LESSON 8: MATERIAL REQUIREMENTS
PLANNING
Outline
•
•
•
•
•
•
Hierarchy of Production Decisions
MRP and its importance
Input and Output of an MRP system
MRP Calculation
Lot Sizing
Lot Sizing with Capacity Constraint
1
Hierarchy of Production Decisions
• The next slide presents a schematic view of the aggregate
production planning function and its place in the hierarchy
of the production planning decisions.
• Forecasting: First, a firm must forecast demand for
aggregate sales over the planning horizon.
• Aggregate planning: The forecasts provide inputs for
determining aggregate production and workforce levels
over the planning horizon.
• Master production schedule (MPS): Recall, that the
aggregate production plan does not consider any “real”
product but a “fictitious” aggregate product. The MPS
translates the aggregate plan output in terms of specific
production goals by product and time period. For example,
2
Hierarchy of Production Decisions
Forecast of Demand
Aggregate Planning
Master Production Schedule
Inventory Control
Operations Scheduling
Vehicle Routing
3
Hierarchy of Production Decisions
suppose that a firm produces three types of chairs: ladderback chair, kitchen chair and desk chair. The aggregate
production considers a fictitious aggregate unit of chair and
find that the firm should produce 550 units of chairs in April.
The MPS then translates this output in terms of three
product types and four work-weeks in April. The MPS
suggests that the firm produce 200 units of desk chairs in
Week 1, 150 units of ladder-back chair in Week 2, and 200
units of kitchen chairs in Week 3.
• Material Requirements Planning (MRP): A product is
manufactured from some components or subassemblies.
For example a chair may require two back legs, two front
legs, 4 leg supports, etc. While forecasting, aggregate plan
4
Hierarchy of Production Decisions
Master Production Schedule
April
1
Ladder-back chair
2
3
Aggregate
production plan
for chair family
4
5
150
6
7
8
150
200
Kitchen chair
Desk chair
May
200
120
120
200
550
200
790
5
Hierarchy of Production Decisions
and MPS consider the volume of finished products, MRP
plans for the components, and subassemblies. A firm may
obtain the components by in-house production or
purchasing. MRP prepares a plan of in-house production or
purchasing requirements of components and
subassemblies.
• Scheduling: Scheduling allocates resource over times in
order to produce the products. The resources include
workers, machines and tools.
• Vehicle Routing: After the products are produced, the firm
may deliver the products to some other manufacturers, or
warehouses. The vehicle routing allocates vehicles and
prepares a route for each vehicle.
6
Hierarchy of Production Decisions
Materials Requirement Planning
Back
legs
Back slats
Seat cushion
Leg supports
Seat-frame
boards
Front
legs
7
Material Requirements Planning
• The demands for the finished goods are obtained
from forecasting. These demands are called
independent demand.
• The demands for the components or subassemblies
depend on those for the finished goods. These
demands are called dependent demand.
• Material Requirements Planning (MRP) is used for
dependent demand and for both assembly and
manufacturing
• If the finished product is composed of many
components, MRP can be used to optimize the
inventory costs.
8
Importance of an MRP System
• Next two slides explain the importance of an MRP
system. The first one shows inventory levels when an
MRP system is not used. The next one shows the same
when an MRP system is used.
• The chart at the top shows inventory levels of the
finished goods and the chart on the bottom shows the
same of the components.
• If the production is stopped (like it is at the beginning of
the chart), the finished goods inventory level decreases
because of sales. However, the component inventory
level remains unchanged. When the production
resumes, the finished goods inventory level increases,
but the component inventory level decreases.
9
Importance of an MRP System
Inventory
without
an MRP
System
10
Importance of an MRP System
Inventory
with an
MRP
System
11
Importance of an MRP System
• Without an MRP system:
– Component is ordered at time A, when the inventory
level of the component hits reorder point, R
– So, the component is received at time B.
– However, the component is actually needed at time C,
not B. So, the inventory holding cost incurred
between time B and C is a wastage.
• With an MRP system:
– We shall see in this lesson that given the production
schedule of the finished goods and some other
information (see the next slide), it is possible to
predict the exact time, C when the component will be
required. Order is placed carefully so that it is
received at time C.
12
MRP Input and Output
• MRP Inputs:
– Master Production Schedule (MPS): The MPS of the
finished product provides information on the net
requirement of the finished product over time.
– Bill of Materials: For each component, the bill of
materials provides information on the number of units
required, source of the component (purchase/
manufacture), etc. There are two forms of the bill of
materials:
• Product Structure Tree: The finished product is
shown at the top, at level 0. The components
assembled to produce the finished product is
shown at level 1 or below. The sub-components
used to produce the components at level 1 is
13
MRP Input and Output
Master
Production
Schedule
Forecasts
Bill of
Materials
file
MRP
computer
program
Inventory
file
To Production
Reports
Orders
To Purchasing
14
MRP Input and Output
shown at level 2 or below, and so on.
The number in the parentheses shows the
requirement of the item. For example, “G(4)”
implies that 4 units of G is required to produce 1
unit of B.
The levels are important. The net requirements of
the components are computed from the low levels
to high. First, the net requirements of the
components at level 1 is computed, then level 2,
and so on.
15
MRP Input and Output
• Bill of Materials: For each item, the name, number,
source, and lead time of every component required
is shown on the bill of materials in a tabular form.
– Inventory file: For each item, the number of units on
hand is obtained from the inventory file.
• MRP Output:
– Every required item is either produced or purchased.
So, the report is sent to production or purchasing.
