Aggregation in Dynamic Programming Yu Ashwin Introduction n

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Aggregation in Dynamic Programming Yu & Ashwin

Aggregation in Dynamic Programming Yu & Ashwin

Introduction n State aggregation done to reduce computation burdens Since information is lost through

Introduction n State aggregation done to reduce computation burdens Since information is lost through state identification, this does not always lead to optimal solution Value = Reduced computation Vs Increased error

Introduction (contd. ) Aggregation Approach n Aggregate the nodes n Solve the aggregated problem

Introduction (contd. ) Aggregation Approach n Aggregate the nodes n Solve the aggregated problem n Disaggregate the solution

Original Problem n n Goal: find min-length path from node 0 to node m

Original Problem n n Goal: find min-length path from node 0 to node m with respect to specified arc lengths Cij for fi: the length of a shortest path from node 0 to node i, for i=0, 1, 2, …, m, so we actually looking for fm

Aggregation n n Notations Entering node Sequential Aggregation Dis-aggregation (SADA) Algorithm Error and bound

Aggregation n n Notations Entering node Sequential Aggregation Dis-aggregation (SADA) Algorithm Error and bound

Notations n n n Macro-nodes: sets of micro-nodes that form a partition of the

Notations n n n Macro-nodes: sets of micro-nodes that form a partition of the nodes of original problem Macro-arc: directs from macro-node J to J’ if and only if a Macro-network: Require this to be acyclic so that when we number the macro-nodes 0 through M, a macro-arc is directed from J to J’ only if J<J’

Entering Node n Entering node and set of entering node n Entering node of

Entering Node n Entering node and set of entering node n Entering node of J from a specific I n Fixed entering node of J

SADA n n Goal: Min-length path from macro-node 0 to macro-node M with respect

SADA n n Goal: Min-length path from macro-node 0 to macro-node M with respect to specified macro-arc lengths Fi is the length of a shortest Macro-path from macro-node 0 to macro-node i, for i=0, 1, 2, …, m, so we actually looking for FM

SADA (Con’d) Macro-Node 2 Macro-Node 1 C 12 C 27 Macro-Node 3 C 01

SADA (Con’d) Macro-Node 2 Macro-Node 1 C 12 C 27 Macro-Node 3 C 01 Macro-Node 0 S C 37 t C 03 C 35 C 57 C 04 Macro-Node 4 C 45 Macro-Node 5

Concern of Error n n SADA finds the micro-path by construction a shortest path

Concern of Error n n SADA finds the micro-path by construction a shortest path through the micro-network, with the restriction that it must pass through the fixed entering micro-nodes. This may not give the optimal result for the original problem

Analysis of Error Bound n Now we will prove

Analysis of Error Bound n Now we will prove

Analysis of Error Bound (Con’d)

Analysis of Error Bound (Con’d)

Error Bound (con’d) We can set the error bound to n the longest path

Error Bound (con’d) We can set the error bound to n the longest path length among any set of paths known to contain the path through N* n the longest path length in the macro-network with arc lengths

Benefits of SADA n Computational savings if not all entering micro-nodes of a macro-node

Benefits of SADA n Computational savings if not all entering micro-nodes of a macro-node are accessible from the fixed entering micro -node of that macro-node

Corollary #3 Let I є N be any macronode and let J є N

Corollary #3 Let I є N be any macronode and let J є N be any adjacent macronode, so that (I, J) є A. Suppose for any i є GI, there is a j є GI U GJ, with f(i, j) ≤ εI , for some εI ≥ 0 and dk. J ≥ dj. J for all k є GI Then F* – f*≤ ε, where ε is the maximum path length in the macroarc network with zero arc lengths and node penalties εJ for J є N.

i εI j dj. J dk. J k I J

i εI j dj. J dk. J k I J

Remark If all pairs of nodes are mutually accesible then εI would be bounded

Remark If all pairs of nodes are mutually accesible then εI would be bounded by the diameter p. I of macronode I , i. e, εI ≤ p. I = maxi, j є I f(i, j)

Block staging We can avoid the task solving a longest path problem in finding

Block staging We can avoid the task solving a longest path problem in finding error bounds in the important special case that the network contains a single macropath from 0 to node M.

Applications to infinite Horizon otpimization We have a acyclic infinite network (N, A) with

Applications to infinite Horizon otpimization We have a acyclic infinite network (N, A) with time ti associated with each node i. The nodes are numbered from 0 to N-1, such that (i, j) є A only if ti<tj, with t 0 = 0. Let C(i, j) arc cost for each arc (i, j).

n n No. of arcs along every path is infinite Arc cost tends to

n n No. of arcs along every path is infinite Arc cost tends to zero. Every C(i, j) emerging out of node i will be multiplied by a discount factor e-rti for some constant r >0. Time diverges to infinity Capacity expansion, production and inventory conrol, etc.

n n n Such problems are solved by truncating them to a finite length

n n n Such problems are solved by truncating them to a finite length problem of T periods. The resulting problem is solved by standard dynamic programming Under mild regulations, this procedure for large enough T will result in optimal initial arc decision that is optimal for infinite horizon problem

n n A sufficiently large T is difficult to identify. All data must be

n n A sufficiently large T is difficult to identify. All data must be first forecasted over infinite horizon For an ε error the planner willing to accept, then an approximate horizon length Tε* can be determined

a) Bounded reaching

a) Bounded reaching

b) Declining future cost

b) Declining future cost

Let F* denote the minimum discounted cost of the macronetwork, is obtained by solving

Let F* denote the minimum discounted cost of the macronetwork, is obtained by solving a sequence of T period finite horizon problems

Rolling horizon procedure 1. 2. 3. 4. 5. Solve the finite horizon problem over

Rolling horizon procedure 1. 2. 3. 4. 5. Solve the finite horizon problem over [0, T] Implement the optimal sequence of decision obtained in 1 Beginning with micronode state obtained from solution 1, solve the problem over [T, 2 T] Implement the optimal sequence of decisions in 3 Repeat steps 1 -4 indefinitely

Theorem 5

Theorem 5

Proof

Proof

Proof (Contd. )

Proof (Contd. )