Lecture 8 - Time Study PDF

Lecture 8 – Time Study Methods Engineering Process ✓ Select the project ✓ Gather & summarize existing information about the operation, and present ✓ Analyze the information ✓ Develop ideal method ✓ Present ideal, and install method ✓ Job analysis  Establish time standards  Follow-up Methods f...

Lecture 8 – Time Study Methods Engineering Process ✓ Select the project ✓ Gather & summarize existing information about the operation, and present ✓ Analyze the information ✓ Develop ideal method ✓ Present ideal, and install method ✓ Job analysis  Establish time standards  Follow-up Methods for establishing time standards  Estimation  Records  Work measurement techniques ◼ Time study ◼ Standard data ◼ Time formulas ◼ Fundamental motion data ◼ Work sampling Work Measurement  Objective, data-driven process for establishing time standards  Uses:  planning  comparing alternative work methods  determining capacity  justify purchase of new equipment  develop pay system  i.d. employee training needs  ….. Time Standards  Appropriate and reasonable amount of time to perform a specific work task.  Based on concept of a fair day’s work: “...amount of work that can be produced by a qualified employee when working at a normal pace and effectively utilizing his time where work is not restricted by process limitations.” Time Study  A common method for establishing time standards  Steps:  Measure work content of prescribed method  Evaluate studied operator’s performance  Add in allowances for PFD (personal delays , fatigue, and unavoidable delays)  Calculate the Standard time for the task: Standard = Work x Perf. x (1+PFD) Time Content Rating Time Total Job Time  Work content: the amount of work contained in a product or process, in person-hrs or machine-hrs. Basic work Total content (WC) work Excess WC from Total content design or spec. defects operation Excess WC from poor time methods Total Due to management shortcomings ineffective Factors under worker time control Time Study Procedural Requirements  standardized method in place  study an operator who is proficient in the standardized method  preparation for time study  personnel  equipment  materials  knowledgeable analyst Time Study Equipment  The basics:  Stopwatch  Clipboard  Time study form  Calculator sc507_m Time Study Form  Record pertinent information that completely describes the job as studied that day.  operator  analyst  date  tools  work conditions  department  travel and reach distances  work station layout sketch  … Time study form Time study form Time Study Preparation Decisions  Stopwatch operation:  snapback V.  continuous  Define the elements of the task  Determine required number of work cycles to study Measuring Work Content  Observe job, and divide into elements with distinct start and stop points Cycle 1 Cycle 2 1 2 3 1 2 3 Time Defining Work Elements  Cohesive groups of therbligs  As fine as needed  Definite start and stop points  Stop for Elementi is start for Elementi+1  Repeated and necessary  Separate machine and operator elements  Separate constant and variable elements  Assign labels and record in sequence on time study form Measuring Work Content Stopwatch Readings Cycle 1 1 2 3 t1 t2 t3 Cycle 2 1 2 3 t4 t5 t6 Cycle 3 1 2 3 t7 t8 t9 Cycle 4 1 2 3. t10 t11 t12.. How many cycles to study? How many cycles to study?  Basis is the confidence interval: X- Zs    X+ Zs n>30 n1/2 n1/2 X- ts    X+ ts n30 n1/2 n1/2 X- ts    X+ ts n /2=.025)  you decide you need the element sample mean time to be within  10% of the true mean time, the population mean (so set k to 10%, or 0.1) How many cycles to study? Example #1 - Single element Based on your pilot study of 25 work cycles you determine that:  sample mean element time(x) = 0.3 min  sample s.d (s) = 0.09 min  You’ve decided how confident and how precise you want to be about your estimates of the time for the work element:  You decide you need a 95% confidence interval (=5% --> /2=.025)  you decide you need the element sample mean time to be within  10% of the true mean time, the