Pull System Simulation 2
(Lean Production)
By: Ana Rita Almeida Marques
Course: Lean Production
Índice
STEP 1: Define the pacemaker process ............................................................... 3
STEP 2 : Define the planned leveling period ......................................................... 4
STEP 3 : Define the runner typologies e exotics .................................................... 4
Step 4: Analyze customer behavior for missed withdrawals .................................. 4
Step 5: Capacity Analysis ................................................................................... 5
Step 5a: Define nº change-over (per day) .................................................................... 5
STEP 5b: Verify the supply capacity of pre -processes .................................................. 6
STEP 6: Define EPEI for each runner .................................................................... 7
STEP 7: Define the batch size for each runner ...................................................... 8
STEP 8: Define the leveling production plan and sequence ................................... 8
Step 8a: Leveling Production Plan............................................................................... 8
Step 8b: Leveling Sequence ....................................................................................... 8
STEP 9: Calculation of required decoupling ......................................................... 9
Calculate RE, LO and WI. .................................................................................................... 9
The SA Factors ................................................................................................................. 10
STEP 10: Design the heijunka board .................................................................. 12
STEP 10a: Distribution of “ lot sizes” ......................................................................... 12
STEP 10b: define the daily setting of SA1 on overflow................................................. 13
PRE-PROCESSES............................................................................................. 14
AMPLIFIER PROCESS:.............................................................................................. 14
GREEN BODY PROCESS: ......................................................................................... 16
STEP 1: Define the pacemaker process
The products we need to create are:
Our Value Stream includes:
We have chosen the assembly process as our pacemaker. We made this decision to position the
pacemaker as close to the customer as possible, aiming to minimize ineJiciencies in our system.
Ideally, the pacemaker should be at the start of the Value Stream to keep inventory low. However,
given that we are still in the early stages and lack extensive knowledge of the processes, this setup
is more practical for now.
Let's consider the customer's request:
STEP 2 : Define the planned leveling period
1 week
2 weeks ( avg .)
3 weeks ( avg .)
4 weeks ( avg .)
CW1
CW2
CW3
CW4
CW5
CW6
3280
3540
3533
3430
3800
3540
3533
3430
3520
3320
3533
3430
3120
3320
2960
3430
2760
2880
2960
3000
2880
2960
Let's examine the diJerent alternatives:
• 1 week: Ideally, this would be best, but the period is too short. We wouldn't have enough
time to set up the system because, after 5 days, we would already need to change it. If we
had the capacity, we might choose 1 week as the leveling period, but we would need to
adjust the line from 3280 pieces to 3800 pieces to 3520 pieces, requiring very high
flexibility in terms of personnel.
• 2 weeks: This is the optimal leveling period.
• 3 weeks: This would result in excessively high inventory levels because we would be too far
removed from the actual quantity requested.
STEP 3 : Define the runner typologies e exotics
All types are classified as runners, indicating they are high-volume products with regular
withdrawals (AX category). This simplifies the planning process as the focus will be on ensuring
steady production rates for these products.
Step 4: Analyze customer behavior for missed withdrawals
Does the costumer at the end of the leveling period pickup the same quantity that he ordered?
- If not: it has to exist a strategy to react if costumer takes more (ex: overtime, other line…) if
costumer takes less (ex: flexibility in other lines, products…)
-
If Yes: the costumer order volume can be used to plan capacity (POT) and nº of changeovers for the period.
