Example: Receiving goods to warehouse Detailed view Receive Goods Inspect Goods (30) Match order? (10) Supervisor Report (5) Quality Check (45) Accept?

Example: Receiving goods to warehouseExample: Receiving goods to warehouseDetailed viewDetailed view

Receive Goods

Inspect Goods (30)

Match order?

(10)

Supervisor Report (5)

Quality Check (45)

Accept? (2)

Goods 4 pick up

Inform Purchasing

Yes

No

YesNo

What is the throughput time for good items?

What is the capacity of each station?

Where is the bottleneck?

What is the cycle time?

What is the throughput rate?

If there are 15 orders coming in an 8 hr day, what would each stations utilization rate be?

7 – 2

Solution – part 1 (single QC station)Solution – part 1 (single QC station) Throughput Time for good items 30 + 10 + 45 + 2 = 87 minutes

Throughput Time for bad orders 30 + 10 + 5 = 45 minutes

Throughput Time for bad quality 30 + 10 + 45 + 2 + 5 = 92 minutes

Cycle time is 45 minutes

Throughput rate for good orders is 10 orders / day

Inspection capacity 2 orders/ hr = 16 orders / day (assuming 8 hr day)utilization rate = 15/16 = 94%

Matching orders capacity 6 orders / hr = 48 orders / day utilization rate = 15/48 = 31%

Supervisor Reports capacity= 12 orders / hr = 96 orders / dayMAX utilization rate = 15/96 = 16%

Quality Check capacity = 480 / 45 = 10 orders / day MAX utilization rate = 15/10 = 150%

Acceptance processing capacity 30 orders / hr = 240 orders / day MAX utilization rate = 15/240= 6%

Bottleneck for good orders is quality check

Bottleneck for bad orders is inspect goods

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

7 – 3

Example: Receiving goods to warehouseExample: Receiving goods to warehouseDetailed viewDetailed view

Receive Goods

Inspect Goods (30)

Match order?

(10)

Supervisor Report (5)

Quality Check (45)

Accept? (2)

Goods 4 pick up

Inform Purchasing

Yes

No

Yes

No

If we get 15 orders in an 8 hr day, what would the utilization rate be for each station?

Quality Check (45)

What is the throughput time for good items?

What is the capacity of each station?

Where is the bottleneck?

What is the cycle time?

What is the throughput rate?

7 – 4

Solution – part 2 (Two QC stations)Solution – part 2 (Two QC stations)

Throughput Time for good items 30 + 10 + 45 + 2 = 87 minutes

Throughput Time for bad orders 30 + 10 + 5 = 45 minutes

Throughput Time for bad quality 30 + 10 + 45 + 2 + 5 = 92 minutes

Throughput rate is 16 orders/ day

Cycle time is 30 minutes

Inspection capacity 2 orders/ hr = 16 orders / day (assuming 8 hr day)utilization rate = 15/16 = 94%

Matching orders capacity 6 orders / hr = 48 orders / day utilization rate = 15/48 = 31%

Supervisor Reports capacity= 12 orders / hr = 96 orders / day MAX utilization rate = 15/96 = 16%

Quality Check capacity = 480 / 22.5 = 21 orders / day MAX utilization rate for each station = 15/21 = 71%

Acceptance processing capacity 30 orders / hr = 240 orders / day MAX utilization rate = 15/240= 6%

Bottleneck for ALL orders is inspect goods


7 – 5

Chapter 3: Problem 2Chapter 3: Problem 2

a. For Type A customers, step T2 can process (60/13) = 4.62 customers per hour. T3 has three work stations and a capacity of (60/14) + (60/10) + (60/11) = 15.74 customer per hour. Step T4 can process (60/18) = 3.33 customers per hour. The bottleneck for type A customers is the step with the highest processing time per customer, T4.

TypeA or B?

Type A

Type B

T1(12)

T2(13)

T3-a(14)

T3-b(10)

T3-c(11)

T4(18)

T7(10)

T6(22)

T5(15)

7 – 6


b. The bottleneck for Type B customers is T6 since it has the longest processing time per customer. The capacity for Type B customers is (60/22) = 2.73 customers per hour. Thus the average capacity is 0.3(3.33) + 0.7(2.73) = 2.9 customers per hour. This of course also the expected throughput rate of the entire process (Theoretically of course since we can’t have partial customers)..

