Kanban Calculation July 7, 2006
Posted by Lawrence Loucka in : Lean, Lean Sigma, Supply Chain , trackback
"That’s not the formula I use" was the start of the debate. So looking at a few refernce books and searching the web here’s what I’ve come up with. How can something so simple have so many variations? Let us count the ways …
1. No. of kanban = (DD*LT+SS*SQRT(LT/TB))/KB+(DD*EPEI)/KB
- Where: DD = Daily demand (units)
- LT = Replenishment leadtime (days)
- SS = Statistically calculated safety stock (units)
- SQRT = Square root
- TB = Time bucket of the safety stock data points (days)
- KB = Quantity per kanban (units)
- EPEI = Supplier’s replenishment interval (days)
2. #KB = (DD*(LT+SS))/KBS +1
- #KB = Number of Kanbans
- DD = Daily Demand
- LT = Lead Time
- SS = Safety Stock
- KBS = Kanban Size
3. Total Req’d Inventory = (Average period demand * Replenishment time) + 1 or 2sigma + safety stock
4. Total Req’d Inventory = (Average period demand * Replenishment time) * 1.X {where X= 20-40%} and the # of bins = TRI / container or bin size
5. # Kanban = ((AD * RT) + (SF * SD))/SCQ
- AD = average period demand
- RT = replenishment time (in the same time bucket as AD)
- SF = the Z factor, typically 1.28 for 95% (sic, should be 1.645 for 95%)
- SD = demand standard deviation
- SCQ = the standard container quantity
6. # Kanban = (average demand during lead tiime + safety stock) / container quantity 6. N = (dL + S)/C
- N = number of kanban
- d = average demand per hour
- L = lead time in hrs
- S = safety
- C = container quantity
7. K=((RT * AC)/Cont) * (SF + C)
- K = number of kanban
- Cont = contents per kanban
- RT = replenishment lead time per kanban
- AC = average consumption per time period
- SF = safety factor
- C = constant, default = 1
Got any more?
Comments»
I’ve posted a response on this topic on my blog, http://www.leanblog.org or
http://kanban.blogspot.com/2006/08/its-not-formula-that-matters.html
i have a problem with kanban calculation. i have no idea how to figure out the correct solution the following question. if you able to answer this questions, your help will be highly appreaciate. i am the beginner of learning kanban calculation. hope can get your reply soon. your help will be highly appreciate.
.
1. In a lean manufacturing environment a ‘pull’ system is used for material control. This means splitting the day’s production into small transfer batches each controlled with a kanban card.
Consider the time required to process 150 units of a product through four sequential processes with each process taking 2 minutes - use the situation to list and explain the advantages of small transfer batches.
2.
A Kanban system is used to control the manufacture of valves in a feeder cell for a main pump assembly line.
• 2 valves are required to assemble each pump.
• 1850 Pumps are manufactured each day.
• Valve processing time is 0.001days/valve.
• The total authorised inventory for the cell is 5060 and each container holds 253 spools.
• On average a Kanban spends 0.65 days queuing and 0.12 days moving around the system.
How many kanbans are used in the cell?
What factor is being applied to control safety stock in the system?
1. If 150 units are processed in a batch it will take 300 minutes to be worked in the first process, then 300 minutes in the second and so on. The time to process all 150 units through four processes will be 1200 minutes.
If the batch size is one, then it will take eight minutes for the first piece to be processed through all four operations. The last piece will take 308 minutes.
2. Problem states pumps, values, and spools. 2 values per pump. How are spools related to values or pumps?
#K = (AD * R)/C = ((1850 pumps/day * 2 values/pump) * (0.65 days 0.12 days))/253 = 11.26, round up = 12
Safety might be considered the 0.65 days waiting in queue. Maybe don’t need so much.
Thanks Mr Loucka. You information has been very helpful.
In formula 5.
SF = the Z factor, typically 1.28 for 95% <<< I think it should read 90% instead of 95%.
It would be 1.65 times the SD for 95%, ie Z = 1.65 for 95%
and 2.33 for 98% and so on…
Pls correct me if I’m wrong.
Ren, right you are. When looking at customer demand we are interested in protecting from unusually high demand. If demand is less than the average we should have enough stock or in-process to cover. It’s when demand spikes that things get interesting. So we use the one sided normal probability distribution. 95% coverege is the then 50% below the mean and 45% above the mean. So the z-value for 0.05% is 1.645.