### Quantitative Review II

```Quantitative Review II
Spring 2013
Quantitative Review II
Goods and Service Design (Chapter 5 )
Process Selection (Chapter 7 )
Lean Manufacturing (Chapter 16 )
Reliability (Chapter 17)
Facility Location + Facility & Work Design (Chapters 8 & 9)
Value Chains and Supply Chain Design (Chapters 11)
Key Concept: Taguchi Loss Function
• Taguchi measures quality as the variation from the target
value of a design specification and then translats that
variation into an economic “loss function” that expresses
the cost of variation in monetary terms
L( x)  k ( x  T )2
Where :
x = is the actual value of the product dimension
T = target limit
L(x) = the monetary value of the loss associated with deviating from the
target limit “T”
k = the constant that translates the deviation into dollars
Designing a product with a smaller design tolerance = better quality
Example: A quality engineer has a manufacturing specification (in cm)
of 0.200 plus or minus 0.05. Historical data indicates that if the quality
characteristic takes on values larger than .25 cm or smaller than .15
cm, the product fails and a cost of \$75 is incurred. Determine the
Taguchi Loss Function and estimate the loss for a dimension of 0.135
cm.
T = 0.2 (cm);
X< 0.15 (cm) or X> 0.25 (cm); L (X) = \$75
L( x)  k ( x  T )2
75 = k (0.15-0.2)2 or 75 = k (0.25-0.2)2
K = 30,000
75= k (± 0.05)2
Taguchi Loss Function: L(X) =30,000 (x-T)2
L(.135)  30,000(0.135  0.200) 2  \$126.75
Example: A quality characteristic has a specification (in inches)
of 0.200  0.020. If the value of the quality characteristic
exceeds 0.200 by the tolerance of 0.020 on either side, the
product will require a repair of \$150. Develop the appropriate
Taguchi loss function (k).
The Taguchi Loss Function is:
L(x) = k(x - T)2
(x-T)2 = ( 0.020)2
150 = k( 0.02)2 or k = 375,000
Example: Suppose that the specifications for a part (in inches) are 7.00
± 0.25, and that the Taguchi Loss Function is estimated to be L(x) =
8,500(x-T)2. Determine the estimated loss if the quality characteristic
under study takes on a value of 7.50 inches. Is the loss greater at 7.50
as opposed to 7.25?
T= 7.00 ; 6.75 < specification range < 7.25,
Since the upper limit of specification for the part should be <7.25, a part
with x=7.5 is way exceeded that limt.
L(x) = 8500 (7.50-7.00)2 = \$2125
when x=7.5 inches
L(x) = 8500 (7.25 -7)2 = \$531.25
when x=7.25 inches
Key Concept: Reliability Management
• Reliability is defined as the probability (often expressed
as a percentage) that a manufactured good, piece of
equipment, or system performs its intended function for a
stated period of time under specified operating
conditions.
• A probability of .97 indicates that on average, 97 out of
100 times the item will perform its function for a given
period of time under specified operating conditions
• Reliability of a manufactured item depends on the
reliability of each component of that system
Components in a Series
• A system is composed of a series of individual
interrelated components. If any one component fails to
perform, the overall system can fail.
R1
Component
1
R2
Component
2
R3
Component
3
Formula for series: The whole system reliability Rs = R1 x R2 x R3
R: Reliability of a component
Components in Parallel
• To increase the reliability of a system, redundancy is added.
The technique is to “back up” with the additional components,
putting units in parallel. Only if both components in parallel
fail, the entire system will then fail.
Component 1
Component 2
Component 3
Component 1
Component 2
R3
R1
R2
Formula for parallel: R1 = 1 – (1- p1) (1- p2)(1-p3)…(1-pn) -if there are n components
in parallel
In this case, each component 1’s reliability is p1, each component 2’s reliability is p2
Then the total system reliability Rs = R1 x R2 xR3
Rs = [1 – (1- p1) (1- p1)] x [1 – (1- p2) (1- p2)] x R3
Only two parallel components in R1
Only two parallel components in R2
Example: Series Product Reliability
The manufacturing of compact disks requires four sequential
steps. The reliability of each of the steps is 0.96, 0.87, 0.92, and
0.89 respectively. What is the reliability of the process?