TypeA or B?

Type A

Type B

T1(12)

T2(13)

T3-a(14)

T3-b(10)

T3-c(11)

T4(18)

T7(10)

T6(22)

T5(15)

7 – 7


c. Type A customers would wait before T2 and T4 because the activities immediately preceding them have a higher rate of output. Type B customers would wait before T5 and T6 because of the same reasons. This assumes that new customers are always arriving.

TypeA or B?

Type A

Type B

T1(12)

T2(13)

T3-a(14)

T3-b(10)

T3-c(11)

T4(18)

T7(10)

T6(22)

T5(15)

7 – 8Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Another type of problem involving Another type of problem involving Parallel Processes Parallel Processes

Bill’s Car Wash offers two types of washes: Standard and Deluxe. The process flow for both types of customers is shown in the following chart. Both wash types are first processed through steps A1 and A2. The Standard wash then goes through steps A3 and A4 while the Deluxe is processed through steps A5, A6, and A7. Both offerings finish at the drying station (A8). The numbers in parentheses indicate the minutes it takes for that activity to process a customer.

A8(10)

A7(12)

A6(20)

A5(5)Deluxe

A4(15)

A3(12)Standard

Standard or Deluxe

A2(6)

A1(5)



a. Which step is the bottleneck for the Standard car wash process? For the Deluxe car wash process?

b. What is the capacity (measured as customers served per hour) of Bill’s Car Wash to process Standard and Deluxe customers? Assume that no customers are waiting at step A1, A2, or A8.

c. If 60 percent of the customers are Standard and 40 percent are Deluxe, what is the average capacity of the car wash in customers per hour?

d. Where would you expect Standard wash customers to experience waiting lines, assuming that new customers are always entering the shop and that no Deluxe customers are in the shop? Where would the Deluxe customers have to wait, assuming no Standard customers?



SOLUTION

a. Step A4 is the bottleneck for the Standard car wash process, and Step A6 is the bottleneck for the Deluxe car wash process, because these steps take the longest time in the flow.

b. The capacity for Standard washes is 4 customers per hour because the bottleneck step A4 can process 1 customer every 15 minutes (60/15). The capacity for Deluxe car washes is 3 customers per hour (60/20). These capacities are derived by translating the “minutes per customer” of each bottleneck activity to “customers per hour.”

c. The average capacity of the car wash is (0.60 4) + (0.40 3) = 3.6 customers per hour.



d. Standard wash customers would wait before steps A1, A2, A3, and A4 because the activities that immediately precede them have a higher rate of output (i.e., smaller processing times). Deluxe wash customers would experience a wait in front of steps A1, A2, and A6 for the same reasons. A1 is included for both types of washes because the arrival rate of customers could always exceed the capacity of A1.

7 – 12

In class Exercise: Emergency Room In class Exercise: Emergency Room CaseCase

The first thing patients do when they arrive into an emergency room is register with the front desk unless it is a life threatening condition that requires immediate attention. Those are taken immediately inside to one of the exam rooms where they receive care from the ER doctor and nurses to stabilize them. Almost all of these patients will be admitted to the hospital for further tests, observation, or surgery. The non-critical patients have to wait in a lounge until one of the exam rooms empties at which time a nurse invites them in, takes down their vital signs (blood pressure, temperature, heart rate) and then she documents their ailment. At their leisure the ER doctor comes in examines the patient and either orders more tests or prescribes medication and releases the patient. Those requiring more tests have to wait for the test results from the lab or radiology before receiving further treatment. Some of those patients are released while others are admitted to the hospital. Patients released have to settle their bills before heading home.

REGISTER

ER CARE

NC

C

WAIT WAITVITALSIGNS

MEDEXAM

MORETESTS

?

HOSP ADMIT END

DISCHARGE

END

YES

1

1

BLOOD WORKXRAY, OTHERS

WAIT

SERIOUSAILMENT

?

