Lean Systems Characteristics of Lean Systems JustinTime Pull

Lean Systems

Characteristics of Lean Systems: Just-in-Time · Pull method of materials flow · Consistently high quality · Small lot sizes · Uniform workstation loads · Standardized components and work methods · Close supplier ties · Flexible workforce · Line flows · Automated production · Preventive maintenance

Lot Size and Cycle Inventory

Lot Size and Cycle Inventory 100 – On-hand inventory Average cycle inventory 75 – Lot size = 100 50 – 25 – 0– Figure 16. 1 5 10 15 Time (hours) 20 25 30

Lot Size and Cycle Inventory 100 – On-hand inventory Average cycle inventory 75 – Lot size = 100 50 – 25 – 0– Figure 16. 1 5 10 15 Time (hours) 20 25 30

Lot Size and Cycle Inventory 100 – On-hand inventory Average cycle inventory 75 – Lot size = 100 50 – Lot size = 50 25 – 0– Figure 16. 1 5 10 15 Time (hours) 20 25 30

Lot Size and Cycle Inventory 100 – On-hand inventory Average cycle inventory 75 – Lot size = 100 50 – Lot size = 50 25 – 0– Figure 16. 1 5 10 15 Time (hours) 20 25 30

Lot Size and Cycle Inventory 100 – On-hand inventory Average cycle inventory 75 – Lot size = 100 50 – Lot size = 50 25 – 0– Figure 16. 1 5 10 15 Time (hours) 20 25 30

Continuous Improvement with Lean Systems Scrap Figure 16. 2 Unreliable suppliers Capacity imbalance

Single-Card Kanban System

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Receiving post Kanban card for product 1 Kanban card for product 2 Storage area Empty containers Assembly line 1 O 2 O 1 Fabrication cell O 2 O 3 Assembly line 2 Full containers Figure 16. 3

Single-Card Kanban System Location: Part Number: 6 Aisle 5 Bin 47 1234567 Z KANBAN Lot Quantity: WS 83 ü ü Supplier: ü WS 116 ü 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 Customer: ü ü

Number of Containers

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 08 day c = 22 units k= d( w + p )( 1 + a ) c Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 08 day c = 22 units k= 2000( 0. 08 + 0. 02 )( 1 + 0. 10 ) 22 Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 08 day c = 22 units k= 2000( 0. 08 + 0. 02 )( 1 + 0. 10 ) 22 Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 08 day c = 22 units k = 10 containers Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 06 day c = 22 units k = 10 containers k= d( w + p )( 1 + a ) c Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 06 day c = 22 units k = 10 containers k= 2000(0. 06 + 0. 02)(1. 10) 22 Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 06 day c = 22 units k = 10 containers k= 2000(0. 06 + 0. 02)(1. 10) 22 Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 06 day c = 22 units k = 10 containers k = 8 containers Example 16. 1

Number of Containers Westerville Auto Parts d = 2000 units/day p = 0. 02 day a = 0. 10 w = 0. 06 day c = 22 units k = 10 containers k = 8 containers Figure 16. 4

Lean Systems in Services · Consistently high quality · Uniform facility loads · Standardized work methods · Close supplier ties · Flexible workforce · Automation · Preventive maintenance · Pull method of materials flow · Line flows

Operational Benefits · Reduce space requirements · Reduce inventory investment · Reduce lead times · Increase labor productivity · Increase equipment utilization · Reduce paperwork and simple · · · planning systems Valid priorities for scheduling Workforce participation Increase product quality