1. Measurement Of Efficiency
In Manufacturing Cycle Time
(How well do we manage waste ?)

2. Background
Manufacturing site was given an uphill task to improve overall cycle time
with a general goal of 2x theoretical cycle time.
Current practice
There is a common cycle time goal set for the site to achieve. Each
individual process along the flow will have it’s own goal to match
the common goal.

3. Advantage of this approach
This approach is good to track each process progress and contribution
towards the common goal.
Disadvantage of this approach
When there is slip up, every individual process will need to be dwell
in details and find the main process contributor to the slip up. From
there, actions taken to revert the situation. This approach will usually
end up with some processes will fail it’s goal marginally, but not enough
to explain the overall slips. Furthermore, time taken to dwell into each
process during presentation is lengthy.
Suggestion for a better approach
Find a measurement method to track cycle time performance and
process improvement at macro level which is easily understand by all
levels. This method should be able to separate entitlement or ideal time
from actual.

4. Expanding the process flow with time incurred :
Assembly EOLLoad Test Pack Ship
PTprinting
Queuetime
Allocationtime
Queuetime
Processtime
Inspectiontime(3rdopt)
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Holdlottime
Holdlottime
Inspectiontime(Buyoff,IRreflow)
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Processtime
Queuetime
Total cycle time
The typical macro process flow :
Theoretical/entitlement fixed time (value added)
Inspection time or must have to survive for quality reason (value added)
Mainly non value added or waste time that need improvement

5. Comments
The non value added and waste time is the main target for cycle time
improvement.
Proposed measurement or tracking method
Manufacturing cycle efficiency or MCE
= Value added time / Total cycle time
= (Theoretical process time + Inspection time) /
(Theoretical process time + Inspection time + Queue time +
allocation time + down time + Hold time + etc)
Ideal MCE should be 1.0 or 100% efficient.
The following 2 examples demonstrate based on 6.5 days theoretical/ideal
cycle time, even though we can achieve 2x = 13 days ( as set ), using
the above formulae will show how efficient is our process towards
the ideal cycle time.

6. Typical example 1 ( not efficient )
Assumptions :
A = Total process time from all processes = 5 days
B = Total inspection time from all processes = 1.5 days
C = Total queue time from all processes = 0.5 days
D = Total hold time from all processes = 3 days
E = Total down time from all processes = 0.5 days
F = Total allocation/execution time from all processes = 1.5 days
Manufacturing Cycle Efficiency (MCE)
= ( A + B) / ( A + B + C + D + E + F )
= ( 5 + 1.5) / (5 + 1.5 + 0.5 +3 + 0.5 + 1.5)
= 6.5 / 12
= 0.54
Conclusion : For the above example, we only able to achieve 54% of
entitlement or ideal cycle time. Although we can meet
2x goal, the efficiency is still low.

7. Typical example 2 ( 75% efficient )
Assumptions :
A = Total process time from all processes = 5 days
B = Total inspection time from all processes = 1.5 days
C = Total queue time from all processes = 0.5 days
D = Total hold time from all processes = 1 days
E = Total down time from all processes = 0.2 days
F = Total allocation/execution time from all processes = 0.5 days
Manufacturing Cycle Efficiency (MCE)
= ( A + B) / ( A + B + C + D + E + F )
= ( 5 + 1.5) / (5 + 1.5 + 0.5 +1 + 0.2 + 0.5)
= 6.5 / 8.7
= 0.75
Conclusion : For the above example, we are able to achieve 75% of
entitlement or ideal cycle time and meet the 2x goal
at the same time.

8. Suggestion for a start
Find out the current performance on MCE, set an internal goal to improve
50% from the ideal cycle time.
Example :
To date, our MCE = 50% efficiency
our Goal = 75% efficiency
If we really can achieved 75% efficiency, we have improved 25%. All these
will contribute to plant cycle time.

9. Takt Time
A meaningful measurement gauge

10. Takt Time = Total available time/Customer demand
Typical scenario 1 (More applicable)
Given customer demand of 1M/week, one week consist of
22 hr X 7 days X 60 mins = 9240 mins. (taking into account of efficiency)
Takt time per unit = 9240/1000000 = 0.00924 mins
What does this mean ?
It means we must produce one single unit out in every 0.00924 mins in
order to service the fixed customer demand per week. Lets say
average lot size is 12k, we need (12000 X 0.00924) = 110.88 mins =
1.848 hour to complete whole lot. Based on our simplify process from
start to end (refer to next page), our model process should looks and
perform in that manner.