Rsystem = (R1)(R2)(R3)……(Rn)
RaRbRcRd = (0.96)(0.87)(0.92)(.89) = 0.6839
Example: Parallel Product Reliability
The system reliability for a two-component parallel system is
0.99968. If the reliability of the first component is 0.992,
determine the reliability of the second component.
Rs 1 (1 p1 )(1 p2 )(1 p3 )........(
1 pn )
0.99968 = 1 – (1 – 0.992)(1 – p2)
0.99968=1-(.008-.008p2)
0.99968-1= -.008+.008p2
p2 = 0.96
Redundancy
.91
B
A
.98
C
B
.99
.91
We have both series and parallel components here in this system
1) RB = 1- (1-0.91)(1-0.91) = 0.9919 (*parallel Rs is always larger than the
each individual R in parallel situation)
2) Rsystem = (0.98) (0.9919)(0.99) = 0.962 or 96.2% (*series Rs is always
smaller than each individual R in series situation)
Example: Given the diagram below, determine the system
reliability if the individual component reliabilities are: A =
0.94, B = 0.92, C = 0.97, and D = 0.94.
A
C
B
D
A and B are parallel components
RAB = 1 - (1 - 0.94)(1 - 0.92) =.9952
C and D are also parallel components
RCD = 1 - (1 - 0.97)(1 - 0.94) = .9982
A-B and C-D are sequentially arranged
Rsystem = RABRCD= (0.9952)(0.9982) = 0.9934
Example: What is the reliability of this system? If you could
add one process (must be one of the existing processes) to
best improve reliability what would be the improved
reliability?
A
0.987
B
C
D
0.965
0.912
0.988
C
0.912
The lowest reliability component in this system is C, so we first improve Rc by
adding another C in parallel fashion
Parallel Reliability
= 1- (1-0.912)(1-0.912) = 0.992256
Total System Reliability
= (0.987)(0.965)(0.992256)(0.988) =0.9337
Key Concepts
• Lean Operations/Just In Time
The manufacturing and service operations that uses approaches of
focusing on the elimination of waste in all forms, and smooth,
efficient flow of materials and information throughout the value
chain to obtain faster customer response (JIT), higher quality, and
lower costs.
• Lean concepts were initially developed and implemented by the
Toyota Motor Corporation.
• Toyota production system (TPS) emphasizes continuous
improvement, respect for people and standard work practices.
• Lean production supplies the customer with their exact wants when
the customer wants it wants it without waste.
Five Basic Lean Principles
1. Elimination of Waste: Eliminate any activities that do not add
value in an organization. Includes overproduction, waiting time,
processing, inventory, and motion.
2. Increased Speed and Response(throughput): Better process
designs allow efficient responses to customers’ needs and the
competitive environment. Lean/JIT is market driven
3. Improved Quality (reducing variability): Poor quality creates
waste, so improving quality is essential to the lean environment.
4. Reducing Cost: Simplifying processes and improving efficiency
translates to reduced costs.
5. Maintaining respect for all workers: people, not machines make
things happen.
Key concepts: Push Production/Distribution Systems
•
A push system produces finished goods inventory in
advance of customer demand using a forecast of sales.
•
Parts and subassemblies are “pushed” through the
operating system based on a predefined schedule that
is independent of actual customer demand.
•
A traditional automobile factory and distribution system
is a good example of a push system.
Just-in-Time Systems (JIT) - Market Driven
•
In a pull system, employees at a given operation (work
station) go to the source of the required parts, such as
machining or subassembly, and withdraw the units as
they need them
•
By pulling parts from each preceding workstation, the
entire manufacturing process is synchronized to the
final-assembly schedule.
•
Finished goods are made to coincide with the actual
rate of customer demand, resulting in minimal
inventories and maximum responsiveness.
Just-in-Time Systems (JIT)
•
JIT systems are sometimes called a Kanban system.
•
A Kanban is a flag or a piece of paper that contains all
relevant information for an order.
•
Slips, called Kanban cards, are circulated within the
system to initiate withdrawal and producing items
through the production process.
•
The Kanban cards are simple visual controls.
Just-in-Time Systems (JIT)
•
The withdraw Kanban authorizes the material handler to
transfer empty containers to the storage area. Next, a
production Kanban triggers production of parts. Finally,
the full container is delivered to the material handler.
Number of Kanban Cards Required
K= d(p + w)(1 + )
C
Where: K = the number of Kanban cards in the operating system.
d = the average production rate.
w = the waiting time of Kanban cards (that is, the time it takes to
deliver the container).
p = the processing time per container.
