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PROCESS DESIGN & ANALYSIS
Pokok Bahasan: I. Water Tank Analogy II. Peta Proses dengan Flow Chart III. Process Blocking & Starving IV. Process Performances & Examples V. Littles Law.
Kuliah ke-3: Jumat, 27 Februari 2015 HW #2: Lihat copy soal (Chase, 2011)!
Sispro/Sesi-3/27 Feb_2015/yad
Manajemen Operasi
Baca Chapter 11. Process Design & Analysis-
Operations and Supply Chain Management;
Chase et.al; Irwin McGraw-Hill; 14th Ed., 2014
SALEMBA
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Learning objectives:
Mengenal dan menerapkan process flowcharting Mengenal dan menerapkan tiga jenis proses:
serial process, parallel process, dan logistic process
Menganalisa proses dengan Littles Law Mengukur dan menghitung process
performance.
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If you can't describe what you
are doing as a process, you
don't know what you're doing.
W. Edwards Deming
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I. Water Tank Analogy
Pekerjaan
memasuki proses
Pekerjaan yang dihasilkan
Pekerjaan menunggu
untuk diselesaikan
Throughput time
Proses diperlukan utk menyelesaikan pekerjaan
What goes in a Process
must come out of the
Process.
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II. Peta Proses dengan Flowchart
Tugas atau
operasi
Contoh: Pemasangan
mesin mobil, tiket
diserahkan ke calon
penumpang, dll.
Simpul
keputusan
Contoh: Berapa
banyak perubahan
desain dilakukan?
Personil mana yang
akan dipilih?, dll.
Semua proses dapat dipetakan dengan flowchart
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Tempat
penyimpanan,
atau antrian
Contoh: Pelanggan
menunggu dilayani,
dll.
Aliran material
atau pelanggan
Contoh: Pelanggan
berpindah tempat,
mekanik mengambil
alat, dll.
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Contoh: Process Flow Chart Pasokan Material
Inspeksi cacat
material
Kembalikan
ke pemasok
Material
diterima dari
pemasok
Ditemukan
cacat?
Ya
Tidak
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Contoh: Process flowchart Las Vegas slot machine
Masukkan
coin ke
slot mesin
Tarik
lengan
slot mesin
Win or
lose
Play
Again? Quit
Is bucket
full?
Pindahkan
coin ke
winnings
bucket
Pindahkan
coin ke payout
bucket
Winnings
bucket
Activate
payout
Payout
bucket
Pay
winnings
Aktivitas di
slot machine
Aktivitas pemain Win
Yes
Lose No
No
Yes
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Jenis Proses
Proses 1 Proses 2 Proses 3
Proses 1
Single stage process
Multi stage process
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Alternatif lain: Simbol Process Flowchart
Operations
Inspection
Transportation
Delay
Storage
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Ste
p
Op
era
tio
n
Tra
nsp
ort
Insp
ect
De
lay
Sto
rag
e
Dis
tan
ce
(fe
et)
Tim
e
(min
) Description
of
process
1
2
3
4
5
6
7
8
9
10
11
Unload apples from truck
Move to inspection station
Weigh, inspect, sort
Move to storage
Wait until needed
Move to peeler
Apples peeled and cored
Soak in water until needed
Place in conveyor
Move to mixing area
Weigh, inspect, sort
Total Page 1 0f 3 480
30
5
20
15
360
30
20
190 ft
20 ft
20 ft
50 ft
100 ft
Date: 9-30-02
Analyst: TLR
Location: Graves Mountain
Process: Apple Sauce
Example:
Process
Flowchart
of Apple
Processing
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Contoh: Making Hamburgers at Mc Donalds, Burger King & Wendys
A. McDonalds Old Process
B. Burger King
C. Wendys
D. McDonalds New Process
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Service Blueprinting
- The process of flowcharting for services that includes the customer:
Identifying (mapping) processes
Isolating fail points
Establishing a time frame
Analyzing profitability
Line of visibility
Above the line: stages in process, in direct contact with customer, that focus on providing good service.
Below the line: stages in the process, not in contact with the customer, that focus on process efficiency.
Failsafing
Creating a control condition where the customer, server, or process can take only the correct (or desired) action while engaged in a service process.
Process Analysis in Services
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Contoh: Express Mail Delivery Service
Customer
Calls Customer
Gives Package
Receive
Package
Truck Packaging Forms Hand-held Computer Uniform
Truck Packaging Forms Hand-held Computer Uniform
Deliver
Package
Driver
Picks
Up Pkg.
