Chapters 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.

1. Chapters 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

2. 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

3. 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

4. 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

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

6. 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

7. 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.

8. 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 SequenceJob Work (Processing) TimeFlow TimeJob Due
DateJob LatenessB 2 2 6 0A 6 8 8 0D 3
11 15 0C 8 19 18 1E 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 SequenceJob Work (Processing) TimeFlow TimeJob
Due DateJob LatenessB 2 2 6 0A 6 8 8
0D 3 11 15 0C 8 19 18 1E 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 SequenceJob Work (Processing)
TimeFlow TimeJob Due DateJob LatenessB 2 2 6
0A 6 8 8 0D 3 11 15 0C 8 19 18
1E 9 28 23 5 28 68 6

9. 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)

10. 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.

11. 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.

12. 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

13. 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 =
=

14. =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

15. 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
Chapters 15 & 16 Homework (SCM 488)
1. 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 _______________
2. Finite loading depends on knowing capacity and actual work
center performance to achieve Input-Output Control. Please
complete the below Input-Output Chart. What problems are
apparent to you?

16. _____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
______________________
week
1
2
3
4
planned input
200
200
200
200
actual input
180
190
170
180
cumulative deviation
-20
?
?
?
planned output
200
200
200
200

17. actual output
180
170
160
150
cummulative deviation
-20
?
?
?
cumulative change in backlog
0
?
?
?
3. Gantt Load & Schedule Charts are effective ways to show
loading, idle times & jobs in process. Based on the below,
when does job A need to finish_____________? What days of
the week could we do maintenance in the electronics work
center.
4. The assignment method allocates tasks to minimize cost (or
time). You have 3 operators who can each do 3 different jobs.
Which jobs should be assigned to which operators?
Operator 1
Operator 2
Operator 3

18. Job A
10
12
4
Job B
12
4
9
Job C
8
8
5
5. You are managing a production shop which involves 4
operations. You decide to sequence the jobs by due. What is
the average completion time__________,
utilization_____________, average # of jobs in the
system____________,and average job lateness_____________.
Job
Process time (days)
Job Due Date (Days)
1
3
6
2
5
5
3
2
18
4

19. 9
7
6. For the above table, your production manager has a different
idea. She’d like to sequence by shortest processing time. What
is the average completion time__________,
utilization_____________, average # of jobs in the
system____________,and average job lateness_____________.
7. For the two methods, which performed best for
utilization?_________________________
8. Today is the 7th. You are a plant manager in a plant making
parts for an automotive customer. If your parts are late you are
charged a financial penalty. For your plant, jobs are scheduled
to minimizing average job lateness using a critical ratio (CR).
Job
Date Due
Days needed to fill order
1
15
2
2
19
5
3
13
3

20. What is the CR of each job. Which needs to be done 1st?
9. You would like to use a kanban system. Wait time + material
handling time + process time is 7 days. Daily demand is 500
units and your EOQ is 250 units. How many bins do you need?
10. You are value stream mapping the following 3 step
production process. What is the time needed for each step to
produce 1000 pieces? Using total production time, process time
and available work time what is the value added time
%______________?