C = the capacity of a standard container in the proper units of
measure (parts, items, etc.). JIT practice is to set the lot size or
container size equal to about 5% to 20% of a days demand or between
20 to 90 minutes worth of demand. Also, the container is sized to be
able to be moved by the operator without help, if possible.
 = safety stock as a %, usually ranging from 0 to 1.
Example: Anna works on an assembly line where it takes her 20
minutes to produce 53 units of a product needed to fill a container. It
takes her an additional 5 minutes to transport the container to Josh,
who works at the next station. The company uses a safety stock of
15%. The current assembly line uses 5 kanbans between Anna's and
Josh's stations. What is the demand for the product?
P=20 min, W= 5 min, C=53,  =.15, K=5
Calculate d
K= d(p + w)(1 + )
C
d(p + w)(1+ ) = KC
d(20+5)(1+.15)=5*53
d= 9.217 /minute
Example: Computing the Number of Kanbans: an aspirin
manufacturer has converted to JIT manufacturing using Kanban
containers. They wish to determine the number of containers at the
bottle filling operation which fills at a rate of 200 per hour. Each
container holds 25 bottles, it takes 30 minutes to receive more
bottles (processing plus delivery time) and safety stock is set at
10% .
Solut ion :
d  200 bot t les per hour
P  W  30 minut es .5 hour
C  25 bot t les per cont ainer
  0.10
d(P  W)(1  ) (200)(.5)(
1.1)
K

 4.4 kanban cont ainers
C
25
Key Concepts
Location analysis methods
1. Factor rating method (where to locate a production facility)
Six steps:
• Develop a list of relevant factors called key success factors
• Assign a weight to each factor
• Develop a scale for each factor
• Score each location for each factor
• Multiply score by weights for each factor for each location
• Recommend the location with the highest point score
Center-of Gravity Method
• Finds location of distribution center that minimizes
distribution costs
• Considers
1. Location of markets
2. Volume of goods shipped to those markets
3. Shipping cost (or distance)
Center-of Gravity Method
• Place existing locations on a coordinate grid
Grid origin and scale is arbitrary
Maintain relative distances
• Calculate X and Y coordinates for ‘center of gravity’
Assumes cost is directly proportional to distance
and volume shipped
Center-of Gravity Method
d Q
X-coordinate =
Q
ix
•
i
i
i
dix: x-coordinate of location i
i
diy: y-coordinate of location i
d Q
Y –coordinate =
Q
iy
•
i
i
i
i
Qi: Quantity of goods moved to
or from location i
Factor Rating Example
The owners of Speedy Logistics, a company that provides overnight
delivery of documents, are considering where to locate their new
facility in the Midwest. They have narrowed their search down to
two locations and have decided to use factor rating to make their
decision. They have listed the factors they consider important and
assigned a factor score to each location based on a 5-point scale 1
lowest, 5 highest). The information is shown here.
What is the best location, and what is its factor rating score?
Factor Score at Each Location
Proximity to Airport
Factor
Weight
20
30
4
3
Proximity to Labor Source
10
3
5
Size of Facility
40
5
4
Factor
100
Location 1
Location 2
5
4
Factor Score at Each Location
Weighted Score
Factor
Factor
Weight
Location 1
Location 2
Proximity to Airport
20
5
4
5*20=100 4*20=80
30
4
3
4*30=120 3*30=90
Proximity to Labor Source
10
3
5
3*10=30 5*10=50
Size of Facility
40
5
4
5*40=200 4*40=160
100
Location 1 Location 2
450
380
Choose location 1 as its factor rating score is 450, higher than that of location 2.
Center-of Gravity Method
Example: find the optimal warehouse location from which
deliveries will be made to four locations: Chicago, New York,
Pittsburgh and Atlanta.
New York (130,130)
North-South
Chicago (30,120)
120
Chicago (90,110)
90
60
Atlanta (60,40)
30
East-West
30
60
90
120
150
Center-of Gravity Method
Store Locations
Chicago (30, 120)
Pittsburgh (90,110)
New York (130, 130)
Atlanta (60,40)
No. of Containers Shipped per Month
2,000
1,000
1,000
2,000
Let’s find out the location of the warehouse by calculating its x and y
Coordinates:
(30)(2000)+(90)(1000)+(130)(1000)+(60)(2000) = 66.7
x=
2000+1000+1000+2000
y=
(120)(2000)+(110)(1000)+(130)(1000)+(40)(2000)
2000+1000+1000+2000
= 93.3
Answer: the warehouse location is at (66.7, 93.3) or somewhere very close to this.