Step 5: Capacity Analysis
Step 5a: Define nº change-over (per day)
Data:
Customer fluctuations in the period
OEE Improvement after System CIP projects
OEE(Final assembly): 91.7 % = 1320 min (24*60*0.917)
- (1320-4*60)/1440=75%
OEE (Actuator module production): 84.4 % = 1215 min (24*60*0.844)
- (1215-2*60)/1440=76%
OEE (Amplifier module production): 76.4 % = 1100 min (24*60*0.764)
- (1100-0*60)/1440=76.4%
OEE (Body production): 88.4% = 1273 min (24*60*0.884)
- (1273-3*60)/1440=76%
POT (planned operation time): 24 h = 1440 minutes
Change Over Time: 60 min
Cycle Time/pc: 1.5 min
NPK: 40 pcs
Time available for our process:
𝑇𝑖𝑚𝑒 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 = 𝑃𝑂𝑇 ∗ 𝑂𝐸𝐸 = 1440 𝑚𝑖𝑛 ∗ 0.917 = 1320 𝑚𝑖𝑛
Time due to OEE loss:
𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 = 𝑃𝑂𝑇 – 𝑡𝑖𝑚𝑒 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 = 1440 𝑚𝑖𝑛 – 1320 𝑚𝑖𝑛 = 120 𝑚𝑖𝑛
Nº pieces produced per day:
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 𝑝𝑒𝑟 𝑑𝑎𝑦 =
𝑇𝑜𝑡𝑎𝑙 𝑛º 𝑜𝑓 𝑝𝑖𝑐𝑒𝑠 𝑖𝑛 2 𝑤𝑒𝑒𝑘𝑠
3280 + 3800 𝑝𝑐𝑠
=
= 708 𝑝𝑧
𝑁° 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑑𝑎𝑦𝑠
5 𝑑𝑎𝑦𝑠
NPT (Net Production Time)
𝑁𝑃𝑇 = 𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 𝑝𝑒𝑟 𝑑𝑎𝑦 ∗ 𝑇𝐶 = 708 𝑝𝑐𝑠 ∗ 1.5 𝑚𝑖𝑛 = 1062 𝑚𝑖𝑛
Available time for change over
𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑡𝑖𝑚𝑒 𝑓𝑜𝑟 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑣𝑒𝑟 = 𝑃𝑂𝑇 – 𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 − 𝑁𝑃𝑇 = 1440 𝑚𝑖𝑛 – 120 𝑚𝑖𝑛 − 1062 𝑚𝑖𝑛
= 258 𝑚𝑖𝑛
Maximum number of possible change overs
𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑡𝑖𝑚𝑒 𝑓𝑜𝑟 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑣𝑒𝑟
258 𝑚𝑖𝑛
𝑁° 𝐶𝑂 =
=
= 4.3
𝑇 𝐶𝑂
60 𝑚𝑖𝑛
We now must choose between 4 or 5 change overs per day.
Our goal is to produce the same pieces every day in small batches so we must try reducing the
change over time as much as possible. The change over time is a loss that is measured and not
calculated by diJerence.
So, we choose 4 change overs.
Total time we will allocate to the change over
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝐶𝑂 = 𝑇𝐶𝑂 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟 𝐶𝑂 = 60 𝑚𝑖𝑛 ∗ 4 = 240 𝑚𝑖𝑛
NPT (Net production time)
𝑁𝑃𝑇 = 𝑃𝑂𝑇 − 𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝐶𝑂 = 1440 𝑚𝑖𝑛 − 120 min − 240 min = 1080 𝑚𝑖𝑛
Nº pieces produced per day
𝑁𝑃𝑇
1080 𝑚𝑖𝑛