Customer
Service
Order
Load
On Truck
Unload
&
Sort
Fly to Destination
Load on
Airplane
Sort
Packages
Fly to
Sort
Center
Airport
Receives
& Loads
Dispatch
Driver
PHYSICAL EVIDENCE
CUSTOMER
(On Stage)
CONTACT PERSON
(Back Stage)
SUPPORT PROCESS
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Starving 15 detik
III. Blocking & Starving Process
Proses 1 Proses 2
Proses 1
Proses 2
Cycle time 30 detik Cycle time 45 detik
Blocking 15 detik
Cycle time 30 detik Cycle time 45 detik
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Contoh Analisa Multi-Proses
Perhitungan: 1. Pada proses 1, 100 unit selesai dalam 3000 detik. 2. Pada proses 2, dalam 3000 detik, output-nya 66 unit, berasal dari (3000
detik 30 detik)/(45 detik/unit)= 66 unit. 3. Jadi persediaan antara proses 1 dan proses 2 selama 3000 detik (WIP) =
(100 66) = 34 unit. 4. Semua unit selesai dalam = 3000 detik + (34 unit X 45 detik/unit) = (3000 + 1530) detik = 4530 detik. Proses tahap 2 adalah bottleneck.
Proses 1
Proses 2
Cycle time 30 detik Cycle time 45 detik
Blocking 15 detik
Produksi 100 unit
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Terminologi Proses Bertahap
1. Tertahan (blocking) Aktivitas tahap berikut berhenti (stop), karena kekurangan kapasitas.
2. Kekosongan (starving) Aktivitas tahap berikut berhenti (stop) karena kekurangan aktivitas yang akan dikerjakan.
3. Penyangga (buffer) Penyimpanan antar tahap proses, agar tahap proses tersebut independen terhadap proses sebelumnya.
4. Bottleneck Tahap proses yang membatasi kapasitas, karena output proses sebelumnya lebih besar.
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Kondisi Multi-Proses dengan Buffer
Proses 1 Proses 2
Buffer
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IV. Pengukuran Process Performance & Contoh
Run time = Batch size x time/unit
Littles Law: Inventory =Throughput rate X Flow time
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Contoh Bread making
Contoh 1: Single line production Hitung troughput time-nya! JAWAB:
Bread making adalah bottleneck dl proses, bila bread making dan packing beroperasi sama-sama.
Kapasitas bakery 100 loaves/hr, tapi krn kapasitas pack 100 loaves/ hr, utilisasi pack hanya 75%. Tdk terjadi inventory pada proses bread making dan packing.
Jadi, throughput time = 1 jam + jam = 1 jam
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Contoh Bread making (lanjutan)
Contoh 2: Parallel line production Hitung throughput time-nya bila bread making diparalelkan dua line! JAWAB: Ini kasus Littles Law dimana terbentuk WIP krn pack menjadi
bottleneck. Bila bread making dioperasikan 2 shifts per day dan packing 3 shifts per
day, pada 2 shifts pertama akan terbentuk WIP, sbb: - Pada bread making 2 X 100 loaves/hr X 2 X 8 hr = 3,200 loaves - Pada pack 100 loaves/0.75 hr X {(2 X 8 hr)-1hr} = 2,000 loaves Sehingga WIP yg terbentuk= 3,200 2,000 = 1,200 loaves WIP rata-rata = 1,200/2 = 600 loaves. Maka, dengan throughput rate 100/0.75 = 133.3 loaves/hr.
Dengan Littles Law didapat throghput time loaves ada di WIP = 600 loaves/(133.3 loaves/hr) = 4.5 hrs. Jadi, total throughput time yaitu waktu loaves di WIP plus operations
time utk bread making dan packaging processes = 1 hr utk bread making + 4.5 hrs utk inventory + 0.75 hr utk packaging = 6.25 hrs.
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Contoh Bread making
Contoh 1. Pembuatan Roti Satu Lintasan
Contoh 2. Pembuatan Roti Dua Lintasan Paralel
Dalam 2 shift inventory terbentuk 0-1200 loaves. WIP rata2=600 loaves. Throughput rate di WIP = 133.3 loaves/hour (dari 100 loaves/ hour).
