Decision theoretic troubleshooting Ji r Vomlel Academy of - PowerPoint PPT Presentation

Decision theoretic troubleshooting Jiˇ r´ ı Vomlel Academy of Sciences of the Czech Republic 11th July, 2007 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 1 / 26

Light Print Problem Your trouble : “The page that came out of your printer is light.” Our trouble-shooter : “Perform these steps that will help you solve the trouble.” Problem description: problem causes C ∈ C actions A ∈ A - troubleshooting steps that may solve the problem questions Q ∈ Q - troubleshooting steps that help identify the problem cause. every action and question has assigned a cost: c A ... cost of an action A c Q ... cost of a question Q J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 2 / 26

Light Print Problem - causes, actions and questions Causes of light print p ( C i ) C 1 : Distribution problem 0.4 C 2 : Defective toner 0.3 C 3 : Corrupted dataflow 0.2 C 4 : Wrong driver setting 0.1 Actions and questions c i A 1 : Remove, shake and reseat toner 5 A 2 : Try another toner 15 A 3 : Cycle power 1 Q 1 : Is the printer configuration page printed light? 2 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 3 / 26

Light Print Problem - Bayesian Network Actions Causes A 1 C 1 Problem A 2 C 1 C 2 C 2 A 3 C 3 C 4 C 3 Questions XOR C 4 Q 1 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 4 / 26

Light Print - conditional probability tables (CPT) for every action A i and for every parent cause C j an expert provides a CPT for p ( A i = yes | C j ) for every answer q k to every question Q k and for every parent cause C j the expert provides a CPT for p ( Q k = q k | C j ) C j p ( A 2 = yes | C j ) C j p ( Q 1 = yes | C j ) C 1 0.9 C 1 1 C 2 0.9 C 2 1 C 3 - C 3 0 C 4 - C 4 0 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 5 / 26

Troubleshooting strategy A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 6 / 26

Expected Cost of Repair (ECR) A strategy may terminate: by giving up (e.g. if there are no further steps left) J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 7 / 26

Expected Cost of Repair (ECR) A strategy may terminate: by giving up (e.g. if there are no further steps left) a penalty function c ( e ℓ ) applies J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 7 / 26

Expected Cost of Repair (ECR) A strategy may terminate: by giving up (e.g. if there are no further steps left) a penalty function c ( e ℓ ) applies can be interpreted as a cost of calling service J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 7 / 26

Expected Cost of Repair (ECR) A strategy may terminate: by giving up (e.g. if there are no further steps left) a penalty function c ( e ℓ ) applies can be interpreted as a cost of calling service by solving the problem: c ( e ℓ ) def = 0 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 7 / 26

Expected Cost of Repair (ECR) Example A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 8 / 26

Expected Cost of Repair (ECR) Example A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes Strategy Expected Cost of Repair (ECR) p ( Q 1 = no , A 1 = yes ) · ( c Q 1 + c A 1 + 0 ) � A 1 Q 1 A 2 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 8 / 26

Expected Cost of Repair (ECR) Example A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes Strategy Expected Cost of Repair (ECR) p ( Q 1 = no , A 1 = yes ) · ( c Q 1 + c A 1 + 0 ) � A 1 + p ( Q 1 = no , A 1 = no ) · ( c Q 1 + c A 1 + c CS ) Q 1 A 2 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 8 / 26

Expected Cost of Repair (ECR) Example A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes Strategy Expected Cost of Repair (ECR) p ( Q 1 = no , A 1 = yes ) · ( c Q 1 + c A 1 + 0 ) � A 1 + p ( Q 1 = no , A 1 = no ) · ( c Q 1 + c A 1 + c CS ) Q 1 A 2 + p ( Q 1 = yes , A 2 = yes ) · ( c Q 1 + c A 2 + 0 ) J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 8 / 26

Expected Cost of Repair (ECR) Example A 1 = yes A 1 = no Q 1 = no A 2 = no Q 1 = yes A 2 = yes Strategy Expected Cost of Repair (ECR) p ( Q 1 = no , A 1 = yes ) · ( c Q 1 + c A 1 + 0 ) � A 1 + p ( Q 1 = no , A 1 = no ) · ( c Q 1 + c A 1 + c CS ) Q 1 A 2 + p ( Q 1 = yes , A 2 = yes ) · ( c Q 1 + c A 2 + 0 ) + p ( Q 1 = yes , A 2 = no ) · ( c Q 1 + c A 2 + c CS ) J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 8 / 26

Expected Cost of Repair (ECR) � � ( A = yes / no ) A ∈ { performed actions } , Node n �→ e n = ( Q = yes / no ) Q ∈ { performed questions } J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 9 / 26

Expected Cost of Repair (ECR) � � ( A = yes / no ) A ∈ { performed actions } , Node n �→ e n = ( Q = yes / no ) Q ∈ { performed questions } �→ p ( e n ) ... probability of getting to node n J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 9 / 26