population mean (so set k to 10%, or 0.1) How many cycles to study? Example #1 - Single element  Sol’n: 2 ◼ s = std dev = n= ts ◼ k = tolerance = kx ◼ x = mean = ◼ t24,=.05 = t-statistic = (two-tail test) n= How many cycles to study? Example #1 - Single element  Sol’n: ◼ s = 0.09 minutes 2 ◼ k = 0.1 n= ts ◼ x = 0.3 minutes kx ◼ t24,=.05 = 2.064 (two-tail test) n = { (2.064 * 0.09)/(0.1 * 0.3)}2 n = 38.3, therefore, 39 cycles should be studied How many cycles to study? Example #2 - Multi-element task A task consists of three elements. Data from your pilot study of 10 work cycles: x1 = 0.3 min; s1 =.12 min x2 = 1.3 min; s2 =.35 min 2 x3 = 0.5 min; s3 =.05 min n= ts kx You require: - a confidence interval of 95% - your estimate of the element times to be within +/- 10% of their true values How many observations (cycles) are needed in your time study? How many cycles to study? Example #2 - Multi-element task A task consists of three elements. Data from your pilot study of 10 work cycles: x1 = 0.3 min; s1 =.12 min x2 = 1.3 min; s2 =.35 min 2 x3 = 0.5 min; s3 =.05 min n= ts kx  Sol’n: {(.12*2.262)/(0.10*0.3)}2 = 81.9 cycles {(.35*2.262)/(0.10*1.3)}2 = 37.1 cycles {(.05*2.262)/(0.10*0.5)}2 = 5.1 cycles Conducting time Study Conducting a Time Study  Enter job description on form  Enter element names on form  Keep track of time, from beginning to end  Write start time on form  Record stop watch times on form ◼ W: watch reading here, if using continuous (split) ◼ OT: observed element time ◼ NT: normal time  Rate operator performance ◼ is the performance normal? Adjust if not… ◼ how: ◼ element by element, observation by observation, or ◼ average rating for each element, or ◼ for whole performance Conducting a Time Study Each element should be recorded in its proper sequence, including a basic division of work terminated by a distinctive sound or motion. For example, the element “up part to manual chuck and tighten” would include the following basic divisions: reach for part, grasp part, move part, position part, reach for Chuck wrench, grasp chuck wrench, move chuck wrench, position chuck wrench, turn chuck wrench, and release Chuck wrench. The termination point of this element would be the chuck wrench being dropped on the head of the lathe, as evidenced by the accompanying sound. The element “start machine” could include reach for lever, grasp lever, move lever, and release lever. The rotation of the machine, with the accompanying sound, would identify the termination point so that readings could be made at exactly the same point in each cycle. Enter operator rating here (R) or here Enter continuous time (W) Enter element time (OT) Recording problems  Foreign element  unnecessary, non-productive, may be added by worker, not part of normal assignment ◼ if occurs between elements, and can be measured, mark in NT column of following element, use letter label, note ◼ if too short to time, include in elemental time, circle to designate as wild  Missed element  Analyst’s error ◼ designate with M in W column Recording problems  Foreign element  unnecessary, non-productive, may be added by worker, not part of normal assignment ◼ if occurs between elements, and can be measured, mark in NT column of following element, use letter label, note ◼ if too short to time, include in elemental time, circle to designate as wild  Missed element  Analyst’s error ◼ designate with M in W column 95 12 114 90 33A297 90 14 126 85 38 323 Time Study Form Foreign elements are described here Recording problems  Foreign element  unnecessary, non-productive, may be added by worker, not part of normal assignment ◼ if occurs between elements, and can be measured, mark in NT column of following element, use letter label, note ◼ if too short to time, include in elemental time, circle to designate as wild  Missed element  Analyst’s error ◼ designate with M in W column 95 12 114 90 33 297 Dropped tool 90 14 126 56 Recording problems  Foreign element  unnecessary, non-productive, may be added by worker, not part of normal assignment ◼ if occurs between elements, and can be measured, mark in NT column of following element, use letter label, note ◼ if too short to time, include in elemental time, circle to designate as wild  Missed element  Analyst’s error ◼ designate with M in W column 95 12 114 M 90 14 126 85 38 323 Recording problems, cont.  