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 =
=
= 720 𝑝𝑐𝑠
𝑇𝐶
1.5 𝑚𝑖𝑛
𝐍º 𝐊𝐚𝐧𝐛𝐚𝐧𝐬 𝐜𝐚𝐫𝐝𝐬 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐬 𝐩𝐞𝐫 𝐝𝐚𝐲
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠
720 𝑝𝑐𝑠
𝑁° 𝑘𝑎𝑛𝑏𝑎𝑛 𝑝𝑒𝑟 𝑑𝑎𝑦 =
=
= 18
𝑁𝑃𝐾
40 𝑝𝑐𝑠
STEP 5b: Verify the supply capacity of pre -processes
ACTUATOR
NPT (Net Production time)
𝑁𝑃𝑇 = 𝑃𝑂𝑇 − 𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝐶𝑂 = 𝑃𝑂𝑇 − d𝑃𝑂𝑇 ∗ (1 − 𝑂𝐸𝐸)g − (𝑁° 𝐶𝑂 ∗ 𝑇 𝐶𝑂)
= 1440 min − (1440 min∗ (1 − 0.844)) − (2 ∗ 60 𝑚𝑖𝑛) = 1095 𝑚𝑖𝑛
Total nº pieces per day
!"#
%&'( *+,
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 =
=
= 730.2 𝑝𝑖𝑒𝑐𝑒𝑠
#$
%.( *+,
AMPLIFIER
NPT (Net Production time)
𝑁𝑃𝑇 = 𝑃𝑂𝑇 − 𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝐶𝑂 = 𝑃𝑂𝑇 − d𝑃𝑂𝑇 ∗ (1 − 𝑂𝐸𝐸)g − (𝑁° 𝐶𝑂 ∗ 𝑇 𝐶𝑂)
= 1440 min − (1440 min∗ (1 − 0.764)) − (0 ∗ 60 𝑚𝑖𝑛) = 1100 𝑚𝑖𝑛
Total nº pieces per day
𝑁𝑃𝑇
1100 𝑚𝑖𝑛
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 =
=
= 733 𝑝𝑖𝑒𝑐𝑒𝑠
𝑇𝐶
1.5 𝑚𝑖𝑛
BODY
NPT (Net Production time)
𝑁𝑃𝑇 = 𝑃𝑂𝑇 − 𝑂𝐸𝐸 𝑙𝑜𝑠𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝐶𝑂 = 𝑃𝑂𝑇 − d𝑃𝑂𝑇 ∗ (1 − 𝑂𝐸𝐸)g − (𝑁° 𝐶𝑂 ∗ 𝑇 𝐶𝑂)
= 1440 min − (1440 min∗ (1 − 0.884)) − (3 ∗ 60 𝑚𝑖𝑛) = 1092 𝑚𝑖𝑛
𝑁° 𝑝𝑖𝑒𝑐𝑒𝑠 =
𝑁𝑃𝑇
1092 𝑚𝑖𝑛
=
= 728.6 𝑝𝑖𝑒𝑐𝑒𝑠
𝑇𝐶
1.5 𝑚𝑖𝑛
STEP 6: Define EPEI for each runner
𝑉𝑊 =
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑞𝑢𝑒𝑠𝑡 𝑤𝑒𝑒𝑘 1 + 𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑞𝑢𝑒𝑠𝑡 𝑤𝑒𝑒𝑘 2
1120 𝑝𝑖𝑒𝑐𝑒𝑠 + 1160 𝑝𝑖𝑒𝑐𝑒𝑠
=
𝑁𝑢𝑚𝑏𝑒𝑟 𝑑𝑎𝑦𝑠 𝑖𝑛 2 𝑤𝑒𝑒𝑘𝑠
10 𝑑𝑎𝑦𝑠
= 228 𝑝𝑖𝑒𝑐𝑒𝑠
𝑅𝑒𝑛𝑎𝑢𝑙𝑡 =
960 + 1440
= 240 𝑝𝑖𝑒𝑐𝑒𝑠
10
𝐷𝑎𝑖𝑚𝑙𝑒𝑟 =
400 + 400
= 80 𝑝𝑖𝑒𝑐𝑒𝑠
10
𝐵𝑀𝑊 =
400 + 400
= 80 𝑝𝑖𝑒𝑐𝑒𝑠
10
𝐴𝑢𝑑𝑖 =
160 + 240
= 40 𝑝𝑖𝑒𝑐𝑒𝑠
10
𝑃𝑒𝑢𝑔𝑒𝑜𝑡 =
240 + 160
= 40 𝑝𝑖𝑒𝑐𝑒𝑠
10
We necessarily must produce some types of pieces on some days and other types on other days
and thus we define the EPEI.
The types of parts produced in large quantities (VW and Renault), will have EPEI = 1, therefore they
will be produced every day.
The parts produced in smaller quantities (Daimler, BMW, Audi, Peugeot) will have EPEI = 2, they will
be produced every other day.
The system takes the largest EPEI, so the system has EPEI 2.
STEP 7: Define the batch size for each runner
Average nº kanban cards for each customer
𝑉𝑊 =
Average daily demand
228 𝑝𝑖𝑒𝑐𝑒𝑠
=
= 5.7 = 6
𝑁𝑃𝐾
40 𝑝𝑖𝑒𝑐𝑒𝑠
𝑅𝑒𝑛𝑎𝑢𝑙𝑡 =
240
=6
40
𝐷𝑎𝑖𝑚𝑙𝑒𝑟 =
80
=2
40
𝐵𝑀𝑊 =
80
=2
40
𝐴𝑢𝑑𝑖 =
40
=1
40
𝑃𝑒𝑢𝑔𝑒𝑜𝑡 =
40
=1
40
In total we will therefore produce 18 cards.
