Download presentation

Presentation is loading. Please wait.

Published byAlicia Pick Modified over 2 years ago

1
On the Complexity of Allocation Problems with Probabilistic Players Rishab Nithyanand Research Proficiency Examination Summer 2012 TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A AAAA A

2
Presentation Outline Introduction The Password Allocation Problem The Weapon-Target Allocation Problem Conclusions and Future Work 2On the Complexity of Allocation Problems with Probabilistic Players

3
Traditional Allocation Problems Given: – resources (r 1, r 2, …, r n ) and tasks (t 1, t 2, …, t k ) – objective function F Goal: – Find allocation for which F is optimal Constraint: – at most one task per resource 3On the Complexity of Allocation Problems with Probabilistic Players

4
Allocation Problems with Probabilistic Players Given: – resources (r 1, r 2, …, r n ) and tasks (t 1, t 2, …, t k ) – resource r i completes task t j with probability p ij – objective function F Goal: – Find allocation for which E[F] is optimal Constraint: – at most one task per resource 4On the Complexity of Allocation Problems with Probabilistic Players

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., ) are gateway accounts. Problem: – What allocation results in minimum expected loss? The Password Allocation Problem 5 On the Complexity of Allocation Problems with Probabilistic Players

6
I dont care. Im super secure, phish-proof, and use 40 char long passwords! – People do stupid things! July 12, 2012: Yahoo lost 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 On the Complexity of Allocation Problems with Probabilistic Players6

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? 7On 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? 8On the Complexity of Allocation Problems with Probabilistic Players

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) [ ], Athans (Bell Labs) [ ] Approximation (heuristics): 1977 – today Best approximations: Ahuja (UF), Orlin (MIT) [2007] (Existence of) Constant-factor approximations: ?? 9On the Complexity of Allocation Problems with Probabilistic Players

10
Presentation Outline Introduction The Password Allocation Problem The Weapon-Target Allocation Problem Conclusions and Future Work 10On the Complexity of Allocation Problems with Probabilistic Players

11
The PA Problem: Definition Problem Instance: – n accounts: a 1, a 2, …, a n – k passwords: PW 1, PW 2, …, PW k – a i has value v i and compromise probability q i = (1-p i ) i.e., compromise probability is independent of password strength Compromise of one account 2 PW j ) compromise of all accounts 2 PW j Constraint: Every account receives exactly one password. Goal: Minimize expected loss through password compromise – Equivalent to maximizing expected survival value (or, Expected Gain (EG)). On the Complexity of Allocation Problems with Probabilistic Players11

12
Allocation matrix: X = {x ij } – x ij = 1 ) account a j is allocated password PW i – x ij = 0, otherwise Objective Function (Expected Gain): [to be maximized] Constraint: – Every account is allocated exactly one password. The PA Problem: Mathematical Formulation On the Complexity of Allocation Problems with Probabilistic Players12

13
Complexity of PA Theorem: PA with 2 passwords PA 2 2 NP Complete. Proof: Part I: Formulating the Decision Version (PA 2 ) Instance: – P = {p 1,…,p n } where p i 2 (0,1) – V = {v 1,…,v n } –r–r Is there a partition of N ={1,…,n} into S 1 and S 2 such that: Clearly PA 2 2 NP. On the Complexity of Allocation Problems with Probabilistic Players13

14
Complexity of PA Theorem: PA with 2 passwords PA 2 2 NP Complete. Proof: Part II: Finding the known hard problem The Partition Problem: – Instance: Q = {q 1, …, q n }, q i 2 Z + – Is there a partition of Q into Q 1 and Q 2 such that: On the Complexity of Allocation Problems with Probabilistic Players14

15
Complexity of PA Theorem: PA with 2 passwords PA 2 2 NP Complete. Proof: Part III: Making the Transformation Convert Partition instance to PA 2 instance in poly-time. Given: Q = {q 1, …, q n } Construct PA 2 instance as follows: What is x? – For now, just a rational 2 (0,1) On the Complexity of Allocation Problems with Probabilistic Players15

