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```Quantitative Review II
Taguchi Loss Function (p.199)
• Design the product or service so that it
will not be sensitive to variations during
the manufacturing or delivery process
• For example, design a manufactured
good with a smaller design tolerance =
better quality
Taguchi Loss Function
L( x)  k ( x  T )
2
where
L(x) =
k
=
x
T
=
=
the monetary value of the loss
associated with deviating from the
target limit “T”
the constant that translates the
deviation into dollars
the actual value of the dimension
target limits
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).
L( x)  k ( x  T ) 2
150  k (0.200 0.220or0.180) 2  k (0.020) 2
150  k /(0.020)
k  375,000
2
A quality engineer has a manufacturing specification (in
cm) of 0.200 plus or minus 0.050. Historical data
indicates that if the quality characteristic takes on values
larger than .250 cm or smaller than .150 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.
L( x)  \$75
( x  T )  (0.200 0.250or0.150)  0.050
k  (75) /(0.050)
k  30,000
2
L( x)  30,000( x  T )2
L(0.135)  30,000(0.200 0.135)2  \$126.75
Reliability Management (pp.651-654)
• Series product components
Rs  ( p 1 )( p2 )( p3 ).......(pn )
• Parallel product components
R p  1  (1  p1 )(1  p 2 )(1  p3 )........(
1  pn )
The manufacturing of compact disks requires
four sequential steps. The reliability of each of
the steps is 0.96, 0.87, 0.92, and 0.88
respectively. What is the reliability of the
process?
Rs  ( p 1 )( p2 )( p3 ).......(pn )
Rs  (0.96)(0.87)(0.92)(0
.88) 0.6762
The system reliability for a two-component
parallel system is 0.99968. If the reliability of
the first component is 0.99, determine the
reliability of the second component.
R p  1  (1  p1 )(1  p 2 )(1  p3 )........(
1  pn )
0.99968
0.99968
0.99968–1
p2
=
=
=
=
1 – (1 – 0.99)(1 – p2)
1 – (0.01 – 0.01p2)
-0.01 + 0.01p2
0.968
Redundancy
B
.91
A
C
.98
.97
B
.91
RB  1  (1  0.91)(1  0.91)  0.9919
RS  (0.98)(0.9919)(0.97)  0.9429or94.29%
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 = 0.93.
A
C
B
D
RaRb = 1 - (1 - 0.94)(1 - 0.92) = 0.9952
RcRd = 1 - (1 - 0.97)(1 - 0.93) = 0.9979
RabRcd = (0.9952)(0.9979) = 0.9931
Kanban (pp.632-634)
d ( p  w)(1   )
K
C
where:
K = the number of Kanban cards
d = the average production rate OR demand of product
p = the processing time
w = the waiting time of Kanban cards
α = safety stock as a %, usually ranging from 0 to 1
C = the capacity of a standard container
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 400 per hour. Each container
holds 35 bottles, it takes 30 minutes to receive more
bottles (processing plus delivery time) and safety stock
is set at 10%.
d = 400 bottles per hour
p+w = 30 minutes or 0.5 hour
C = 35 bottles per container
α = 0.10
d ( p  w)(1   ) 400 (0.5)(1  .1) 220
K


