WSD1 2012 06 Manual Work Systems1
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Transcript of WSD1 2012 06 Manual Work Systems1
Manual Work Systems
Work Systems and How They WorkPart - 2
Harsh V Bhasin
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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