Overslaan en naar de inhoud gaan

Hi David,

Thank you so much for reaching out to me and look into my numbers in such detail.

a) The maximum stock that a FIFO lane can hold is a choice that you can make. There are some ideas for minimal buffer sizes that you can calculate, which I will copy below.
The goal is, that the size of the FIFO is big enough to prevent the bottleneck from slowing down, but small enough to minimize waiting time in the overall production lead time. So this depends on the situation of the FIFO and whether the bottleneck is in front of it, behind it, or not at all next to it.

For instance situation 1:

- when you have a fifolane between machine A and machine B, 
- machine A is the bottlenek
- Machine B has a C/O time of one hour
You want your FIFO lane to be longer than one hour, so that machine A -the bottleneck- never has to idle

For instance situation 2:
- if you have a fifolane between machine A and machine B
- Machine B is the bottleneck
- Machine A has a delivery time of 24 hours for raw materials
You could decide the FIFO lane to contain at least 24 hours of buffer to make sure the machine does not starve when the delivery is late/cancelled

 

About the 333 in the FIFO lane... In my first draft of this example for future state VSM, I had a planned cycle time of 10.8 seconds because I had not included the difference between Takt time and Planned cycle time yet.

Since Molding was the bottleneck, and the FIFO lane was not linked to the bottleneck, I got to choose a random size of the buffer for this example. I chose a buffer of one hour, which meant 3600 seconds / 10.8 planned cycle time = 333 products in the buffer.

So, continuing this same thinking, in the current example in which I did include the difference between the planned cycle time (10s) and te takt time (10.8s), the correct number should be 3600/10 = 360 parts for a one hour of buffer.

Thank you for helping me find this mistake.  I have to correct this in both my blog and my book. :)

 

b) My thinking on the yield problem is as follows: if step Assembly 1 has 80% (so 20% is taken out) and Assembly 2 has 80% of the second part of the process (20% taken out of the remaining 80%),  the total yield of both assemblies together is 0.8 * 0.8 = 64%

If we want to combine the two steps together, the total assembly yield has to go up from 64% to about 75% as you write. we therefore have an almost infinite number of options again

option 1:  .95 *.80 = >.75
option 2:  .80 * .95 = > .75
option 3:  .90 * .90 = > 0.75

I simply chose the third option, starting a project to get both parts of assembly up to 90%, without writing it down in the article (which maybe I should).

 

Thijs