 6.29 kanbans
C
35
35
Location Analysis Methods (Chapter 8)
Factor Rating Method (pp.319-320):
Σ (Factor Weighti * Factor Scorei)
5*10=
4*20=
2*30=
5*10=
3*30=
2*10=
2*20=
5*30=
3*10=
5*30=
Location Analysis Methods
Center-of-Gravity Method (pp.322-323):
where
dix = x-coordinate of location i
diy = y-coordinate of location i
Qi = Quantity of goods moved to or from location i
Center-of-Gravity Method
X
Y
coordinate
coordinate
Number of Containers
Shipped per Week
Chicago
30
120
2,000
Pittsburgh
90
110
1,000
New York
130
130
1,000
Atlanta
60
40
2,000
Location
Where would be the best place to put the warehouse?
Location Analysis Methods
Find load distance score by: Calculate the rectilinear
distance and multiply by the number of loads
 Calculate Rectilinear Distance
DAB  X A  X B  YA  YB
DAB  30  10  40  15
DAB  20  25
DAB  45m iles
Load Distance Score for AB = 45*4 = 180
TT Logistics Co. has just signed a contract to deliver
products to three locations, and they are trying to
decide where to put their new warehouse. The three
delivery locations are A, B, and C. The two potential
sites for the warehouse are D and E. The total quantity
to be delivered to each destination is: 200 to A, 100 to
B, and 300 to C. The x, y coordinates for the delivery
locations and warehouses are as follows:
Where to locate
warehouse, D or E?
X
Y
coordinate
coordinate
Location A
92
42
Location B
80
40
Location C
90
35
Warehouse D
90
45
Warehouse E
90
40
Location
Location
 Warehouse D
Location A
Location B
Location C
X
coordinate
Y
coordinate
Location A
92
42
Location B
80
40
Location C
90
35
Warehouse D
90
45
Warehouse E
90
40
Distance
2+3=5
10+5=15
0+10=10
200
100
300
Score
1000
1500
3000
5500
Distance
2+2=4
10+0=10
0+5=5
200
100
300
Score
800
1000
1500
3300
 Warehouse E
Location A
Location B
Location C
Designing Process Layouts (Chapter 9)
 Step 1: Gather information
 Space needed, space available, importance of proximity
between various units
 Step 2: Develop alternative block plans
 Using trial-and-error or decision support tools
 Step 3: Develop a detailed layout
 Consider exact sizes and shapes of departments and
work centers including aisles and stairways
 Tools like drawing, 3-D models, and CAD software are
available to facilitate this process.
Process Layout (Step 1: Gather information)
 Recovery First Sports Medicine Clinic Layout
(total space 3750 sq.ft.)
A
400 sq.ft.
D
800 sq.ft.
B
300 sq.ft.
E
900 sq.ft.
C
300 sq.ft.
F
1050 sq.ft.
Process Layout (Step 2: Develop a block layout)
Current
Proposed
A
400 sq.ft.
B
300 sq.ft.
C
300 sq.ft.
A
400 sq.ft.
D
800 sq.ft.
C
300 sq.ft.
D
800 sq.ft.
E
900 sq.ft.
F
1050 sq.ft.
E
900 sq.ft.
B
300 sq.ft.
F
1050 sq.ft.
Proposed layout
would require less
walking.
 What is the load distance for this layout?
B
A
D
C
E
F
Trips between departments
Dept.
A
B
C
E
A
B
10
C
30
D
10
30
20
E
0
15
15
F
10
15
5
25
B
A
D
Dept.
C
E
F
A
Depts. Trips
Distance
Score
B
C
D
E
F
10
30
10
0
10
B
30
15
15
20
15
5
AB
10
1
10
C
AC
30
2
60
E
10
1
10
AF
10
2
20
BD
30
2
60
BE
15
2
30
BF
15
3
45
CD
20
3
60
CE
15
1
15
CF
5
2
10
EF
25
1
25
345
A
25
Assembly Line Balancing (Chapter 9)
 Step 1: Identify task & immediate predecessors
 Step 2: Calculate the cycle time
 Step 3: Determine the output rate
 Step 4: Compute the theoretical minimum number of
workstations
 Step 5: Assign tasks to workstations (balance the line)
 Step 6: Compute efficiency, idle time & balance delay
Assembly Line Balancing
(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 Calculation
 Step 2: Determine cycle time (The amount of time
each workstation is allowed to complete its tasks.)
 Cycle time = Station A (50 seconds) -- the bottleneck
 Step 3: Determine output rate
AvailableT ime 3600 sec/ hour
MaximumOut put 

 72 Pizzas / hour
Bottleneck
50 sec/ unit
 Step 4: Compute the theoretical minimum number of
workstations (number of station needed to achieve
100% efficiency)
TM 

 3.30Stations
CycleTim e
50sec
Assembly Line Balancing
(Step 5: Balance the line)
3 Work Stations
(A,B), (C,D,G), (E,F,H,I)
55 sec
55 sec
55 sec
Assembly Line Balancing
(Step 6: Compute efficiency, idle time & balance delay)
 Efficiency
Efficiency(%) 
(100%)
( Num berOfStations)(NewCycleTim e)
165 sec 165
Efficiency (%) 

 100 %
3 * 55 165
 Balance Delay
 Balance Delay = 1 – Assembly Line Efficiency
BalanceDelay  1  1  0IdleTime
Line Balancing Problem
 What is the bottleneck?
 What is the maximum production per hour?
 What is efficiency and balance delay?
 How to minimize work stations?
 How should they be groups?
 New efficiency?
3.4 mins
B
E
2.3 mins
A
C
2.7 mins
4.1 mins
D
F
G
1.6 mins 3.3 mins 2.6 mins
Line Balancing Problem
 What is the bottleneck?
 4.1 minutes
 What is the maximum production per hour?
 60/4.1 = 14.63 units/hour
 What is efficiency and balance delay?
 Efficiency = 20/(7*4.1) = 69.69%
 Balance Delay = 1-.6969 = 30.31%
 How to minimize work stations?
TM 

