Chapter 09 Layout Strategy 8th Ed 2011 - Faculty Websites 09...9 Layout Strategies. Outline ... •...
Transcript of Chapter 09 Layout Strategy 8th Ed 2011 - Faculty Websites 09...9 Layout Strategies. Outline ... •...
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SCM 352
9 Layout Strategies
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Outline
• Global Company Profile: McDonald’s• The Strategic Importance of Layout
Decisions• Process-Oriented and Repetitive Layout• Assembly Line Balancing
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McDonald’s New Layout
• Redesigning all 30,000 outlets around the world• Three separate dining areas
• Linger zone with comfortable chairs and Wi-Fi connections
• Grab and go zone with tall counters• Flexible zone for kids and families
• Facility layout is a source of competitive advantage
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McDonald’s
Savings of $100,000,000 per year in food costs
New Kitchen Layout
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Drive-Up Grocery
In France, a Drive-Up Grocery Takes Off, Wall Street Journal, Jan 14, 2010
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Facility Layout Strategy
• The objective of layout strategy is to develop a cost-effective layout that will meet a firm’s competitive needs
• Location or arrangement of everything within & around a facility to achieve:– Higher utilization of space, equipment, and people– Improved flow of information, materials, or people– Improved employee morale and safer working
conditions– Improved customer/client interaction– Flexibility
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Surgery
Radiology
ER Triage Room
ER Beds Pharmacy
Emergency room admissions
Billing/exit
Laboratories
Patient A - broken leg
Patient B - erratic heart pacemaker
Figure 9.3
Process-Oriented Layout
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Minimize cost = ∑ ∑ Xij Cij
n
i = 1
n
j = 1
where n = total number of work centers or departments
i, j = individual departmentsXij = number of loads moved from
department i to department jCij = cost to move a load between
department i and department j
Process-Oriented Layout
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Process Layout Example
Arrange six departments in a factory to minimize the material handling costs. Each department is 20 x 20 feet and the building is 60 feet long and 40 feet wide.• Construct a “from-to matrix”• Determine the space requirements• Develop an initial schematic diagram• Determine the cost of this layout • Try to improve the layout• Prepare a detailed plan
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Department Assembly Painting Machine Receiving Shipping Testing(1) (2) Shop (3) (4) (5) (6)
Assembly (1)
Painting (2)
Machine Shop (3)
Receiving (4)
Shipping (5)
Testing (6)
Number of loads per week
50 100 0 0 20
30 50 10 0
20 0 100
50 0
0
Process Layout Example
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Process Layout Example
Room 1 Room 2 Room 3
Room 4 Room 5 Room 660’
40’
Receiving Shipping TestingDepartment Department Department
(4) (5) (6)
Assembly Painting Machine ShopDepartment Department Department
(1) (2) (3)
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100
50
50
10100
30
Interdepartmental Flow Graph
1 2 3
4 5 6
Process Layout Example
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Cost = $50*20 + $100*40 + $20*60(1 and 2) (1 and 3) (1 and 6)
+ $30*20 + $50*40 + $10*20(2 and 3) (2 and 4) (2 and 5)
+ $20*60 + $100*20 + $50*20(3 and 4) (3 and 6) (4 and 5)
= $13,200
Cost = ∑ ∑ Xij Cij
n
i = 1
n
j = 1
Process Layout Example
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Room 1 Room 2 Room 3
Room 4 Room 5 Room 660’
40’
Receiving Shipping TestingDepartment Department Department
(4) (5) (6)
Painting Assembly Machine ShopDepartment Department Department
(2) (1) (3)
Process Layout Example
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30
50
50
50 100
100
Revised Interdepartmental Flow Graph
2 1 3
4 5 6
Process Layout Example
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Cost = $50*20 + $100*20 + $20*40(1 and 2) (1 and 3) (1 and 6)
+ $30*40 + $50*20 + $10*40(2 and 3) (2 and 4) (2 and 5)
+ $20*60 + $100*20 + $50*20(3 and 4) (3 and 6) (4 and 5)
= $10,600
Cost = ∑ ∑ Xij Cij
n
i = 1
n
j = 1
Process Layout Example
Lower Cost Layout
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Assembly Line Balancing
• Objective is to minimize the imbalance between machines or personnel while meeting required output
• Starts with the precedence relationships– Determine cycle time– Calculate theoretical minimum number of
workstations– Balance the line by assigning specific tasks to
workstations– Calculate efficiency
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Cycle time = Production time available
Demand per day
Minimum number of work stations
Σ Task timesCycle time
Efficiency =
=
Σ Task times(Actual number of work stations)*(Largest cycle time)
Assembly Line Balancing
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This means that tasks B and E cannot be done until task A has been completed
Performance Task Must FollowTime Task Listed
Task (minutes) BelowA 10 —B 11 AC 5 BD 4 BE 12 AF 3 C,DG 7 FH 11 EI 3 G,HTotal time 66
Example
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Performance Task Must FollowTime Task Listed
Task (minutes) BelowA 10 —B 11 AC 5 BD 4 BE 12 AF 3 C,DG 7 FH 11 EI 3 G,HTotal time 66
I
GF
C
D
H
B
E
A10
1112
5
4 3
711 3
Example
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I
GF
C
D
H
B
E
A10
1112
5
4 3
711 3
Performance Task Must FollowTime Task Listed
Task (minutes) BelowA 10 —B 11 AC 5 BD 4 BE 12 AF 3 C, DG 7 FH 11 EI 3 G, HTotal time 66
480 available mins per day
40 units required
Cycle time =Production time available per day
Units required per day= 480 / 40= 12 minutes per unit
Minimum number of
workstations=
∑ Time for task i
Cycle time
n
i = 1
= 66 / 12= 5.5 or 6 stations
(ROUND UP!)
Example
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I
GF
C
D
H
B
E
A10
1112
5
4 3
711 3
Copier Example
Performance Task Must FollowTime Task Listed
Task (minutes) BelowA 10 —B 11 AC 5 BD 4 BE 12 AF 3 C, DG 7 FH 11 EI 3 G, HTotal time 66
480 available mins per day
40 units requiredCycle time = 12 mins
Minimum workstations = 5.5 or 6
Line-Balancing Heuristics
1. Longest task time Choose the available task with the longest task time
2. Most following tasks Choose the available task with the largest number of following tasks
3. Ranked positional weight
Choose the available task for which the sum of following task times is the longest
4. Shortest task time Choose the available task with the shortest task time
5. Least number of following tasks
Choose the available task with the least number of following tasks
Primary Rule
Secondary Rule
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480 available mins per day
40 units requiredCycle time = 12 mins
Minimum workstations = 5.5 or 6
Performance Task Must FollowTime Task Listed
Task (minutes) BelowA 10 —B 11 AC 5 BD 4 BE 12 AF 3 C, DG 7 FH 11 EI 3 G, HTotal time 66Station
1
Station 2
Station 4
Station 5
Station 3
Station 6
Example
5
I
GF
H
C
D
B
E
A10 11
12
4
3 7
11
3
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480 available mins per day
40 units requiredCycle time = 12 mins
Minimum workstations = 5.5 or 6
Efficiency =∑ Task times
(actual number of workstations) x (largest cycle time)
= 66 minutes / (6 stations) x (12 minutes)= 91.7%
Example
I
GF
C
D
H
B
E
A10
1112
5
4 3
711 3
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Thank You
Questions? ?