16
Bill of Materials: Product Structure Tree
Level 0
Level 1
Level 2
Level 3
17
Bill of Materials
Item
B
C
D
E
BILL OF MATERIALS
Product Description: Ladder-back chair
Item: A
Component
Quantity
Source
Required
Description
Ladder-back
1
Manufacturing
Front legs
2
Purchase
Leg supports
4
Purchase
Seat
1
Manufacturing
18
Bill of Materials
Item
H
I
BILL OF MATERIALS
Product Description: Seat
Item: E
Component
Quantity
Required
Description
Seat frame
1
Seat cushion
1
Source
Manufacturing
Purchase
19
On Hand Inventory and Lead time
Component
Units in
Inventory
Lead
time
(weeks)
Seat
Subassembly
25
2
Seat frame
50
3
Seat frame
boards
75
1
20
MRP Calculation
• Now, the MRP calculation will be demonstrated with an
example.
• Suppose that 150 units of ladder-back chair is required.
• The previous slide shows a product structure tree with
seat subassembly, seat frames, and seat frame boards.
For each of the above components, the previous slide
also shows the number of units on hand.
• The net requirement is computed from top to bottom.
Since 150 units of ladder-back chair is required, and
since 1 unit of seat subassembly is required for each unit
of ladder-back chair, the gross requirement of seatsubassembly is 1501 =150 units. Since there are 25
units of seat-subassembly in the inventory, the net
requirement of the seat-subassembly is 150-25 = 125
21
MRP Calculation
units. Since 1 unit of seat frames is required for each unit
of seat subassembly, the gross requirement of the seat
frames is 1251 = 125 units. (Note that although it
follows from the product structure tree that 1 unit of seat
frames is required for each unit of ladder-back chair, the
gross requirement of seat frames is not 150 units
because each of the 25 units of seat-subassembly also
contains 1 unit of seat frames.) Since there are 50 units
of seat frames in the inventory, the net requirement of
the seat frames is 125-50 = 75 units. The detail
computation is shown in the next two slides.
• A similar logic is used to compute the time of order
placement.
22
MRP Calculation
Quantity of ladder-back chairs to be produced
Gross requirement, seat subassembly
Less seat subassembly in inventory
Net requirement, seat subassembly
Gross requirement, seat frames
Less seat frames in inventory
Net requirement, seat frames
Gross requirement, seat frame boards
Less seat frame boards in inventory
Net requirement, seat frame boards
Units
150
25
50
75
Assume that 150 units of ladder-back chairs are to be produced
at the end of week 15
23
MRP Calculation: Time of Order Placement
Week
Complete order for seat subassembly
Minus lead time for seat subassembly
Place an order for seat subassembl y
Complete order for seat frames
Minus lead time for seat frames
Place an order for seat frames
Complete order for seat frame boards
Minus lead time for seat frame boards
Place an order for seat frame boards
14
2
3
1
Assume that 150 units of ladder-back chairs are to be
produced at the end of week 15 and that there is a one-week
lead time for ladder-back chair assembly
24
MRP Calculation: Some Definitions
• Scheduled Receipts:
– Items ordered prior to the current planning period
and/or
– Items returned from the customer
• Lot-for-lot (L4L)
– Order quantity equals the net requirement
– Sometimes, lot-for-lot policy cannot be used. There
may be restrictions on minimum order quantity or
order quantity may be required to multiples of 50, 100
etc.
25
MRP Calculation
Example 1: Each unit of A is composed of one unit of B, two
units of C, and one unit of D. C is composed of two units of
D and three units of E. Items A, C, D, and E have on-hand
inventories of 20, 10, 20, and 10 units, respectively. Item B
has a scheduled receipt of 10 units in period 1, and C has
a scheduled receipt of 50 units in Period 1. Lot-for-lot (L4L)
is used for Items A and B. Item C requires a minimum lot
size of 50 units. D and E are required to be purchased in
multiples of 100 and 50, respectively. Lead times are one
period for Items A, B, and C, and two periods for Items D
and E. The gross requirements for A are 30 in Period 2, 30
in Period 5, and 40 in Period 8. Find the planned order
releases for all items.
26
MRP Calculation
Level 0
Level 1
Level 2
27
MRP Calculation
Period
1
Item Gross
Requirements
A Scheduled
receipts
LT= On hand from
prior period
Net
requirements
Q= Time-phased Net
Requirements
Planned order
releases
Planned order
delivery
2
3
4
5
6
7
8
9
10
28
MRP Calculation
Period
1 2 3
Item Gross
30
Requirements
A Scheduled
receipts
LT= On hand from
20
prior period
1
Net
WK requirements
Q= Time-phased Net
Requirements
L4L Planned order
releases
Planned order
delivery
All the information above are given.
4
5
30
6
7
8
9
10
40
29
MRP Calculation
Period
1 2 3 4 5 6
Item Gross
30
30
Requirements
A Scheduled
receipts
LT= On hand from
20 20
prior period
1
Net
WK requirements
-Q= Time-phased Net
Requirements
L4L Planned order
releases
Planned order
delivery
20 units are just transferred from Period 1 to 2.
7
8
9
10
40
30
MRP Calculation
Period
1 2 3 4 5 6 7 8 9 10
Item Gross
30
30
40
Requirements
A Scheduled
receipts
LT= On hand from
20 20
prior period
1
Net
WK requirements
-- 10
Q= Time-phased Net
10
Requirements
L4L Planned order
10
releases
Planned order
10
delivery
31
The net requirement of 30-20=10 units must be ordered in week 1.
MRP Calculation
Period
1 2 3 4 5 6
Item Gross
30
30
Requirements
A Scheduled
receipts
LT= On hand from
20 20 0 0 0
prior period
1
Net
WK requirements
-- 10
Q= Time-phased Net
10
Requirements
L4L Planned order
10
releases
Planned order
10
delivery
On hand in week 3 is (20+10)-30=0 unit.
7
8
9
10
40
32
MRP Calculation
Period
1 2 3 4 5 6 7 8 9 10
Item Gross
30
30
40
Requirements
A Scheduled
receipts
LT= On hand from
20 20 0 0 0
prior period
1
Net
30
WK requirements
-- 10
Q= Time-phased Net
30
10
Requirements
L4L Planned order
30
10
releases
Planned order
10
30
delivery
33
The net requirement of 30-0=30 units must be ordered in week 4.
