileither :

(i) all nodes have been pruned from the tree

(in which case the problem has no feasible solutions)

(ii) all active nodes have bounds no greater than the value of the incumbent

(in which case the incumbent represents an optimal solution)


Branch and Bound for KP - principles

The solution set is partitioned into subsetsS1, . . . , SM

each subset corresponds to a partial (feasible) solution,

i.e. some of the variables have been fixed

When abranchingwith respect to a variablexj is done, 2 new subsets

are created fromSk :

S0k where we set xj = 0, S1

k where we set xj = 1

An upper boundUBk corresponding toSk can be computed by the

greedy algorithm on the CKP corresponding toSk (integer part)

A lower boundLBk corresponding toSk can be computed by

rounding down the fractional variable, if it exists.

If UBk = LBk the subproblemk is solved exactly.


Branch and Bound for KP - principles

When we continue branching ?

Let z∗ be the value of the best feasible solution found so far:

current best solution: z∗ ≥ max{LBk, ∀k}

UBk > z∗ : the subsetSk may contain a feasible solution with value> z∗.

We continue working onSk. (Sk, UBk) is active ⇒ Branching

UBk ≤ z∗ : the subsetSk cannot contain a solution with value> z∗.

The subset is of no interest anymore.⇒ Fathoming


Branch and Bound for KP - steps

Initialization

construct a list L containing all active pairs(Sk, UBk) :

initially, the original problem

initial Upper BoundUB : solution of the CKP via the greedy algorithm

initial Lower BoundLB : a feasible solution

(rounding down the value of the fractional variable in the solution of CKP)


Branch and Bound for KP - steps

As long asL is not empty :

choose among the active pairs the one with the maximalUB : (Sk, UBk)

split Sk in S0k andS1

k according to a branching criterion

(ex. : the fractional variablexi in the continuous solution)

computeUB0k andUB1

k (greedy algorithm for CNP)

computeLB0k andLB1

k (feasible solution for KP)

if LB0k > z∗, then set z∗ ← LB0

k and updatex∗ accordingly

if LB1k > z∗, then set z∗ ← LB1

k and updatex∗ accordingly

if UB0k > z∗, then add (S0

k, UB0k) to L

if UB1k > z∗, then add (S1

k, UB1k) to L

updateL deleting the elements(Sh, UBh) such thatUBh < z∗


Branch and Bound for KP - algorithm

Sort values/volumes by non-increasing order.Let i j be indexes such thatui j /vi j with j = 1 is the maximum ratio

xi ← 2 ∀ik← 1 // iteration counter

L← (Sk, UBk) // original problem

z∗ ← LBk

while (L not empty) do

ν ← min{i j |j ∈ {1, ..n} ∧ xν = 2}x0ν ← 0, x1

ν ← 1 // two branches

for p = 0, 1 do

compute UBpk and LBp

k

if(∑ν

j=1 vi j ≤ V)

then

if (UBpk > z∗ and LBp

k 6= UBpk) then

add (Spk, UBp

k) to L by non-increasing order forUB


Branch and Bound for KP - algorithm (cont.)

if (LBpk > z∗) then

z∗ ← LBpk // current best solution

x∗ ← corresponding solution

end if

end for

L← L \ (Sk, UBk)k← k + 1

end while


Example

max z = 12x1 + 11x− 2 + 8x3 + 6x4

s.t. 8x1 + 8x2 + 6x3 + 5x4 ≤ 20

xi ∈ {0, 1} ∀i

u1

v1>

u2

v2>

u3

v3>

u4

v4


branching on the variable with lowest index www.win.tue.nl/ wscor/OW/2V300


branching on fractional variables www.win.tue.nl/ wscor/OW/2V300


Designing a Branch and Bound algorithm

Branch and Bound is a very general framework.

To completely specify how the process is to proceed, you alsoneed to define

How to obtain bounds?

How to branch?

In what order to examine nodes?


Branch and Bound: Obtaining bounds

For a maximization problem

- the (current) tightest lower bound, provided by the best solution we have found,is also referred to as theprimal bound.