STEP 8: Define the leveling production plan and sequence
Step 8a: Leveling Production Plan
Customer
VW
Renault
Daimler
BMW
Audi
Peugeot
Total
Day 1
6
6
4
0
2
0
18
Day 2
6
6
0
4
0
2
18
Day 3
6
6
4
0
2
0
18
Day 4
6
6
0
4
0
2
18
Day 5
6
6
4
0
2
0
18
Day 6
6
6
0
4
0
2
18
Day 7
6
6
4
0
2
0
18
Day 8
6
6
0
4
0
2
18
Day 9
6
6
4
0
2
0
18
Day 10
6
6
0
4
0
2
18
Step 8b: Leveling Sequence
We can decide how to schedule different types of products in our leveling production plan. Here's our
approach:
1. Night Shift: We schedule the simplest types of products at night because they are easier to
handle.
2. Day Shift: More complex types that need more frequent changes are scheduled during the
central part of the day. This period typically has more disturbances due to the presence of the
entire staff.
Floating Runner:
- Renault: Chosen as the floating runner due to high demand and flexibility in production.
-
VW: Not chosen because the production is already higher than customer demand ( I
approximated the number of kanban cards to produce from 4.6 to 5) , leading to potential
overproduction and large inventory.
STEP 9: Calculation of required decoupling
The sum of Renault's kspots is 6 + 6 + 14 = 26 kanbans.
The sum of BMW kspots is 8 + 4 + 0 = 12 kanbans.
Calculate RE, LO and WI.
WI= kspot1+kspot2+kspot3-RE-LO
RE=number of cards to cover the replenishment time (time from the moment the customer picks
up a piece to the moment this piece returns to the supermarket, this time interval is called RTloop)
𝑅𝐸 =
𝑅𝑇𝑙𝑜𝑜𝑝
𝑇𝑇 ∗ 𝑁𝑃𝐾
RTloop = sum of a series of times = 3 days
TT = Takt Time
𝑇𝑇 =
𝑃𝑂𝑇
𝑃𝑅
𝑅𝐸 =
𝑅𝑇𝑙𝑜𝑜𝑝 ∗ 𝑃𝑅
𝑝𝑜𝑡 ∗ 𝑁𝑃𝐾
𝑅𝐸 =
𝑅𝑇𝑙𝑜𝑜𝑝 ∗ 𝑃𝑅
3 𝑑𝑎𝑦𝑠 ∗ (960 𝑝𝑖𝑒𝑐𝑒𝑠 + 1440 𝑝𝑖𝑒𝑐𝑒𝑠)
=
= 18 𝑘𝑎𝑛𝑏𝑎𝑛
𝑝𝑜𝑡 ∗ 𝑁𝑃𝐾
10 𝑑𝑎𝑦𝑠 ∗ 40 𝑝𝑖𝑒𝑐𝑒𝑠
𝐿𝑂 =
𝑙𝑜𝑡 𝑠𝑖𝑧𝑒
240
−1=
−1=5
𝑁𝑃𝐾
4
When we talk about a leveled process, however, LO is equal to 0 because we have the kanbans in
the leveling plan. We know, in fact, that it is not the customer who determines the type to produce
but our level plan.
WI is the number of cards needed to handle the customer's scheduled swings.
𝑊𝐼 = 𝑘𝑠𝑝𝑜𝑡(𝑡𝑜𝑡) − 𝑅𝐸 − 𝐿𝑂 = 26 − 18 = 8
The SA Factors
The SA factors are the factors that are used to guarantee continuity in production and delivery to
our customer
We have three SAs: SA1, SA2 and SA3.
SA1 is the number of cards to cover the problems generated by the production process as each
process has fluctuations in performance and therefore we will calculate those fluctuations to put in
our supermarket.
SA2 is the number of cards to cover customer-generated fluctuations that occur when the
customer does not meet scheduled withdrawals.
SA3 is not taken into account as it is an additional factor based on experience. It is usually used
when we know neither our processes nor our customers.