Out-of-sequence elements  why... ◼ operator varies order ◼ lack of training, practice; “sabotage” study  what to do about it if you are using continuous timing method... Element 1 Element 2 Element 3 Element 4 w w w w 332 380 350 410 350 332 380 Concluding a Time Study  Record finishing time  Provide operator with performance rating  discuss/explain as needed  Calculate OT if continuous method was used  Complete the Summary:  # of observations per element  incorporate Performance Rating in Summary if not done observation by observation  % Allowance: ◼ extra time, since time study conditions do not occur throughout the day ◼ three categories (PFD)  Time check - all accounted for? Using Standard Time to Determine Operator Efficiency Efficiency = 100 x Standard hrs earned Hrs on clock Using Standard Time to Determine Operator Efficiency Efficiency = 100 x Standard hrs earned Hrs on clock Operator worked 4 hours Completed 8 complete work cycles Standard times for 6 elements that make up each work cycle are: element 1: 8.15 minutes element 2: 15.24 minutes element 3: 2.66 minutes element 4: 4.33 minutes element 5: 3.50 minutes What was her efficiency? element 6: 2.45 minutes  Using Standard Time to Determine Operator Efficiency Efficiency = 100 x Standard hrs earned Hrs on clock Operator worked 4 hours Completed 8 complete work cycles Standard times for 6 elements that make up each work cycle are : e1: 8.15 minutes e2: 15.24 minutes e3: 2.66 minutes e4: 4.33 minutes 36.33 min * 8 cycles = 290.64 min e5: 3.50 minutes e6: 2.45 minutes Efficiency = 100*290.64/240 = 121.1% cycle time: 36.33 mins Other Needs or Types of Standards  Temporary standards:  Apply when operator is learning (new operator or new operation).  Specify upper limit on production level.  Set-up Standards  either distribute time across production time (if production lots are consistent, or  treat separately  Partial Set-up  may choose to provide whole set-up time Time study resources Time study packages… Training materials… Adjustments to Measured Times from Time Study  Performance rating  learning curves  individual differences  Allowances  personal  fatigue  delay Performance Rating Performance Rating  Observed time may need to be adjusted, to represent performance of a normal operator. Normal Time = Perf Rating x Observed Time  Accounts for differences between and within operators.  Critical element of time study  Subjective  Requires clear definition of “normal” operator Differences between operators...  Reasons:  Work-related ◼ ◼  Personal ◼ ◼ ◼ ◼ ◼ Differences between operators...  Reasons:  Work-related ◼ training ◼ experience  Personal ◼ strength ◼ coordination ◼ skill ◼ intelligence ◼ vision ◼ aptitude ◼ attitude Performance Rating System  Characteristics of a sound system:  Accurate, consistent.  Easy to use.  Keyed to benchmarks.  When to use:  During time study, for ◼ Elemental rating ◼ longer cycles, tasks ◼ machine time rating = 100 ◼ Overall rating ◼ short cycle repetitive tasks Performance Rating System  Characteristics of a sound system:  Accurate, consistent.  Easy to use.  Keyed to benchmarks.  When to use:  During time study, for ◼ Elemental rating ◼ longer cycles, tasks ◼ machine time rating = 100 ◼ Overall rating ◼ short cycle repetitive tasks Performance Rating System  Characteristics of a sound system:  Accurate, consistent.  Easy to use.  Keyed to benchmarks.  