 4.88WorkStations
CycleTim e
4.1
 Should we use 4 or 5 work stations?
4 Work Stations
Efficiency = 20/(4*6) = 20/24 = 83.3%
Balance delay = 1-.833 = 16.7%
Maximum production/hour = 60/6 = 10 units/hour
5.7 mins
3.4 mins
B
6 mins
E
2.3 mins
A
C
2.7 mins
4.1 mins
5.7 mins
D
F
G
1.6 mins
3.3 mins
2.6 mins
2.6 mins
5 Work Stations
Efficiency = 20/(5*5.7) = 20/28.5 = 70.18%
Balance delay = 1-.7018 = 29.82%
Maximum production/hour = 60/5.7 = 10.52 units/hour
5.7 mins
3.4 mins
B
E
2.3 mins
A
C
4.1 mins
4.9 mins
2.7 mins
2.7 mins
4.1 mins
D
F
G
1.6 mins
3.3 mins
2.6 mins
2.6 mins
Should we use 4 or 5 or 7 work stations?
 4 Work Stations
 Efficiency = 83.3%
 Balance delay = 16.7%
 Maximum production/hour = 10 units/hour
 5 Work Stations
 Efficiency = 70.18%
 Balance delay = 29.82%
 Maximum production/hour = 10.52 units/hour
 7 Work Stations
 Efficiency = 69.69%
 Balance delay = 30.31%
 Maximum production/hour = 14.63 units/hour
Supply Chain Efficiency
 Measuring Cash to Conversion Cycle
 Inventory Turnover (IT)
 Inventory Days’ Supply (IDS)
 Accounts Receivable Turnover (ART)
 Accounts Receivable Days’ Supply (ARDS)
 Accounts Payable Turnover (APT)
 Accounts Payable Days’ Supply (APDS)
Cash-to-Cash Conversion Cycle = IDS + ARDS - APDS
Inventory Ratios
 Inventory Turnover (IT):
# of times you turn your inventory annually
 Inventory Days’ Supply (IDS):
how many days inventory you keep
Accounts Receivable Ratios
 Accounts Receivable Turnover (ART):
# of times you turn your accts. rec. annually
 Accounts Receivable Days’ Supply (ARDS):
how long it takes to get \$ owed paid to you
Accounts Payable Ratios
 Accounts Payable Turnover (APT):
# of times you turn your accts. payable annually
 Accounts Payable Days’ Supply (APDS):
how long you take to pay your bills
Dell’s Financial data
 Revenue
 Cost of goods sold
 Average Inventory Value
 Average Accounts Receivable
 Average Accounts Payable
\$35.40 billions
\$29.10 billions
\$0.306 billions
\$2.586 billions
\$5.989 billions
Dell’s Example
 Dell’s Inventory Turnover
 Dell’s Inventory Days’ Supply
Dell’s Example
 Dell’s Accounts Receivable Turnover
 Dell’s Accounts Receivable Days’ Supply
Dell’s Example
 Dell’s Accounts Payable Turnover
 Dell’s Accounts Payable Days’ Supply
Dell’s Example
 Dell’s Cash-to-Cash Conversion Cycle
= IDS + ARDS – APDS
= 3.84 days + 26.66 days – 61.76 days
= -31.26 days
 The negative value means that Dell receives
customers’ payments (accounts receivable) 31.26
days, on average, before Dell has to pay its
suppliers (accounts payable).
 This means that Dell’s value chain is a selffunding cash model.
Dell’s Negative Cash-to-Cash Conversion Cycle
 Total Cost of Outsourcing:
 Total Cost of Insourcing:
 Indifference Point:
The Bagel Shop Problem
 Jim & John plan to open a small bagel shop.
 The local baker has offered to sell them
bagels at 50 cents each. However, they will
need to invest \$2,000 in bread racks to
transport the bagels back and forth from the
bakery to their store.
 Alternatively, they can bake the bagels at
their store for 20 cents each if they invest
\$20,000 in kitchen equipment.
 They expect to sell 80,000 bagels each year.
 What should they do?
The Bagel Shop Problem
 Indifference Point Calculation:
The Bagel Shop Problem
 Make vs Buy Decision at 80,000 bagels
\$2,000+(\$.5*80,000)
= \$42,000
In House (Make)
\$20,000+(\$.2*80,000)
= \$36,000
 Make vs Buy Decision at 50,000 bagels