MRP Calculation
Period
1 2 3 4 5 6 7 8 9 10
Item Gross
30
30
40
Requirements
A Scheduled
receipts
LT= On hand from
20 20 0 0 0
0 0 0
prior period
1
Net
30
40
WK requirements
-- 10
Q= Time-phased Net
30
40
10
Requirements
L4L Planned order
30
40
10
releases
Planned order
10
30
40
delivery
34
The net requirement of 40-0=30 units must be ordered in week 7.
MRP Calculation
Period
1 2 3 4 5 6 7 8 9 10
Item Gross
30
30
40
Requirements
A Scheduled
receipts
LT= On hand from
20 20 0 0 0
0 0 0 0 0
prior period
1
Net
30
40
WK requirements
-- 10
Q= Time-phased Net
30
40
10
Requirements
L4L Planned order
30
40
10
releases
Planned order
10
30
40
delivery
35
The net requirement of 40-0=30 units must be ordered in week 7.
MRP Calculation
Period
1
Item Gross
Requirements
B Scheduled
receipts
LT= On hand from
prior period
Net
requirements
Q= Time-phased Net
Requirements
Planned order
releases
Planned order
delivery
2
3
4
5
Exercise
6
7
8
9
10
36
MRP Calculation
Period
1
Item Gross
Requirements
C
Scheduled
receipts
LT= On hand from
prior period
Net
requirements
Q= Time-phased Net
Requirements
Planned order
releases
Planned order
delivery
2
3
4
5
Exercise
6
7
8
9
10
37
MRP Calculation
Period
1
Item Gross
Requirements
D
Scheduled
receipts
LT= On hand from
prior period
Net
requirements
Q= Time-phased Net
Requirements
Planned order
releases
Planned order
delivery
2
3
4
5
Exercise
6
7
8
9
10
38
MRP Calculation
Period
1
Item Gross
Requirements
E
Scheduled
receipts
LT= On hand from
prior period
Net
requirements
Q= Time-phased Net
Requirements
Planned order
releases
Planned order
delivery
2
3
4
5
Exercise
6
7
8
9
10
39
READING AND EXERCISES
Lesson 8
Reading:
Section 7.1 pp. 355-364 (4th Ed.), pp. 346-358 (5th
Ed.)
Exercise:
4 and 9 pp. 364-366 (4th Ed.), pp. 356-358 (5th Ed.)
40
LESSON 9: MATERIAL REQUIREMENTS
PLANNING: LOT SIZING
Outline
• Lot Sizing
• Lot Sizing Methods
– Lot-for-Lot (L4L)
– EOQ
– Silver-Meal Heuristic
– Least Unit Cost (LUC)
– Part Period Balancing
41
Lot-Sizing
• In Lesson 21
– We employ lot for lot ordering policy and order
production as much as it is needed.
– Exception are only the cases in which there are
constraints on the order quantity.
– For example, in one case we assume that at least 50
units must be ordered. In another case we assume that
the order quantity must be a multiple of 50.
• The motivation behind using lot for lot policy is minimizing
inventory. If we order as much as it is needed, there will
be no ending inventory at all!
42
Lot-Sizing
• However, lot for lot policy requires that an order be placed
each period. So, the number of orders and ordering cost
are maximum.
• So, if the ordering cost is significant, one may naturally try
to combine some lots into one in order to reduce the
ordering cost. But then, inventory holding cost increases.
• Therefore, a question is what is the optimal size of the lot?
How many periods will be covered by the first order, the
second order, and so on until all the periods in the
planning horizon are covered. This is the question of lot
sizing. The next slide contains the statement of the lot
sizing problem.
43
Lot-Sizing
• The lot sizing problem is as follows: Given net
requirements of an item over the next T periods, T >0, find
order quantities that minimize the total holding and
ordering costs over T periods.
• Note that this is a case of deterministic demand. However,
the methods learnt in Lessons 11-15 are not appropriate
because
– the demand is not necessarily the same over all
periods and
– the inventory holding cost is only charged on ending
inventory of each period
44
Lot-Sizing
• Although we consider a deterministic model, keep in mind
that in reality the demand is uncertain and subject to
change.
• It has been observed that an optimal solution to the
deterministic model may actually yield higher cost
because of the changes in the demand. Some heuristic
methods give lower cost in the long run.
• If the demand and/or costs change, the optimal solution
may change significantly causing some managerial
problems. The heuristic methods may not require such
changes in the production plan.
• The heuristic methods require fewer computation steps
and are easier to understand.
• In this lesson we shall discuss some heuristic methods.
The optimization method is discussed in the text,
Appendix 7-A, pp 406-410 (not included in the course).45
Lot-Sizing
• Some heuristic methods:
– Lot-for-Lot (L4L):
• Order as much as it is needed.
• L4Lminimizes inventory holding cost, but maximizes
ordering cost.
– EOQ:
• Every time it is required to place an order, lot size
equals EOQ.
• EOQ method may choose an order size that covers
partial demand of a period. For example, suppose
that EOQ is 15 units. If the demand is 12 units in
period 1 and 10 units in period 2, then a lot size of
15 units covers all of period 1 and only (15-12)=3
units of period 2. So, one does not save the
ordering cost of period 2, but carries some 3 units in
46
Lot-Sizing
• Some heuristic methods:
the inventory when that 3 units are required in period 2. This is
not a good idea because if an order size of 12 units is chosen,
one saves on the holding cost without increasing the ordering
cost!
• So, what’s the mistake? Generally, if the order quantity covers
a period partially, one can save on the holding cost without
increasing the ordering cost. The next three methods, SilverMeal heuristic, least unit cost and part period balancing avoid
order quantities that cover a period partially. These methods
always choose an order quantity that covers some K periods,
K >0.