When to use:  During time study, for ◼ Elemental rating ◼ longer cycles, tasks ◼ machine time rating = 100 ◼ Overall rating ◼ short cycle repetitive tasks Rating Systems Examples  Westinghouse  skill, effort, conditions, consistency  Synthetic  based on fundamental motion times  P = Expected time/Observed time  Speed rating  focus on work rate  Objective rating  judge pace relative to benchmark task, then modify for difficulty  Regardless of the system, training and practice are essential for reliable, accurate application Westinghouse System  Westinghouse System  Once each factor rated  Sum the four values to give allowance in %  Or add to 1 to give performance factor  Best for whole job, not individual elements Allowances  The last step in determination of Standard Time  The need for allowances:  Normal time is still idealized, best case scenario.  No accounting for interruptions, problems, breaks, etc.  Allowances might be applied to:  total cycle time ◼ ex: personal needs delays, workstation clean-up, minor machine maintenance  machine time only ◼ ex: tool maintenance, power variance  effort time only ◼ ex: fatigue, unavoidable delays Allowances, cont  Determined through:  production study  work sampling  Types of:  constant ◼ personal - applied to total cycle time ◼ basic fatigue - applied to total cycle time  variable ◼ variable fatigue ◼ unavoidable delays - supervisor talk, machine interference,... ◼ extra allowances ◼ variable material quality ◼ clean and oil machine ◼ shutdown and tool maintenance - apply to machine time ◼ policy allowances - new employee, light duty,... Personal Allowance (P)  Restroom and water breaks  Affected by environmental conditions (heat, cold)  5% is typical  ex. 5% of 480 min = 24 min Fatigue Allowances (F)  Basic  accounts for variability in work pace  4% is typical for light work, under good conditions  Variable*  less than optimal work conditions stress a worker physiologically, psychologically, or both ways  conditions include: ◼ heavy lifting requirements (Tables) ◼ high cardiovascular demands (RA = (HR/40 - 1)*100) ◼ high mental demands ◼ high visual demands ◼ poor sensory conditions (excess noise, poor lighting, etc) ◼ high physical stress (fine, detailed work, continuous standing, awkward postures, etc) ◼ monotonous or tedious work *Table 11-2 & pg 435-447 in N&F Allowances for Unavoidable Delays (D)  Factors that impede operator effort:  interruptionsfrom supervisor, engineer, etc  problems with tolerances, specifications; materials  machine interference -- machines waiting for operator Unavoidable delay due to multiple machine interference  Recall from chapter 2, calculation of machine interference time:  Example: pg 53, Fig. 2-18 and Wright’s formula  Both provide “I”, interference as % of mean servicing time  Machine interference time = I x mean service time  % Allowance = machine interference time (machine run time+service time)  Stand. Time/Prod. Unit = (machine run time + service time + machine interference time) production units per run time Or, (machine run time + service time) x (1+allowance) production units per run time Unavoidable delay due to multiple machine interference  Example: If machine runs for 150 min, requires operator attention for 10 min, operator oversees 5 machines, and each machine produces 1 unit per run, calculate interference allowance.  Unavoidable delay due to multiple machine interference  Example: If machine runs for 150 min, requires operator attention for 10 min, operator oversees 5 machines, and each machine produces 1 unit per run, calculate interference allowance.  Sol’n: N  6, so from graph (fig 2-19*), I = 20% *Figure 10-4, N&F, p 398 (ed 10)  Unavoidable delay due to multiple machine interference  Example: If machine runs for 150 min, requires operator attention for 10 min, operator oversees 5 machines, and each machine produces 1 unit per run, calculate interference allowance.  Sol’n: N  6, so from graph (fig 2-19*), I = 20% machine interference time = 20% x 10 minutes = 2.0 minute *Figure 10-4, N&F, p 398 (ed 10)  Unavoidable delay due to multiple machine interference  Example: If machine runs for 150 min, requires operator attention for 10 min, operator oversees 5 machines, and each machine produces 1 unit per run, calculate interference allowance.  