11. Expand the process flow with time incurred :
Assembly EOLLoad Test Pack Ship
PTprinting
Queuetime
Allocationtime
Queuetime
Processtime
Inspectiontime(3rdopt)
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Holdlottime
Holdlottime
Inspectiontime(Buyoff,IRreflow)
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Processtime
Queuetime
Takt = 1.85 hr
The typical macro process flow :
Takt = 1.85 hr Takt = 1.85 hr
Takt = 1.85 hr Takt = 1.85 hr Takt = 1.85 hr
Continue next page ->
Each lot must be processed within 1.85 hour for each process step above

12. How does this work in actual situation ?
Since we have six processes , for the very first lot, it took 6x1.848 = 11.088 hr
to move from start until end. For every subsequent lots, if we can control
each process to take < takt time, we will achieve the goal of 1.848 hour to
ship each lot out. Total lot produced is (1000000/12000) = 83.33 lots.
This is equivalent to a total cycle time of
11.088 + 82.33(1.848) = 163.23 hours to process total of 83.33 lots
[1st
lot CT] + [(Total lot-1st
lot)TakT]
Converted to days, equivalent to 163.23/24 = 6.80days
WE CAN MEET CUSTOMER DEMAND FOR EVERY WEEK IF WE CAN
CONTROL TO PROCESS A LOT WITHIN THE CALCULATED Takt TIME.

13. Takt Time = Total available time/Customer demand
Typical scenario 2 (Less applicable)
Given a goal of 2.5 days cycle time and demand of 1kk/week, equivalent
to [(1kkx2.5)/7] = 357143 pieces per every 2.5 days. Assuming each lot
consists of 10k, equivalent to 35.7143 lots per every 2.5 days.
Takt time per lot = 2.5/35.7143 = 0.07 day = 1.68 hour
What does this mean ?
It means we must produce one single lot out in every 1.68 hour in order
not to jeopardize the total cycle time (2.5 days). Based on our simplify
process from start to end (refer to next page), our model process should
looks and perform in that manner.

14. Expand the process flow with time incurred :
Assembly EOLLoad Test Pack Ship
PTprinting
Queuetime
Allocationtime
Queuetime
Processtime
Inspectiontime(3rdopt)
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Holdlottime
Holdlottime
Inspectiontime(Buyoff,IRreflow)
Allocationtime
Queuetime
Processtime
Downtime
Holdlottime
Processtime
Queuetime
Takt = 1.68 hr
The macro process flow :
Takt = 1.68 hr Takt = 1.68 hr
Takt = 1.68 hr Takt = 1.68 hr Takt = 1.68 hr
Continue next page ->

15. How does this work in actual situation ?
Since we have six processes , for the very first lot, it took 6x1.68 = 10.08 hr
to move from start until end. For every subsequent lots, if we can control
each process to take < takt time, we will achieve the goal of 1.68 hour to
ship each lot out. This is equivalent to a total cycle time of
10.08 + 34.713(1.68) = 68.40 hours to process total of 35.713 lots
Converted to days, equivalent to 68.40/24 = 2.85 days ( 0.3 days extra
for the very first lot processed)

16. Production Kanban
A visual control for Lean manufacturing

17. A
M/C 1 M/C 2 M/C 3
A B A BA B B B
2 bin before
process, B (white)
under processed
After
processed
Material handler
OperatorOperator Operator
Oper place card B
at Kanban board
(1)
MH get card B
and tag to
new lot (3)
A
A
B
Inventory from
upstream process
B
MH send
new lot to
m/c 1
(4)
Note :
• Sequence (1) to (4) happen
to m/c 2 and 3 as well if the
processes are stand alone.
If this is a full conveyor line,
only apply to M/C1.
• Kanban system ensure no
disruption to value added time,
reduce downtime and improve
productivity with Lean line.
Signal to
MH (2)
B
B
Initial stage

18. Lot Size
An easy task for early win engagement
if properly managed

19. Time
(min)
Start Step 1 Step 2 Step 3 Finish
(Step 4)
Start Step
1
Step
2
Step
3
Finish (Step 4)
1 a,b,c,d,e a,b,c,d,e
2 a
3 b a
4 c b a
5 a,b,c,d,e d c b a
6 e d c b
7 e d c
8 e d
9 e + (a,b,c,d)
10 a,b,c,d,e
11
12
13
14
15 a,b,c,d,e
16
17
18
19
20 a,b,c,d,e
Assumptions : Each alphabet contained 100 pcs, total 4 process steps, each step required 1 min to complete 100 pcs,
we only have one machine for each process.
If we build a big lot of 500 pcs, total time to complete whole process is 20 minutes.
In contrast, when we build in small lot size (100 pcs each), total time to complete whole process is only 9 minutes.