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Lean Systems

Chapter 6

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What is a Lean System?

Lean Systems

Operations systems that maximize the value added by each of a company’s activities by removing waste and delays from them.

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Continuous Improvement Using a Lean Systems Approach

Just-in-time (JIT) philosophy

The belief that waste can be eliminated by cutting unnecessary capacity or inventory and removing non-value-added activities in operations.

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Eight Types of Waste or Muda

Overproduction
Inappropriate Processing
Waiting
Transportation

Motion
Inventory
Defects
Underutilization of Employees

Table 6.1

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Continuous Improvement �with Lean Systems

Figure 6.1

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Supply Chain Considerations �in Lean Systems

Close Supplier Ties

Small Lot Sizes

Single-digit setup

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Process Considerations �in Lean Systems

Pull Method of Workflow (LEAN)

A method in which customer demand activates the production of the service or item.

Push Method of Workflow (NOT LEAN)

A method in which production of the item begins in advance of customer needs.

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Process Considerations �in Lean Systems

Quality at the Source

Jidoka

Automatically stopping the process when something is wrong and then fixing the problems on the line itself as they occur.

Poka-Yoke

Mistake-proofing methods aimed at designing fail-safe systems that minimize human error.

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Process Considerations �in Lean Systems

Uniform Workstation Loads

Takt time
Heijunka
Mixed-model assembly

Standardized Components and Work Methods

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Process Considerations �in Lean Systems

Flexible Workforce

Automation

5S

Total Preventative Maintenance

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5S

Table 6.2

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Toyota Production System

All work must be completely specified as to content, sequence, timing, and outcome.
All customer-supplier connections should be direct and unambiguous.
All pathways should be simple and direct.
All improvements should be made under the guidance of a teacher using the scientific method.

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House of Toyota

Figure 6.3

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One-Worker, Multiple Machines

Figure 6.4

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Group Technology

Jumbled Flows without GT

Lines Flows with 3 GT cells

Figure 6.5

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What is a Value Stream Mapping?

Value Stream Mapping

A widely used qualitative lean tool aimed at eliminating waste or muda.

Figure 6.6

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VSM Icons

Figure 6.7

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VSM Metrics

Takt Time

Daily Availability/Daily Demand

Cycle Time
Setup Time
Per Unit Processing Time

Cycle Time + Setup Time

Capacity

Availability/Time at bottleneck

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Example 6.1

Jensen Bearings, Inc makes two types of retainers that are packaged and shipped in returnable trays with 40 retainers in each tray. The operations data is on the following slides.

Create a VSM for Jensen Bearings
What is the takt time?
What is the lead time at each cell?
What is the total processing time?
What is the capacity?

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Example 6.1

Table 6.3

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Example 6.1

Table 6.3

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Example 6.1

a.

Figure 6.8

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Example 6.1

Daily Demand

[(1,000 + 2,200) pieces /week]/5 days =

640 pieces per day

Daily Availability

(7 hours/day) x (3,600 seconds per hour) =

25,200 seconds per day

Takt Time = Daily availability/Daily Demand =

25,200/640 =

39.375 seconds per piece

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Example 6.1

Production Lead time = Inventory/Daily Demand

Raw Material Lead Time - 5 days

WIP between Press and Pierce/Form =

(2,250/640) = 3.5 days

WIP between Pierce/Form and Finish/Grind =

(3,350/640) = 5.2 days

WIP between Finish/Grind and Shipping=

(1,475/640) = 2.3 days

Total Production Lead Time =

(5 + 3.5 + 5.2 + 2.3) = 16 days

d. Total Processing Time = Sum of the Cycle Times

(12 + 34 + 35) = 81 seconds

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Example 6.1

e.

Pierce and Form is the bottleneck

Capacity = 25,200/38.5 = 654 units/day

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Application 6.1

Gilman Inc. makes vending machines. The operations data is on the following slides.

What is the cell’s current inventory level?
What is the takt time?
What is the lead time at each cell?
What is the total processing time?
What is the capacity?

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Application 6.1

Overall Process Attributes	Average demand: 200/day Batch size: 20 Number of shifts per day: 2 Availability: 8 hours per shift with a 45-minute break

Customer Shipments	One shipment of 1,000 units each week
Information Flow	All communications with the customer are electronic There is a weekly order release to Cutting All material is pushed

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Application 6.1

Processing Step 1	Cut	Cycle time = 160 seconds Setup time = 3 minutes Up time = 100% Operators = 1 WIP = 600 units (Before Cut)
Processing Step 2	Grind	Cycle time = 120 seconds Setup time = 1 minute Up time = 99% Operators = 1 WIP = 800 units (Before Grind)
Processing Step 3	Bend	Cycle time = 240 seconds Setup time = none Up time = 100% Operators = 1 WIP = 400 units (Before Bend) WIP = 600 units (After Bend)

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Application 6.1

Current Inventory Level

(600 + 800 + 400 + 600) = 2400 units

b. Daily Demand

200 units per day

Daily Availability

(8 hours/day x 60 min) – 45 minutes = 435 min

435 min x 2 shifts/day=

870 minutes per day

Takt Time = Daily availability/Daily Demand = 870/200 =

4.35 minutes per unit

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Application 6.1

Production Lead time = Inventory/Daily Demand

Raw Material Lead Time - (600/200) = 3 days

WIP between Cut and Grind =

(800/200)= 4 days

WIP between Grind and Bend =

(400/200) = 2 days

Finished Goods Lead Time after Bend =

(600/200) = 3 days

Total Production Lead Time = (3 + 4 + 2 + 3) = 12 days

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Application 6.1

d. Total Processing Time = Sum of the Cycle Times

(160 + 120 + 240) = 520 seconds

e. Bending is the bottleneck

Availability at Bending = 870 min/day

Time at bottleneck = (240 + 0)/240 sec/unit = 4 min/unit

Capacity = 870/4 = 217.5 units/day

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What is a Kanban?