SA1 CALCULATION
When I go to calculate the WI I take into consideration the data from the future, when I calculate the
SA I take the data from the past. How big is my past period? I decide it based on the client, if it is
stable I will take the last period into consideration, if it is unstable I will take a longer period (6-8
weeks). In this case we have the two previous periods, we take into consideration a period equal to
RTloop and therefore 3 days. My biggest loss is days 8, 9 and 10. My three best days instead are 4, 5
and 6.
We will divide SA1 into two further factors: SA1 supermarket (when I have an OEE lower than
average) and SA1 overflow (when I have an OEE lower than average)
Let's therefore calculate the OEE fluctuations that we have had in terms of Kanban cards by making
a simple proportion:
𝐷𝑎𝑦: 𝑃𝑙𝑎𝑛𝑛𝑒𝑑 𝑂𝐸𝐸: 𝑃𝑙𝑎𝑛𝑛𝑒𝑑 𝐾𝑎𝑛𝑏𝑎𝑛 = 𝑅𝑒𝑎𝑙𝑂𝐸𝐸: 𝑅𝑒𝑎𝑙𝐾𝑎𝑛𝑏𝑎𝑛
To calculate SA1 supermarket I consider the three worst days
18 ∗ 0.7
𝐷𝑎𝑦 8: 75%: 18 = 70% ∶ 𝑥 → 𝑥 =
= 16.8 𝐾
0.75
18 ∗ 0.71
𝐷𝑎𝑦 9: 75%: 18 = 71% ∶ 𝑥 → 𝑥 =
= 17 𝐾
0.75
18 ∗ 0.71
𝐷𝑎𝑦 10: 75%: 18 = 71% ∶ 𝑥 → 𝑥 =
= 17 𝐾
0.75
On the first day I lost 1.2 cards, on the second 1 card and on the third 1 card, for a total of 3.2 cards
lost.
SA1 supermarket: 3
To calculate SA1 overflow I consider the three best days
18 ∗ 0.8
𝐷𝑎𝑦 4: 75%: 18 = 80% ∶ 𝑥 → 𝑥 =
= 19.2 𝐾
0.75
𝐷𝑎𝑦 5: 75%: 18 = 79% ∶ 𝑥 → 𝑥 =
18 ∗ 0.79
= 19 𝐾
0.75
𝐷𝑎𝑦 6: 75%: 18 = 78% ∶ 𝑥 → 𝑥 =
18 ∗ 0.78
= 18.7 𝐾
0.75
On the first day I increased by 1.2 cards, on the second by 1 and on the third by 0.7, for a total of 2.9
cards.
SA1 overflow: 3
SA1= SA1supermarket + SA1overflow = 3 + 3 = 6
SA2 CALCULATION
SA2 Supermarket is when the customer takes more than planned.
SA2 Overflow is when the customer withdraws less and therefore it is an unplanned stock. It's the
opposite of SA1
The only day in which the customer withdrew more than planned was day 4 when he withdrew 95=4 more kanban cards.
SA2supermarket = 4 kanbans
The three days on which he withdraws the least are the 7th, 8th and 9th.
SA2overflow = 1 + 1 + 1 = 3 kanbans
SA2 = SA2supermarket + SA2overflow = 4 + 3 = 7 kanban
The total number of Kanban cards related to Renault in our process will be:
K (Renault) = RE + LO + WI + SA1 + SA2 = 18 + 0 + 8 + 6 + 7 = 39 kanban
STEP 10: Design the heijunka board
STEP 10a: Distribution of “ lot sizes”
1.
2.
3.
4.
5.
In the first hour, we perform a type change.
After that, we produce 6 VW kanban cards and then change the type again.
OEE losses, totaling 120 minutes, are distributed across all product types.
.
For VW, the losses are calculated as 120 ∗ %/ , where 6 represents the number of VW cards
As a result, instead of taking 360 minutes to produce the batch of 6 VW cards (6 cards × 60
minutes each), we take 460 minutes.
.
o This includes 60 minutes for type change and 40 minutes of OEE losses 120 ∗ = 40
%/
min of losses
o Thus, the total time is 360+40+60=460360 + 40 + 60 = 460360+40+60=460 minutes
Same thing for each type.