EMERGENCY ROOM FLOWCHARTNO

YES

EVALUATE RESULTS

PRESCRIBE MEDS

NO

REG

MEDEXAM

MORETESTS

END

NO

IDEAL “ER” SCENARIO!

DISCHARGE

PERFORM TESTS

PRESCRIBE MEDS

WAIT

EVALUATE RESULTS

SERIOUSAILMENT

?

HOSP ADMIT

END

YES

CHECK VITALSIGNS YES

NO

Process Throughput Time ReductionProcess Throughput Time Reduction

Perform activities in parallel

Change the sequence of activities

Reduce interruptions

7 – 16

7 Key Principles of Bottleneck analysis7 Key Principles of Bottleneck analysis

1. The focus is on balancing flow, not on balancing capacity.

2. Maximizing output and efficiency of every resource will not maximize the throughput of the entire system.

3. An hour lost at a bottleneck or constrained resource is an hour lost for the whole system.

An hour saved at a non-constrained resource does not necessarily make the whole system more productive.

7 – 17


4. Inventory is needed only in front of the bottlenecks to prevent them from sitting idle, and in front of assembly and shipping points to protect customer schedules. Building inventories elsewhere should be avoided.

5. Work should be released into the system only as frequently as the bottlenecks need it. Bottleneck flows should be equal to the market demand. Pacing everything to the slowest resource minimizes inventory and operating expenses.

7 – 18


6. Activation of non-bottleneck resources cannot increase throughput, nor promote better performance on financial measures.

7. Every capital investment must be viewed from the perspective of its global impact on overall throughput (T), inventory (I), and operating expense (OE).

7 – 19

How Operational Measures How Operational Measures Relate to Financial MeasuresRelate to Financial Measures

Relationship to Financial Measures

Description

Utilization (U)

Operating Expense (OE)

Throughput (T)

Inventory (I)

Operational Measures

A decrease in OE leads to an increase in net profit, ROI, and cash flows

An increase in U at the bottleneck leads to an increase in net profit, ROI, and cash flows

An increase in T leads to an increase in net profit, ROI, and cash flows

A decrease in I leads to an increase in net profit, ROI, and cash flow

The degree to which equipment, space, or labor is currently being used, and is measured as the ratio of average output rate to maximum capacity, expressed as a %

All the money the system spends to turn inventory into throughput

Rate at which system generates money through sales

All the money invested in the system in purchasing things that it intends to sell

Constraint Management – Making Constraint Management – Making product choices using bottleneck product choices using bottleneck analysisanalysis



Identifying the BottleneckIdentifying the BottleneckMulti-Product Multi-stationsMulti-Product Multi-stations

EXAMPLE

Diablo Electronics manufactures four unique products (A, B, C, and D) that are fabricated and assembled in five different workstations (V, W, X, Y, and Z) using a small batch process. Each workstation is staffed by a worker who is dedicated to work a single shift per day at an assigned workstation. Batch setup times have been reduced to such an extent that they can be considered negligible. Figure 7.2 is a flowchart of the manufacturing process. Diablo can make and sell up to the limit of its demand per week, and no penalties are incurred for not being able to meet all the demand.

Which of the five workstations (V, W, X, Y, or Z) has the highest utilization, and thus serves as the bottleneck for Diablo Electronics?

Identifying the BottleneckIdentifying the Bottleneck

Product A

$5Raw materials

Purchased parts

Product: APrice: $75/unitDemand: 60 units/wk

Step 1 at workstation V

(30 min)

Finish with step 3at workstation X

(10 min)

Step 2 atworkstation Y

(10 min)

$5

Product C

Raw materialsPurchased parts

Product: CPrice: $45/unitDemand: 80 units/wk

Finish with step 4at workstation Y

(5 min)

Step 2 atworkstation Z

(5 min)

Step 3 at workstation X

(5 min)

Step 1 atworkstation W

(5 min)

$2

$3

Product B


Product: BPrice: $72/unitDemand: 80 units/wk

Finish with step 2at workstation X

(20 min)


(10 min)

$3

$2

Product D


Product: DPrice: $38/unitDemand: 100 units/wk

$4 Step 2 atworkstation Z

(10 min)

Finish with step 3at workstation Y

(5 min)


(15 min)

$6

Flowchart for Products A, B, C, and DOverhead Costs: $8,500; Labor Costs: $18/hr (8hrs/day; 40 hrs/week)



SOLUTION

Because the denominator in the utilization ratio is the same for every workstation, with one worker per machine at each step in the process, we can simply identify the bottleneck by computing aggregate workloads at each workstation.

