Ch02 Manual Work 1

Work Systems and the Methods, Measurement, and Management of Workby Mikell P. Groover, ISBN 0-13-140650-7.

©2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Work Systems and How They Work

Chapters:2. Manual Work and Worker-Machine

Systems3. Work Flow and Batch Processing4. Manual Assembly Lines5. Logistics Operations6. Service Operations and Office Work7. Projects and Project Management

Part I



Manual Work & Worker-Machine Systems

Sections:1. Manual Work Systems2. Worker-Machine Systems3. Automated Work Systems4. Determining Worker and Machine

Requirements5. Machine Clusters

Chapter 2



Three Categories of Work Systems

1. Manual work systemWorker performs one or more tasks without the aidof powered tools (e.g. hammers, screwdrivers,shovels)

2. Worker-machine systemHuman worker operates powered equipment (e.g.a machine tool)

Physical effort (less)Machine power(more)

3. Automated work systemProcess performed without the direct participationof a human worker



A Work System as a Physical Entity



Manual Work System



Worker-Machine System



Automated System



Manual Work SystemsMost basic form of work in which human bodyis used to accomplish some physical taskwithout an external source of power

With or without hand toolsEven if hand tools are used, the power to operatethem is derived from the strength and stamina of ahuman workerHairbrush vs hair dryer

Of course other human faculties are alsorequired, such as hand-eye coordination andmental effort



Pure Manual Work

Involves only the physical and mentalcapabilities of the human worker withoutmachines or tools.

Material handler moving cartons in a warehouseWorkers loading furniture into a moving van withoutthe use of dolliesDealer at a casino table dealing cardsOffice worker filing documentsAssembly worker snap-fitting two parts together



Manual Work with Hand Tools

Manual tasks are commonly augmented by useof hand tools.

Tool is a device for making changes to objects(formally work units) such as cutting,grinding,striking, sequeezing

Scissor, screwdriver, shovel

Tools can also be used for measurementand/or analysis purposes

Workholder to grasp or poisiton work units



Manual Work with Hand Tools

Machinist filing a partAssembly worker using screwdriverPainter using paintbrush to paint door trimQC inspector using micrometer to measure thediameter of a shaftMaterial handling worker using a dolly to movefurnitureOffice worker writing with a pen



Repetitive vs. Nonrepetitive Tasks

Repetitive TaskWork cycle is relatively short (usually a few minutesor less)High degree of similarity from one cycle to the next

Nonrepetitive TaskWork cycle takes a long timeWork cycles are not similar

In either case, the task can be divided intowork elements that consist of logical groupingsof motions



Work Element Example:

Grommet : sealant like ring



REMINDER: The Pyramidal Structure of Work

Tasks consist of work elements

Work elements consist of basic motion elements

Required time



Cycle Time Analysis

Cycle time Tc

whereTek= time of work element k, where k is used toidentify the work elements (min)ne = number of work elements into which acycle is divided.

1

en

c ekk

T T=

= ∑



Example 2.1: A repetitive Manual Task

Current method: An assembly worker performs arepetitive task consisting of inserting 8 pegs into 8 holesin a board. A sightly interference fit is involved in eachinsertion. The worker holds the board in one hand andpicks up the pegs from a tray with other hand andinserts them into the holes, one peg at a time.



Current method and current layout:




Improved method and improved layout:Use a work-holding device to hold and position theboard while the worker uses both handssimultaneously to insert pegs.Instead of picking one peg at a time, each hand willgrab four pegs to minimize the number of times theworker’s hands must reach the trays.




Improved method

The cycle time is reduced from 0.62 min to 0.37 min.

% cycle time reduction=(CTcurrent-CTimproved)/CTcurrent

=(0.62-0.37)/0.62=%40




Production ratecurrent=1/0.62 min=1.61 units per min(throughput)Production rateimproved=1/0.37 min=2.70 units per min

% increase in R=(Rimproved-Rcurrent)/Rcurrent

=(1.61-2.70)/1.61=%68

It is important to design the work cycle so as to minimizethe time required to perform it.

Of course there are many alterantive ways to perform agiven task. Our focus is on the best one.




One Best Method Principle

Of all the possible methods that can be used to perform agiven task, there is one optimal method that minimizes thetime and effort required to accomplish it

Attributed to Frank Gilbreth

A primary objective in work design is to determine the one

best method for a task, and then to standardizeit!