16
Complexity of PA Theorem: PA with 2 passwords PA 2 2 NP Complete. Proof: Part IV: Why it works Solving equations: Gives us the following solutions: On the Complexity of Allocation Problems with Probabilistic Players16

17
Complexity of PA Theorem: PA with 2 passwords PA 2 2 NP Complete. Proof: Part IV: Why it works We will eliminate the solutions where V S 1 V S 2. – As a result our solver will return that the constructed PA 2 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. On the Complexity of Allocation Problems with Probabilistic Players17

18
Complexity of PA Theorem: PA with 2 passwords PA 2 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 ) PA 2 2 NP Complete. On the Complexity of Allocation Problems with Probabilistic Players18

19
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. On the Complexity of Allocation Problems with Probabilistic Players19

20
The case of identical accounts We have The problem reduces to: where x i = number of accounts allocated to p i 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. Efficiently Solvable Cases On the Complexity of Allocation Problems with Probabilistic Players20

21
On the Complexity of Allocation Problems with Probabilistic Players21 PW 1 PW 2 Passwords: Number of accounts (x i ) f i (x i ) = x i p x i Threshold point: f i (x i +1)- f i (x i ) < f i (x i ) – f i (x i -1) x 1 = 1, ± (f 1 ) = p x 1 = 2, ± (f 1 ) = 2p 2 - p x 1 = 1, ± (f 2 ) = p x 2 = 2, ± (f 2 ) = 2p 2 - p x 1 = j, ± (f 1 ) = jp j – (j-1)p j-1 x 2 = j, ± (f 2 ) = jp j – (j-1)p j-1

22
A Special Case The Case of Correlated Values and Probabilities We have: and where. Property of Optimal Solution: Proof (Sketch): On the Complexity of Allocation Problems with Probabilistic Players22 a1a1 aiai ajaj akak anan PW 1 PW l PW m PW k EG(PW l ) = (v i + v k + …) p i p k ….. EG(PW m ) = (v j + …) p i ….. If we have: p i > p j > p k, v i > v j > v k, and p i /q i > v j /v i then… EG(PW l ) = (v i + v j + …) p i p j …..EG(PW m ) = (v k + …) p k ….. EG(PW l ) + EG(PW m ) < EG(PW l ) + EG(PW m )

23
Presentation Outline Introduction The Password Allocation Problem The Weapon-Target Allocation Problem Conclusions and Future Work 23On the Complexity of Allocation Problems with Probabilistic Players

24
The Single Round WTA Problem: Definition Problem Instance: – n targets: t 1, t 2, …, t n – k weapons: w 1, w 2, …, w k – w i destroys t j with probability q ij 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. On the Complexity of Allocation Problems with Probabilistic Players24

25
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. – q ij (kill probability) = 1-p ij (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 On the Complexity of Allocation Problems with Probabilistic Players25

26
Allocation matrix: X = {x ij } – x ij = 1 ) weapon w j is allocated to target t i – x ij = 0, otherwise Objective Function (Survival Value): [to be minimized] Constraint: – Every weapon is allocated to exactly one target. The SWTA Problem: Mathematical Formulation On the Complexity of Allocation Problems with Probabilistic Players26

27
Complexity of SWTA Theorem: SWTA with 2 targets (SWTA 2 ) 2 NP Complete. Proof: Part I: Formulating the Decision Version Instance: – P = {p ij } where p ij 2 (0,1) –r–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 SWTA 2 2 NP. On the Complexity of Allocation Problems with Probabilistic Players27

28
Complexity of SWTA Theorem: SWTA with 2 targets (SWTA 2 ) 2 NP Complete. Proof: Part II: Finding the known hard problem The Rational Product Dichotomy (Fractional Subset Product): – Instance: Q = {q 1, …, q n }, q i 2 (0,1) – Is there a partition of N={1, …, n} into S 1 and S 2 such that: On the Complexity of Allocation Problems with Probabilistic Players28

