TUGAS
TEKNIK OPTIMASI
“EMPLOYEE
SCHEDULING”
Oleh
:
RISDAWATI
HUTABARAT
NPM
: 1215031064
Kelas : B
JURUSAN
TEKNIK ELEKTRO
FAKULTAS
TEKNIK
UNIVERSITAS
LAMPUNG
2014
SOAL
Employed has a 24-hour-day, 7-days-a-week toll free
hotline that is being set up to answer questions regarding a new product. The
following table summarizes the number of full-time equivalent employees (FTEs)
must be on duty in a each time block.
Interval
|
Time
|
FTEs
|
1
|
0-4
|
15
|
2
|
4-8
|
10
|
3
|
8-12
|
40
|
4
|
12-16
|
70
|
5
|
16-20
|
40
|
6
|
20-0
|
35
|
Constraints for Employee Scheduling
·
Macrosoft may
hire both full-time and part time employees. The former work 8-hour shifts and
the latter work 4-hour shifts ; their respective hourly wages are $15.20 and
$12.95. Employees may start work only at the beginning of 1 of the 6 intervals.
·
Part-time
employees can only answer 5 calls in the time a full-time employee can answer 6
calls. (i,e.,a part-time employee is only 5/6 a full-time employee.)
·
At least
two-thirds of the employee working at any one time must be full-time employees.
Formulate an LP to determine how to
staff the hotline at minimum cost.
Solution :
Formulate an LP to determine how to
staff the hotline at minimum cost.
Decision variables:
Xt = # of full time employes that
begin the day at the start of interval t to work
For 8 hours
Yt = # of part time employes that
are signed interval t.
Dari tabel diatas maka diperoleh
nilai dari masing-masing interval waktu untuk memulai bekerja adalah :
X1= 0-8
X2= 4-12
X3=8-16
X4=12-20
X5=16-0
X6=20-4
Maka untuk mendapatkan nilai minimum cost adalah
Part-time employees can only answer
5 calls in the time a full-time employee can answer 6 calls.
(8
x $ 15,2) (4 x $12,95)
|
Min
$ 121,6 (X1+X2+…X6) + $
51,8 (Y1+Y2+…Y6 =
 |
X1+X6 +
(5/6) Y1
 15
|
X1+X2 + (5/6) Y2
10
|
X2+X3 + (5/6) Y3
40
|
X3+X4 + (5/6) Y4
70
|
X4+X5 + (5/6) Y5
40
|
X5+X6 + (5/6) Y6
35
|
At least two-thirds of the employee
working at any one time must be full-time employees.
More constraints
X1+X6 (2/3) (X6+XI+Y1) -->
(1/3) X1 + (1/3)X6 – (2/3)YI 0
(2/3) (X6+XI+Y1) -->
(1/3) X1 + (1/3)X6 – (2/3)YI 0
X1+X2 (2/3)
(X1+X2+Y2) -->
(1/3) X1 + (1/3)X2 – (2/3)Y2 0
(2/3)
(X1+X2+Y2) -->
(1/3) X1 + (1/3)X2 – (2/3)Y2 0
X2+X3 (2/3)
(X2+X3+Y3) -->
(1/3) X2 + (1/3)X3 – (2/3)Y3 0
(2/3)
(X2+X3+Y3) -->
(1/3) X2 + (1/3)X3 – (2/3)Y3 0
X3+X4 (2/3)
(X3+X4+Y4) -->
(1/3) X3 + (1/3)X4 – (2/3)Y4 0
(2/3)
(X3+X4+Y4) -->
(1/3) X3 + (1/3)X4 – (2/3)Y4 0
X4+X5 (2/3)
(X4+X5+Y5) --> (1/3) X4 + (1/3)X5 – (2/3)Y5 0
(2/3)
(X4+X5+Y5) --> (1/3) X4 + (1/3)X5 – (2/3)Y5 0
X5+X6 (2/3)
(X5+X6+Y6) --> (1/3) X5 + (1/3)X6 – (2/3)Y6 0
(2/3)
(X5+X6+Y6) --> (1/3) X5 + (1/3)X6 – (2/3)Y6 0
Xt 0, Yt 0
t=1,2,3,4…..6
0, Yt 0
t=1,2,3,4…..6
Hasil print screen dari ms.excel dengan
menggunakan Jensen.lib
Dari
gambar diatas maka dapat diketahui bahwa nilai value dari :
X1= 7,1 Y1
= 0 Dengan nilai profit= 12800,19
X2
= 0 Y2
= 3,5
X3=
40 Y3 =
0
X4
=12,9 Y4 =
20,6
X5=
27,1 Y5 =
0
X6
=7,1 Y6 = 0