Chap 7 Constraint Management

8/3/2019 Chap 7 Constraint Management

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7 1CONSTRAINT MANAGEMENT

Constraint ManagementConstraint Management

7


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7 2

CONSTRAINTCONSTRAINT

Any factor that limits the performance of asystem and restricts its output.

Constraints can occur up or down the supplychain, with either the firms suppliers or

customers, or within one of firmsprocesses like service/productdevelopment or order fulfillment.

CONSTRAINT MANAGEMENT


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KINDS OF CONSTRAINTSKINDS OF CONSTRAINTS

Three kinds of constraints can generally beidentified:

Physical (usually machine, labor orworkstation capacity or material shortages)

Market ( demand is less than capacity)

Managerial ( policy, metrics, or mind-setsthat create constraints which impede workflow)



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BOTTLENECKBOTTLENECK

A bottleneck is a special type of constraintthat relates to:

The capacity shortage of a process, & isdefined as

Any resource whose available capacitylimits the organizations ability to meet theservice or product volume, product mix, orfluctuating requirements demanded by the

market place



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MANAGING CONSTRAINT ACROSS THEMANAGING CONSTRAINT ACROSS THEORGANIZATIONORGANIZATION

Firms must manage their constraints andmake appropriate capacity choices at theindividual-process level, as well as at theorganization level.

Hence this process involves inter-functionalcooperation

A bottleneck could be the sales departmentnot getting enough sales or the loan

department not processing loans fastenough as in many banks.

The constraint could be a lack of capital orequipment or it could be planning or

schedulingCONSTRAINT MANAGEMENT


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THEORY OF CONSTRAINTTHEORY OF CONSTRAINT

A systematic management approach thatfocuses on actively managing thoseconstraints that impede a firms progresstoward its goals

TOC methods increase the firms profitsmore effectively by focusing on makingmaterials flow rapidly through the entiresystem

They help firms look at the big picturehowprocesses can be improved to increaseoverall work flows, & how inventory andworkforce levels can be reduced while still

effectively utilizing critical resources.CONSTRAINT MANAGEMENT


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Theory of ConstraintsTheory of Constraints

Bottlenecks can both be internal or external to thefirm and are typically a process or step with thelowest capacity

Throughput time is the total elapsed time from thestart to the finish of a job or a customer being

processed at one or more work centers

A bottleneck can be identified in several differentways

1. If it has the highest total time per unit processed

2. If it has the highest average utilization and total workload3. If a reduction of processing time would reduce the

average throughput time for the entire process


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Identifying the BottleneckIdentifying the Bottleneck

EXAMPLE 7.1

Managers at the First Community Bank are attempting to shorten

the time it takes customers with approved loan applications to

get their paperwork processed. The flowchart for this process,

consisting of several different activities, each performed by a

different bank employee, is shown in Figure.

Approved loan applications first arrive at activity or step 1, where

they are checked for completeness and put in order.


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At step 2, the loans are categorized into different classes

according to the loan amount and whether they are being

requested forpersonalorcommercial reasons.

While credit checking commences at step 3, loan application

data are entered in parallel into the information system for

record-keeping purposes at step 4.

Finally, all paperwork for setting up the new loan is finished at

step 5. The time taken in minutes is given in parentheses.



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Complete paperworkfor new loan

(10 min)


Check for credit rating(15 min)

Enter loan applicationinto the system

(12 min)

Categorize loans(20 min)

Check loan documentsand put them in order

(15 min)

Figure: Processing Credit Loan Applications at First Community Bank

Which single step is the bottleneck? The management is alsointerested in knowing the maximum number of approved loansthis system can process in a 5-hour work day.


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SOLU

TION

We define the bottleneck as step 2, where a single-minutereduction in its time reduces the average throughput time of theentire loan approval process. The throughput time to complete anapproved loan application is 15 + 20 + max(15, 12) + 10 = 60minutes. Although we assume no waiting time in front of any

step, in practice such a smooth process flow is not always thecase. So the actual time taken for completing an approved loanwill be longer than 60 minutes due to non uniform arrival ofapplications, variations in actual processing times, and therelated factors.