Key Concepts
Facility Layout Ch 9 p. 344
Specific arrangement of physical facilities
Good layouts consider
•Supporting firm’s core competencies and competitive
priorities
•Considers material handling equipment, and minimizing
delays in materials handling
•Capacity and space requirements
•Environment and aesthetics
•Flow of information
•Cost of moving between various work areas
•Promote high employee morale and customer satisfaction
Types of Layout
Office
Supermarket / Retail
Warehouse (storage)
Project
Job Shop (Process oriented)
Work Cell (Product families)
Repetitive / Continuous (Product oriented)
1.Process layout design
- Grouping of similar activities or consisting of similar
functional equipments
Pros
•General purpose & flexible resources; Flexible and
capable of handling a wide variety of products or services
•Lower capital/equipment cost
•Higher worker satisfaction
Cons
•Higher labor cost – workers more skilled
•Processing rates are slower
•Material handling costs are higher/more WIP
•Scheduling resources & work flow is more complex and
difficulty
•Space requirements are higher
e.g. Legal offices, Hospitals
1.Process layout design
3- step process
• Gather information: space needed, space available,
importance of proximity between various units
• Develop alternative block plans: using trial-and-error
or decision support tools
• Develop a detailed layout: consider sizes, shapes of
departments and centers; utilize tools such as 3-D
2. Product layout design
-Arrangement based on the sequence of operations
that is performed
Pros
• Processing rates are faster
• Material handling and labor costs are lower
•Less space required for inventories
Cons
• High capital cost and wide use of automation
•Less volume or design flexibility
•Specialized equipment required
•Support staff expensive
e.g. automobile assembly lines, wine-making
industry
Key Concepts
1.Study process layout design
Using load-distance score to choose a layout plan
2. Study product layout
Using Assembly line balancing to design work stations
Product Layout Design
Grouping tasks among workstations so that each workstation has –in the
ideal case- the same amount of work
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Calculate the cycle time
Determine the output rate
Compute the theoretical minimum number
of workstations
Assign tasks to workstations (balance the
line)
Compute efficiency, idle time & balance
delay
Process Layout Example
The following from-to matrix shows daily customer trips between departments of
Fresh Foods Grocery
Dept.
A
B
C
D
E
F
A
Trips Between Departments
B
C
D
E
9
45
25
90
30
22
25
34
96
40
Their proposed layout looks like the following
F
70
25
20
10
36
A
C
D
F
B
E
What is the load-distance for Fresh Foods proposed layout?
1
A
C
D
2
F
Dept.
A
B
C
D
E
F
B
A
distance for A-C is 1
distance for C-E is 2
distance for A-E is 3
E
Trips Between Departments
B
C
D
E
9
45
25
90
30
22
25
34
96
40
F
70
25
20
10
36
A-B
2
9*2=18
A-C
1
45*1=45
A-D
2
25*2=50
A-E
3
90*3=270
A-F
1
70*1=70
B-C
1
30*1=30
B-D
2
22*2=44
B-E
1
25*1=25
B-F
1
25*1=25
C-D
1
34*1=34
C-E
2
96*2=192
C-F
2
20*1=20
D-E
1
40*1=40
D-F
3
10*3=30
E-F
2
36*2=72
965
To evaluate multiple layout plans, choose the one with smallest
load-distance score- a good proposed layout would require less walking
Product Layout Example
The following table shows the tasks required to assemble an aluminum storm
door and the length of time needed to complete each task.
A
B
C
D
E
F
G
H
Immediate Predecessor
None
A
A
A
B, C, D
E
None
F, G
32
55
12
23
15
70
20
5
1. Calculate the cycle time
2. What is the maximum output per hour for this line with an operator at
each station?
3. Calculate the minimum number of work stations required and round up
4. What is the efficiency of the line if the theoretical minimum number of
stations is used? Round down. Use the combination of stations shown.