Kanban

A Japanese word meaning “card” or “visible record” that refers to cards used to control the flow of production through a factory

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The Kanban System

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Storage area

Empty containers

Full containers

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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The Kanban System

Storage area

Empty containers

Full containers

Receiving post

Kanban card for product 1

Kanban card for product 2

Fabrication cell

O₁

O₂

O₃

O₂

Assembly line 1

Assembly line 2

Figure 6.9

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General Operating Rules

Each container must have a card.
Assembly always withdraws from fabrication (pull system).
Containers cannot be moved without a kanban.
Containers should contain the same number of parts.
Only good parts are passed along.
Production should not exceed authorization.

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Determining the �Number of Containers

Two determinations

Number of units to be held by each container
Number of containers

Little’s Law

Average work-in-process inventory equals the average demand rate multiplied by the average time a unit spends in the manufacturing process

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Determining the �Number of Containers

Work in Process (WIP) =

(average demand rate) × (average time a container spends

in the manufacturing process) + safety stock

WIP = kc

k =

where

k = number of containers

d = expected daily demand for the part

w = average waiting time

p = average processing time

c = number of units in each container

α = policy variable

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Example 6.2

The Westerville Auto Parts Company produces rocker-arm assemblies
A container of parts spends 0.02 day in processing and 0.08 day in materials handling and waiting
Daily demand for the part is 2,000 units
Safety stock equivalent of 10 percent of inventory

If each container contains 22 parts, how many containers should be authorized?
Suppose that a proposal to revise the plant layout would cut materials handling and waiting time per container to 0.06 day. How many containers would be needed?

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Example 6.2

k =

2,000(0.08 + 0.02)(1.10)

22

= = 10 containers

220

22

b. Figure 8.10 from OM Explorer shows that with reduced waiting time, the number of containers drops to 8.

Figure 6.10

if

d = 2,000 units/day,

p = 0.02 day,

α = 0.10,

w = 0.082 day, and

c = 2,000 units

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Application 6.2

Item B52R has an average daily demand of 1,000 units. The average waiting time per container of parts (which holds 100 units) is 0.5 day. The processing time per container is 0.1 day. If the policy variable is set at 10 percent, how many containers are required?

= 6.6, or 7 containers

=

1,000(0.5 + 0.1)(1 + 0.1)

100

k =

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Other Kanban Signals

Container System

Using the container itself as a signal device.
Works well with containers specifically designed for parts.

Containerless System

Using visual means in lieu of containers as a signal device.
Examples: a painted square on a workbench = one unit.

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Organizational Considerations

The Human Costs of Lean Systems

Cooperation and Trust

Reward Systems and Labor Classification

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Process Considerations

Inventory and Scheduling

Schedule Stability

Setups

Purchasing and Logistics

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Solved Problem 1

Metcalf, Inc makes brackets for two major automotive customers. The operations data is on the following slides.

Create a VSM for Metcalf Bearings
What is the takt time?
What is the lead time at each cell?
What is the total processing time?
What is the capacity?

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Solved Problem 1

Table 6.4

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Solved Problem 1

Figure 6.11

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Solved Problem 1

Daily Demand

2,700 units per day

Daily Availability

(7.5 hours/day) x (3,600 seconds per hour) x (2 shifts/day)=

54,000 seconds per day

Takt Time = Daily availability/Daily Demand =

54,000/2,700 =

20 seconds per unit

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Solved Problem 1

Production Lead time = Inventory/Daily Demand

Raw Material Lead Time – (4,000/2,700) = 1.48 days

WIP between Forming and Drilling =

(5,000/2,700)= 1.85 days

WIP between Drilling and Grinding =

(2,000/2,700) = .74 day

WIP between Grinding and Packaging=

(1,600/2,700) = .59 day

Finished Goods Lead Time before Shipping =

(15,700/2,700) = 5.81 days

Total Production Lead Time =

(1.48 + 1.85 + .74 + .59 + 5.81) = 10.47 days

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Solved Problem 1

d. Total Processing Time = Sum of the Cycle Times

(11 + 10 + 17 + 15) = 53 seconds

e.

Grinding is the bottleneck

Capacity = 54,000/17 = 3,176 units/day

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Solved Problem 2

A company using a kanban system has an inefficient machine group. For example, the daily demand for part L105A is 3,000 units. The average waiting time for a container of parts is 0.8 day. The processing time for a container of L105A is 0.2 day, and a container holds 270 units. Currently, 20 containers are used for this item.

a. What is the value of the policy variable, α?

b. What is the total planned inventory (work-in-process and finished goods) for item L105A?

c. Suppose that the policy variable, α, was 0. How many containers would be needed now? What is the effect of the policy variable in this example?

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Solved Problem 2

a. We use the equation for the number of containers and then solve for α:

α = 1.8 – 1 = 0.8

20 =

3,000(0.8 + 0.2)(1 + α)

270

(1 + α) = = 1.8

20(270)

3,000(0.8 + 0.2)

k =

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Solved Problem 2

With 20 containers in the system and each container holding 270 units, the total planned inventory is

20(270) = 5,400 units

c. If α = 0

k =

= 11.11, or 12 containers

3,000(0.8 + 0.2)(1 + 0)

270

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