Muhammad Rusman ST MT

  Konsep Dasar Simulasi

Muhammad Rusman, ST. MT Muhammad Rusman ST MT

  Sistem Antrian Sederhana  

  Sistem Sistem 

  Sebagian Model Simulasi melibatkan antrian sebagai bagunan dasar.

   Oleh karena itu digambarkan suatu kasus a antrian sederhana yang menggambarkan a sede a a ya g e gga ba a 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 Jika Komponen datang pada saat mesin dalam kondisi idle maka akan langsung diproses

   Jika Mesin dalam Keadaan Busy, maka komponen harus menunggu dalam antrian FIFO

  Sistem  Kita harus menentukan aspek numerik dari  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)

   Satuan yang kita gunakan Menit

   Satuan yang kita gunakan Menit

  Pa r t N u m be r Ar r iv a l Tim e I n t e r va l Tim e Se r v ice Tim e

  1 0 . 0 0 6 . 8 4 4 . 5 8 2 6 . 8 4 2 . 4 0 2 . 9 6 3 9 . 2 4 2 . 7 0 5 . 8 6 4 1 1 . 9 4 2 . 5 9 3 . 2 1 5 1 4 . 5 3 0 . 7 3 3 . 1 1

   Simulasi akan dilaksanakan selama 15 menit

  Tujuan Studi 

  Output yang akan dianalisis:

   Total Produksi :

   Jumlah komponen yang dapat diselesaikan dalam waktu 15 menit simulasi.

  

  Rata-rata waktu tunggu 

  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 i l i Mi lk Q(t) d l h j l h k t i d saat t maka jumlah rata-rata antrian selama simulasi adalah luas _T 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. 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 b h k j k l i d t k t i k j i l i 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

  T dengan panjang waktu simulasi :

  B t dt ( )

   U

  

T

  Grafik Jumlah Antrian Q(t) t=T

  Grafik Status Mesin B(t) t=T

   Satuan yang kita gunakan Menit

   Satuan yang kita gunakan Menit

  Pa r t N u m be r Ar r iv a l Tim e I n t e r va l Tim e Se r v ice Tim e

  1 0 . 0 0 6 . 8 4 4 . 5 8 2 6 . 8 4 2 . 4 0 2 . 9 6 3 9 . 2 4 2 . 7 0 5 . 8 6 4 1 1 . 9 4 2 . 5 9 3 . 2 1 5 1 4 . 5 3 0 . 7 3 3 . 1 1

   Simulasi akan dilaksanakan selama 15 menit

  Kurva Q(t)

3 Q( t )

  2

  1

  2

  3

  4

  5

  6

  7

  8

  9 1 0 1 1 1 2 1 3 1 4 1 5 Kurva B(t)

  1

3 B( t )

  1

  2

  3

  4

  5

  6

  7

  8 9 1 0

1 1

1 2 1 3 1 4 1 5

  2

  1

   Event  Event  Menggambar keajadian yang baru terjadi. 

  D* = the maximum time in queue observed so far

  

  Q* = the maximum value of Q(t) so far

  

  the area under the Q(t) curved so far

  

  = the area under the Q(t) curved so far

  

  F* = the maximum flowtimes observed so far

  

  F = the sum of the flowtimes that have been observed so far

  

  

  Arr = Arrival, Dep= Departure  Variables

  D = the sum of the queue times that have been observed so far

  

  N = the number of the entities that have passed throught the queue so far

  

  P = the total number of the parts produced so far

  

  Statistical Acumulators

  

  If Part is in service at the machine, its arrival time is underlined

   Attributes 

   B(t)  Server busy system

   Q(t)  Jumlah parts yang antri dalam waktu t

  = the area under the B(t) curve so far Simulation by Hand 

  Manually track state variables statistical 

  10

  20

  15

  10

  5

  2

  1

  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, ...

  20

  15

  5

  Manually track state variables, statistical accumulators 

  4

  3

  2

  1

  Q(t) graph

  completed waiting times in queue waiting times in queue Q(t) B(t)

  Simulation by Hand: Setup

  custs. in queue Number of Total of Area under Area under

  Keep track of event calendar  “Lurch” clock from one event to the next  Lurch clock from one event to the next  Will omit times in system, “max” computations here (see text for complete details)

  Use “given” interarrival, service times 

2 B(t) graph

  System Clock

  4

  Number of Total of Area under Area under

  Simulation by Hand: t = 0.00, Arrival of Part 1

  1

  completed waiting times in queue 1 waiting times in queue

  0.00 Q(t)

  0.00 B(t)

  0.00 Q(t) graph

  1

  2

  3

  5

  1 custs. in queue <empty>

  10

  15

  20

  2 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

  5

  10

  15

  [2, 1.73, Arr] [1, 2.90, Dep] [–, 20.00, End]

  0.00

  0.00 B(t) Q(t) Arrival times of custs. in queue <empty>

  4

  Event calendar [1, 0.00, Arr] [–, 20.00, End]

