Tugas Teknik Optimasi "Employee Scheduling"

Diposting oleh Unknown
Selasa, 04 November 2014
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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
X1+X2   (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
X3+X4   (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
X5+X6   (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




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





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