The capacity for loan completions is derived by translating theminutes per customer at the bottleneck step to customer perhour. At First Community Bank, it is 3 customers per hourbecause the bottleneck step 2 can process only 1 customer every20 minutes (60/3).


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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 smallestnumber of workstations

Achieving the goal is similar to the theory ofconstraints but it differs in how it addressesbottlenecks

Precedence diagram AON network


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Precedence DiagramPrecedence Diagram

EXAMPLE

Green Grass, Inc., a manufacturer of lawn and gardenequipment, is designing an assembly line to produce a newfertilizer spreader, the Big Broadcaster. Using the followinginformation on the production process, construct a precedence

diagram for theBig Broadcaster.

WorkElement

Description Time(sec)

ImmediatePredecessor(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


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Precedence DiagramPrecedence Diagram

SOLU

TION

Figure shows the complete diagram. We begin with work elementA, which has no immediate predecessors. Next, we add elementsB 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 simplifiesinterpretation. Work element F,for example, can be doneanywhere on the line afterelement C is completed.However, element I must

await completion ofelements F and G.

D

40

I

18

H

20

F

25

G

15

C

50

E

6

B

30

A

40

Figure: Precedence Diagram for

Assembling the Big Broadcaster


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CYCLE TIMECYCLE TIME

Cycle time is the maximum time allowed forwork on a unit at each station.

c =1

r

wherec = cycle time in hours

r = desired output rate


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THEORETICAL MINIMUMTHEORETICAL MINIMUM

A benchmark or goal for the smallestnumber of stations possible, where the totaltime required to assemble each unit isdivided by the cycle time.

TM =7t

c

where

7t= total time required to assembleeach unit


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Idle time, efficiency, and balance delay

Idle time = nc 7t

where

n = number of stations

Efficiency (%) = (100)7t

nc

Balance delay (%) = 100 Efficiency


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

EXAMPLE

Green Grasss plant manager just received marketings latestforecasts ofBig Broadcaster sales for the next year. She wantsits production line to be designed to make 2,400 spreaders perweek for at least the next 3 months. The plant will operate 40

hours per week.

a. What should be the lines cycle time?

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

c.S

uppose that she finds a solution that requires only fivestations. What would be the lines efficiency?


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SOLUTION

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

c = 1/r =

b. Now calculate the theoretical minimum for the number ofstations by dividing the total time, 7t, by the cycle time,c = 60 seconds. Assuming perfect balance, we have

TM =7t

c

244 seconds

60 seconds= = 4.067 or 5 stations

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


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

Efficiency = (100) =7t

nc244

5(60)= 81.3%


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The theoretical minimum number of workstationsis 5 and the cycle time is 60 seconds, so Figure7.5 represents an optimal solution to the problem

Figure 7.5 Big Broadcaster Precedence Diagram Solution

D40

I

18

H

20

F

25

C

50

E

6

B

30

A

40

G

15


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Solved ProblemSolved Problem

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

Work Element Time (sec) Immediate Predecessor(s)

A 40 NoneB 80 A

C 30 D, E, F

D 25 B

E 20 B

F 15 B

G 120 AH 145 G

I 130 H

J 115 C, I

Total 720


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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 yoursolution on a precedence diagram.

d. What are the efficiency and balance delay of the solutionfound?

SOLUTION

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

c = =1

r

8 hours

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


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

TM =7t

c

= = 4.8 or 5 stations720 sec/unit

150 sec/unit-station


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c. The precedence diagram is shown in Figure. Each row in thefollowing table shows work elements assigned to each of thefive 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 Precedence Diagram

WorkElement

ImmediatePredecessor(s)

A None

BA

C D, E, F

D B

E B

F B

G A

H G

I H

J C, I


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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


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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


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Solved ProblemSolved Problem

d. Calculating the efficiency, we get

Thus, the balance delay is only 4 percent (10096).

Efficiency (%) = (100)7t

nc

=720 sec/unit

5(150 sec/unit)

= 96%

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Chap 7 Constraint Management