• Be careful when you compute EOQ. Express both holding cost
and demand over the same period. If the holding cost is
annual, use annual demand. If the holding cost is weekly, use
weekly demand.
47
Lot-Sizing
• Some heuristic methods:
– Silver-Meal Heuristic
• As it is discussed in the previous slide, Silver-Meal heuristic
chooses a lot size that equals the demand of some K periods
in future, where K>0.
• If K =1, the lot size equals the demand of the next period.
• If K =2, the lot size equals the demand of the next 2 periods.
• If K =3, the lot size equals the demand of the next 3 periods,
and so on.
• The average holding and ordering cost per period is computed
for each K=1, 2, 3, etc. starting from K=1 and increasing K by
1 until the average cost per period starts increasing. The best
K is the last one up to which the average cost per period
decreases.
48
Lot-Sizing
• Some heuristic methods:
– Least Unit Cost (LUC)
• As it is discussed before, least unit cost heuristic
chooses a lot size that equals the demand of some
K periods in future, where K>0.
• The average holding and ordering cost per unit is
computed for each K=1, 2, 3, etc. starting from K=1
and increasing K by 1 until the average cost per unit
starts increasing. The best K is the last one up to
which the average cost per unit decreases.
• Observe how similar is Silver-Meal heuristic and
least unit cost heuristic. The only difference is that
Silver-Meal heuristic chooses K on the basis of
average cost per period and least unit cost on
average cost per unit.
49
Lot-Sizing
• Some heuristic methods:
– Part Period Balancing
• As it is discussed before, part period balancing
heuristic chooses a lot size that equals the demand
of some K periods in future, where K>0.
• Holding and ordering costs are computed for each
K=1, 2, 3, etc. starting from K=1 and increasing K
by 1 until the holding cost exceeds the ordering
cost. The best K is the one that minimizes the
(absolute) difference between the holding and
ordering costs.
• Note the similarity of this method with the SilverMeal heuristic and least unit cost heuristic. Part
period balancing heuristic chooses K on the basis of
the (absolute) difference between the holding and
ordering costs.
50
Lot-Sizing
• Some important notes
– Inventory costs are computed on the ending inventory.
– L4L minimizes carrying cost
– Silver-Meal Heuristic, LUC and Part Period Balancing
are similar
– Silver-Meal Heuristic and LUC perform best if the costs
change over time
– Part Period Balancing perform best if the costs do not
change over time
– The problem extended to all items is difficult to solve
51
Lot-Sizing
Example 2: The MRP gross requirements for Item A are
shown here for the next 10 weeks. Lead time for A is three
weeks and setup cost is $10. There is a carrying cost of
$0.01 per unit per week. Beginning inventory is 90 units.
Week
1
2
3
4
5
Gross requirements Week
30
6
50
7
10
8
20
9
70
10
Determine the lot sizes.
Gross requirements
80
20
60
200
50
52
Lot-Sizing: Lot-for-Lot
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
53
Use the above table to compute ending inventory of various periods.
Lot-Sizing: Lot-for-Lot
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
20
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
54
Week 4 net requirement = 20 > 0. So, an order is required.
Lot-Sizing: Lot-for-Lot
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
20
Requirements
Planned order
20
Release
Planned
20
Deliveries
Ending
60 10 0
Inventory
55
A delivery of 20 units is planned for the 4th period..
Lot-Sizing: Lot-for-Lot
Exercise
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0 0
Inventory
Net
0 0 0 20 70
Requirements
Time-phased Net
20
Requirements
Planned order
20
Release
Planned
20
Deliveries
Ending
60 10 0 0
Inventory
56
The net requirement of the 5th period is 70 periods.
Lot-Sizing: EOQ
• First, compute EOQ
– Annual demand is not given. Annual demand is
estimated from the known demand of 10 weeks.
Estimated annual demand, 
Total demand over 10 weeks

 52 weeks/yea r
10
30  50  10  20  70  80  20  60  200  50

 52
10
590

 52
10
 3,068 units/year
– Compute annual holding cost per unit
h  $0.01/unit/week  $0.52/unit /year
57
Lot-Sizing: EOQ
• First, compute EOQ
  3,068 units/year
K  $10 /order
h  $0.52 /unit/year
2 K
2  10  3,068
EOQ 

 343.51  344 units
h
0.52
• Therefore, whenever it will be necessary to place an
order, the order size will be 344 units. This will now be
shown in more detail.
58
Lot-Sizing: EOQ
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
59
Use the above table to compute ending inventory of various periods.
Lot-Sizing: EOQ
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net 20
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
60
Week 4 net requirement = 20 > 0. So, an order is required.
Lot-Sizing: EOQ
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net 20
Requirements
Planned order
344
Release
Planned
344
Deliveries
Ending
60 10 0
Inventory
61
Order size = EOQ = 344, whenever it is required to place an order.
Lot-Sizing: EOQ
Exercise
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0 324
Inventory
Net
0 0 0 20
Requirements
Time-phased Net 20
Requirements
Planned order
344
Release
Planned
344
Deliveries
Ending
60 10 0 324
Inventory
62
Week 5 b. inv=344-20=324>70= gross req. So, no order is required.
Lot-Sizing: Silver-Meal-Heuristic
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
Per
H. Ord. Period
Cost Cost Cost
The order is placed for K periods, for some K>0. Use the above table
to find K.
63
Lot-Sizing: Silver-Meal-Heuristic
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
20
Per
H. Ord. Period
Cost Cost Cost
0.00 10 10.0
If K=1, order is placed for 1 week and the order size = 20. Then, the
ending inventory = inventory holding cost =0. The order cost = $10.
64
Average cost per period = (0+10)/1=$10.