Calculation
Time window
6
= 460 min (7.67 h)
18
4
60 + 60 ∗ 4 + 120 ∗
= 327 min (5.44 h)
18
2
60 + 60 ∗ 2 + 120 ∗
= 193 min (3.22 h)
18
6
60 + 60 ∗ 6 + 120 ∗
= 460 min (7.67 h)
18
VW
00:00 – 07:40
60 + 60 ∗ 6 + 120 ∗
(LS=6), C/O=60 min , 1K=60 min
Daimler/BMW
(LS=4)
Audi/Peugeot
(LS=2)
Renault
(LS=6)
07.40 – 13.07
1.07pm – 4.20pm
4.20pm – 12.00am
We therefore have the distribution of " lot sizes".
VW:
6h + (6*0.11)h
c/o
0
1
2
3
4
5
Daimler:
4h + (4*0.11)h
c/o
6
7
8
9
10
11
Audi:
c/o
c/o
2h + (2*0.11)h
12
13
14
15
16
17
Renault:
6h + (6*0.11)h
18
19
20
21
22
23
24
STEP 10b: define the daily setting of SA1 on overflow
We need to determine the optimal placement for the 3 SA1 cards.
By analyzing the data from our process, we see that the best performance was on day 4, with an OEE of
80%, exceeding the target of 75%. This resulted in an additional 1.2 cards.
On the first day, we will place 1 kanban card in the OEE performance and 2 additional cards in Overflow.
If we perform better and manage to produce the extra kanban card, we would halt production at that
point, as we would have reached our maximum limit. Producing significantly more pieces than
requested by the customer would shift us into a push production mode, which we want to avoid.
The final situation will be this:
Customer
VW
Renault
Daimler
BMW
Audi
Peugeot
Today
Tomorrow
6
6
4
0
2
0
0
0
0
0
0
0
Overflow
14 + 2
4
OEE
Performance
0
1
0
0
0
0
Backlog
0
0
0
0
0
0
The minimum stock we need to put in the supermarket for each runner before starting the leveling
period is SA1supermarket + Kspot2 + 50%WI.
For Renault therefore the minimum stock to be placed in the supermarket will be 3 + 6 + 0.5 x 8 = 16
kanbans.
When customer knowledge improves we will try to lower this inventory at the beginning of the period.
PRE-PROCESSES
In comparison to the pacemaker, the preprocess (operated with a pure pull system) has a shorter
replenishment time. This is because, unlike the 3-day RT loop used in leveling, we have a
today/tomorrow management approach where the cards produced today are scheduled for production
tomorrow.
A process managed with pure pull has a shorter cycle time than one managed with leveling. Specifically,
a card takes one day to be released from the supermarket and another day to be produced. Therefore,
the RT loop is 2 days.
AMPLIFIER PROCESS:
Let's calculate the kanban formula factors for the AMPLIFIER process:
180
𝑅𝑇𝑙𝑜𝑜𝑝 ∗ 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐷𝑒𝑚𝑎𝑛𝑑 2 ∗ 10
=
= 36
𝑃𝑂𝑇 ∗ 𝑁𝑃𝐾
1∗1
They are 160 pcs/day because we have 16 kanbans for 10 days. Let's consider 1 kanban=1 piece
(just for convenience, because initially we know that 1 kanban is 40 pieces)
WA = peak withdrawal in the RT loop. In this case RT loop is 2 days and every day the customer
requires the same amount, this is because the customer is the pacemaker. The pacemaker must
follow the leveling, so the client's withdrawal will be constant over the period. So the peak
withdrawal is 36.
WA ( Amplifier ) = 36 kanbans
set LO equal to 0 since we consider the type change time of the preprocesses corresponding to the
type change time of the pacemaker (which is very diJicult in reality).
WI ( Amplifier ) = 36 – 36 – 0 = 0 kanban
We calculate SA1 supermarket by considering the performance on days 4 and 5.
We calculate SA1 overflow considering the performance on days 2 and 3.