29
Complexity of SWTA Theorem: SWTA with 2 targets (SWTA 2 ) 2 NP Complete. Proof: Part III: Making the Transformation Convert RPD instance to SWTA 2 instance in poly-time. Given: Q = {q 1, …, q n } Construct SWTA 2 instance as follows: Where p ij is the survival probability of target i after a strike by weapon j. On the Complexity of Allocation Problems with Probabilistic Players29

30
Complexity of SWTA Theorem: SWTA with 2 targets (SWTA 2 ) 2 NP Complete. Proof: Part IV: Why it works Our SWTA 2 solver will return yes iff Where q i is the ith rational in the given RPD instance. By AGMI: – Therefore, can never occur. – SWTA 2 solver returns yes iff By AGMI: – SWTA 2 solver returns yes iff which is a yes instance of RPD. SWTA 2 2 NP Complete On the Complexity of Allocation Problems with Probabilistic Players30

31
Efficiently Solvable Cases The Case of Identical Weapons and Targets We have all weapon-target pairs with same survival probability p i.e., The problem reduces to: subject to where x i is the number of weapons allocated to target i. Optimal Strategy: Divide weapons as evenly as possible. On the Complexity of Allocation Problems with Probabilistic Players31

32
On the Complexity of Allocation Problems with Probabilistic Players32 If dividing k weapons evenly is not optimal. Then: Target iTarget j x i weapons x j = d+x i weapons But, switching one of the weapons target gives us: 1+x i weapons x j = d-1+x i weapons Since p 2 (0,1) and x i < d+x i - 1 Therefore, switching targets strictly decreases the net survival value ) solution is not optimal

33
Efficiently Solvable Cases The Case of Equal Weapons We have one type of weapon – so all weapons destroy target i with the same probability – p i. Problem reduces to: Optimal Strategy: Assign weapons to the target for which the objective function (i.e., p i x i ) decreases the most. Proof of Optimality: By induction. – When allocating one weapon to n weapons trivially true. On the Complexity of Allocation Problems with Probabilistic Players33

34
On the Complexity of Allocation Problems with Probabilistic Players34 Assume X k is the optimal solution for k weapons to n targets X k = Let X k+1 be the solution returned for k+1 weapons to n targets X k+1 = Where ± m · ± i 8 i 2 {1, …, n} ± = p m x m £ (p m -1) Z k+1 = Let Z k+1 be any other solution Since Z k+1 X k+1, there is a j where z k+1 (j) > x k+1 (j) ¸ x k (j) Z k = Let Z k be the same solution with one less weapon for target j. ± = p j (zj -1) £ (p j -1) X* k+1 = Let X* k+1 be X k with one more weapon allocated to target j. ± = p j x j £ (p j -1) · · Since x j < z j · ·

35
Efficiently Solvable Cases The Case of One Weapon per Target We have each of the n targets getting at most one weapon – i.e., As a result: (1)is true since x ij 2 {0,1} (2)is true since there is only one x ij = 1 for each target. Therefore: On the Complexity of Allocation Problems with Probabilistic Players35

36
Efficiently Solvable Cases The Case of One Weapon per Target This can now be written as: Which is the transportation problem with: – cost ij = -q ij – k supply nodes with supply = 1 – n demand nodes with demand = 1 On the Complexity of Allocation Problems with Probabilistic Players36

37
SWTA Approximation Heuristics Three 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. On the Complexity of Allocation Problems with Probabilistic Players37

38
Presentation Outline Introduction The Password Allocation Problem The Weapon-Target Allocation Problem Conclusions and Future Work 38On the Complexity of Allocation Problems with Probabilistic Players

39
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 On the Complexity of Allocation Problems with Probabilistic Players39 Variability of passwords Variability of accounts Equal accounts and PWs Equal #accounts and PWs Correlated accounts 2 P 2 NP Complete 2 ?

40
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. On the Complexity of Allocation Problems with Probabilistic Players40 Variability of weapons Variability of targets All equal targets Equal weapons and targets All equal weapons One weapon per target 2 P 2 NP Complete

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google