Expected Cost of Repair (ECR) � � ( A = yes / no ) A ∈ { performed actions } , Node n �→ e n = ( Q = yes / no ) Q ∈ { performed questions } �→ p ( e n ) ... probability of getting to node n �→ t ( e n ) ... total cost of actions and questions per- formed (to get to node n ) J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 9 / 26

Expected Cost of Repair (ECR) � � ( A = yes / no ) A ∈ { performed actions } , Node n �→ e n = ( Q = yes / no ) Q ∈ { performed questions } �→ p ( e n ) ... probability of getting to node n �→ t ( e n ) ... total cost of actions and questions per- formed (to get to node n ) � ECR ( s ) = p ( e ℓ ) · [ t ( e ℓ ) + c ( e ℓ ) ] ℓ ∈{ terminal nodes of s } J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 9 / 26

Optimal strategy Optimal strategy s ⋆ ⇐ ⇒ s ⋆ = arg min s ∈{ all possible strategies } ECR ( s ) J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 10 / 26

Troubleshooting with dependent actions is NP-hard Theorem (1) Assume decision-theoretic troubleshooting problem with fixed costs and dependent actions. The decision whether there exists a troubleshooting sequence with ECR ≤ K for a given constant K is NP-complete problem for both single fault assumption and independent faults. Proof: The problem is NP: if we guess a good sequence we calculate ECR and compare whether ECR ≤ K . It takes polynomial time to calculate ECR of a sequence. J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 11 / 26

Troubleshooting with dependent actions is NP-hard Theorem (1) Assume decision-theoretic troubleshooting problem with fixed costs and dependent actions. The decision whether there exists a troubleshooting sequence with ECR ≤ K for a given constant K is NP-complete problem for both single fault assumption and independent faults. Proof: The problem is NP: if we guess a good sequence we calculate ECR and compare whether ECR ≤ K . It takes polynomial time to calculate ECR of a sequence. The problem is NP-hard: We reduce the Exact cover by 3-sets to troubleshooting. J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 11 / 26

Exact cover by 3-sets Definition (Exact cover by 3-sets) We are given a family F = { S 1 , . . . , S n } of subsets of a set U , such that | U | = 3 m for some integer m , and | S i | = 3 for all i . We are asked if there are m sets in F that are disjoint and have U as their union. J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 12 / 26

Exact cover by 3-sets Definition (Exact cover by 3-sets) We are given a family F = { S 1 , . . . , S n } of subsets of a set U , such that | U | = 3 m for some integer m , and | S i | = 3 for all i . We are asked if there are m sets in F that are disjoint and have U as their union. The proof of NP-completeness is for example in: Christos H. Papadimitriou. Computational complexity . Addison-Wesley Publishing Company, 1994. J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 12 / 26

COVER BY 3-SETS � Troubleshooting U = { 1 , 2 , 3 · · · , 12 } C 1 A 1 S 1 = { 1 , 2 , 3 } C 2 C 3 S 2 = { 2 , 3 , 4 } A 2 C 4 S 3 = { 1 , 5 , 9 } p ( C i ) uniform C 5 A 3 p ( A j | C i ) ∈ { 0 , 1 } C 6 S 4 = { 6 , 7 , 8 } A 4 c A = 1 C 7 c ( e ℓ ) = 2 · ( m + 1 ) 2 S 5 = { 7 , 8 , 9 } C 8 A 5 C 9 A 6 S 6 = { 5 , 10 , 11 } C 10 A 7 C 11 S 7 = { 10 , 11 , 12 } C 12 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 13 / 26

COVER BY 3-SETS � Troubleshooting U = { 1 , 2 , 3 · · · , 12 } C 1 A 1 S 1 = { 1 , 2 , 3 } C 2 C 3 S 2 = { 2 , 3 , 4 } A 2 C 4 S 3 = { 1 , 5 , 9 } p ( C i ) uniform C 5 A 3 p ( A j | C i ) ∈ { 0 , 1 } C 6 S 4 = { 6 , 7 , 8 } A 4 c A = 1 C 7 c ( e ℓ ) = 2 · ( m + 1 ) 2 S 5 = { 7 , 8 , 9 } C 8 A 5 C 9 A 6 S 6 = { 5 , 10 , 11 } C 10 A 7 C 11 S 7 = { 10 , 11 , 12 } C 12 The exact cover by 3-sets exists iff ECR ≤ m + 1 for some sequence. 2 J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 13 / 26

Proof of NP-hardness - 1 Lemma (1) If we have exact 3-sets cover V = { S j 1 , . . . , S j l } then the ECR of corresponding action sequence A j 1 , . . . , A j l (in any order) has the ECR = m + 1 2 . J. Vomlel ( ´ UTIA AV ˇ CR) Troubleshooting 11th July, 2007 14 / 26