Dg Littles Law, throughput time di WIP = 600/133.3 = 4 jam
RM WIP Bread making
Cycle time:
1 hr/100 loaves
Packing
Cycle time:
hr/100 loaves
Finished
Goods
RM
Bread
making
Cycle time:
1 hour/100 loaves
Bread
making
Cycle time:
1 hour/100 loaves
WIP Pack
Cycle time:
hr/100 loaves
Finished
Goods
Total Throughput Time = 1 hrs
Total Throughput Time (1 hr pembuatan roti +
4 hr in inventory+ hr pengepakan) = 6 hrs
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Contoh 3: Restaurant Operation
Diketahui: Restaurant: Buffet arrangement Non-steady state process. Pre-prepared items (salad, desserts, dll) guna mempercepat proses pelayanan. Diasumsikan waktu rata-rata pelanggan mendapatkan makanan sampai selesai
30 menit/meja. Meja yang tersedia adalah 40 meja untuk 4 orang. Problem bagi restoran ini setiap orang ingin makan pada waktu berbarengan. Berapa kapasitas restoran? JAWAB: Kapasitas pelanggan maksimum adalah 4 x 40 = 160 kursi? (kenyataannya tidak)
Bila rata-rata meja ditempati 2.5 orang, maka utilisasi restoran 2.5 kursi : 4 kursi = 62.5% Cycle time restoran bila beroperasi pada kapasitas: Cycle time = menit (= 45 seconds) (yaitu dari 30 menit/meja : 40 meja = menit)
Dengan Cycle Time restoran 45 seconds (3/4 menit), meja akan kosong tiap menit. Artinya tiap interval waktu 15 menit, meja yang kosong = 15 : = 20 meja
Jadi, tiap 1 jam, restoran dapat menangani = 60 menit : min/kelompok = 80 kelompok pelanggan/jam.
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Problem bagi restoran adalah pelanggan datang
berbarengan antara jam 11.30 am1.00 pm.
Data yang dikumpulkan, sbb:
Time Parties arriving
11.30-11.45 15
11.45-12.00 35
12.00-12.15 30
12.15-12.30 15
12.30-12.45 10
12.45-13.00 5
Total parties 110
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Restaurant analysis
Data untuk interval 15 menit (dari 2 jam operasi restoran)
Time period
(1)
Parties
arriving
during
period (2)
Parties
departing
during period
(3)
Parties either at
table or waiting
to be served
(4)=(2)-(3)
Tables
used
(5)
Customer
parties
waiting
(6)=(4)-(5)
Expected
waiting
time
(7)=0.75x(6)
11-30-11-45 15 0 15 15 --- -----
11.45-12.00 35 (50) 0 50 40 10 7.5 min
12.00-12.15 30 (80) 15 65 40 25 18.75 min
12.15-12.30 15 (95) 20 (35) 60 40 20 15 min
12.30-12.45 10 (105) 20 (55) 50 40 10 7.5 min
12.45-13.00 5 (110) 20 (75) 35 35
13.00-13.30 0 (110) 35 (110)
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0
10
20
30
40
50
60
70
11.45 12.00 12.15 12.30 12.45 13.00 13.15 13.30
Parties waiting
for tables
Tables available in the restaurant
Time
Customer in the Restaurant
35
15
50
65
60
50
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Kesimpulan Restaurant Analysis
1. Jam 12.00 siang, ada 10 kelompok customers yang menunggu dilayani.
2. Bahkan meningkat menjadi 25 kelompok s/d jam 12.15.
3. Penurunan terjadi s/d jam 12.45 menjadi 10 kelompok.
4. Apa langkah mengurangi problem waiting line?:
- Menurunkan cycle time utk 1 meja (tapi apakah
bisa lunch hour customer masih bisa dipercepat,
misalnya menjadi kecil dari 30 menit?)
- Menambah meja, misalnya dengan 25 meja
- Meningkatkan utilisasi meja, mis. menjadi 2 X lipat.
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Contoh 4. Planning A Transit Bus Operation
Non-steady state process Bus route: Notre-Dame, Louvre, Concorde, Champs Elysees, Arc de
Triomphe, Eiffel Tower, and others. The route has 60 stops. Seating capacity 50 passengers; and 30 passengers can stand. Assume 2 hours needed to traverse the route during the peak traffic Key measure of service: How long customer waits before arrival of the
bus? Jadi: cycle time 2 jam. Dengan 1 bus rata-rata menunggu 1 jam. Bila ada 2 bus, menunggu rata-rata 30 menit. Bila ingin rata-rata menunggu 2 menit, maka cycle time 4 menit, dan
dibutuhkan 30 bus (120 menit : 4 menit/bus = 30 bus) Bila ada 30 bus, penumpang yang bisa diangkut 1500 duduk, dan 2400
duduk & berdiri. Berapa jumlah bus yang dibutuhkan bila semua duduk, atau duduk & berdiri?