Sol’n: N  6, so from graph (fig 2-19*), I = 20% machine interference time = 20% x 10 minutes = 2.0 minute Stnd time/Prod Unit = (150 +10+2.0)/5units = 32.4min/unit *Figure 10-4, N&F, p 398 (ed 10)  Unavoidable delay due to multiple machine interference  Example: If machine runs for 150 min, requires operator attention for 10 min, operator oversees 5 machines, and each machine produces 1 unit per run, calculate interference allowance.  Sol’n: N  6, so from graph (fig 2-19*), I = 20% machine interference time = 20% x 10 minutes = 2.0 minute Stnd time/Prod Unit = (150 +10+2.0)/5units = 32.4min/unit % Allowance = [2.0/(150+10)] x 100= 1.25% *Figure 10-4, N&F, p 398 (ed 10) Other Allowances  Attention time allowance  when watching process is required, and machine run time >> service time  Workstation clean-up, machine upkeep  if not provided with time allotment at end of shift  Power feed  added to machine time to cover shutdowns and tool maintenance  Policy allowances  Applying Allowances  Two ways to do this: - One way is to just apply allowance to work time: ST = NT (1 + PFD)  Alternative is to apply to total work day: total hours x PFD = allowance time ST=Standard Time; NT = Normal Time Applying Allowances Example  just apply allowance to work time: ST = NT (1 + PFD) If NT = 2 min and PFD totaled 10%, then ST = 2 min ( 1+ 0.10) = 2.2 min / unit Applying Allowances Example  just apply allowance to work time: ST = NT (1 + PFD) If NT = 2 min and PFD totaled 10%, then ST = 2 min ( 1+ 0.10) = 2.2 min / unit (480 min/day) / (2.2 min/unit) = 218.2 units/day 480 min = (2 +.2 min/unit) x 218.2 units = 436.4 min + 43.6 min Applying Allowances Example, cont  alternatively, apply to total work day: total hours x PFD = allowance time example: 480 min x 0.10 = 48 min 480 min = 432 min + 48 min Maintaining Standard Times  Method must match standard  Conduct periodic audits  Compliance  operators  supervisor  analyst Learning Curve Learning Curves  Practice affects performance.  Skills improve with practice.  Learning curves quantify decrease in task performance time, as function of task repetition. Learning Curve - theory  Theory: as total quantity produced doubles, time/unit declines at a constant rate (expressed as a percentage).  Example: Total Time Percent- Total Avg Units for age time time/ Produced unit unit 1 100 min - 100 min 100 min 2 90 90% 190 95 3 84.6 - 274.6 91.5 4 81 90% 355.6 88.9. 8 72.9 90% ……... Learning Curve - theory Learning curve 120 Production time of each unit, minutes 100 80 60 90% 40 20 0 0 20 40 60 80 100 Num ber of units producted Characteristic Equation of Learning Curve YX = KX N or log YX = log K + N log X where: YX = predicted production time for Xth unit K = time req. to produce first unit X = total units produced N = exponent = tan  (angle from log-log plot of data) 2N = rate Characteristic Equation of Learning Curve Learning curve 120 Production time of each unit, minutes YX = KX N 100 80 90% 2N=.90 60 80% 2N=.80 40 20 0 0 20 40 60 80 100 Num ber of units producted Learning Curve Example 1 Q1. How many cycles are anticipated before a new operator reaches a standard production time of 45 min, if  firstunit required 94 min.  the learning curve rate is 82% Learning Curve Example 1 Q1. How many cycles are anticipated before a new operator reaches a standard production time of 45 min, if  firstunit required 94 min.  the learning curve rate is 82% 2n =.82 Yx = K Xn Learning Curve Example 1 Q1. How many cycles are anticipated before a new operator reaches a standard production time of 45 min, if  firstunit required 94 min.  the learning curve rate is 82% 2n =.82 => n log102 = log100.82 => n = -.2863 Yx = K Xn Learning Curve Example 1 Q1. How many cycles are anticipated before a new operator reaches a standard production time of 45 min, if  firstunit required 94 min.  the learning curve rate is 82% 2n =.82 => n log102 = log100.82 => n = -.2863 Yx = K Xn 45 = 94 X-0.2863 Learning Curve Example 1 Q1. How many cycles are anticipated before a new operator reaches a standard production time of 45 min, if  firstunit required 94 min.  the learning curve rate is 82% 2n =.82 => n log102 = log100.82 => n = -.2863 Yx = K Xn 45 = 94 X-0.2863 45/94 = X-0.2863 log10(45/94) = -0.2863 log10X 1.1174 = log10X X=13.1 cycles Learning Curve Example 1, cont. Q2. How much cumulative production time to reach a cycle time of 45 min, if  first unit required 94 min.  the learning curve rate is 82% Learning Curve Example 1, cont. Q2. How much cumulative production time to reach a cycle time of 45 min, if  first unit required 94 min.  the learning curve rate is 82% Determine by calculating the area under the learning curve: Yx => k{(x2+1/2)n+1 – (x1 –1/2)n+1}/(n+1) Where: k=time for first cycle x2=the cycle that takes 45 minutes x1=the first cycle (x1 =1) n=exponent that represents the slope of the learning curve Learning Curve Example 1, cont. Q2. How much production time to reach a cycle time of 45 min, if  first unit required 94 min.  the learning curve rate is 82% Determine by calculating the area under the learning curve: Yx => k{(x2+1/2)n+1 – (x1 –1/2)n+1}/(n+1) Where: k=time for first cycle = 94 x2=the cycle that takes 45 minutes = 13 x1=the first cycle (x1 =1) = 1 n=exponent that represents the slope of the learning curve = -.2863 Learning Curve Example 1, cont. Q2. How much production time to reach a cycle time of 45 min, if  first unit required 94 min.  the learning curve rate is 82% Determine by calculating the area under the learning curve: Yx => k{(x2+1/2)n+1 – (x1 –1/2)n+1}/(n+1) Where: k=time for first cycle = 94 x2=the cycle that takes 45 minutes = 13 x1=the first cycle (x1 =1) = 1 n=exponent that represents the slope of the learning curve = -.2863 Yx = 768 minutes of production time Learning Curve Example 1, cont. Q3. What is the cumulative average cycle time? Learning Curve Example 1, cont. Q3. What is the cumulative average cycle time? 768 min/ 13 cycles = 58.6 min/cycle Learning Curve Example 2  How many cycles are anticipated before a new operator reaches a standard production time of 20 min, if  first unit required 30 min.  the learning curve rate is 93%  How long (in minutes) will it take to reach this level of productivity?  Learning Curve Example 3  How many cycles are anticipated before a new operator reaches a standard production time of 20 min, if  firstunit required 30 min.  the learning curve rate is 93% 2n =.93 => n log102 = log100.93 => n = -.1047 Yx = K Xn 20 = 30 X-0.1047 20/30 = X-0.1047 log10(20/30) = -0.1047 log10X 1.682= log10X X=48.1 therefore, 48 (or 49) cycles Learning Curve Example 3  How long (in minutes) will it take to reach this level of productivity? Yx => k{(x2+1/2)n+1 – (x1 –1/2)n+1}/(n+1) 30{(48+1/2)(-.1047+1) – (1 –1/2)(-.1047+1)} (-.1047+1) 1065.9 minutes ~ 17.8 hours Learning Remission  Performance decrement due to time away from task (s.a. 1-10 days)  Amount of remission is a function of location on the learning curve prior to interruption  Prediction of time to produce first unit after interruption:  Learning Remission  Performance decrement due to time away from task (s.a. 1-10 days)  Amount of remission is a function of location on the learning curve prior to interruption  Prediction of time to produce first unit after interruption: yx = k + (k-s) (x -1) 1-xs where: s = stnd time xs = # cycles to reach stnd time x = # of cycles completed +1 (x refers to the first unit produced upon return from break) k = time for first cycle Note: yx becomes revised K in learning curve equation, YX = KX N Learning Remission Example  How long would it take for an operator to perform the first cycle after returning from vacation, if the last cycle before vacation was the 10th cycle? Assuming conditions for job in Example 1... Learning Remission Example  How long would it take for an operator to perform the first cycle after returning from vacation, if the last cycle before vacation was the 10th cycle? y = 94 + ( 94 - 45 ) ( 11 - 1 ) 1-13 y = 53.5 minutes Learning Remission Example  How long would it take for an operator to perform the first cycle after returning from vacation, if the last cycle before vacation was the 10th cycle? y = 94 + ( 94 - 45 ) ( 11 - 1 ) 1-13 y = 53.5 minutes If no break had occurred, the 