The firm wants to satisfy as much of the product demand in a week as it can. Each week consists of 2,400 minutes of available production time. Multiplying the processing time at each station for a given product with the number of units demanded per week yields the workload represented by that product. These loads are summed across all products going through a workstation to arrive at the total load for the workstation, which is then compared with the others and the existing capacity of 2,400 minutes.


WorkstationLoad from Product A

Load from Product B

Load from Product C

Load from Product D

Total Load (min)

V

W

X

Y

Z




These calculations show that workstation X is the bottleneck, because the aggregate work load at X exceeds the available capacity of 2,400 minutes per week.

60 30 = 1800 0 0 0 1,800

0 0 80 5 = 400100 15 =

1,5001,900

60 10 = 600 80 20 = 1,600 80 5 = 400 0 2,600

60 10 = 600 80 10 = 800 80 5 = 400 100 5 = 500 2,300

0 0 80 5 = 400100 10 =

1,0001,400

WorkstationLoad from Product A

Load from Product B

Load from Product C

Load from Product D

Total Load (min)

V

W

X

Y

Z


Determining the Product MixDetermining the Product Mix

Pedro Rodriguez, the newly hired production supervisor, is knowledgeable about the theory of constraints and bottleneck-based scheduling. He believes that profitability can indeed be improved if bottleneck resources were exploited to determine the product mix. What is the change in profits if, instead of the traditional method used by Diablo Electronics, the bottleneck method advocated by Pedro is used to select the product mix?

SOLUTION

Decision Rule 1: Traditional Method

Select the best product mix according to the highest overall contribution margin of each product.



Step 1: Calculate the contribution margin per unit of each product as shown here.

A B C D

Price

Raw material and purchased parts

= Contribution margin



Step 1: Calculate the contribution margin per unit of each product as shown here.

When ordered from highest to lowest, the contribution margin per unit sequence of these products is B, A, C, D.

$75.00 $72.00 $45.00 $38.00

–10.00 –5.00 –5.00 –10.00

$65.00 $67.00 $40.00 $28.00

A B C D

Price

Raw material and purchased parts

= Contribution margin



Step 2: Allocate resources V, W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation X) is encountered. Subtract minutes away from 2,400 minutes available for each week at each stage.

Work Center

Minutes at the Start

Minutes Left After Making 80 B

Minutes Left After Making 60 A

Can Only Make 40 C

Can Only Make 100 D

V

W

X

Y

Z




The best product mix according to this traditional approach is then 60 A, 80 B, 40 C, and 100 D.

2,400 2,400 600 600 600

2,400 2,400 2,400 2,200 700

2,400 800 200 0 0

2,400 1,600 1,000 800 300

2,400 2,400 2,400 2,200 1,200

Work Center


Minutes Left After Making 80 B

Minutes Left After Making 60 A

Can Only Make 40 C

Can Only Make 100 D

V

W

X

Y

Z



Step 3: Compute profitability for the selected product mix.

Profits

Revenue

Materials

Labor

Overhead

Profit




Manufacturing the product mix of 60 A, 80 B, 40 C, and 100 D will yield a profit of $1,560 per week.

(60 $75) + (80 $72) + (40 $45) + (100 $38) = $15,860

Profits

Revenue

Materials

Labor

Overhead

Profit

(60 $10) + (80 $5) + (40 $5) + (100 $10) = –$2,200

(5 workers) (8 hours/day) (5 days/week) ($18/hour) = –$3,600

= –$8,500

= $1,560



Decision Rule 2: Bottleneck Method

Select the best product mix according to the dollar contribution margin per minute of processing time at the bottleneck workstation X. This method would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck.