A
B
C
D
E
F
G
H
Immediate Predecessor
None
A
A
A
B, C, D
E
None
F, G
32
55
12
23
15
70
20
5
Bottleneck:
70 seconds for
each unit
locate near E
Draw the layout first based on the information given:
B
A
F
C
D
E
H
G
1. Cycle time – is determined by the bottleneck or station that takes the
longest: 70 sec.
2.
3.
Ma xi m u m o u tp u t
a va i l a b l eti m e 3 6 0 0se c./h r.

 5 1 .4d o o rs p e r h o u r
b o ttl e n e ck
7 0 se c./u n i t
ta sk ti m e s  2 3 2 se co n d s 3 .3 1sta ti o n s
TM  
cycl eti m e
7 0 se co n d s
You need 4 workstations
55
55
B
A
32
12
15
70
5
E
F
H
C
G
55
75
20
D
47
23
Balance the line: (A,D), (B), (C,E,G), (F,H), assign these tasks to 4 work stations
4. After the activities are grouped and assigned to 4 work stations, the new
cycle time becomes 75 seconds.
Efficency% 

number of stationsnew cycletime
232
100  0.7733
475
100
Step 1: Identify Tasks & Immediate Predecessors
Example 10.4 Vicki's Pizzeria and the Precedence Diagram
Immediate
Predecessor
(seconds
A
B
C
D
E
F
G
H
I
Roll dough
Place on cardboard backing
Sprinkle cheese
Shrinkwrap pizza
Pack in box
None
A
B
C
D
D
D
E,F,G
H
50
5
25
15
12
10
15
18
15
165
Layout Calculations
• Step 2: Determine cycle time
– Cycle time is determined by the bottleneck or station that takes
the longest
– Cycle time = Station A (50 seconds) since it is the bottleneck
• Step 3: Determine output rate per hour
(note: always make the units for both available time and bottleneck consistent)
Maxi m umoutput
avai l abl eti m e 3600sec./hr.

 7 2 pi zzas p e r hour
bottl eneck
5 0 sec./uni t
• Step 4: Compute the theoretical minimum number of stations
– TM = number of stations needed to achieve 100% efficiency
(every second is used)
– Round up or down depending on circumstances
ta sk ti m e s  1 6 5 se co n d s 3 .3 0 sta ti o n s
TM  
cycl eti m e
5 0 se co n d s
55
55
55
Balance the line: (A,B), (C,D,G), (E,F,H,I), assign these tasks to 3 work stations
Layout Calculations
• Step 6: Compute efficiency, idle time & balance
delay
Efficency% 
number of stationsnew cycletime
Efficiency % 
100
1 6 5 se co n d s
 100%
55 x 3
Balance Delay = 1 - Assembly Line Efficiency
In this case Balance Delay= 1-1.00 = “0” idle time
Another Line Balancing Problem
3.4 mins
Total: 20 mins
B
E
2.2 mins
A
2.7 mins
C
4.1mins
D
F
G
1.7mins
3.3
mins
2.6 mins
Before balancing the line, the efficiency is:
Efficency% 
number of stationsnew cycletime
100
= (2.2+3.4+4.1+2.7+1.7+3.3+2.6) (100)
4.1x7
=69.7%
Balance Delay: 1- 69.7% =30.3%
Questions?
1. What is the bottleneck? 4.1 minutes
2. What is maximum production per hour? (convert to
minutes for the available time as the unit for bottleneck is in minutes)
60 min per hour /4.1 min = 14.63 units
3. How to minimize work stations? Balance the line!
  ta sk ti m e s 2 0 m i n u te s
TM 

 4 .8 7 sta ti o n s
cycl eti m e
4 .1 m i n u te s


4. How should they be grouped?
4 or 5 workstations.
5. New efficiency?
If use 4 Stations
(5.6 min)
3.4
Station 1
B
2.2
E
A
2.7
Station 3
(6.0 min)
C
Station 4
4.1
Station 2
(5.8 min)
D
1.7
F
G
3.3
2.6
All under 6 minutes, so Maximum output = (60 min/hr)/ 6min/unit =10 units /hour
Efficency%  

number of stationsnew cycle time
100   20 /(4 * 6)  83.3%
If use 5 Stations
Max.output = 60/5.6 =10.7 units/hour
Efficiency = 20/(5.6x5 )= 20/28 = 71.4%
5.6
3.4 mins
B
E
2.2 mins
A
2.7 mins
C
4.1mins
D
F
G
1.7mins
3.3
mins
2.6 mins
5.0
```