11th cycle would have been predicted to take only… 47.3 minutes = YX = KX N = 94x11-.2863 Learning Remission Example  How long would it take for an operator to perform the first cycle after returning from vacation, if the last cycle before vacation was the 10th cycle? y = 94 + ( 94 - 45 ) ( 11 - 1 ) 1-13 y = 53.5 minutes If no break had occurred, the 11th cycle would have been predicted to take only 47.3 minutes = YX = KX N = 94x11-.2863 Now, 11th cycle is new “first cycle” of new learning curve: YX = (53.5)X-.2863 Unavoidable Delays Multiple machine interference  Alternative suggestion when N > 6  Use values from tables based on queuing theory models (see A3-13, N&F, p 702-3), and relationships below: C = T 1 + T2 + T3 k = T2 T1 where: T1 is machine run time /cycle T2 is attention time per cycle/ machine T3 is lost production time due to interference C is cycle time Confidence Intervals An interval within which the value of the parameter of interest (Ф) is expected to lie. P (L ≤ Ф ≤ U ) = 1-  Where (L ≤ Ф ≤ U ) is the 100(1-)% confidence interval for parameter Ф Interpretation: if a large number of intervals were constructed, from repeated random samplings, 100(1-)% of those intervals would contain the true value of Ф An online stopwatch… http://www.shodor.org/interactivate/activities/stopwatch/ 00:00:20.5 00:00:25.7 00:00:31.7 00:00:36.7 00:00:41.5 00:00:49.5 00:00:58.2 though units are seconds Unavoidable delay due to multiple machine interference Suggestion when N  6  Figure 10-4, N&F, p 398 (ed 10)  Figure 2-19, N&F, p 54 (ed 11) ex. If machine runs for 150 min, requires operator attention for 5 min, and operator oversees 5 machines, calculate interference allowance.  Unavoidable delay due to multiple machine interference  Suggestion when N  6  Figure 10-4, N&F, p 398 (ed 10)  Figure 2-19, N&F, p 54 (ed 11) ex. If machine runs for 150 min, requires operator attention for 5 min, and operator oversees 5 machines, calculate interference allowance. From graph, I = 9.5% machine interference time = 0.095x5 minutes =.48 minute Stnd time/Prod Unit = (150 +5+.48)/5units = 31.1min/unit % Allowance = [.48/(150+5)] x 100= 0.31% Unavoidable delay due to multiple machine interference  Suggestion when N > 6  use the following equation to predict interference time: I = 50 { [ (1+X-N)2 + 2N]1/2 - (1+X-N)} where: I= interference as % of attention time X= machine run time = MRT machine attention time req. MAT N= number of machines/operator Standard time / unit = [MRT+MAT+(I x MAT)] / N % Allowance for interference = (I x MAT)/(MRT+MAT) Unavoidable delay due to multiple machine interference Example, N>6 machines: Known: machine runs for 150 min, requires operator attention for 5 min (for unloading and loading, etc), and operator oversees 55 machines. Calculate: interference allowance as percentage of attention time; standard time per unit.  Unavoidable delay due to multiple machine interference Example, continued: MRT = 150 min MAT = 5 min N=55; X=MRT/MAT=30 I = 50 { [ (1+X-N)2 + 2N]1/2 - (1+X-N)} I = 50 { [ (1+ 30 – 55)2 + 110]1/2 – (1+30-55)} I = 2510%  Unavoidable delay due to multiple machine interference Example, continued: MRT = 150 min MAT = 5 min N=55 I = 50 { [ (1+X-N)2 + 2N]1/2 - (1+X-N)} I = 50 { [ (1+ 30 – 55)2 + 110]1/2 – (1+30-55)} I = 2510% 25.1x5 minutes = 125.5 minutes = machine interference time  Unavoidable delay due to multiple machine interference Example , continued : MRT = 150 min MAT = 5 min N=55 I = 50 { [ (1+X-N)2 + 2N]1/2 - (1+X-N)} I = 50 { [ (1+ 30 – 55)2 + 110]1/2 – (1+30-55)} I = 2510% 25.1x5 minutes = 125.5 minutes = machine interference time Stnd time/Prod Unit = (150 +5+125.5)/55units = 5.1 min/unit  Unavoidable delay due to multiple machine interference Example , continued : MRT = 150 min MAT = 5 min N=55 I = 50 { [ (1+X-N)2 + 2N]1/2 - (1+X-N)} I = 50 { [ (1+ 30 – 55)2 + 110]1/2 – (1+30-55)} I = 2510% 25.1x5 minutes = 125.5 minutes = machine interference time Stnd time/Prod Unit = (150 +5+125.5)/55units = 5.1 min/unit % Allowance = [125.5/(150+5)] x 100= 80.97% End of Lecture