Step 1: Calculate the contribution margin/minute of processing time at bottleneck workstation X:



Product A Product B Product C Product D

Contribution margin

Time at bottleneck

Contribution margin per minute



When ordered from highest to lowest contribution margin/ minute at the bottleneck, the manufacturing sequence of these products is D, C, A, B, which is reverse of the earlier order. Product D is scheduled first because it does not consume any resources at the bottleneck.

$65.00 $67.00 $40.00 $28.00

10 minutes 20 minutes 5 minutes 0 minutes

$6.50 $3.35 $8.00 Not defined

Product A Product B Product C Product D

Contribution margin

Time at bottleneck

Contribution margin per minute




Work Center


Minutes Left After Making

Minutes Left After Making

Min Left After Making

Min Left After Making

V

W

X

Y

Z




The best product mix according to this bottleneck based approach is then 60 A, 70 B, 80 C, and 100 D.

2,400 2,400 2,400 600 600

2,400 900 500 500 500

2,400 2,400 2,000 1,400 0

2,400 1,900 1,500 900 200

2,400 1,400 1,000 1,000 1,000

Work Center


Minutes Left After Making 100 D

Minutes Left After Making 80 C

Min Left After Making 60 A

Can Only Make 70 B

V

W

X

Y

Z




Profits

Revenue

Materials

Labor

Overhead

Profit




Manufacturing the product mix of 60 A, 70 B, 80 C, and 100 D will yield a profit of $2,490 per week.

(60 $75) + (70 $72) + (80 $45) + (100 $38) = $16,940

(60 $10) + (70 $5) + (80 $5) + (100 $10) = –$2,350

(5 workers) (8 hours/day) (5 days/week) ($18/hour) = –$3,600

Profits

Revenue

Materials

Labor

Overhead

Profit

= –$8,500

= $2,490


In classIn class - Example - Example

O’Neill Enterprises manufactures three unique products (A, B, C) that are fabricated and assembled in four different workstations (W, X, Y, Z) using a small batch process. Each of the products visits every one of the four workstations, though not necessarily in the same order. Batch setup times are negligible. A flowchart of the manufacturing process is shown below. O’Neill can make and sell up to the limit of its demand per week, and there are no penalties for not being able to meet all the demand. Each workstation is staffed by a worker dedicated to work on that workstation alone, and is paid $12 per hour. Variable overhead costs are $8000/week. The plant operates one 8-hour shift per day, or 40 hours/week.

Which of the four workstations W, X, Y, or Z has the highest total workload, and thus serves as the bottleneck for O’Neill Enterprises?



Product B

Raw materials

Purchased part

Product: BPrice: $85/unitDemand: 70 units/wk

Finish with step 4at workstation Z

(13 min)


(10 min)

Step 3 at workstation Y

(10 min)

Step 1 atworkstation X

(12 min)

$9

$5

Product A

Raw materials

Purchased part

Product: APrice: $90/unitDemand: 65 units/wk


(16 min)


(15 min)

Step 3 at workstation X

(9 min)


(10 min)

$7

$6

Product C

Raw materials

Purchased part

Product: CPrice: $80/unitDemand: 80 units/wk


(10 min)

Step 2 atworkstation X

(10 min)

Step 3 at workstation W

(12 min)


(5 min)

$10

$5

Flowchart for Products A, B, and C

Flowchart for Products A, B, and COverhead Costs: $8,000; Labor Costs: $12/hr (as much as worker works)


Work Station

Load from Product A

Load from Product B

Load from Product C

Total Load (minutes)

W

X

Y

Z




These calculations show that workstation Z is the bottleneck, because the aggregate work load at Z exceeds the available capacity of 2400 minutes per week.