Lot-Sizing: Silver-Meal-Heuristic
Exercise
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
20
90 70
Per
H. Ord. Period
Cost Cost Cost
0.00 10 10.0
0.70 10 5.35
If K=2, order is placed for 2 weeks and the order size = 20+70=90.
Then, inventory at the end of week 4 = 90-20=70 and holding cost
65
=70 0.01. = 0.70. Average cost per period = (0.70+10)/2=$5.35.
Lot-Sizing: Silver-Meal-Heuristic
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
66
Use the above table to compute ending inventory of various periods.
Lot-Sizing: Silver-Meal-Heuristic
Exercise
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net 20
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
67
Week 4 net requirement = 20 > 0. So, an order is required.
Lot-Sizing: Least Unit Cost
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
H. Ord. Unit
Cost Cost Cost
The order is placed for K periods, for some K>0. Use the above table
to find K.
68
Lot-Sizing: Least Unit Cost
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
20
H. Ord. Unit
Cost Cost Cost
0.00 10 .500
If K=1, order is placed for 1 week and the order size = 20. Then, the
ending inventory = inventory holding cost =0. The order cost = $10.
69
Average cost per unit = (0+10)/20=$0.50
Lot-Sizing: Least Unit Cost
Exercise
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70 80 20
60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
20
90 70
H. Ord. Unit
Cost Cost Cost
0.00 10 .500
0.70 10 .119
If K=2, order is placed for 2 weeks and the order size = 20+70=90.
Then, inventory at the end of week 4 = 90-20=70 and holding cost
70
=70 0.01. = 0.70. Average cost per unit = (0.70+10)/90=$0.119.
Lot-Sizing: Least Unit Cost
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
71
Use the above table to compute ending inventory of various periods.
Lot-Sizing: Least Unit Cost
Exercise
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net 20
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
72
Week 4 net requirement = 20 > 0. So, an order is required.
Lot-Sizing: Part Period Balancing
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70
80 20 60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
H. Ord.
Cost Cost
The order is placed for K periods, for some K>0. Use the above table
to find K.
73
Diff
Lot-Sizing: Part Period Balancing
j
rj
Order for weeks
1 week, week 4
2 weeks, weeks 4 to 5
3 weeks, weeks 4 to 6
4 weeks, weeks 4 to 7
5 weeks, weeks 4 to 8
6 weeks, weeks 4 to 9
7 weeks, weeks 4 to 10
1
2
3
4
5
6
7
20 70
80 20 60 200 50
Units in the inventory at the end of Week
Q
4
5
6
7
8
9 10
20
90
170
190
250
450
1 week, week 9
200
2 weeks, weeks 9 to 10 250
70
150 80
170 100 20
230 160 80 60
430 360 280 260 200
NOT COMPUTED
50
H. Ord.
Cost Cost
0.00
0.70
2.30
2.90
5.30
15.30
10
10
10
10
10
10
Diff
10.0
9.30
7.70
7.10
4.70
5.30
0.00 10 10.0
0.50 10 9.50
The above computation is similar to that of the Silver-Meal heuristic.
The primary difference is that the (absolute) difference between
74
holding and ordering cost is shown in the last column.
Lot-Sizing: Part Period Balancing
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0
Inventory
Net
0 0 0 20
Requirements
Time-phased Net
Requirements
Planned order
Release
Planned
Deliveries
Ending
60 10 0
Inventory
75
Use the above table to compute ending inventory of various periods.
Lot-Sizing: Part Period Balancing
Period
1 2 3 4 5 6 7 8 9 10
Gross
30 50 10 20 70 80 20 60 200 50
Requirements
Beginning
90 60 10 0 230 160 80 60 0 50
Inventory
200
Net
0 0 0 20
Requirements
Time-phased Net 20
200
Requirements
Planned order
250
250
Release
Planned
250
250
Deliveries
Ending
60 10 0 230 160 80 60 0 50 0
Inventory
The computation is similar to that of the Silver-Meal heuristic. 76
Cost Comparison
• Lot-for-Lot
– See the last slide entitled “lot-sizing: lot-for-lot”
– Number of orders: 7
– Ordering cost = 7  $10/order = $70
– Holding cost = (60+10)  $0.01/unit/week = $0.70
– Total cost = 70+0.70 =$70.70
• EOQ
– See the last slide entitled “lot-sizing: EOQ”
– Number of orders: 2
– Ordering cost = 2  $10/order = $20
– Holding cost = (60 +10 +324 +254 +174 +154 +94
+237 +187)  $0.01/unit/week = $14.94
– Total cost = 20+14.94 =$34.94
77
Cost Comparison
• Silver-Meal Heuristic
– See the last slide entitled “lot-sizing: Silver-Meal
heuristic”
– Number of orders: 2
– Ordering cost = 2  $10/order = $20
– Holding cost = (60 +10 +230 +160 +80 +60 +50) 
$0.01/unit/week = $6.50
– Total cost = 20+6.50 =$26.50
• Least Unit Cost
– See the last slide entitled “lot-sizing: least unit cost”
– Number of orders: 2
– Ordering cost = 2  $10/order = $20
– Holding cost = (60 +10 +430 +360 +280 +260 +200) 
$0.01/unit/week = $16.00
78
– Total cost = 20+16.00 =$36.00
Cost Comparison
• Part-Period Balancing
– See the last slide entitled “lot-sizing: part-period
balancing”
– Number of orders: 2
– Ordering cost = 2  $10/order = $20
– Holding cost = (60 +10 +230 +160 +80 +60 +50) 
$0.01/unit/week = $6.50
– Total cost = 20+6.50 =$26.50
• Conclusion: In this particular case, Silver-Meal heuristic
and part period balancing yield the least total holding and
ordering cost of $26.50 over the planning period of 10
weeks.
