WIP controlling - Example line
Unbalanced line
Jesse Uitto

Unbalanced line simulation
• Source: Factory Physics by Wallace Hopp &
Mark Spearman - Chapter 7, Intuition building
exercise 1
• Unlike in the book, in this example there is
only 5, not 6, machines in the third phase.
• More about the theory can be found at
http://jesseuitto.fi/?p=464 (example line 2)

2 h 5 h 10 h 3 h
Phase 1
- 1 machine
- 2 hour raw process time
- Capacity 0,5 products/hour

2 h 5 h 10 h 3 h
Phase 2
- 2 machines
- 5 hour raw process time
- Capacity 0,4 products/hour

2 h 5 h 10 h 3 h
Phase 3
- 5 machines
- 10 hour raw process time
- Capacity 0,5 products/hour

2 h 5 h 10 h 3 h
Phase 4
- 2 machines
- 3 hour raw process time
- Capacity 0,67 products/hour

Experiment #1
• First we try this line with WIP limit of 7 jobs
• Red balls are jobs
• Each slide simulates the beginning of that
particular hour.
– Example at the beginning of the hour 1 first job is
released and it is finished when hour 2 changes
into hour 3 (slide = hour 3)
– The number below each work station tells what
process hour is going with that part.

2 h 5 h 10 h 3 h
Hour 1
1

2 h 5 h 10 h 3 h
Hour 2
2

2 h 5 h 10 h 3 h
Hour 3
1
1

2 h 5 h 10 h 3 h
Hour 4
2
2

2 h 5 h 10 h 3 h
Hour 5
1
3
1

2 h 5 h 10 h 3 h
Hour 6
2
4
2

2 h 5 h 10 h 3 h
Hour 7
1
5
3

2 h 5 h 10 h 3 h
Hour 8
2
1
4
1

2 h 5 h 10 h 3 h
Hour 9
1
2
5
2

2 h 5 h 10 h 3 h
Hour 10
2
3
1
3
1

2 h 5 h 10 h 3 h
Hour 11
1
4
2
4
2

2 h 5 h 10 h 3 h
Hour 12
2
5
3
5
3

2 h 5 h 10 h 3 h
Hour 13
1
1
4
6
4
1

2 h 5 h 10 h 3 h
Hour 14
2
2
5
7
5
2

2 h 5 h 10 h 3 h
Hour 15
3
1
8
6
3
1

2 h 5 h 10 h 3 h
Hour 16
4
2
9
7
4
2

2 h 5 h 10 h 3 h
Hour 17
5
3
10
8
5
3

2 h 5 h 10 h 3 h
Hour 18
1
4
9
6
4
1
1

2 h 5 h 10 h 3 h
Hour 19
2
5
10
7
5
2
2

2 h 5 h 10 h 3 h
Hour 20
3
8
6
3
3
1
1

2 h 5 h 10 h 3 h
Hour 21
4
9
7
4
2
2
1

2 h 5 h 10 h 3 h
Hour 22
5
10
8
5
3
3
2

2 h 5 h 10 h 3 h
Hour 23
1
9
6
4
1
1
1

2 h 5 h 10 h 3 h
Hour 24
2
10
7
5
2
2
2

2 h 5 h 10 h 3 h
Hour 25
3 1
8
6
1 3
3

2 h 5 h 10 h 3 h
Hour 26
4 2
9
7
2
1
4

2 h 5 h 10 h 3 h
Hour 27
5 3
10
8
3
2
5

2 h 5 h 10 h 3 h
Hour 28
1
1
1
9
4
1
6

2 h 5 h 10 h 3 h
Hour 29
2
2
2
10
5
2
7

2 h 5 h 10 h 3 h
Hour 30
3
1 3
1
1
3
8

2 h 5 h 10 h 3 h
Hour 31
4
2
1
2
2
4
9

2 h 5 h 10 h 3 h
Hour 32
5
3
2
3
3
5
10

2 h 5 h 10 h 3 h
Hour 33
1
4
1
1
4
6
1

2 h 5 h 10 h 3 h
Hour 34
2
5
2
2
5
7
2

2 h 5 h 10 h 3 h
Hour 35
3
1
3
1
6
8
3

2 h 5 h 10 h 3 h
Hour 36
4
2
4
2
7
91

2 h 5 h 10 h 3 h
Hour 37
5
3
5
3
8
102

2 h 5 h 10 h 3 h
Hour 38
1
4
6
4
9
1
1

Notices of simulation with WIP = 7
• This hour 38 is exactly the same as hour 18 so
it starts to go round between hours 18 -> 37
• Notice that after the beginning there is no
queue in front of the phase 2.
– Actually during the hours 20-24 both stations at
phase 2 aren’t even fully utilized.
– Bottleneck station should be fully utilized, right ?
• So let’s try this with an WIP level of 8 jobs

