Antrian pada stasiun kerja Tunggal
Konsep Dasar Simulasi Simulasi Sistem Industri
Dr. Eng. Muhammad Rusman, ST. MT Sistem Antrian Sederhana
Sistem n
Sebagian Model Simulasi melibatkan n antrian sebagai bagunan dasar.
Oleh karena itu digambarkan suatu kasus n antrian sederhana yang menggambarkan satu stasiun kerja dalam suatu lantai produksi. Antrian pada stasiun kerja Tunggal
Machine
(Server)
Arriving Departing
7
6
5
4
Blank Parts Finished Parts Queue (FIFO)Part in Service Sistem Jika Komponen datang pada saat mesin n dalam kondisi idle maka akan langsung diproses Jika Mesin dalam Keadaan Busy, maka n
komponen harus menunggu dalam antrian FIFO Sistem n Kita harus menentukan aspek numerik dari berkaitan sistem yang kita bangun
Konsisten dalam menggunakan satuan waktu yang ¨ akan dipakai Dalam contoh ini kita menggunakan satuan waktu
¨ MENIT Sistem ini kita asumsikan mulai dari waktu 0 menit,
¨ dimana tidak ada satupun komponen dalam sistem (Sistem Idle)
Waktu Kedatangan dan Proses
Satuan yang kita gunakan Menit n
Part Arrival Interval Service Number Time Time Time
1
0.00
6.84
4.58
2
6.84
2.40
2.96
3
9.24
2.70
5.86
4
11.94
2.59
3.21
5
14.53
0.73
3.11 Simulasi akan dilaksanakan selama 15
n menit
Tujuan Studi Output yang akan dianalisis: n
Total Produksi : ¨ n Jumlah komponen yang dapat diselesaikan dalam waktu 15 menit simulasi.
Rata-rata waktu tunggu
¨ n Waktu yang diperlukan komponen sampai bisa di proses. Jika D
i menyatakan waktu delay tiap komponen ke-i dalam antrian dan N jumlah komponen yang telah selesai menunggu maka rata-rata
N waktu tunggu adalah:
Di ∑ i
=1
N
Waktu Maksimum dalam antrian
¨
Tujuan Studi
Rata-rata jumlah komponen yang menunggu dalam antrian
¨ selama simulasi: Misalkan Q(t) adalah jumlah komponen yang antri pada saat t maka jumlah rata-rata antrian selama simulasi T adalah luas daerah dibawah kurva Q(t) di bagi panjang waktu simulasi.
D i ∫
N Waktu ini menunjukkan beberapa rata-rata panjang antrian yang mungkin dapat digunakan dalam pengambilan keputusan luas stasiun kerja.
Panjang antrian maksimum: Ukuran ini merupakan indikator ¨ yang lebih baik untuk menentukan berapa luas lantai produksi yang perlukan. Tujuan Studi Rata-rata dan Maksimum Flow Time : Waktu yang dibutuhkan oleh
¨ sebuah komponen sejak mulai datang ke stasiun kerja sampai selesai diproses dan keluar. Untuk tiap komponen, flow time adalah rentang waktu antara kedatangan sampai selesai diproses sehingga sama dengan jumlah antara waktu tunggu dalam antrian dan waktu proses.
Dalam sistem antrian, semakin kecil indikator ini maka semakin baik.
Utilisasi Mesin : Proporsi waktu dimana mesin dalam keadaan sibuk ¨
(busy). Misalkan B(t) adalah suatu fungsi yang menyatakan staus dari mesin dari waktu ke waktu dimana:
1 Jika mesin dalam keadaan sibuk pada saat t
§ B(t)
§
0 Jika Mesin dalam keadaan idle pada saat t
§ Maka Utilisasi mesin tersebut adalah luas area dibawah kurva B(t) dibagi
¨ dengan panjang waktu simulasi : T
B t dt ( )
∫
U= T
Grafik Jumlah Antrian
Q(t) t=T Grafik Status Mesin B(t) t=T
Waktu Kedatangan dan Proses
2.70
3.11
0.73
14.53
5
3.21
2.59
11.94
4
5.86
n Satuan yang kita gunakan Menit n
Simulasi akan dilaksanakan selama 15 menit Part Number Arrival Time Interval Time Service Time
3
2.96
2.40
6.84
2
4.58
6.84
0.00
1
9.24 Kurva Q(t)
2
1
9
10
12
13
14
15
5
7
8
11
1
2
3
4
1
10
12
13
14
15
8
9
11
2
3
4
5
6
7
1
Event
n
Arr = Arrival, Dep= Departure n
Variables
¨ Q(t) à Jumlah parts yang antri dalam waktu t
¨ B(t) à Server busy system n
Attributes
¨ If Part is in service at the machine, its arrival time is underlined
¨ Menggambar keajadian yang baru terjadi. ¨
Statistical Acumulators
Σ F = the sum of the flowtimes that have been observed so far n
F* = the maximum flowtimes observed so far = the area under the Q(t) curved so far n
Q* = the maximum value of Q(t) so far = the area under the B(t) curve so far
D* = the maximum time in queue observed so far n
N = the number of the entities that have passed throught the queue so far n
n P = the total number of the parts produced so far n
Σ D = the sum of the queue times that have been observed so far n Simulation by Hand n
Manually track state variables, statistical accumulators n
Use “given” interarrival, service times n
Keep track of event calendar n
“Lurch” clock from one event to the next n
Will omit times in system, “max” computations here (see text for complete details)
Simulation by Hand: Setup
System Clock B(t) Q(t) Arrival times of Event calendar Number of completed waiting times in queue
Total of waiting times in queue Area under
Q(t)
Area under
B(t) Q(t) graph B(t) graph
Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ... Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
1 2 3 4 5 10 15 20 1 2 5 10 15 20 System Clock
