on the complexity of allocation problems with probabilistic players rishab nithyanand research...

On the Complexity of Allocation Problems with Probabilistic Players

Rishab Nithyanand

Research Proficiency ExaminationSummer 2012

On the Complexity of Allocation Problems with Probabilistic Players 2

Presentation Outline

• Introduction

• The Password Allocation Problem

• The Weapon-Target Allocation Problem

• Conclusions and Future Work

3

Traditional Allocation Problems

• Given:– resources (r1, r2, …, rn) and tasks (t1, t2, …, tk)– objective function F

• Goal:– Find allocation for which F is optimal

• Constraint:– at most one task per resource


4

Allocation Problems with Probabilistic Players

• Given:– resources (r1, r2, …, rn) and tasks (t1, t2, …, tk)

– resource ri completes task tj with probability pij

– objective function F

• Goal:– Find allocation for which E[F] is optimal

• Constraint:– at most one task per resource


5

• Users have a large set of accounts– some are very valuable

– and some are less valuable

• Passwords are hard to remember– [Vu, 2006]: Average users remember upto 6 unique passwords.

• [Perito, 2011]: Internet accounts are easily linkable by pseudonyms.– Compromise of one account ) compromise of all accounts allocated the same

password.– Some accounts (eg., email) are gateway accounts.

• Problem: – What allocation results in minimum expected loss?

The Password Allocation Problem



• I don’t care. I’m super secure, phish-proof, and use 40 char long passwords!– People do stupid things!

• July 12, 2012: Yahoo lost 45000 unhashed passwords.– All passwords are equal.

• Compromise probability is only server dependent.

• June 5, 2012: 6.5 million hashed passwords stolen.– Some passwords are uncrackable.

• Compromise probability is server and password dependent.

The Password Allocation Problem

7

PA as a Parallel Job Allocation Problem

• Given a set of programs to be executed and a (smaller) set of machines.

• Each program may cause a system failure with some probability.– This may be machine independent (i.e., all machines are the same).

• Parallel Processing Constraint: Failure of one of the programs ) failure of all programs on the system.

• Problem:– How should programs be allocated to machines to maximize expected

throughput?On the Complexity of Allocation Problems with Probabilistic Players

8

The Weapon-Target Allocation Problem

• Military offense allocation problem.

• Given a set of weapons and a set of enemy targets.

• Not all weapons destroy their targets– Enemy interception– Mechanical failures

• Probability of failure depends on the weapon-target pair– Placement of defenses against weapons– Distance from allocated weapon

• Problem: – What allocation maximizes expected damage to the enemy targets?


9

The Weapon-Target Allocation Problem

Research Timeline:

• Formulation: Allan Manne (Stanford) [1958]

• NP-Completeness: Lloyd and Witsenhausen (Bell Labs) [1988]

• Analysis, Variants: Hosein (MIT) [1987-1992], Athans (Bell Labs) [1989-1992]

• Approximation (heuristics): 1977 – today

• Best approximations: Ahuja (UF), Orlin (MIT) [2007]

• (Existence of) Constant-factor approximations: ??On the Complexity of Allocation Problems with Probabilistic Players



• Introduction





The PA Problem: Definition

• Problem Instance:– n accounts: a1, a2, …, an

– k passwords: PW1, PW2, …, PWk

– ai has value vi and compromise probability qi = (1-pi)• i.e., compromise probability is independent of password strength

• Compromise of one account 2 PWj ) compromise of all accounts 2 PWj

• Constraint: Every account receives exactly one password.

• Goal: Minimize expected loss through password compromise– Equivalent to maximizing expected survival value (or, Expected Gain (EG)).


• Allocation matrix: X = {xij}– xij = 1 ) account aj is allocated password PWi

– xij = 0, otherwise

• Objective Function (Expected Gain): [to be maximized]

• Constraint: – Every account is allocated exactly one password.

The PA Problem: Mathematical Formulation

kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A

kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A (1)xi j =

½1 if account j is allocated to password i0 otherwise

maximize: F =kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A

subj ect to :kX

i=1

xi j =1; 8j 2 f1;:::;ng

F is theexpected gain (EG).

F =P ki=1

³ Q nj =1((pj )

xi j )P nj =1(xi j vj )

´F =

P ki=1

³ Q nj =1((pj )


´


Complexity of PA

Theorem: PA with 2 passwords PA2 2 NP Complete.

Proof: Part I: Formulating the Decision Version (PA2)

• Instance: – P = {p1,…,pn} where pi 2 (0,1)

– V = {v1,…,vn}– r

• Is there a partition of N ={1,…,n} into S1 and S2 such that:

• Clearly PA2 2 NP.