So SA1 supermarket :
18 ∗ 0.72
= 16.9 𝐾
0.764
18 ∗ 0.72
𝐷𝑎𝑦 5: 76,4%: 18 = 72% ∶ 𝑥 → 𝑥 =
= 16.9 𝐾
0.764
𝐷𝑎𝑦 4: 76,4%: 18 = 72% ∶ 𝑥 → 𝑥 =
We lost 18-16.9=1.1 K and 18-16.9=1.1 K, so the sum is 2.2 k which we round to 2 kanbans.
SA1supermarket = 2 kanbans .
SA1 overflow:
%/∗&./1
𝐷𝑎𝑦 2: 76.4%: 18 = 82% ∶ 𝑥 → 𝑥 = &.2.3 = 19.32 𝐾
%/∗&./1
𝐷𝑎𝑦 3: 76.4%: 18 = 82% ∶ 𝑥 → 𝑥 = &.2.3 = 19.32 𝐾
We have 19.32-18=1.32 K and 19.32-18=1.32 K, so the sum is 2.6 k which we round to 2 kanbans.
SA1overflow = 2 kanbans .
SA1 ( Amplifier ) = SA1supermarket + SA1overflow = 2 + 2 = 4 kanban
SA2supermarket = 4 kanbans
SA2overflow = 1 + 1 = 2 kanbans
SA2( Amplifier ) = 4 + 2 = 6 kanbans
The final situation will therefore be:
RE
LO
WI
SA1 (OEE)
SA1supermarket
SA1overflow
SA2 (Customer)
SA2supermarket
SA2overflow
SA
Total
AMPLIFIER
36k
0k
0k
2k
2k
4k
2k
10 k
46 k
GREEN BODY PROCESS:
𝑹𝑬 (𝑮𝒓𝒆𝒆𝒏 𝒃𝒐𝒅𝒚) =
60 𝑘𝑎𝑛𝑏𝑎𝑛 + 100 𝑘𝑎𝑛𝑏𝑎𝑛
2 𝑑𝑎𝑦𝑠 ∗
𝑅𝑇𝑙𝑜𝑜𝑝 ∗ 𝑃𝑅
2
=
= 16 𝑘𝑎𝑛𝑏𝑎𝑛
𝑝𝑜𝑡 ∗ 𝑁𝑃𝐾
1 𝑑𝑎𝑦 ∗ 1 𝑘𝑎𝑛𝑏𝑎𝑛
WA (Body) = 16 kanbans
WI (Green body) = 16 – 16 – 0 = 0 kanban
We calculate SA1 supermarket by considering the performance on days 8 and 9.
We calculate SA1 overflow considering the performance on days 1 and 2.
SA1 supermarket:
18 ∗ 0.73
= 17.3 𝐾
0.759
18 ∗ 0.74
𝐷𝑎𝑦 9: 75,9%: 18 = 74% ∶ 𝑥 → 𝑥 =
= 17.55 𝐾
0.759
We lost 18-17.3 = 0.7k and 18-17.55 = 0.45k, so the sum is 1.15 kanban cards which we round to 1.
SA1supermarket = 1 kanban .
𝐷𝑎𝑦 8: 75,9%: 18 = 73% ∶ 𝑥 → 𝑥 =
SA1 overflow:
%/∗&.2'
𝐷𝑎𝑦 1: 75.9%: 18 = 79% ∶ 𝑥 → 𝑥 = &.2(' = 18.7 𝐾
%/∗&.2'
𝐷𝑎𝑦 2: 75.9%: 18 = 79% ∶ 𝑥 → 𝑥 = &.2(' = 18.7 𝐾
We have 18.7-18=0.7 k and 18.7-18=0.7 k, so the sum is 1.4 k which we round to 1 kanban.
SA1overflow = 1 kanban .
SA1( Green body ) = SA1supermarket + SA1overflow = 1 + 1 = 2 kanban.
SA2(Green body)= SA2(Amplifier) = 4 + 2 = 6 kanbans
RE
LO
WI
SA1 (OEE)
SA1supermarket
SA1overflow
SA2 (Customer)
SA2supermarket
SA2overflow
SA
Total
Black
16 k
0k
0k
BODY
Red
4k
0k
0k
Green
16 k
0k
0k
0k
0k
0k
0k
1k
1k
0k
0k
0k
16 k
0k
0k
0k
4k
4k
2k
8k
24 k