8/9/2019 Lab 5 Best First Search for Scheduling

1/3

BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE

PILANI-333031 (RAJASTHAN) INDIA

Second Semester 2007-2008

Course: EA C461 (Artificial Intelligence) Lab 5: Best-First Search for Scheduling

Prepared By: Dr. Mukesh Kumar Rohil rohil@bits-pilani.ac.in

This Lab is based on the book by Ivan Bratko, Prolog Programming for Artificial Intelligence, ThirdEdition, 2001, Pearson Education Ltd., Second Impression 2007 (in India).

We can formulate a scheduling problem as a state-space search problem as follows:

States are partial schedules;

A successor state of some partial schedule is obtained by adding a not yet scheduled task to thisschedule; another possibility is to leave a processor that has completed its empty schedule;

The start state is the empty schedule;

Any schedule that includes all the tasks in the problem is a goal state;

The cost of a solution (which is to be minimized) is the finish time of a goal schedule;

Accordingly, the cost of a transition between two (partial) schedules whose finishing times are F1,and F2 respectively is the difference F2 F1.

Some refinements are done to the above approach as follows:

We fill the schedule according to increasing time.

Each time a task is added, the precedence constraint has to be checked.

If a task is waiting for resource (i.e. processor or server) then the task is assigned to any idle

resource.

We need the following information:

List of waiting tasks and their execution times

Current engagement of the processors We add (for convenience) the finish time of the (partial) schedule

The list of waiting tasks and their execution times are represented as: [Task1/Duration1,

Task3/Duration3, Task3/Duration3, .]

The current engagement of resources (or processors) represented by a list of tasks currently being

processed: Task/FinishTime

Three components of partial schedule are represented as:

WaitingList*ActiveTasks*FinishingTime

A best-first program along these lines is shown in Table 1.

Do the following:1. Find the time taken by the method to execute the query given at the end of Table 1. [Hint: Read

and Use time/2 predicate given in the LPA Win-Prolog help]2. Comparing the program in Table 1 (of Lab 4) and Table 1 (of this Lab), make a table of

predicates of Table 1 and Table 3 by stating which predicate(s) in Table 3 does the analogous or

similar work of which predicate(s) in Table 1.3. Change the program for any other scheduling problem involving at least 10 tasks and 15

precedence relations.

4. Name your files as L5_AI_YourID_pN and zip them as L5_AI_YourID.zip and transmit to

instructors email id. Replace the text YourID by your BITS IDNumber and N in pN by problemnumber 1, 2, or 3 as the case may be.

mailto:rohil@bits-pilani.ac.inmailto:rohil@bits-pilani.ac.in

8/9/2019 Lab 5 Best First Search for Scheduling

2/3

% Problem-specific relations for the task-scheduling problem.% The particular scheduling problem of Figure 12.8 is also defined by its% precedence graph and an initial (empty) schedule as a start node for search.

/* Problem-specific relations for task scheduling

Nodes in the state space are partial schedules specified by:

[ WaitingTask1/D1, WaitingTask2/D2, ...] * [ Task1/F1, Task2/F2, ...] * FinTime

The first list specifies the waiting tasks and their durations; the second list specifies the currently executed tasks and their finishing times, ordered so that F1 =< F2, F2 =< F3 ... . Fintime is the latest completion time of current engagements of the processors.*/

% s( Node, SuccessorNode, Cost)

s(Tasks1*[_/F|Active1]*Fin1,Tasks2*Active2*Fin2,Cost) :- del(Task/D,Tasks1,Tasks2), % Pick a waiting task not(member(T/_,Tasks2),before(T,Task)), % Check precedence not(member(T1/F1,Active1),F

8/9/2019 Lab 5 Best First Search for Scheduling

3/3

insertidle( A, L, L1).

del( A, [A | L], L). % Delete item from list

del( A, [B | L], [B | L1] ) :-del( A, L, L1).

goal( [] * _ * _). % Goal state: no task waiting

% Heuristic estimate of a partial schedule is based on an% optimistic estimate of the final finishing time of this% partial schedule extended by all the remaining waiting tasks.

h( Tasks * Processors * Fin, H) :-totaltime( Tasks, Tottime), % Total duration of waiting taskssumnum( Processors, Ftime, N), % Ftime is sum of finishing times

% of processors, N is their numberFinall is ( Tottime + Ftime)/N,

( Finall > Fin, !, H is Finall - Fin;H = 0

).

totaltime( [], 0).

totaltime( [_/D | Tasks], T) :-totaltime( Tasks, T1),T is T1 + D.

sumnum( [], 0, 0).

sumnum([_/T|Procs],FT,N) :- sumnum(Procs,FT1,N1), NisN1+1, FTisFT1+T.

% A task-precedence graph prec(TaskX, TaskY) means TaskX precedes TaskY

prec(t1,t4). prec(t1,t5). prec(t2,t4). prec(t2,t5).prec(t3,t5). prec(t3,t6). prec(t3,t7).

% A start node

start([t1/4,t2/2,t3/2,t4/20,t5/20,t6/11,t7/11]*[idle/0,idle/0,idle/0]*0).

% An example query: start( Problem), bestfirst( Problem, Sol).

Table 1