Quantitative Review II

• 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

L ( x )
 k ( x

T )
2 where
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 x = the actual value of the dimension
T = 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 .
220 or 0 .
180 )
2  k (

0 .
020 )
2
150
 k /( 0 .
020 )
2 k

375 , 000

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 .
250 or 0 .
150 )
 
0 .
050 k k

( 75 ) /( 0 .
050 )
2

30 , 000
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
R
 s
( p
1
)( p
2
)( p
3
).......( p n
)
• Parallel product components
R p

1

( 1
 p
1
)( 1
 p
2
)( 1
 p
3
)........( 1
 p n
)

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?
R
 s
( p
1
)( p
2
)( p
3
).......( p n
)
R s

(0.96)(0.8
7)(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
 p
1
)( 1
 p
2
)( 1
 p
3
)........( 1
 p n
)
0.99968
= 1 – (1 – 0.99)(1 – p2)
0.99968
= 1 – (0.01 – 0.01p2)
0.99968–1 = -0.01 + 0.01p2
p2 = 0.968

B
.91
A
.98
C
.97
B
.91
R
B

1

( 1

0 .
91 )( 1

0 .
91 )

0 .
9919
R
S

( 0 .
98 )( 0 .
9919 )( 0 .
97 )

0 .
9429 or 94 .
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
D B
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

K
 d ( p
 w )( 1
 
)
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
K
 d ( p
 w )( 1
 
)
C

400 ( 0 .
5 )( 1

.
1 )

35
220

6 .
29 kanbans
35

Location Analysis Methods (Chapter 8)
Factor Rating Method (pp.319-320):
Σ (Factor Weight i
* Factor Score i
)
5*10=
4*20=
2*30=
5*10=
3*30=
2*10=
2*20=
5*30=
3*10=
5*30=

Center-of-Gravity Method (pp.322-323): where d ix
= x-coordinate of location i d iy
Q i
= y-coordinate of location i
= Quantity of goods moved to or from location i

Location
Chicago
Pittsburgh
New York
Atlanta
X coordinate
30
90
130
60
Y coordinate
120
110
130
40
Number of Containers
Shipped per Week
2,000
1,000
1,000
2,000
Where would be the best place to put the warehouse?

Load Distance Model:
Find load distance score by: Calculate the rectilinear distance and multiply by the number of loads

 Calculate Rectilinear Distance
D
AB

X
A

X
B
D
AB

D
D
AB
AB
30

10

20

25

45 miles


Y
A

Y
B
40

15
 Identify Loads, i.e., 4 loads from A to B
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?
Location
Location A
Location B
Location C
Warehouse D
Warehouse E
X coordinate
92
80
90
90
90
Y coordinate
42
40
35
45
40

Load Distance Score
 Warehouse D
Location A
Location B
Location C
Distance
2+3=5
10+5=15
0+10=10
Location
Location A
Location B
Location C
Warehouse D
Warehouse E
Loads
200
100
300
 Warehouse E
Location A
Location B
Location C
Distance
2+2=4
10+0=10
0+5=5
Loads
200
100
300
X coordinate
92
80
90
90
90
Score
1000
1500
3000
5500
Y coordinate
42
40
35
45
40
Score
800
1000
1500
3300

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.

(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
A
400 sq.ft.
B
300 sq.ft.
C
300 sq.ft.
Proposed
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
A
B
C
E
Trips between departments
Dept.
A B C D E F
10 30 10 0 10
30 15 15
20 15 5
25

B
C
A
E
D
F
Dept.
Depts. Trips Distance Score
AB 10 1 10
AC
AD
AF
BD
BE
30
10
10
30
15
1
2
2
2
2
60
10
20
60
30
BF
CD
CE
CF
EF
15
20
15
5
25
1
3
3
1
2 10
25
345
45
60
15
C
E
A
B
A B C D E F
10 30 10 0 10
30 15 15
20 15 5
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 Task Time
Work Element Task Description
A
B
Roll dough
Place on cardboard backing
F
G
H
C
D
E
I
Sprinkle cheese
Spread Sauce
Add pepperoni
Add sausage
Add mushrooms
Shrinkwrap pizza
Pack in box
Predecessor
None
A
B
C
D
D
D
E,F,G
H
Total task time
(seconds
50
5
25
15
12
10
15
18
15
165

 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
MaximumOut put

AvailableT ime

Bottleneck
3600 sec/ hour
50 sec/ unit

72 Pizzas / hour
 Step 4: Compute the theoretical minimum number of workstations (number of station needed to achieve
100% efficiency)
TM

TotalTaskT imes
CycleTime

165 sec
50 sec

3 .
30 Stations

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 (%)

TotalTaskT imes
( NumberOfSt ations )( NewCycleTi me )
( 100 %)
Efficiency (%)

165 sec
3 * 55

165
165

100 %
 Balance Delay
 Balance Delay = 1 – Assembly Line Efficiency
BalanceDel ay

1

1

0 IdleTime

 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?
3.4 mins
 New efficiency?
B
2.3 mins
A C
4.1 mins
E
2.7 mins
D
1.6 mins
F G
3.3 mins 2.6 mins

 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

TotalTaskT imes
CycleTime

20
4 .
1

4 .
88 WorkStatio ns
 Should we use 4 or 5 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
2.3 mins
A C
4.1 mins
6 mins
E
2.7 mins
5.7 mins
D
1.6 mins
F
3.3 mins
G
2.6 mins
2.6 mins

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
2.3 mins
A
4.1 mins
C
4.1 mins
E
2.7 mins
2.7 mins
4.9 mins
D
1.6 mins
F
3.3 mins
G
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

 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 Turnover (IT):
# of times you turn your inventory annually
 Inventory Days’ Supply (IDS): how many days inventory you keep

 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 Turnover (APT):
# of times you turn your accts. payable annually
 Accounts Payable Days’ Supply (APDS): how long you take to pay your bills

 Revenue
 Cost of goods sold
 Average Inventory Value
$35.40 billions
$29.10 billions
$0.306 billions
 Average Accounts Receivable $2.586 billions
 Average Accounts Payable $5.989 billions

 Dell’s Inventory Turnover
 Dell’s Inventory Days’ Supply

 Dell’s Accounts Receivable Turnover
 Dell’s Accounts Receivable Days’ Supply

 Dell’s Accounts Payable Turnover
 Dell’s Accounts Payable Days’ Supply

 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

(Make/Buy Decision)
 Total Cost of Outsourcing:
 Total Cost of Insourcing:
 Indifference Point:

 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?

 Indifference Point Calculation:

 Make vs Buy Decision at 80,000 bagels
Outsource (Buy) In House (Make)
$2,000+($.5*80,000)
= $42,000
$20,000+($.2*80,000)
= $36,000
 Make vs Buy Decision at 50,000 bagels
Outsource (Buy) In House (Make)
$2,000+($.5*50,000)
= $27,000
$20,000+($.2*50,000)
= $30,000
 If the demand is lower than the indifference point, outsourcing is a cheaper alternative, and vice versa.