79
READING AND EXERCISES
Lesson 9
Reading:
Section 7.2-7.3 pp. 366-375 (4th Ed.), pp. 358-366
(5th Ed.)
Exercise:
17 and 25 pp. 371-373, 375 (4th Ed.), pp. 363, 366
80
LESSON 10: MATERIAL REQUIREMENTS
PLANNING: LOT SIZING WITH CAPACITY
CONSTRAINTS
Outline
• Lot Sizing with Capacity Constraints
– Order Partial Requirements
– Checking Feasibility
– Lot Shifting Technique
– An Improvement Procedure
81
Lot Sizing with Capacity Constraints
• In Lessons 21 and 22 we have assumed that there is
no capacity constraint on production. However, often,
the production capacity is limited.
• In this lesson we assume that it is required to develop
a production plan (i.e., production quantities of
various periods) that minimizes total inventory
holding and ordering costs.
• Capacity constraints make the problem more
realistic.
• At the same time, capacity constraints make the
problem difficult.
82
Lot Sizing with Capacity Constraints
• We shall discuss
– the lot shifting technique, a heuristic procedure
that constructs a production plan, and
– another procedure that improves a given
production plan.
• At times, capacity may be so low that it may not be
possible to meet the demand of all periods. We shall
discuss a procedure to check feasibility.
• First, a property that is new for the problems with
capacity constraints.
83
Order Partial Requirements
• Recall from Lesson 22, that if a lot-sizing solution
includes an order size that covers a period only
partially, then we can reduce the holding cost without
increasing the ordering cost. So, Silver-Meal
heuristic, least unit cost and part-period balancing
consider order sizes that equals demand of K periods
in future, for some K>0.
• As it is shown by the next example, the above does
not hold if there are some capacity constraints. It may
be essential to order partial requirement of a period.
84
Order Partial Requirements
Example 3
Production Requirement
Production Capacity
June
10
12
July
10
8
• Without capacity constraint, June production quantity
must include either all or none of the July production
requirement
– If production is ordered only in June, produce all in
June
– If production is ordered in both June and July,
produce June requirement in June and July
requirement in July
85
Order Partial Requirements
• With capacity constraint, June production quantity
may include a part of the July production requirement
– If production is ordered only in June, produce all in
June
– If production is ordered in both June and July,
June production quantity must include all of the
June requirement and 2 units of the July
requirement
86
Checking Feasibility
• As it was discussed before, sometimes capacity may
be so low that it may not be possible to meet the
demand of all periods.
• So, given demand over the planning horizon and the
corresponding capacity constraints, we ask if there
exists a feasible solution that meets all the demand.
This is the feasibility problem.
• The procedure is stated in the next slide and then the
procedure is illustrated with an example.
87
Checking Feasibility
• Procedure to check feasibility:
– For every period, compute the cumulative
requirement and the cumulative capacity.
• If for every period, the cumulative capacity is
larger than (or equal to) the cumulative
requirement, then there exists a feasible
solution.
• Else, if there is a period in which cumulative
capacity is smaller than the cumulative
requirement, then there will be a shortage in
that period, and, therefore, there is no feasible
solution.
88
Checking Feasibility
Example 4
Production
Requirement
June
10
July
14
August
15
September 16
Production
Capacity
15
11
12
17
Question: Is it possible to meet the production
requirements of all the months?
89
Checking Feasibility
Production
Production
Requirement Cumulative Capacity Cumulative
June
10
10
15
15
July
14
24
11
26
August
15
39
12
38
September 16
55
17
55
Answer: The August requirement cannot be met even
after full production in June, July and August. Hence,
it is not possible to meet the production requirements
of all the months.
90
Lot Shifting Technique
• Lot shifting technique constructs a feasible
production plan, if there exists one or provides a
proof that there is no feasible solution.
• Lot shifting method is a heuristic. The production plan
obtained from the lot shifting technique is not
necessarily optimal. It is possible to improve the
production plan.
• An improvement procedure will be discussed after
the discussion on the lot shifting technique.
91
Lot Shifting Technique
• The lot shifting method repeatedly does the following:
– Find the first period with less capacity.
• If possible, back-shift the excess capacity to
some prior periods. Continue.
• If it is not possible to back-shift the excess
capacity to some prior periods, stop. There is
no feasible solution.
92
Lot Shifting Technique
Example 5
Production
Requirement
June
10
July
14
August
15
September 16
Production
Capacity
30
13
13
17
Question: Find a feasible production plan
93
Lot Shifting Technique
Production
Requirement
June
10
July
14
August
15
September 16
Production
Capacity
30
13 (less capacity)
13
17
Rule: Find the first period with less capacity.
The first period with shortage is July when the
capacity = 13 < 14 = production requirement.
94
Lot Shifting Technique
Production
Requirement
June
10
July
14 13
August
15
September 16
Production
Capacity
30
13 (less capacity)
13
17
July production requirement is 14 units which is 1 unit
more than the capacity of 13 unit. So, this 1 unit must
be produced in some earlier month. There is only one
month before July.
95
Lot Shifting Technique
Production
Requirement
June
10 11
July
14 13
August
15
September 16
Production
Capacity
30 (back-shift)
13 (less capacity)
13
17
Rule: Back-shift the excess requirement to prior periods.
One unit excess demand of July is back-shifted to
June. So, the June production is 10+1=11 units.
96
Lot Shifting Technique
Production
Requirement
June
10 11
July
14 13
August
15 13
September 16
Production
Capacity
30
13
13 (less capacity)
17
Rule: Find the first period with less capacity
97
Lot Shifting Technique
Production
Requirement
June
10 11 13
July
14 13
August
15 13
September 16
Production
Capacity
30 (back-shift)
13
13 (less capacity)
17
Rule: Back-shift the excess requirement to prior periods
98
An Improvement Procedure
• Now, a procedure will be discussed that can
sometimes find an alternative production plan that
may reduce the total holding and ordering cost over
the planning horizon.