(65x10)= 650 (7010)= 700 (8012)= 960 2310

(659)= 585 (7012)= 840 (8010)= 800 2225

(6515)= 975 (70x10)= 700 (80x5)= 400 2075

(6516)= 1040 (7013)= 910 (8010)= 800 2750

Work Station

Load from Product A

Load from Product B

Load from Product C

Total Load (minutes)

W

X

Y

Z



The senior management at O’Neill Enterprises wants to improve the profitability of the firm by accepting the right set of orders. Currently, decisions are made to accept as much of the highest contribution margin product as possible (up to the limit of its demand), followed by the next highest contribution margin product, and so on until no more capacity is available. Since the firm cannot satisfy all the demand, the product mix must be chosen carefully. Jane Hathaway, the newly hired production supervisor, is knowledgeable about the theory of constraints and bottleneck based scheduling. She believes that profitability can indeed be approved if bottleneck resources were exploited to determine the product mix. What is the change in profits if instead of the traditional method that O’Neill has used thus far; a bottleneck based approach advocated by Jane is used instead for selecting the product mix?



SOLUTION

Decision rule 1: Traditional method - Select the best product mix according to the highest overall profit margin of each product.

Step 1: Calculate the profit margin per unit of each product as shown below

A B C

Price

Raw Material & Purchased Parts

Labor

= Contribution Profit Margin



SOLUTION

Decision rule 1: Traditional method - Select the best product mix according to the highest overall profit margin of each product.

Step 1: Calculate the profit margin per unit of each product as shown below

When ordering from highest to lowest, the profit margin per unit order of these products is ABC.

$90.00 $85.00 $80.00

–13.00 –14.00 –15.00

A B C

Price

Raw Material & Purchased Parts

Labor

= Contribution Profit Margin

–10.00 –9.00 –7.40

$67.00 $62.00 $57.60



Step 2: Allocate resources W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation Z) is encountered. Subtract minutes away from 2400 minutes available for each week at each stage.

Work Center Starting After 65 A After 70 B Can Only Make 45 C

W

X

Y

Z




The best product mix is 65 A, 70 B, and 45 C

2400 1750 1050 510

2400 1815 975 525

2400 1425 725 500

2400 1360 450 0

Work Center Starting After 65 A After 70 B Can Only Make 45 C

W

X

Y

Z




Profits

Revenue

Materials

Overhead

Labor

Profit


Profits

Revenue

Materials

Overhead

Labor

Profit



Manufacturing the product mix of 65 A, 70 B, and 45 C will yield a profit of $2980.

$15400

–$2500

–$8000

–$1920

$2980



Decision rule 2: Bottleneck-based approach - Select the best product mix according to the dollar contribution per minute of processing time at the bottleneck workstation Z. This rule would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck.

Step 1: Calculate the contribution/minute of processing time at bottleneck workstation Z:

Product A Product B Product C

Contribution Margin

Time at Bottleneck

Contribution Margin per minute



Decision rule 2: Bottleneck-based approach - Select the best product mix according to the dollar contribution per minute of processing time at the bottleneck workstation Z. This rule would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck.

Step 1: Calculate the contribution/minute of processing time at bottleneck workstation Z:

When ordering from highest to lowest contribution margin/minute at the bottleneck, the manufacturing sequence of these products is CBA, which is reverse of the traditional method order.

$67.00 $62.00 $57.60

16 minutes 13 minutes 10 minutes

4.19 4.77 5.76

Product A Product B Product C

Contribution Margin

Time at Bottleneck

Contribution Margin per minute




Work Center Starting After 80 C After 70 B Can Only Make 43 A

W

X

Y

Z




The best product mix is 43A, 70B, and 80C

2400 1440 740 310

2400 1600 760 373

2400 2000 1300 655

2400 1600 690 2

Work Center Starting After 80 C After 70 B Can Only Make 43 A

W

X

Y

Z



Step 3: Compute profitability for the selected product mix. The new profitability figures are shown below based on the new production quantities of 43A, 70B, and 80C.

Profits

Revenue

Materials

Overhead

Labor

Profit


Profits

Revenue

Materials

Overhead

Labor

Profit


Step 3: Compute profitability for the selected product mix. The new profitability figures are shown below based on the new production quantities of 43A, 70B, and 80C.

Manufacturing the product mix of 43 A, 70 B, and 80 C will yield a profit of $3561.

