Lean Systems, Process & Operations class need within next 6 hrs
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Chapters15_161.pptxChapters 15 & 16
Short Term Scheduling & Kanban
The 3-18 month aggregate plan is used to make the weekly master production plan.
The short term schedule assigns work to specific work centers.
Scheduling goals are:
Minimize completion time
Maximize equipment utilization
Minimize work-in-process (WIP) inventory
In manufacturing, we typically schedule the final operation 1st to meet the due date. Prior steps in the sequence of work (i.e. routing) are scheduled in reverse order.
In finite loading jobs are assigns to work centers only up to their capacity (customer due dates may need to be pushed out).
Finite loading depends on knowing capacity and actual work center performance to achieve Input-Output Control
Below is the input portion of an Input-Output Chart. We were supposed to receive 280 parts but actually only received 270. So our cumulative input deviation to plan is -10 pieces at the end of week 6/6.
The following week (6/13) we are supposed to receive 280 parts but only 250 show up. So, the cumulative input deviation from plan by week 6/13 is -40 (i.e. the 10 we were short in 6/6 week plus the 30 we were short in 6/13 week)
Of the 270 we actually receive, we output 270 (0 cumulative change in backlog)
Now lets look at the output portion of the chart. The plan was to make 320 parts. We only made 270 parts. So our cumulative output deviation to plan is -50 pieces at the end of week 6/6.
The following week (6/13) we are supposed to make 320 parts but only 270 were made. So, the cumulative output deviation from plan by week 6/13 is -100 (i.e. the 50 we were short in 6/6 week plus the 50 we were short in 6/13 week)
Now, let’s put the input and output charts together. In week 6/6 actual input (270 parts) matched out actually output (270), so the cumulative change in backlog was 0.
In week 6/13 actual input (250 parts) was exceeded by actually output (270), so the cumulative change in backlog reduced by -20.
In week 6/20, we received 280 parts and only output 270 parts. So our backlog increased +10. Considering our backlog was -20, the cumulative change -20..+10 is now -10
It is apparent from our Input-Output Chart,
Incoming delivery is supposed to be 280 every week but is actually sporadic (e.g. 250 to 285)
We consistently output 270 – but it’s not the 320 we always planned
One way to improve input- output performance is by using Gantt Load & Schedule Charts. These show loading, idle times & jobs in process
The assignment method allocates tasks to minimize cost (or time)
In this example, we have 3 typesetters (A-C) who can each do 3 different jobs (R-34, S-66 and T-50). Which way is the least costly?
Step 1a: Subtract the smallest number from each row
Step 1b: Subtract the smallest number from each column
-6 from 11,14 & 6
-8 from 8,10 & 11
-7 from 9,12 & 7
-0 from 5,0 & 2
-2 from 8,2 & 5
-0 from 0,3 & 0
3+2=5
Because three lines are needed, the solution is optimal and assignments can be made
A B C
Job
R-34 $ 3 $ 4 $ 0
S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
Assignments
A B C
Job
R-34 $ 3 $ 4 $ 0
S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
Start by assigning R-34 to worker C as the this is the only “0” value for that job.
Since worker C cannot take T-50, the only other “0” value is for worker A
This leaves S-66 to be done by worker B.
From the original cost table
Minimum cost = $6 + $10 + $9 = $25
Step 4 - Assignments
A B C
Job
R-34 $ 3 $ 4 $ 0
S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
A B C
Job
R-34 $11 $14 $ 6
S-66 $ 8 $10 $11
T-50 $ 9 $12 $ 7
Typesetter
Now that we know what jobs will be done in which work centers, the next task is sequencing the jobs.
In manufacturing we typically sequence by due date (earliest first) or processing time (longest or shortest first)
Let’s look at an example:
Sequencing by end due date (EDD)
| Job Sequence | Job Work (Processing) Time | Flow Time | Job Due Date | Job Lateness |
| B | 2 | 2 | 6 | 0 |
| A | 6 | 8 | 8 | 0 |
| D | 3 | 11 | 15 | 0 |
| C | 8 | 19 | 18 | 1 |
| E | 9 | 28 | 23 | 5 |
| 28 | 68 | 6 |
Flow time is the amount of time a unit spends in the process. We can see it took 2 days to complete Job B; then another 6 days to complete job A. So on the 8th day (6+2) job A was done
| Job Sequence | Job Work (Processing) Time | Flow Time | Job Due Date | Job Lateness |
| B | 2 | 2 | 6 | 0 |
| A | 6 | 8 | 8 | 0 |
| D | 3 | 11 | 15 | 0 |
| C | 8 | 19 | 18 | 1 |
| E | 9 | 28 | 23 | 5 |
| 28 | 68 | 6 |
On the 11th day until job D, and so on. So the total processing time is 28 days & total flow time for all jobs to get through the work cell are 68 days.
| Job Sequence | Job Work (Processing) Time | Flow Time | Job Due Date | Job Lateness |
| B | 2 | 2 | 6 | 0 |
| A | 6 | 8 | 8 | 0 |
| D | 3 | 11 | 15 | 0 |
| C | 8 | 19 | 18 | 1 |
| E | 9 | 28 | 23 | 5 |
| 28 | 68 | 6 |
How effective is this sequencing method?