• Keep in mind that it is guaranteed that whenever,
there is an improved plan, the procedure will be able
to identify that plan. Sometimes, the procedure may
fail to identify an improved plan, although there may
actually exist one.
99
An Improvement Procedure
Example 6
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity
30
13
13
17
Question: Is it possible to improve the plan if K= $50,
h=$2/unit/month?
100
An Improvement Procedure
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity
30
13
13
17
Improvement procedure: Start from the last period and work
backwards. In each iteration, back-shift all units of the period
under consideration to one or more previous periods if
additional holding cost is less than the ordering cost saved. 101
An Improvement Procedure
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity Excess
30
17
13
0
13
0
17
1
Improvement procedure: Start from the last period and work
backwards. In each iteration, back-shift all units of the period
under consideration to one or more previous periods if
additional holding cost is less than the ordering cost saved. 102
An Improvement Procedure
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity Excess
30
17
13
0
13
0
17
1
Back-shift September production to June?
Additional holding cost = (16)(2)(3) =$96 > 50 = K
So, do not back-shift September production to June.
103
An Improvement Procedure
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity Excess
30
17
13
0
13
0
17
1
Back-shift August production to June?
Additional holding cost = (13)(2)(2) =$52 > 50 = K
So, do not back-shift August production to June.
104
An Improvement Procedure
Production
Requirement
June
13
July
13
August
13
September 16
Production
Capacity Excess
30
17
13
0
13
0
17
1
Back-shift July production to June?
Additional holding cost = (13)(2)(1) =$26 < 50 = K
So, yes, back-shift July production to June.
105
An Improvement Procedure
Production
Requirement
June
26
July
0
August
13
September 16
Production
Capacity Excess
30
4
13
13
13
0
17
1
Final Plan
The above is the result of the improvement procedure.
106
READING AND EXERCISES
Lesson 10
Reading: Section 7.4 pp. 375-379 (4th Ed.), pp. 366-369
(5th Ed.)
Exercise: 28 p. 380 (4th Ed.), p. 369 (5th Ed.)
Lesson 11
Reading: Section 7.6 pp. 387-395 (4th Ed.) pp. 377-384
(5th Ed.)
Exercise: 37 p. 395 (4th Ed.), p. 384 (5th Ed.)
107
LESSON 11: JUST-IN-TIME
Outline
• Just-in-Time (JIT)
• Examples of Waste
• Some Elements of JIT
108
Just-in-time
•
Producing only what is needed and when it is needed
•
A philosophy
•
An integrated management system
109
Just-in-time
• Theme: eliminate all waste including the ones caused by:
– inventory management
– supplier selection
– defective parts
– scheduling of production and delivery
– information system
110
Examples Of Waste
•
•
•
•
•
•
•
•
•
Watching a machine run
Waiting for parts
Counting parts
Overproduction
Moving parts over long
distances
Storing inventory
Looking for tools
Machine breakdown
Rework
111
Some Elements Of JIT
1.
2.
3.
4.
5.
6.
7.
8.
9.
Focused factory networks
Grouped Technology: Cellular layouts
Quality at the source
Flexible resources
Pull production system
Kanban production control
Small-lot production and purchase
Quick setups
Supplier networks
112
Cellular Layouts
•
•
•
•
Group dissimilar machines in a cell to produce a family
of parts
Reduce setup time and transit time
Send work in one direction through the cell (resembling
a small assembly line)
Adjust cycle time by changing worker paths
113
Cellular Layouts
Machines
Enter
Worker 2
Worker 1
Worker
3
Exit
Key:
Product route
Worker route
114
Original Process Layout
Assembly
4
6
7
5
8
2
1
A
10
3
B
9
12
11
C Raw materials
115
Part Routing Matrix
Parts
A
B
C
D
E
F
G
H
1 2 3 4
x x
x
x
x x
x
x
x
x
Machines
5 6 7 8 9 10 11 12
x
x
x
x
x x
x
x
x
x
x x
x
x
x
x
x
x
x x
116
Part Routing Matrix - Reordered
Machines
Parts 1 2 4 3 5 6 7 8 9 10 11 12
A
x x x
x
x
B
x
x
x x
C
x
x
x
D
x x x
x
x
E
x x
x
F
x
x
x
G
x
x
x
x
H
x
x x
117
Part Routing Matrix - Reordered
Machines
Parts 1 2 4 3 5 6 7 8 9 10 11 12
A
x x x
x
x
D
x x x
x
x
B
x
x
x x
C
x
x
x
E
x x
x
F
x
x
x
G
x
x
x
x
H
x
x x
118
Part Routing Matrix - Reordered
Parts
A
D
B
C
E
F
G
H
1
x
x
x
Machines
2 4 8 3 5 6 7 9 10 11 12
x x x
x
x x x
x
x
x
x x
x
x
x
x x
x
x x
x
x
x
x
x
x x
119
Part Routing Matrix - Reordered
Parts
A
D
F
B
C
E
G
H
1 2 4 8
x x x x
x x x x
x
x x
Machines
3 5 6 7 9 10 11 12
x
x
x
x
x
x
x x
x
x
x
x
x
x
x
x
x
x
x
120
Part Routing Matrix - Reordered
Parts
A
D
F
B
C
E
G
H
1
x
x
x
Machines
2 4 8 10 3 5 6 7 9 11 12
x x x x
x x x x
x x
x
x
x x
x
x
x
x x
x
x
x
x
x
x
x x
121
Part Routing Matrix - Reordered
Parts
A
D
F
C
G
B
E
H
1
x
x
x
Machines
2 4 8 10 3 6 9 5 7 11 12
x x x x
x x x x
x x
x x x
x x x
x
x x x x
x
x
x
x x x
122
Cellular Layout Solution
Assembly
8
10
9
12
11
4
Cell 2 6
Cell1
Cell 3
7
2
1
Raw materials
3
A
C
5
B
123
Advantages of Cellular Layouts
• Reduced material handling and transit time
• Reduced setup time
• Reduced work-in-process inventory
• Better use of human resources
• Easier to control
• Easier to automate
124
Disadvantages of Cellular Layouts
• Inadequate part families
• Poorly balanced cells
• Expanded training and scheduling of workers
• Increased capital investment
125
Quality At The Source
•
•
•
•
•
Jidoka is the authority to stop a production line
Andon lights signal quality problems
Undercapacity scheduling allows for planning,
problem solving & maintenance
Visual control makes problems visible
Poka-yoke prevents defects
126
Kaizen
•
•
•
Continuous improvement
Requires total employment involvement
The essence of JIT is the willingness of workers to
• spot quality problems,
• halt production when necessary,
• generate ideas for improvement,
• analyze problems, and
• perform different functions
127
Flexible Resources
•
Multifunctional workers
•
General purpose machines
•
Study operators & improve operations
128
Flexible Resources
129
Flexible Resources
Cell 1
Cell 2
Worker 2
Worker 1
Worker 3
Cell 3
Cell 4
Cell 5
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Pull Production System
•
•
•
In a push system, a schedule is prepared in advance
and as soon as one process completes its work, its
products are sent to the next process.