$16220

–$2739

–$8000

–$1920

$3561


A Line ProcessA Line Process

Line Balancing Assignment of work to stations in a line so as

to achieve the desired output rate with the smallest number of workstations

Achieving the goal is similar to the theory of constraints but it differs in how it addresses bottlenecks

Precedence diagram – AON network



The desired output rate is matched to the staffing or production plan

Cycle time is the maximum time allowed for work at each station is

c =1

r

wherec = cycle time in hoursr = desired output rate



The theoretical minimum number of stations is

TM =tc

where

t =total time required to assemble each unit



Idle time, efficiency, and balance delay

Idle time = nc – t

where

n =number of stations

Efficiency (%) = (100)tnc

Balance delay (%) = 100 – Efficiency


Precedence DiagramPrecedence Diagram

EXAMPLE

Green Grass, Inc., a manufacturer of lawn and garden equipment, is designing an assembly line to produce a new fertilizer spreader, the Big Broadcaster. Using the following information on the production process, construct a precedence

diagram for the Big Broadcaster.

Work Element Description Time

(sec)Immediate

Predecessor(s)

A Bolt leg frame to hopper 40 None

B Insert impeller shaft 30 A

C Attach axle 50 A

D Attach agitator 40 B

E Attach drive wheel 6 B

F Attach free wheel 25 C

G Mount lower post 15 C

H Attach controls 20 D, E

I Mount nameplate 18 F, G

Total 244


Precedence DiagramPrecedence Diagram

SOLUTION

Figure 4 shows the complete diagram. We begin with work element A, which has no immediate predecessors. Next, we add elements B and C, for which element A is the only immediate predecessor. After entering time standards and arrows showing precedence, we add elements D and E, and so on. The diagram simplifies interpretation. Work element F, for example, can be done anywhere on the line after element C is completed. However, element I must await completion of elements F and G.

D

40

I

18

H

20

F

25

G

15

C

50

E

6

B

30

A

40

Figure 4 – Precedence Diagram for Assembling the Big Broadcaster


Calculating Cycle Time, TM, EfficiencyCalculating Cycle Time, TM, Efficiency

EXAMPLE (cont’d)

Green Grass’s plant manager just received marketing’s latest forecasts of Big Broadcaster sales for the next year. She wants its production line to be designed to make 2,400 spreaders per week for at least the next 3 months. The plant will operate 40 hours per week.

a. What should be the line’s cycle time?

b. What is the smallest number of workstations that she could hope for in designing the line for this cycle time?

c. Suppose that she finds a solution that requires only five stations. What would be the line’s efficiency?



SOLUTION

a. First convert the desired output rate (2,400 units per week) to an hourly rate by dividing the weekly output rate by 40 hours per week to get units per hour. Then the cycle time is

c = 1/r =

b. Now calculate the theoretical minimum for the number of stations by dividing the total time, t, by the cycle time, c = 60 seconds. Assuming perfect balance, we have

TM =tc

244 seconds

60 seconds= = 4.067 or 5 stations

1/60 (hr/unit) = 1 minute/unit = 60 seconds/unit



c. Now calculate the efficiency of a five-station solution, assuming for now that one can be found:

Efficiency = (100) =tnc

244

5(60)= 81.3%


Finding a SolutionFinding a Solution

The goal is to cluster the work elements into workstations so that 1. The number of workstations required is minimized

2. The precedence and cycle-time requirements are not violated

The work content for each station is equal (or nearly so, but less than) the cycle time for the line

Trial-and-error can be used but commercial software packages are also available


Finding a SolutionFinding a Solution

The theoretical minimum number of workstations is 5 and the cycle time is 60 seconds, so Figure 5 represents an optimal solution to the problem

Figure 5 – Big Broadcaster Precedence Diagram Solution

D

40

I

18

H

20

F

25C

50

E

6

B

30

A

40

G

15



A plant manager needs a design for an assembly line to assembly a new product that is being introduced. The time requirements and immediate predecessors for the work elements are as follows:

