Exact methods for ALB.ppt

7/29/2019 Exact methods for ALB.ppt

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Exact methods for ALB

ALB problem can be considered as a shortest

path problem

The complete graph need not be developed since

one can stop as soon as in one node all

operations have been assigned, and also because

of pruning the tree by dominance rules

The graph will still be very large -> heuristic thatconsiders only a subgraph with the cost of loosing

guarantee of optimality


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Pinto heuristic

1. Find some good (and feasible w.r.t. precedence) orderings

of the operations using e.g. different priority rules

2. For each of these orderings (permutations) (jl,. ... , jn) of

operations, define nodes (states)Z0 = {}, {jl}, {jl, j2}, ... , Zend = {jl, ... , jn}.

3. Draw an arrow from node ZtoZ' if Z' - Z represents a

feasible assignment of a station in the sense that cycle

time is not exceeded:4. In the resulting graph find the shortest path fromZ0 = {} to

Zend = {jl, ... , jn}.


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Pinto heuristic

Example: c=4

Ordering 1: 2, 1, 4, 5, 3

Ordering 2: 1, 4, 5 ,2, 3

By coincidence the optimal solution is reached!

2 1

3 4 5

1 3

2 3

2


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B&B algorithm by Johnson (FABLE)

DP algorithm can be considered as breadth search

All nodes in a certain stage are considered, before proceeding tothe next stage

The first complete solution is already the optimal one; if thealgorithm is stopped because of time restrictions no feasiblesolution is available.

-> B&B algorithm by Johnson treis to search the correspondingtree in the sense ofdepth search by trying to reach leaves ofthe tree soon (complete solution).

Algorithm is also known as FABLE: fast algorithm for balancing

lines effectively. Like in all B&B algorithms one tries to keep the tree small; in

addition to dominance rules also bounds are used.


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Branching process

In the starting node 0 no operations have been assigned

yet.

In each iteration an additional operation is assigned (or in a

backtracking step an operation is removed). A new station is opened whenever no further operation can

be assigned to the previous one (maximal stations)

In order to obtain a good first solution, the operations are

ordered according to some priority rules -> in FABLE this isdone according to the following rules (considering

precedences):


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Branching process

1. sort according to decreasing operation times tj (i.e.

allocate long operations first)

2. in case of a tie, use decreasing number of immediate

successors 3. in case there is still a tie choose randomly

Example:

We obtain the following ordering:1, 2, 3, 4, 5

2

1

3 4 5

1 3

2 3

2


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Branching process

This leads us to a leaf of the B&B tree:

Feasible solution (3 stations): {1}, {2, 4}, {3, 5}


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Branching process

In an exact method we must check all alternative solutions in principle.

Hence, one has to perform backtracking. We have to go back in the tree to

find the last "crossroad" where not all directions /alternatives have been

investigated yet. 1st branching opportunity: node 3 (operation 5 instead of

operation 3).

In FABLE only increasing operations number within 1 station are permitted

-> Node 6 is not to be considered any further!


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Branching process

Next branching opportunity: Node 1 (operation 4 instead of operation 2)

{4,2} is not allowed for station 2 (see previous slide). {4,5} is ok (no other possibilities). The

only possibility for station 3 is {2} and for station 4 {3}. Already in node 9 we see that this

cannot lead to a new optimal solution since we will not end up with < 3 stations.

Backtracking from node 9next branching opportunity: node 0.


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Branching process

Branching at node 0:

We try to assign operation 2 instead of operation 1. Maximal station 1 is {2} and only maximal station 2 is

{1}. In node 12 one has again 2 possible operations to be chosen, 3 and 4. We start with the lower index,

i.e. operation 3. Station {3} is maximal. In node 13 or even in node 12 one could have already stopped,since it is clear that no solution with less than 3 stations will be found. All branches have been

investigated and the algorithm stops.

The solution {1}, {2, 4}, {3, 5} with 3 stations is optimal.

Exact methods for ALB.ppt

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