Quantitative Review II

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Quantitative Review II

Taguchi Loss Function

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) = 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 .

220 or 0 .

180

0 .

200 )

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 .

250 or 0 .

150

0 .

200 )

 

0 .

050 k k

( 75 ) /(

0 .

050 )

2

30 , 000

L ( x )

30 , 000 ( x

T )

2

L ( 0 .

135 )

30 , 000 ( 0 .

135

0 .

200 )

2 

$ 126 .

75

Reliability Management

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

Parallel/Redundancy

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

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

More Reliability Question

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

0.895

C

0.947

D

0.918

More Reliability Question -- Solution

A

0.987

B

0.895

C

0.947

D

0.918

Reliability = 0.987*0.895*0.947*0.918= 0.768

B

0.895

A

0.987

C

0.947

B

0.895

Parallel B Reliability = 1-(1-0.895)*(1-0.895) = 0.989

System Reliability = 0.987*0.989*0.947*0.918 = 0.849

Improvement = 0.849 – 0.768 = 0.08

D

0.918

Kanban

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

Factor Rating Method:

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=

Location Analysis Methods

Center-of-Gravity Method: 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

Center-of-Gravity Method

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?

Question 3

A Plus 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 Chicago,

Pittsburgh, and New York.

Which town would be the best place to put the warehouse?

Question 3 (Continued)

Question 3 -- Solution

30 2000 + 90 1000 + (130)(1000)

= 70

2000 + 1000 + 1000

120 2000 + 110 1000 + (130)(1000)

= 120

2000 + 1000 + 1000

Therefore, the best place to put warehouse is Insanity!

Location Analysis Methods

Load Distance Model:

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

Load Distance Model

 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

 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

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.

Load Distance Problem

 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

Load Distance Problem

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

 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

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

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

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?

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

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

TotalTaskT imes

CycleTime

20

4 .

1

4 .

88 WorkStatio ns

 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

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

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

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

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

$35.40 billions

$29.10 billions

$0.306 billions

 Average Accounts Receivable $2.586 billions

 Average Accounts Payable $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

Breakeven Analysis

(Make/Buy Decision)

 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

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.

That’s all, folks!

Good Luck!

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