Complexity of PA


Proof: Part II: Finding the known hard problem

• The Partition Problem:– Instance: Q = {q1, …, qn}, qi 2 Z+

– Is there a partition of Q into Q1 and Q2 such that:


Complexity of PA


Proof: Part III: Making the Transformation

• Convert Partition instance to PA2 instance in poly-time.

• Given: Q = {q1, …, qn}

• Construct PA2 instance as follows:

• What is x?– For now, just a rational 2 (0,1)


Complexity of PA


Proof: Part IV: Why it works• Solving equations:

• Gives us the following solutions:


Complexity of PA


Proof: Part IV: Why it works

• We will eliminate the solutions where VS1 VS2

.

– As a result our solver will return that the constructed PA2 instance is a yes instance iff the Partition instance is a yes instance.

• Eliminating solution 1: – Recall our transformation:

– When we have:• Since x < 1

– Therefore, solution 1 can never occur.


Complexity of PATheorem: PA with 2 passwords PA2 2 NP Complete.

Proof: Part IV: Why it works• Eliminating solution 2:

– Recall our transformation:

– We need to ensure that when :

– We will find an x such that

– Therefore, when , solution 2 can never occur ) PA2 2 NP Complete.


Efficiently Solvable Cases

The case of n = k• Optimal Strategy: Allocate exactly one account to each

password.

• Proof of optimality: – Since and , we have:

– This means an account contributes the most to the EG when it has its own password.


The case of identical accountsWe have The problem reduces to:

where xi = number of accounts allocated to pi

• Optimal Strategy: Assign accounts (sequentially) to the password for which the EG increases the most.

• Proof of Optimality: Greedy argument – we always stay on par or ahead of any feasible solution.



A Special Case

The Case of Correlated Values and ProbabilitiesWe have:

and where .

• Property of Optimal Solution:

• Proof (Sketch):

a1 ai aj ak an

PW1 PWl PWm PWk

EG(PWl) = (vi + vk + …) pi pk ….. EG(PWm) = (vj + …) pi …..

If we have: pi > pj > pk , vi > vj > vk, and pi/qi > vj/vi then…

EG’(PWl) = (vi + vj + …) pi pj ….. EG’(PWm) = (vk + …) pk …..

EG(PWl) + EG(PWm) < EG’(PWl) + EG’(PWm)



• Introduction





The Single Round WTA Problem: Definition

• Problem Instance:– n targets: t1, t2, …, tn

– k weapons: w1, w2, …, wk

– wi destroys tj with probability qij

• i.e., kill probability is weapon and target dependent

• Constraint: Each weapon is allocated to exactly one target.

• Goal: Minimize expected survival of enemy targets– Equivalent to maximizing expected damage to enemy targets.


SWTA Problem Assumptions

• Given kill probabilities are independent of other pairs.– Isolates problem from geometric and geographic factors.

• Either a target is destroyed completely or survives completely.– qij (kill probability) = 1-pij (survival probability)

• Damage is surveyed after weapons are fired.– Models short battles with limited ammunition – does not consider

enemy retreats

• No fractional allocations may be made.– A weapon can only be allocated to a single target


• Allocation matrix: X = {xij}– xij = 1 ) weapon wj is allocated to target ti

– xij = 0, otherwise

• Objective Function (Survival Value): [to be minimized]

• Constraint: – Every weapon is allocated to exactly one target.

The SWTA Problem: Mathematical Formulation

kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A

kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A (1)xi j =

½1 if account j is allocated to password i0 otherwise

maximize: F =kX

i=1

0

@nY

j =1

((pj )xi j )nX

j =1

(xi j vj )

1

A

subj ect to :kX

i=1

xi j =1; 8j 2 f1;:::;ng

F is theexpected gain (EG).

F =P ki=1

³ Q nj =1((pj )


´F =

P ki=1

³ Q nj =1((pj )


´


Complexity of SWTA

Theorem: SWTA with 2 targets (SWTA2) 2 NP Complete.

Proof: Part I: Formulating the Decision Version

• Instance: – P = {pij} where pij 2 (0,1) – r

• Is there a 0-1 matrix X such that:– The sum of the survival probabilities of the 2 targets is less than r

– and every weapon is allocated to at least one target.

• Clearly SWTA2 2 NP.


Complexity of SWTA


Proof: Part II: Finding the known hard problem

• The Rational Product Dichotomy (Fractional Subset Product):– Instance: Q = {q1, …, qn}, qi 2 (0,1)

– Is there a partition of N={1, …, n} into S1 and S2 such that:


Complexity of SWTA


Proof: Part III: Making the Transformation

• Convert RPD instance to SWTA2 instance in poly-time.

