Manual Work Systems

Work Systems and How They WorkPart - 2

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

Requirements5. Machine Clusters

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Three Categories of Work Systems1. Manual work system

– Worker performs one or more tasks without the aid of powered tools (e.g. hammers, screwdrivers, shovels)

2. Worker-machine system– Human worker operates powered equipment (e.g. a

machine tool)• Physical effort (less)• Machine power(more)

3. Automated work system– Process performed without the direct participation of a

human worker

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Manual Work System

Worker-Machine System

Automated System

Some Definitions Work unit – the object that is processed by the

work system– Workpiece being machined (production work)– Material being moved (logistics work)– Customer in a store (service work)– Product being designed (knowledge work)

Unit operations – tasks and processes that are treated as being independent of other work activities– As opposed to sequential operations (sequence of

operations required to manufacture a product or deliver a service)

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Manual Work Systems Most basic form of work in which human body is

used to accomplish some physical task without an external source of power

With or without hand tools– Even if hand tools are used, the power to operate them

is derived from the strength and stamina of a human worker

– Hairbrush vs hair dryer

Of course other human faculties are also required, such as hand-eye coordination and mental effort

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Pure Manual Work Involves only the physical and mental

capabilities of the human worker without machines or tools.– Material handler moving cartons in a warehouse– Workers loading furniture into a moving van

without the use of dollies– Dealer at a casino table dealing cards– Office worker filing documents– Assembly worker snap-fitting two parts together

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Manual Work with Hand Tools Manual tasks are commonly augmented by

use of 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 measurement and/or analysis purposes

Workholder to grasp or poisiton work units

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Manual Work with Hand Tools Machinist filing a part Assembly worker using screwdriver Painter using paintbrush to paint door trim QC inspector using micrometer to measure

the diameter of a shaft Material handling worker using a dolly to

move furniture Office worker writing with a pen

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Repetitive vs. Non-repetitive Tasks

Repetitive Task– Work cycle is relatively short (usually a few

minutes or less)– High degree of similarity from one cycle to the next

Non-repetitive Task– Work cycle takes a long time– Work cycles are not similar

In either case, the task can be divided into work elements that consist of logical groupings of motions

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Cycle Time Analysis

Cycle time Tc

where Tek = time of work element k, where k is used to identify the work elements (min) ne = number of work elements into which a cycle is divided.

1

en

c ekk

T T

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Example 2.1: A repetitive Manual Task Current method: An assembly worker performs a

repetitive task consisting of inserting 8 pegs into 8 holes in a board. A sightly interference fit is involved in each insertion. The worker holds the board in one hand and picks up the pegs from a tray with other hand and inserts them into the holes, one peg at a time.

Current method and current layout:Example 2.1: A repetitive Manual Task

Improved method and improved layout: Use a work-holding device to hold and position

the board while the worker uses both hands simultaneously to insert pegs.

Instead of picking one peg at a time, each hand will grab four pegs to minimize the number of times the worker’s hands must reach the trays.

Example 2.1: A repetitive Manual Task

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

Example 2.1: A repetitive Manual Task

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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

minimize the time required to perform it. Of course there are many alterantive ways to

perform a given task. Our focus is on the best one.

Example 2.1: A repetitive Manual Task

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One Best Method Principle Of all the possible methods that can be used to

perform a given task, there is one optimal method that minimizes the time 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 standardize it This one best refers to an average worker with a

moderate level of skill, operating under normal working conditions with nominal material quality and tool/equipment availability

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Cycle Time Variations Once the method has been standardized,

the actual time to perform the task is a variable because of:– Differences in worker performance– Mistakes, failures and errors– Variations in starting work units– Variations in hand and body motions– Extra elements not performed every cycle– Differences among workers– The learning curve phenomenon

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Worker Performance Defined as the pace (tempo) or relative speed

with which the worker does the task. As worker performance increases, cycle time

decreases From the employer’s viewpoint, it is desirable

for worker performance to be high What is a reasonable performance/pace to

expect from a worker in accomplishing a given task?

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Normal Performance (pace) A pace of working that can be maintained by a

properly trained average worker throughout an entire work shift without harmful short-term or long-term effects on the worker’s health or physical well-being

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

Normal performance = 100% performance– Faster pace > 100%, slower pace < 100%

Common benchmark of normal performance: – Walking at 3 mi/hr (~4.83 km/hr)

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Normal Time The time to complete a task when working at normal

performance Actual time to perform the cycle depends on worker

performance

Tc = Tn / Pw where Tc = cycle time,

Tn = normal time,

Pw = worker performance or pace

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Example 2.2: Normal Performance Given: A man walks in the early morning, for

health and fitness. His usual route is 1.85 miles. The benchmark of normal performance = 3 mi/hr.

