A Study on Heuristic Timetabling Method for
Faculty Course Timetable Problem
Bong Chia Lih∗, Sze San Nah†and Noor Alamshah Bolhassan‡
Faculty of Computer Science and Information Technology
Universiti Malaysia Sarawak
Sarawak, Malaysia
Email:∗[email protected], †[email protected], ‡[email protected]
Abstract—This paper studies university course timetabling
problem in a case study related to faculty management system,
which is concerned with assigning students/lecturers to classes
and time-slots. In order to increase flexibility choices of timeslot
for “repeaters”, the problem has become more tight in scheduling.
A two-stage heuristic approach is presented, where the initial
stage groups the courses that are able to conduct simultaneously.
The second stage then assigns the weekly timeslots for each group
of courses, followed by venue for each course. Computational
results are presented for the proposed solution using real data. It
shows that the proposed solution is effective to handle the faculty
course timetabling.
Keywords—automated timetabling; two-stage heuristic
I. INTRODUCTION
Course timetabling is well known as NP-complete [1] and
has been studied extensively over the past decades [2], [3] and
[4]. However, it is “ambitious” to develop a fully automated
timetabling mechanism to meet the needs of all concerned.
Eventually, the timetabling process requires a lot of manual
effort.
In Universiti Malaysia Sarawak (UNIMAS), course
timetabling is managed by two parties: faculty management
and Centre of Applied Learning and Multimedia (CALM).
Each faculty owns a number of lecture/tutorial rooms, while
CALM manages four Centres of Teaching Facilities (CTF).
Every CTF consists of one lecture hall (LH) with capacity of
540 students and six seminar rooms (SR) with capacity of 150
students. In the beginning of the timetabling setup, CALM will
pre-assign each faculty a number of slots at CTFs based on
past experience, to accommodate courses with large number
of students. Together with the pre-assigned rooms allocation
from CALM and own faculty’s facilities, the responsibility
of constructing faculty course timetable falls on each faculty
management. This is mainly because each faculty has different
preference or request from lecturers.
Previously, Faculty of Computer Science and Information
Technology (FCSIT) course timetabling was handled by two
experience planners. The planning process was overwhelmed
with abundance of constraints and time consuming to even
find a feasible solution. This paper presents a real solution to
this faculty course timetabling by using a two-stage heuristic
method.
II. P ROP OSE D APPRO ACH
In this timetabling problem, we consider a set of offered
courses, C={C1, C2, . . . , Cn}will be enrolled by a set of
students (including lecturers), S={S1, S2, . . . , Su}. Each
course ihas a list of potential students and lecturers. Let
define the set of potential students and lecturers as PC=
{PC1, PC2, . . . , PCn}, where PCi⊂Sand Ci∈C.
A two-stage heuristic is proposed to solve this course
timetabling case study. This is due to its simplicity to cope
with different hard and soft constraints in two stages.
A. Stage I (Course Grouping)
All the courses will be divided into a few groups based on
the following constraints:
1) Students can max enrol one course in each group.
2) Lecturers can max enrol one course in each group.
3) Takes into account the repeating students to enable
them to enrol in certain courses.
The main idea of course grouping is to cluster several
courses into one group so that these courses can have lecture
concurrently during the same timeslot in different venues by
different lecturers and students. The task of this stage is to
reduce the problem size, from ncourses to mgroups where
n > m. Let define a set of group, G={G1, G2, . . . , Gm}
where Gi6=∅and G1∩G2∩. . . ∩Gm=C. The course
grouping algorithm is shown as follow:
Step 1 Calculate each course’s pairing ability, Yiwith other
courses
Algorithm 1 Calculate pairing ability
n=number of courses
for i=1 to ndo
Yi=0
for j=1 to ndo
if (PCi∩PCj=∅)then
Xij =1
else
Xij =0
end if
Yi=Yi+Xij
end for
end for