• Given: Q = {q1, …, qn}

• Construct SWTA2 instance as follows:

Where pij is the survival probability of target i after a strike by weapon j.


Complexity of SWTATheorem: SWTA with 2 targets (SWTA2) 2 NP Complete.

Proof: Part IV: Why it works• Our SWTA2 solver will return yes iff

Where qi is the ith rational in the given RPD instance.

• By AGMI: – Therefore, can never occur.– SWTA2 solver returns yes iff

• By AGMI: – SWTA2 solver returns yes iff

which is a yes instance of RPD.

• SWTA2 2 NP Complete



The Case of Identical Weapons and TargetsWe have all weapon-target pairs with same survival probability p

i.e.,The problem reduces to:

subject towhere xi is the number of weapons allocated to target i.

• Optimal Strategy: Divide weapons as evenly as possible.


If dividing k weapons evenly is not optimal. Then:

Target i Target j

xi weapons xj = d+xi weapons

But, switching one of the weapons target gives us:

1+xi weapons xj = d-1+xi weapons

Since p2(0,1) and xi < d+xi - 1

Therefore, switching targets strictly decreases the net survival value ) solution is not optimal



The Case of Equal WeaponsWe have one type of weapon – so all weapons destroy target i

with the same probability – pi.

Problem reduces to:

• Optimal Strategy: Assign weapons to the target for which the objective function (i.e., pi xi) decreases the most.

• Proof of Optimality: By induction.– When allocating one weapon to n weapons trivially true.


Assume Xk is the optimal solution for k weapons to n targets

Xk = <x1, x2, …, xn>

Let Xk+1 be the solution returned for k+1 weapons to n targets

Xk+1 = <x1, x2, …, xm+1, …, xn>

Where ±m · ±i 8 i 2 {1, …, n}

± =

pm

x m £

(pm

-1)

Zk+1 = <z1, z2, …, zn>

Let Zk+1 be any other solution

Since Zk+1 Xk+1, there is a j where zk+1(j) > xk+1(j) ¸ xk (j)

Zk = <z1, z2, …, zj-1, …, zn>

Let Zk be the same solution with one less weapon for target j.

± =

pj (z

j -1) £

(pj-1

)

X*k+1 = <x1, x2, …, xj+1, …, xn>

Let X*k+1 be Xk with one more weapon allocated to target j.

± = pj xj £ (p

j -1)

·

· Since xj < zj

··



The Case of One Weapon per TargetWe have each of the n targets getting at most one weapon – i.e.,

As a result:

(1) is true since xij 2 {0,1}

(2) is true since there is only one x ij = 1 for each target.

Therefore:



The Case of One Weapon per Target• This can now be written as:

• Which is the transportation problem with:– costij = -qij

– k supply nodes with supply = 1– n demand nodes with demand = 1


SWTA Approximation HeuristicsThree main techniques:• Integer constraint relaxation

– Allow fractional allocations of weapons to targets.– Solve resulting LP .– Use randomized rounding to obtain approximate solution to integer problem.

• Modeling as network flow problems– Create a graph of weapons and targets.– Each edge between a weapon and target has a cost approximately equal to the

change in objective function.• Approximate due to non-linear nature

– Set appropriate constraints (eg., supply/demand, capacity).– Solve network flow problem using MCMF, MF, TP algorithms (as is appropriate).

• Localized search– Start with a feasible solution of reasonable quality.– Perform swaps and multi-swaps yielding better solutions.



• Introduction





Conclusions and Future Work• The Password Allocation Problem

– Also models parallel processing allocation problems.

– NP Complete even when all passwords are equal

– Has several efficiently solvable cases

• Analysis for cases with varying passwords.

• Approximation Techniques – Heuristics– Boundable algorithms (??)

• Online version of the problem

Varia

bilit

y of

pas

swor

dsVariability of accounts

Equal accounts and PWs

Equal #accounts and PWs

Correlated accounts

2 P

2 NP Complete2 ?


Conclusions and Future Work• The Weapon-Target Allocation

Problem– NP Complete even for the single-

round version.– There are special poly-time solvable

cases.– General approaches to making

approximate solutions.

• Most current work ignores analysis – too much focus on heuristics (unboundable)!– Existence of constant-factor

bounds?– Almost no analysis for multi-round

variant.

Varia

bilit

y of

wea

pons

Variability of targets

All equal targets

Equal weapons and targets

All equal weapons

One weapon per target

2 P

2 NP Complete

on the complexity of allocation problems with probabilistic players rishab nithyanand research...

Documents

password allocation

allocation matrix

allocation results

pa problem

passwords pa

partition problem

allocated weapon problem

complexity of pa theorem