- the upper bound on the profit obtainable within a given set of solutionsis also referred to as thedual bound.

To obtain adual (upper) bound:

one typically solves a relaxation of the subproblem :

LP relaxation (but also Lagrangian relaxation, etc..)

The bound can also be computed by solving thedual problem.

To obtain aprimal (lower) bound:

one typically computes a feasible solution of the subproblem.


Dual Problem

For aLP problem in standardPrimalform:

(LP)

max cTx x∈ Rn, c ∈ Rn

s.t. Ax≤ b A∈ Rm×n, b ∈ Rm

x≥ 0

TheDual Problemis :

(D)

min bTy y∈ Rm

s.t. ATy≥ c

y≥ 0


Dual Problem

To obtain the Dual Problem:

- write the Lagrangian : L(x, y, s) = cTx− sTx + yT(b− Ax), with s∈ Rn, y ∈ Rm

(using nonnegative Lagrangian multipliers to add the constraints to the objective function)

- write the Lagrangian dual : min max(yb+ (cT − s− yA)x)

- apply the optimality conditions (KKT) to the Lagrangian dual

The solution of the dual problem provides an upper bound

to the solution of the primal problem


Dual Problem : ExampleKnapsack with 2 constraints, relaxed

maxn∑

i=1

xiui

s.t.n∑

i=1

xivi ≤ V

n∑i=1

xipi ≤ P

xi ≤ 1 ∀ixi ≥ 0 ∀i

A =

v1 . . . vn

p1 . . . pn

1 . . . 0...

...0 . . . 1

b =

VP1...1



AT =

v1 p1 1 . . . 0...

... 0 . . . 0vn pn 0 . . . 1

(D)

min bTy =

n+2∑i=1

biyi

s.t. ATy≥ u ⇔ viy1 + piy2 + yi+2 ≥ ui ∀i ∈ {1, ..n}y≥ 0



A (feasible) solution can be obtained by setting :

y1 = y2 = 0, yi+2 = ui ∀i

The solution is :n∑

i=1

ui . This gives an upper bound for the 0-1 KP.

Another solution can be obtained by setting :

y1 = y2 = 1, yi+2 = max{0, ui − pi − vi ∀i}The solution is :

V + P +n∑

i=1

max{0, ui − pi − vi}

that can be better than the previous one ifV + P <

n∑i=1

ui.

So, we can search for

minα∈R+

α(V + P) + max{0, ui − α(pi + vi)}


Branch and Bound: How to partition into subproblems

IP problems :

- choose xj /∈ Z- set Sj

k = Sk ∪ {x : xj ≤ ⌊xkj ⌋} and

Sj+1k = Sk ∪ {x : xj ≤ ⌊xk

j ⌋+ 1}How to choose the fractionary component? Example : choose the component with

fractionary value closest to 0.5

Binary IP problems :

- choose xj

- setS0k = Sk ∪ {x : xj = 0} and

S1k = Sk ∪ {x : xj = 1}


Branch and Bound: How to choose a node

The search tree can be explored using

Depth-first search

allows to find a feasible solution to the IP quickly

Breadth-first-search

may allow to prune more of the search tree sooner

Best-Node First

chooses the node with the best dual bound.

This might lead to finding a good incumbent quickly, thus eliminating the need

to evaluate the other subproblems

Otherheuristics


Bibliography I

L.R. Foulds.Combinatorial Optimization for UndergraduatesSpringer-Verlag, 1984.

C.H. Papadimitriou and K. Steiglitz.Combinatorial Optimization: Algorithms and Complexity.Dover, New York, 1998.

John W. Chinneck.Practical Optimization: a gentle introductionwww.sce.carleton.ca/faculty/chinneck/po.html, 2010.


Documents

Solving the Knapsack Problem Sonia Cae ri Optimisation ...recherche.enac.fr/~cafieri/slides/knapsack-1x2.pdf · Sonia Cae ri (ENAC) Knapsack Problem 2011-2012 5 / 50 Greedy algorithm