Determine: a)how long the route would take at normal

performance b)the man’s performance when he completes

the route in 30 min.

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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 / Tc

Pw = 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/hr

Pw = 3.7 mi/hr / 3.0 mi/hr = 1.233

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

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

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Standard Performance Same as normal performance, but

acknowledges that periodic rest breaks must be taken by the worker

Periodic rest breaks may be allowed during the work 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

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Rest Breaks in a Work Shift A typical work shift is 8 hours (8:00 A.M. to

5:00 P.M. with one hour lunch break)– work time is usually defined as 40 hours a week

(so 8:00 A.M. to 5:00 P.M. with one hour lunch break, provided that workers work for 5 days)

The shift usually includes one rest break The employers allows these breaks, because

they know that the overall productivity of a worker is higher if rest breaks are allowed.

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Standard Performance Of course other interruptions and

delays also occur during the shift– Machine breakdowns– Receiving instructions from the foreman– Telephone calls– Bathroom/toilette breaks etc.

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Personal time, Fatigue, Delay (PFD) Allowance To account for the delays and rest breaks, an

allowance is added to the normal time in order to determine allowed time for the worker to perform 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

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Standard Time Defined as the normal time but with an allowance

added into account for losses due to personal time, fatigue, and delaysTstd = Tn (1 + Apfd)where Tstd = standard time, Tn = normal time, Apfd = PFD allowance factor

Also called the allowed time Now we are confident to say that a worker working at

100% performance during 8 hours can accomplish a task of 8 hour standard time.

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Irregular Work Elements Elements that are performed with a

frequency of less than once per cycle Examples:

– Changing a tool– Exchanging parts when containers become full

Irregular elements are prorated into the regular cycle according to their frequency

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Example 2.3: Determining Standard Time and Standard Output Given: The normal time to perform the regular

work cycle is 3.23 min. In addition, an irregular work element with a normal time = 1.25 min is performed every 5 cycles. The PFD allowance factor is 15%.

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

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Example 2.3: Solution(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-hr shiftQstd = 8.0(60)/4.00 = 120 work units

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

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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 lost per 8-hour shift. Solution:

8.0 hour =(actual time worked) (1+0.15)Actual time worked = 8/ 1.15 = 6.956 hrTime lost = 8.0 – 6.956 = 1.044 hr

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Example 2.5: Production rate when worker performance exceeds 100%

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

Determine daily production rate.

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Example 2.5: Solution Based on normal time Tn=3.48 min, the actual

cycle time with a worker performance of 125%, Tc=3.48 / 1.25 = 2.78 min.

Assuming one work unit is produced each cycle, the corresponding daily production rate, Rp=6.956(60)/2.78=150 work units

OR 125% of 120 units (we know that from Exercise

2.3.b) at 100% performance = 150 units

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Standard Hours and Worker Efficiency Two (three) common measures of worker

productivity used in industry– Standard hours – represents the amount of work

actually accomplished during a given period (shift, week)

– Quantity of work units (in terms of time) producedHstd = Q Tstd where Hstd =standard hours accomplished, hrQ = quantity of work units completed during the period, pcTstd =standard time per work unit, hr/pc

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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 hoursEw = Hstd / Hsh

where Hstd =standard hours accomplished, hr

Ew =worker efficiency, normally expressed as a percentage, hrHsh =number of shift hours, hr

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Example 2.6: Standard hours and worker efficiency Given: The worker performance of 125% in the previous

example. 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 be equal to the worker performance (rate).

What are the reasons for that?– The number of hours actually worked is

consistent with 15% allowance factor.– The entire work cycle consists of manual

labor. • So, worker efficiency=worker performance (rate)

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Example 2.7: Standard hours and worker efficiency as affected by hours actually worked Given: The worker performance of 125%,

actual hours worked 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=160 units (Tn=3.48 min, with a worker performance of 125%, Tc=3.48 / 1.25 = 2.78 min)(b) Hstd=160(4 min)=640 min= 10.67 hr(c) Ew = 10.67hr / 8 hr =133.3 %

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

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

The reason for that – Actual work hours of that labor (7.42–given in the example)

is greater than the allowed time (or anticipated work time, which is found to be 6.956 hr in Example 2.4) , (with an allowance of 15% Actual time worked = 8/ 1.15 = 6.956 hr)

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

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

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More on Worker Efficiency Worker efficiency is commonly used to evaluate

workers in industry. In many incentive wage payment plans, the worker’s

earnings are based on – worker’s efficiency, Ew,

or– the number of standard hours accomplished, Hstd.

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

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Work Systems and the Methods, Measurement, and Management of Work

by Mikell P. Groover, ISBN 0-13-140650-7.

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

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