Work Element Time (sec) Immediate Predecessor

A 12 ―

B 60 A

C 36 ―

D 24 ―

E 38 C, D

F 72 B, E

G 14 ―

H 72 ―

I 35 G, H

J 60 I

K 12 F, J

Total = 435


K


Draw a precedence diagram, complete I, F, J, and K

Work Element Time (sec) Immediate

Predecessor

A 12 ―

B 60 A

C 36 ―

D 24 ―

E 38 C, D

F 72 B, E

G 14 ―

H 72 ―

I 35 G, H

J 60 I

K 12 F, J

Total = 435

F

J

B

E

I

A

C

G

H

D



If the desired output rate is 30 units per hour, what are the cycle time and theoretical minimum?

c = =1r

1

30(3600) = 120 sec/unit

TM =tc

= = 3.6 or 4 stations435

120



Suppose that we are fortunate enough to find a solution with just four stations. What is the idle time per unit, efficiency, and the balance delay for this solution?

Idle time = nc – t

Efficiency (%) = (100)tnc

Balance delay (%) = 100 – Efficiency

= 4(120) – 435 = 45 seconds

= 100 – 90.6 = 9.4%

= (100) = 90.6%435

480


Station

Work Elements Assigned Cumulative Time

Idle Time(c = 120)

1

2

3

4

5


Using trial and error, one possible solution is shown below.



Using trial and error, one possible solution is shown below.

H, C, A 120 0

B, D, G 98 22

E, F 110 10

I, J, K 107 13

A fifth station is not needed

Station

Work Elements Assigned Cumulative Time

Idle Time(c = 120)

1

2

3

4

5


Managerial ConsiderationsManagerial Considerations

Pacing is the movement of product from one station to the next

Behavioral factors such as absenteeism, turnover, and grievances can increase after installing production lines

The number of models produced complicates scheduling and necessitates good communication

Cycle times are dependent on the desired output rate


Solved Problem 2Solved Problem 2

A company is setting up an assembly line to produce 192 units per 8-hour shift. The following table identifies the work elements, times, and immediate predecessors:

Work Element Time (sec) Immediate Predecessor(s)

A 40 None

B 80 A

C 30 D, E, F

D 25 B

E 20 B

F 15 B

G 120 A

H 145 G

I 130 H

J 115 C, I

Total 720



a. What is the desired cycle time (in seconds)?

b. What is the theoretical minimum number of stations?

c. Use trial and error to work out a solution, and show your solution on a precedence diagram.

d. What are the efficiency and balance delay of the solution found?

SOLUTION

a. Substituting in the cycle-time formula, we get

c = =1r

8 hours

192 units(3,600 sec/hr) = 150 sec/unit



b. The sum of the work-element times is 720 seconds, so

TM =tc

= = 4.8 or 5 stations720 sec/unit

150 sec/unit-station

which may not be achievable.



c. The precedence diagram is shown in Figure 7.6. Each row in the following table shows work elements assigned to each of the five workstations in the proposed solution.

J

115

C

30

D

25

E

20

F

15

I

130

H

145

B

80

G

120

A

40

Figure 7.6 – Precedence Diagram

Work Element

Immediate Predecessor(s)

A None

B A

C D, E, F

D B

E B

F B

G A

H G

I H

J C, I


Station Candidate(s) ChoiceWork-Element

Time (sec)Cumulative Time (sec)

Idle Time(c= 150 sec)

S1

S2

S3

S4

S5


J

115

C

30

D

25

E

20

F

15 I

130H

145

B

80

G

120

A

40



J

115

C

30

D

25

E

20

F

15 I

130H

145

B

80

G

120

A

40

A A 40 40 110

B B 80 120 30

D, E, F D 25 145 5

E, F, G G 120 120 30

E, F E 20 140 10

F, H H 145 145 5

F, I I 130 130 20

F F 15 145 5

C C 30 30 120

J J 115 145 5

Station Candidate(s) ChoiceWork-Element

Time (sec)Cumulative Time (sec)

Idle Time(c= 150 sec)

S1

S2

S3

S4

S5



d. Calculating the efficiency, we get

Thus, the balance delay is only 4 percent (100–96).

Efficiency (%) = (100)tnc =

720 sec/unit

5(150 sec/unit)

= 96%

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