ISQS 3344 Introduction to Production and Operations Management Spring 2014

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ISQS 3344 Introduction to Production and Operations Management Spring 2014. 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 - PowerPoint PPT Presentation

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ISQS 3344 Introduction to Production and

Operations Management

Spring 2014

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

2)()( TxkxL 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

Taguchi Loss Function

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).

000,375)020.0(150

)020.0()200.0180.0220.0(150

)()(

2

22

2

kk

kork

TxkxL

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.

000,30)050.0/()75(

050.0)200.0150.0250.0()(75$)(

2

kk

orTxxL

2)(000,30)( TxxL

75.126$)200.0135.0(000,30)135.0( 2 L

Reliability Management

• Series product components

• Parallel product components

)).......()()(( 321 ns ppppR

)1)........(1)(1)(1(1 321 npppppR

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?

)).......()()(( 321 ns ppppR

0.6762 .88)7)(0.92)(0(0.96)(0.8 sR

Parallel/Redundancy

A

B

C

B

.98

.91

.91

.97

9919.0)91.01)(91.01(1 BR

%29.949429.0)97.0)(9919.0)(98.0( orRS

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

D

C

B

RaRb = 1 - (1 - 0.94)(1 - 0.92) = 0.9952RcRd = 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.

)1)........(31)(21)(11(1 npppppR

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 QuestionWhat 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?

A0.987

B0.895

C0.947

D0.918

More Reliability Question -- Solution

Reliability = 0.987*0.895*0.947*0.918= 0.768

A0.987

B0.895

C0.947

D0.918

A0.987

B0.895

C0.947

D0.918

B0.895

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

Kanban

where:K = the number of Kanban cardsd = the average production rate OR demand of productp = the processing timew = the waiting time of Kanban cardsα = safety stock as a %, usually ranging from 0 to 1C = the capacity of a standard container

CwpdK )1)((

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

kanbansCwpdK 29.6

35220

35)1.1)(5.0(400)1)((

Location Analysis Methods

Factor Rating Method: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 MethodsCenter-of-Gravity Method:

where dix = x-coordinate of location i diy = y-coordinate of location i Qi = Quantity of goods moved to or from

location i

Where would be the best place to put the warehouse?

Location X coordinate

Ycoordinate

Number of Containers Shipped per Week

Chicago 30 120 2,000

Pittsburgh 90 110 1,000New York 130 130 1,000

Atlanta 60 40 2,000

Center-of-Gravity Method

Question 3A 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

Therefore, the best place to put warehouse is Insanity!

Location Analysis MethodsLoad Distance Model:

Find load distance score by: Calculate the rectilinear distance and multiply by the

number of loads

Load Distance ModelCalculate Rectilinear Distance

Identify Loads, i.e., 4 loads from A to B

Load Distance Score for AB = 45*4 = 180

milesDD

D

YYXXD

AB

AB

AB

BABAAB

452520

15401030

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 locatewarehouse, D or E?

Location X coordinate

Ycoordinate

Location A 92 42

Location B 80 40Location C 90 35

Warehouse D 90 45

Warehouse E 90 40

Load Distance Score

Warehouse DDistance Loads ScoreLocation A 2+3=5 200 1000Location B10+5=15 100 1500Location C0+10=10 300 30005500

Warehouse EDistance Loads ScoreLocation A 2+2=4 200 800Location B10+0=10 100 1000Location C 0+5=5 300 15003300

Location X coordinate

Ycoordinate

Location A 92 42Location B 80 40

Location C 90 35Warehouse D 90 45

Warehouse E 90 40

Designing Process LayoutsStep 1: Gather information

Space needed, space available, importance of proximity between various units

Step 2: Develop alternative block plansUsing trial-and-error or decision support tools

Step 3: Develop a detailed layoutConsider 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.B

300 sq.ft.C

300 sq.ft.D

800 sq.ft.E

900 sq.ft.F

1050 sq.ft.

Process Layout (Step 2: Develop a block layout)Current Proposed

A400

sq.ft.

B300

sq.ft.

C300

sq.ft.D

800 sq.ft.

E900

sq.ft.

F1050 sq.ft.

A400

sq.ft.

D800

sq.ft.

C300

sq.ft.E

900 sq.ft.

B300

sq.ft.

F1050 sq.ft.

Proposed layout would require less walking.

What is the load distance for this layout?

Trips between departments

B A D

C E F

Dept.

A B C D E F

A 10 30 10 0 10B 30 15 15C 20 15 5E 25

Load Distance Problem

Load Distance Problem

Depts.