One Best Method Principle

This ‘One Best’ refers to:1. an average worker with a moderate level

of skill,2. operating under normal working conditions3. with nominal material quality and4. tool/equipment availability



Cycle Time Variations

Once the method has been standardized, theactual time to perform the task is a variablebecause of:

Differences in worker performanceMistakes, failures and errorsVariations in starting work unitsVariations in hand and body motionsExtra elements not performed every cycleDifferences among workersThe learning curve phenomenon



Worker Performance

Defined as the pace (tempo) or relative speedwith which the worker does the task.

As worker performance increases, cycle timedecreases

From the employer’s viewpoint, it is desirablefor worker performance to be high

What is a reasonable performance/pace toexpect from a worker in accomplishing a giventask?

What is a reasonable performance for a worker?





Normal Performance (normal pace)

A pace of working that can be MAINTAINED by a properlytrained average worker throughout an entire work shiftwithout harmful short-term or long-term effects on theworker’s health or physical well-being




The work shift is usually 8 hours, duringwhich periodic rest breaks are allowed

Normal performance = 100%performance

Faster pace > 100%, slower pace < 100%

Common benchmark of normalperformance:

Walking at 3 mi/hr (~4.83 km/hr)

The normal pace refers to the pace of worker while actually working (does not include or relate to rest periods)






Normal Time

The time to complete a task when working atnormal performance

Actual time to perform the cycle depends onworker performance

Tc = Tn / Pw

whereTc = cycle time,Tn = normal time,Pw = worker performance (or pace)



Example 2.2: Normal Performance

Given: A man walks in the early morning forhealth and fitness. His usual route is 1.85 miles.The benchmark of normal performance = 3mi/hr.

Determine:(a) how long the route would take at normalperformance(b) the man’s performance when he completesthe route in 30 min.



Example 2.2: Solution

(a) At 3 mi/hr, time = 1.85 mi / 3 mi/hr= 0.6167 hr = 37 min

(b) Rearranging equation, Pw = Tn / TcPw = 37 min / 30 min = 1.233 = 123.3 %

or an alternative approach in (b):Using v = 1.85 mi / 0.5 hr = 3.7 mi/hrPw = 3.7 mi/hr / 3.0 mi/hr = 1.233

If worker performance > 100%, then the time required tocomplete the cycle will be less than normal time.

If worker performance < 100%, then the time required tocomplete the cycle will be greater than normal time.

233.10.37.3===

VnVcPw

233.1min30min37

===TnTcPw



Standard Performance

Same as normal performance, butacknowledges that periodic rest breaks mustbe taken by the worker

Periodic rest breaks are allowed during thework shift

Lunch breaks (1/2 or 1 hour)usually not counted as part of work shifts

Shorter rest beraks (15 mins)usually counted as part of work shifts



Rest Breaks in a Work ShiftA typical work shift is 8 hours (8:00 A.M. to 5:00 P.M.with one hour lunch break)

E.g. In one country work time is defined as 45 hoursa week(so 8:00 A.M. to 6:00 P.M. with one hour lunch break,provided that workers work for 5 days)

The shift usually includes one rest break in the morningand another in the afternoon.

The employers allows these breaks, because they knowthat the overall productivity of a worker is higher if restbreaks are allowed.

The rest periods are not included in daily work hours inwhich employers are paid for.



Standard Performance

Of course other interruptions and delays alsooccur during the shift

Machine breakdowns

Receiving instructions from the foreman

Telephone calls

Bathroom/toilette breaks etc.



Personal time, Fatigue, Delay (PFD)Allowance

To account for the delays and rest breaks, anallowance is added to the normal time in orderto determine allowed time for the worker toperform the task throughout a shift

Personal time (P)Bathroom breaks, personal phone calls

Fatigue (F)Rest breaks are intended to deal with fatigue

Delays (D)Interruptions, equipment breakdowns



Standard TimeDefined as the normal time but with an allowance addedinto account for losses due to personal time, fatigue, anddelays

Tstd = Tn (1 + Apfd)whereTstd = standard time,Tn = normal time,Apfd = PFD allowance factor

Also called the allowed time

Now we are confident to say that a worker working at100% performance during 8 hours CAN ACCOMPLİSHa task of 8 hour standard time.