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Solution for Bus Operation
Data dikumpulkan sbb:
Time
(1)
# of
cstmrs/hr
(2)
Ave. time
on bus
(3)
Load
(pssngr hrs)
(4)=(2)X(3)
Min. # of
bus needed
(5)=(4)/80
# Bus for all pssngrs
to be seated
(6)=(4)/50
8-9 a.m 2000 45 min or 3/4 hr 1500 18.75 30
9-10 am 4000 30 min or 1/2 hr 2000 25 40
10-11 am 6000 30 min 3000 37.5 60
11 am-12 5000 30 min 2500 31.25 50
12-1 pm 4000 30 min 2000 25 40
1-2 pm 3500 30 min 1750 21.875 35
2-3 pm 3000 45 min 2250 28.125 45
3-4 pm 3000 45 min 2250 28.125 45
4-5 pm 3000 45 min 2250 28.125 45
5-6 pm 4000 45 min 3000 37.5 60
6-7 pm 3000 45 min 2250 28.125 45
7-8 pm 1500 45 min 1125 14.0625 22.5
Totals 42000 25875
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Kesimpulan Bus Operation Analysis
1. Bila disediakan hanya 30 bus, maka sebagian harus berdiri
2. Dengan 30 bus, maka selama rush hour 10-11 am dan 5-6 pm, tidak semua penumpang bisa terangkut; sehingga lebih beralasan menyediakan 40 bus jam 9 am 7 pm
3. Bila disediakan 40 bus, maka yang terangkut duduk 40 busx12 jamx50 seat/bus=24000 seat-hrs. Jadi utilisasi 25875/24000=107.8%. Artinya rata-ratanya 7.8% berdiri
4. Jadi, design sistem transit adl trade-off antara keenakan layanan, frekuensi bus datang di tiap stop-an, dan utilisasi kapasitas dari bus.
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V. Littles Law (a really quite useful law)
Throughput (TH) = Work In Process (WIP) x Cycle Time (CT)
WIP = 10
Throughput time = ?
Cycle time
= 2 mins
Throughput time = 20 mins
Throughput time = 10 2 mins
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Littles law (a really quite useful law) (Continued)
Throughput (TH) = Work In Process (WIP) x Cycle Time (CT)
Contoh 5: Need to mark 500 exam scripts in 5 days (working 7 hours a
day). Takes 1 hour to mark a script. How many markers are needed?
Throughput time = 5 days 7 hours = 35 hours
35 hours = 500 scripts Cycle times
Cycle time = 35 hours 500 scripts
= 0.07 hours
Number of markers = Work content = 1 hour = 14.29
Cycle time 0.07
Throughput efficiency = Work content
Throughput time 100
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Average cost $45 12 hours to make a car Assembles 200 cars per 8 hour shift
Currently one shift Holds on average 8,000 batteries in raw material inventory. Hitung: a). jumlah battery di WIP dan raw material inventory. b). Nilai battery. c). Days of supply disimpan ory.raw material inven
Contoh 6: Car Batteries
Solution:
a). WIP = Throughput rate x flow time
= 25 batteries/hour x 12 hours = 300 batteries
Total = 8,000 + 300 = 8,300 batteries
b). Value of batteries = 8,300 X $45 = $373,500
c). Days of supply in raw material inventory is the flow time for a battery in raw
material inventory.
Flow time = Inventory / Throughput rate = (8,000 batteries) / (200 batteries/day) = 40 days
Note: 200 cars per 8 hour, atau 200/8 = 25 cars atau batteries/hour
Inventory = Throughput rate X Flow time atau: I = R X T
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Textbook: Operations and Supply Chain Management; Chase
et.al; Irwin McGraw-Hill; 14th Ed., 2014
HW #2:
Kerjakan Practice Exam (p. 293) ... (Chase, 2014) tdk dikumpul
Baca Chapter 11 Process Design & Analysis (Chase, 2014)
Soal no. 1, 2, 3, 5, 6, 7,8,10,11,12,13 (copy terlampir; Chase 2011)
Dikumpul Sabtu, 7 Maret 2015.