Trips Distance

Score

AB 10 1 10AC 30 2 60AD 10 1 10AF 10 2 20BD 30 2 60BE 15 2 30BF 15 3 45CD 20 3 60CE 15 1 15CF 5 2 10EF 25 1 25

345

B A DC E F

Dept.

A B C D E F

A 10 30 10 0 10B 30 15 15C 20 15 5E 25

Assembly Line BalancingStep 1: Identify task & immediate

predecessorsStep 2: Calculate the cycle timeStep 3: Determine the output rateStep 4: Compute the theoretical minimum

number of workstationsStep 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 DiagramImmediate Task Time

Work Element Task Description Predecessor (secondsA Roll dough None 50B Place on cardboard backing A 5C Sprinkle cheese B 25D Spread Sauce C 15E Add pepperoni D 12F Add sausage D 10G Add mushrooms D 15H Shrinkwrap pizza E,F,G 18I Pack in box H 15

Total task time 165

Layout CalculationStep 2: Determine cycle time (The amount of

time each workstation is allowed to complete its tasks.)Cycle time = Station A (50 seconds) -- the

bottleneckStep 3: Determine output rate

Step 4: Compute the theoretical minimum number of workstations (number of station needed to achieve 100% efficiency)

hourPizzasunithour

BottleneckimeAvailableTputMaximumOut /72

sec/50sec/3600

StationsCycleTime

imesTotalTaskTTM 30.3sec50sec165

Assembly Line Balancing(Step 5: Balance the line)

3 Work Stations(A,B), (C,D,G), (E,F,H,I)

55 sec

55 sec55 sec

Assembly Line Balancing(Step 6: Compute efficiency, idle time & balance delay)

Efficiency

Balance DelayBalance Delay = 1 – Assembly Line Efficiency

%)100())((

(%)meNewCycleTiationsNumberOfSt

imesTotalTaskTEfficiency

%100165165

55*3sec165(%) Efficiency

IdleTimeayBalanceDel 011

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?

A

B

C

4.1 mins

D

1.6 mins

E

2.7 mins

F

3.3 mins

G

2.6 mins

2.3 mins

3.4 mins

Line Balancing Problem

Line Balancing ProblemWhat is the bottleneck?

4.1 minutesWhat is the maximum production per hour?

60/4.1 = 14.63 units/hourWhat is efficiency and balance delay?

Efficiency = 20/(7*4.1) = 69.69%Balance Delay = 1-.6969 = 30.31%

How to minimize work stations?

Should we use 4 or 5 work stations?

nsWorkStatioCycleTime

imesTotalTaskTTM 88.41.4

20

4 Work Stations

A

B

C

4.1 mins

D

1.6 mins

E

2.7 mins

F

3.3 mins

G

2.6 mins

2.3 mins

3.4 mins5.7 mins

5.7 mins

6 mins

2.6 mins

Efficiency = 20/(4*6) = 20/24 = 83.3%Balance delay = 1-.833 = 16.7%Maximum production/hour = 60/6 = 10 units/hour

5 Work Stations

A

B

C

4.1 mins

D

1.6 mins

E

2.7 mins

F

3.3 mins

G

2.6 mins

2.3 mins

3.4 mins5.7 mins

4.9 mins

2.7 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

4.1 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 StationsEfficiency = 70.18%Balance delay = 29.82%Maximum production/hour = 10.52 units/hour

7 Work StationsEfficiency = 69.69%Balance delay = 30.31%Maximum production/hour = 14.63 units/hour

Supply Chain EfficiencyMeasuring 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 dataRevenue $35.40

billionsCost of goods sold $29.10

billionsAverage Inventory Value $0.306

billionsAverage Accounts Receivable $2.586

billionsAverage Accounts Payable $5.989

billions

Dell’s ExampleDell’s Inventory Turnover

Dell’s Inventory Days’ Supply

Dell’s ExampleDell’s Accounts Receivable Turnover

Dell’s Accounts Receivable Days’ Supply

Dell’s ExampleDell’s Accounts Payable Turnover

Dell’s Accounts Payable Days’ Supply

Dell’s ExampleDell’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 self-funding 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 ProblemJim & 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 ProblemIndifference Point Calculation:

The Bagel Shop ProblemMake vs Buy Decision at 80,000 bagels Outsource (Buy) In House (Make)

$2,000+($.5*80,000) $20,000+($.2*80,000)= $42,000 = $36,000

Make vs Buy Decision at 50,000 bagels Outsource (Buy) In House (Make)$2,000+($.5*50,000) $20,000+($.2*50,000)= $27,000 = $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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