2 h 5 h 10 h 3 h
Hour 1
1

2 h 5 h 10 h 3 h
Hour 2
2

2 h 5 h 10 h 3 h
Hour 3
1
1

2 h 5 h 10 h 3 h
Hour 4
2
2

2 h 5 h 10 h 3 h
Hour 5
1
3
1

2 h 5 h 10 h 3 h
Hour 6
2
4
2

2 h 5 h 10 h 3 h
Hour 7
1
5
3

2 h 5 h 10 h 3 h
Hour 8
2
1
4
1

2 h 5 h 10 h 3 h
Hour 9
1
2
5
2

2 h 5 h 10 h 3 h
Hour 10
2
3
1
3
1

2 h 5 h 10 h 3 h
Hour 11
1
4
2
4
2

2 h 5 h 10 h 3 h
Hour 12
2
5
3
5
3

2 h 5 h 10 h 3 h
Hour 13
1
1
4
6
4
1

2 h 5 h 10 h 3 h
Hour 14
2
2
5
7
5
2

2 h 5 h 10 h 3 h
Hour 15
3
1
8
6
3
1
1

2 h 5 h 10 h 3 h
Hour 16
4
2
9
7
4
2
2

2 h 5 h 10 h 3 h
Hour 17
5
3
10
8
5
3

2 h 5 h 10 h 3 h
Hour 18
1
4
9
6
4
1
1

2 h 5 h 10 h 3 h
Hour 19
2
5
10
7
5
2
2

2 h 5 h 10 h 3 h
Hour 20
3
8
6
3
3
1
11

2 h 5 h 10 h 3 h
Hour 21
4
9
7
4
2
2
1
2

2 h 5 h 10 h 3 h
Hour 22
5
10
8
5
3
3
2
3

2 h 5 h 10 h 3 h
Hour 23
4
9
6
4
1
1
1
1

2 h 5 h 10 h 3 h
Hour 24
5
10
7
5
2
2
2
2

2 h 5 h 10 h 3 h
Hour 25
1 1
8
6
3 3
3
1

Notices of simulation with WIP = 8
• This hour 25 is exactly the same as hour 20 so
it starts to go round between hours 20 – 24
• So after the start there is no queue in front of
the second phase. But it’s still fully utilized.
– Bottleneck stations are fully utilized, so this is the
best solution?
• Let’s try this once again with WIP limit of 9
jobs

2 h 5 h 10 h 3 h
Hour 1
1

2 h 5 h 10 h 3 h
Hour 2
2

2 h 5 h 10 h 3 h
Hour 3
1
1

2 h 5 h 10 h 3 h
Hour 4
2
2

2 h 5 h 10 h 3 h
Hour 5
1
3
1

2 h 5 h 10 h 3 h
Hour 6
2
4
2

2 h 5 h 10 h 3 h
Hour 7
1
5
3

2 h 5 h 10 h 3 h
Hour 8
2
1
4
1

2 h 5 h 10 h 3 h
Hour 9
1
2
5
2

2 h 5 h 10 h 3 h
Hour 10
2
3
1
3
1

2 h 5 h 10 h 3 h
Hour 11
1
4
2
4
2

2 h 5 h 10 h 3 h
Hour 12
2
5
3
5
3

2 h 5 h 10 h 3 h
Hour 13
1
1
4
6
4
1

2 h 5 h 10 h 3 h
Hour 14
2
2
5
7
5
2

2 h 5 h 10 h 3 h
Hour 15
3
1
8
6
3
1
1

2 h 5 h 10 h 3 h
Hour 16
4
2
9
7
4
2
2

2 h 5 h 10 h 3 h
Hour 17
5
3
10
8
5
3
1

2 h 5 h 10 h 3 h
Hour 18
1
4
9
6
4
1
1
2

2 h 5 h 10 h 3 h
Hour 19
2
5
10
7
5
2
2

2 h 5 h 10 h 3 h
Hour 20
3
8
6
3
3
1
11

2 h 5 h 10 h 3 h
Hour 21
4
9
7
4
2
2
1
2

2 h 5 h 10 h 3 h
Hour 22
5
10
8
5
3
3
2
3

2 h 5 h 10 h 3 h
Hour 23
4
9
6
4
1
1
1
1

2 h 5 h 10 h 3 h
Hour 24
5
10
7
5
2
2
2
2

2 h 5 h 10 h 3 h
Hour 25
1 1
8
6
3 3
3
1

Notices of simulation with WIP = 9
• Notice that here too hour 25 is same than
hour 20. So it goes same round than with WIP
of 8 jobs
• Now after the start there is always 1 job in
queue waiting to be processed at the second
phase

Variability at the phase 2
• If process time at the second phase would be
shorter they could take a new job from the
queue. On the other hand it would mean that
stations have variability and they would also
have longer process times which would
increase the queue.

Differences
• With WIP limit of 7 second work phase
sometimes starves and is not fully utilized. So the
cycle time for the line is same as theoretical raw
process time T0 = 20 h
• With second example and WIP limit of 8 jobs the
second phase is always fully utilized. And still
there is no queue in front of it. So the cycle time
still remains at 20 hours.
• With the WIP of 9 jobs there is always queue in
front of the second phase. After the start when
all the WIP is in use, jobs will wait 2 hours to get
processed at the phase 2.
– So the cycle time is 22 hours

Summarize
• As calculated in post (http://jesseuitto.fi/?p=464) Critical
WIP is W0 = rb* T0 = 0,4 * 20 = 8 jobs
– In simulations we noticed that line works quite
perfectly with this WIP limit.
• Machines at second phase are fully utilized
• No queues in the line
• Cycle time = raw process time T0
• So if we have perfect line with no variability,
no loss, optimum situation would be to limit
the WIP for 8 jobs
– Achieve maximum TH with minimum WIP and CT