  Number of Total of Area under Area under

  Simulation by Hand: t = 0.00, Initialize

  completed waiting times in queue waiting times in queue

  0.00 Q(t)

  0.00 B(t)

  0.00 Q(t) graph

  1

  2

  3

  5

  20

  10

  15

  20

  2 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

  5

  10

  15

  20 System Clock

  1.73 B(t)

  3

  [3, 3.08, Arr] [2, 4.66, Dep] [–, 20.00, End]

  Number of Total of Area under Area under

  Simulation by Hand: t = 2.90, Departure of Part 1

  2

  completed waiting times in queue 2 waiting times in queue

  1.17 Q(t)

  1.17 B(t)

  2.90 Q(t) graph

  1

  2

  4

  2.90

  5

  10

  15

  20

  2 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

  5

  10

  15

  1 custs. in queue <empty>

  20

  1 Q(t)

  2

  1 Arrival times of custs. in queue (1.73)

  Event calendar [1, 2.90, Dep] [3, 3.08, Arr] [–, 20.00, End]

  Number of Total of Area under Area under

  Simulation by Hand: t = 1.73, Arrival of Part 2

  1

  2

  completed waiting times in queue 1 waiting times in queue

  0.00 Q(t)

  0.00 B(t)

  1.73 Q(t) graph

  1

  3

  15

  4

  5

  10

  15

  20

  2 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

  5

  10

  20 System Clock

  3.08 B(t)

  2

  [5, 4.41, Arr] [2, 4.66, Dep] [–, 20.00, End]

  Number of Total of Area under Area under

  Simulation by Hand: t = 3.79, Arrival of Part 4

  2

  3

  4

  completed waiting times in queue 2 waiting times in queue

  1.17 Q(t)

  1.88 B(t)

  3.79 Q(t) graph

  1

  3

  1

  4

  5

  10

  15

  20

  2 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

  5

  10

  15

  2 custs. in queue (3.79, 3.08)

  3.79

  1 Q(t)

  2

  1 Arrival times of custs. in queue (3.08)

  Event calendar [4, 3.79, Arr] [2, 4.66, Dep] [–, 20.00, End]

  Number of Total of Area under Area under

  Simulation by Hand: t = 3.08, Arrival of Part 3

  2

  3

  completed waiting times in queue 2 waiting times in queue

  1.17 Q(t)

  1.17 B(t)

  3.08 Q(t) graph

  1

  3

  20

  4

  5

  10

  15

  20

  2 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

  5

  10

  15

  20 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

  3.12

  4.41

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

2 B(t) graph

  1

  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

  custs. in queue [3, 8.05, Dep]

  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

  3.87

  4.66

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

2 B(t) graph

  1

  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

  10.65

  8.05

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

2 B(t) graph

  1

  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

  custs. in queue [5, 17.03, Dep]

  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

  15.17

  12.57

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

2 B(t) graph

  1

  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

  15.17

  17.03

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

  2

1 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

  custs. in queue [7, 19.39, Arr]

  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

  15.17

  17.03

  4

3 Q(t) graph

  2

  1

  5

  10

  15

  20

  2

2 B(t) graph

  1

  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, ...

  System Clock

  17.73 Q(t) graph

  20

  15

  10

  5

  4

  3

  2

  1

  15.17 B(t)

  2

  15.17 Q(t)

  completed waiting times in queue 6 waiting times in queue

  7

  6

  Simulation by Hand: t = 19.39, Arrival of Part 7

  Number of Total of Area under Area under

  Event calendar [–, 20.00, End] [6, 23.05, Dep] [8, 34.91, Arr]

  1 Arrival times of custs. in queue (19.39)

  1 Q(t)

  19.39 B(t)

2 B(t) graph

  1

  10

  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, ...

  20 B(t) graph

  15

  10

  5

  2

  1

  18.34 Q(t) graph

  15.78 B(t)

  15.17 Q(t)

  completed waiting times in queue 6 waiting times in queue

  2

  20

  15

  5

  5

  4

  3

  2

  1

  Number of Total of Area under Area under

  [6, 23.05, Dep] [8, 34.91, Arr]

  1 custs. in queue (19.39)

  1

  20.00

  7

  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, ...

  20 Simulation by Hand: t = 20.00, The End

  15

  10

  6 Simulation by Hand: Finishing Up 

  Average waiting time in queue: T t l Total of f times ti i in queue

  15 15 .

  17

  17 2 . 53 minutes per part   No. of times in queue

  6 

  Time-average number in queue: Area under Q ( t ) curve 15 .

  78   . 79 part p Final Fi l clock l k value l

  20 

  20

  Utilization of drill press: Area under B ( t ) curve 18 .

  34 . 92 (dimension less)   Final clock value

20 Complete Record of the Hand Simulation

  Tugas  

  Buatlah kembali grafik B(t) dan Q(t) dalam Buatlah kembali grafik B(t) dan Q(t) dalam sebuah grafik demikian juga dengan Hand simulation nya? Dikumpul minggu depan