Another scheduling option is sequencing work by shortest processing time (SPT)
How effective is this sequencing method?
Our scheduler might go another way and sequence jobs by longest processing time (LPT)
How effective is this sequencing method?
Which sequencing method performs best depends on which criteria are most important. In our example:
Average completion time (SPT was best)
Utilization (SPT was best)
# of jobs in the system (LPT was best)
Average job lateness (EDD was best)
In regards to minimizing average job lateness a critical ratio (CR) could be used.
Jobs with lower critical ratios are scheduled first.
CR = =
Due date - Today’s date
Work (lead) time remaining
Time remaining
Workdays remaining
In all these examples schedulers are pushing what should be made in what sequence….
These MRP based systems rely on pushing orders
The downside is work may be moving in larger (or smaller) quantities before (or after) downstream operations need them.
In the 1980’s Toyota developed a pull system called Just-in-Time (JIT) manufacturing.
JIT is designed to produce the exact amount of a product at the exact time downstream operations need it.
JIT uses Kanban cards (or bins) to pull orders.
How?
1-bin (or 1-card) System
An empty bin in a designated location indicates to an upstream operation it needs to be refilled (which schedules the upstream process to begin manufacturing).
2-bin (or card) System
Two bins are full of an item. When the first bin is empty, the empty bin (or a kanban card in the bin) schedules the upstream process to begin the order.
When the new supplies arrive, the newly filled bin is placed behind the bin currently in use.
A 3-bin (or card) System
One bin is placed at the factory where products are made, one in the warehouse (store) where raw materials are held, and one at the raw material supplier.
When the factory runs out of raw materials, it sends its empty bin to the warehouse to be refilled. The warehouse sends its newly emptied bin to the supplier. The supplier then sends a full bin to the store.
The bins (or cards) serve as the scheduling signal that downstream parts of the process need more stock.
A 2-card system
When the customer process (Work Centre B) needs product, it sends a withdrawal kanban card (W) to the upstream process’ storage area (supermarket). The material handler pulls parts out of storage and provides to Work Centre B.
At Work Centre B, the operator removes a withdrawal kanban (W) when using the first item in a container.
This withdrawal kanban card goes in a nearby collection box and is picked up by a material handler when Work Center B needs more parts.
The material handler then returns to the upstream supermarket again; the withdrawal kanban is placed on a new container of parts for delivery to the downstream process.
When withdrawing the container the material handler removes the container’s Production Kanban (P) and places it in a collection box.
The material handler serving the upstream process returns this production kanban to the upstream process, where it signals the need to produce one additional container of parts.
As long as no parts are produced or moved in the absence of a kanban, a true pull system is maintained.
In the simplest situation, a card corresponds to one container of parts, which the upstream process makes for the supermarket ahead of the next downstream process.
In large batch situations—for example, a stamping press with very short cycle times and long changeover times—a signal kanban (often called a triangle kanban) is used when a minimum quantity of containers is reached.
In this example, parts in the #409 location have a signal Kanban in place indicating upstream production is needed.
After 2 containers of #407 are pulled from the supermarket its production Kanban will be displayed.
Let’s look at an example to determine how many bins (inventory) are needed?
Lead time = Wait time + material handling time + process time = 2 days
Daily demand=500 cakes
Safety stock = ½ day
EOQ = 250 cakes
# bins =
=
=5 bins
The benefits of this type of system are it can reduce overproduction, eliminate waiting time between processes, and reduce inventory of materials.
We can use value stream mapping to calculate a ratio of process time to lead time to see how customer-effective is the process.
Define inventory points & process data to be collected at each step
In our example, process data for each step:
C/T = cycle time for a machine to make a part
C/O = change over time to set up a machine
Uptime % = ratio of actual hrs running production to hrs work time available
Work time available (1 shift which nets to 25,200 sec)
In our example production lead time (5+3.5+5.2+2.3) is 16 days
While process time is (3+22+35) is 60 seconds
A low value added time (<5%) indicates opportunities to reduce waste
Lean manufacturing tries to eliminate waste (which reduces lead time and lowers cost)
Over-production
Unnecessary transportation
Extra motion
Waiting
Over-processing
Inventory
Under utilization