In a pull system, workers take the parts or materials
from the preceding stations as needed.Workers at the
preceding stations may produce the next unit only after
their outputs are taken by the workers in the
subsequent processes.
Although the concept of pull production seems simple,
it can be difficult to implement. Kanbans are introduced
to implement the pull system.
131
Kanban Production Control
Part no.:
7412
Description: Slip rings
From :
Machining
M-2
Box capacity
25
Box Type
A
Issue No.
To:
Assembly
A-4
3/5
•
A kanban is a card that indicates quantity of production
•
Kanbans maintain the discipline of pull production
– - A production kanban authorizes production
– - A withdrawal kanban authorizes the movement of
132
goods
The Origin Of Kanban
a. Two-bin inventory system
Bin 1
Reorder
Card
b. Kanban Inventory System
Bin 2
Kanban
Q-R
R
Q = order quantity
R = reorder point
= demand during lead time
133
A Single-Kanban System
134
A Single-Card Kanban System
Consider the fabrication cell that feeds two assembly lines.
1. As an assembly line needs more parts, the kanban card
for those parts is taken to the receiving post and a full
container of parts is removed from the storage area.
2. The receiving post accumulates cards for both assembly
lines and sequences the production of replenishment
parts.
135
A Dual
Kanban
System
136
A Two-Card System
1. When the number of tickets on the withdrawal kanban
reaches a predetermined level, a worker takes these
tickets to the store location.
2. The workers compares the part number on the production
ordering kanban at the store with the part number on the
withdrawal kanban.
3. The worker removes the production ordering kanban from
the containers, places them on the production ordering
kanban post, and places the withdrawal kanbans in the
containers.
4. When a specified number of production ordering kanbans
have accumulated, work center 1 proceeds with
production.
137
5. The worker transports parts picked up at the store to work
center 2 and places them in a holding area until they are
required for production.
6. When the parts enter production at work center 2, the
worker removes the withdrawal kanbans and places them
on the withdrawal kanban post.
138
Kanban Squares
X
X
X
X
X
X
Flow of work
Flow of information
Kanban Racks
407
409
410
408
411
412
Signal Kanban
407
408
409
407
408
409
Kanban Post Office
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
Types Of Kanbans
•
•
•
•
Kanban Square
– marked area designed to hold items
Signal Kanban
– triangular kanban used to signal production at the
previous workstation
Material Kanban
– used to order material in advance of a process
Supplier Kanban
– rotates between the factory and supplier
Determining Number Of Kanbans
avg.demand during lead time + safety stock
No. of kanbans =
container size
DL  w
y
a
where
– y = number of kanbans or containers
– = average demand over some time period
D
– L = lead time to produce parts
– w = safety stock, usually 10% of the demand during
lead time
– a = container size
Kanban Calculation Example
Problem statement:
D = 150 bottles per hour
DL = (150)(0.5) = 75
a = 25 bottles
L = 30 minutes = 0.5 hours
w = 10% of DL
Solution:
DL  w 75  (0.1)( 75)
y

 3.3 Kanban containers
a
25
Round up to 4 (allow some slack)
or down to 3 (force improvement)
Small-Lot Production
•
Requires less space & capital investment
•
Moves processes closer together
•
Makes quality problems easier to detect
•
Makes processes more dependent on each other
146
Small-lot
Production
and Purchase
147
Reducing Setup Time
•
•
•
•
•
•
Preset desired
settings
Use quick fasteners
Use locator pins
Prevent
misalignments
Eliminate tools
Make movements
easier
148
Trends In Supplier Policies
1. Locate near to the customer
2. Use small, side loaded trucks and ship mixed loads
3. Consider establishing small warehouses near to the
customer or consolidating warehouses with other suppliers
4. Use standardized containers and make deliveries according
to a precise delivery schedule
5. Become a certified supplier and accept payment at regular
intervals rather than upon delivery
149
Benefits Of JIT
1.
2.
3.
4.
Reduced inventory
Improved quality
Lower costs
Reduced space
requirements
5. Shorter lead time
6. Increased productivity
7.
8.
Greater flexibility
Better relations with
suppliers
9. Simplified scheduling
and control activities
10. Increased capacity
11. Better use of human
resources
12. More product variety
150
READING AND EXERCISES
Lesson 11
Reading: Section 7.6 pp. 387-395 (4th Ed.) pp. 377-384
(5th Ed.)
Exercise: 37 p. 395 (4th Ed.), p. 384 (5th Ed.)
151