Irregular Work Elements

Elements that are performed with a frequencyof less than once per cycle

Examples:Changing a toolExchanging parts when containers become full

Irregular elements are prorated into the regularcycle according to their frequency



Example 2.3: Determining Standard Timeand Standard Output

Given: The normal time to perform the regularwork cycle is 3.23 min. In addition, an irregularwork element with a normal time = 1.25 min isperformed every 5 cycles. The PFD allowancefactor is 15%.

Determine(a) the standard time(b) the number of work units produced duringan 8-hr shift if the worker's pace is consistentwith standard performance.




(a) Normal time Tn = 3.23 + 1.25/5= 3.48 min

Standard time Tstd = 3.48 (1 + 0.15)= 4.00 min

(b) Number of work units produced during an 8-hrshift

Qstd = 8.0(60)/4.00 = 120 work units

Normal time of a task involves normal timesfor regular and irregular work elements



Example 2.4: Determining Lost Time due to the Allowance Factor

Given: An allowance factor of 15% is used.

Determine the anticipated amount of time lostper 8-hour shift.

Solution:8.0 hour =(actual time worked) (1+0.15)

Actual time worked = 8/ 1.15 = 6.956 hr

Time lost = 8.0 – 6.956 = 1.044 hr



Example 2.5: Production rate when worker performance exceeds 100%

Given: Tsd=4.00 min. The worker’s averageperformance during an 8-hour shift is 125%and the hours actually worked is 6.956 hr(which corresponds to the 15% allowancefactor).

Determine daily production rate.




Based on normal time Tn=3.48 min, the actualcycle time with a worker performance of 125%,Tc=3.48 / 1.25 = 2.78 min.

Assuming one work unit is produced eachcycle, the corresponding daily production rate,Rp=6.956(60)/2.78=150 work unitsOR125% of 120 units (we know that from Exercise2.3.b) at 100% performance = 150 units



Standard Hours and Worker Efficiency

Two (three) common measures of workerproductivity used in industry

Standard hours – represents the amount of workactually accomplished during a given period (shift,week)

Quantity of work units (in terms of time) producedHstd = Q Tstd

whereHstd =standard hours accomplished, hrQ = quantity of work units completed during the

period, pcTstd =standard time per work unit, hr/pc



Standard Hours and Worker Efficiency

Two (three) common measures of worker productivity used in industry

Worker efficiency – work accomplished during the shift expressed as a proportion of shift hours

Ew = Hstd / Hsh

where Hstd =standard hours accomplished, hrEw =worker efficiency, normally expressed as a

percentage, hrHsh =number of shift hours, hr



Example 2.6: Standard hours and worker efficiency

Given: The worker performance of 125% in the previousexample.

Determine:(a) number of standard hours produced(b) worker efficiency

Solution:(a) Hstd=150(4 min)=600 min= 10.0 hr

(Hstd = Q Tstd)(b) Ew = 10hr / 8 hr =125 %

(Ew = Hstd / Hsh)



Example 2.6: Standard hours and worker efficiency

Note that worker efficiency is found to beequal to the worker performance (rate).

What are the reasons for that?

The number of hours actually worked is consistentwith 15% allowance factor.

The entire work cycle consists of manual labor.So, worker efficiency=worker performance (rate)



Example 2.7: Standard hours and worker efficiency as affected by hours actually worked

Given: The worker performance of 125%, actual hoursworked is 7.42 hr.

Determine:(a) number of pieces produced,(b) number of standard hours accomplished,(c) the worker’s efficiency

Solution:(a) Tc= 2.78 (prev. example), Q=7.42(60)/2.78=160units(b) Hstd=160(4 min)=640 min= 10.67 hr(c) Ew = 10.67hr / 8 hr =133.3 %



Example 2.7: Standard hours and worker efficiency as affected by hours actually worked

Note that in this example worker efficiency, Ew, andworker pace, Pw, are not equivalent.

The reason for thatActual work hours of that labor (7.42–given in theexample) is greater than the allowed time (or anticipatedwork time, which is found to be 6.956 hr in Example 2.4)

That is, on that specific day, the worker delays (8-7.42=0.58hr) and the rest breaks are less than theanticipated time loss due to PDF allowances (1.044hr).

It is better to calculate worker efficiency by using actualoutputs (in terms of hour).



More on Worker Efficiency

Worker efficiency is commonly used to evaluateworkers in industry.

In many incentive wage payment plans, the worker’searnings are based on

worker’s efficiency, Ew,orthe number of standard hours accomplished, Hstd.

Either one of these two measures can be derived fromthe other one. Thus, they are equivalent.

Ch02 Manual Work 1

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