0.00 B(t) Q(t) Arrival times of custs. in queue <empty>
Event calendar [1, 0.00, Arr] [–, 20.00, End]
Number of completed waiting times in queue
Total of waiting times in queue
0.00 Area under
Q(t)
0.00 Area under
B(t)
0.00 Q(t) graph
B(t) graph
Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ... Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 0.00, Initialize
1 2 3 4 5 10 15 20 1 2 5 10 15 20 System Clock
Simulation by Hand: t = 0.00, Arrival of Part 1
0.00 B(t)
Event calendar [1, 2.90, Dep] [–, 20.00, End]
Number of completed waiting times in queue
1 Total of waiting times in queue
0.00 Area under
Q(t)
0.00 Area under
B(t)
0.00 Q(t) graph
B(t) graph
Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ... Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
1 2 3 4 5 10 15 20 1 2 5 10 15 20
1
1 Q(t) Arrival times of <empty>
Simulation by Hand: t = 1.73, Arrival of Part 2
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [1, 2.90, Dep]
2
1
1.73 (1.73) [3, 3.08, Arr]
1
1 [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
1
0.00 4 3
0.00
1.73 Q(t) graph 2 1 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 2.90, Departure of Part 1
System Clock B(t) Q(t) Arrival times of Event calendar
2
2.90 1 <empty> [2, 4.66, Dep] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
2
1.17 3 4
1.17
2.90 Q(t) graph 1 1 2 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 3.08, Arrival of Part 3
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [4, 3.79, Arr]
3
2
3.08 (3.08) [2, 4.66, Dep]
1
1 [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
2
1.17 3 4
1.17
3.08 Q(t) graph 1 2 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 3.79, Arrival of Part 4
System Clock B(t) Q(t) Arrival times of Event calendar
4
3
2
3.79
1 2 (3.79, 3.08) [2, 4.66, Dep] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
2
1.17 3 4
1.88
3.79 Q(t) graph 1 2 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 4.41, Arrival of Part 5
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [2, 4.66, Dep]
5
4
3
2
4.41
1 3 (4.41, 3.79, 3.08) [6, 18.69, Arr] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
2
1.17 4 3
3.12
4.41 Q(t) graph 1 2 1 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 4.66, Departure of Part 2
System Clock B(t) Q(t) Arrival times of Event calendar
5
4
3
4.66
1 2 (4.41, 3.79) [6, 18.69, Arr] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
3
2.75 3 4
3.87
4.66 Q(t) graph 1 2 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 8.05, Departure of Part 3
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [4, 12.57, Dep]
5
4
8.05
1 1 (4.41) [6, 18.69, Arr] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
4
7.01 4 3
10.65
8.05 Q(t) graph 2 1 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 12.57, Departure of Part 4
System Clock B(t) Q(t) Arrival times of Event calendar
5
12.57 1 () [6, 18.69, Arr] [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
5
15.17 3 4
15.17
12.57 Q(t) graph 1 1 2 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 17.03, Departure of Part 5
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [6, 18.69, Arr] 17.03 () [–, 20.00, End]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
5
15.17 3 4
15.17
17.03 Q(t) graph 2 1 1 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 18.69, Arrival of Part 6
System Clock B(t) Q(t) Arrival times of Event calendar
6
18.69 1 () [–, 20.00, End] [6, 23.05, Dep]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
6
15.17 3 4
15.17
17.03 Q(t) graph 1 2 1 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 19.39, Arrival of Part 7
System Clock B(t) Q(t) Arrival times of Event calendar custs. in queue [–, 20.00, End]
7
6
19.39
1 1 (19.39) [6, 23.05, Dep] [8, 34.91, Arr]
Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
6
15.17 4 3
15.17
17.73 Q(t) graph 2 1 2 1 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: t = 20.00, The End
System Clock B(t) Q(t) Arrival times of Event calendar
7
6
20.00
1 1 (19.39) [8, 34.91, Arr] Number of Total of Area under Area under completed waiting waiting times in queue Q(t) B(t) times in queue
6 3 4
15.17
15.78
18.34 Q(t) graph 1 1 2 2 5 10 15 20 B(t) graph 5 10 15 20 Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...
Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...
Simulation by Hand: Finishing Up Average waiting time in queue: n
Total of times in queue
15
17 . = =
2
53 minutes per part.
No. of times in queue6 Time-average number in queue: n
Q t ( ) .
Area under curve
15
78 = = .
79 part Final clock value
20
Utilization of drill press: nB t Area under ( ) curve 18 .
34 = =
.
92 (dimension less)Final clock value Tugas Buatlah kembali grafik B(t) dan Q(t) dalam n sebuah grafik demikian juga dengan Hand simulation nya? Dikumpul minggu depan