2.1 (a) A machine has a cycle time of 1/2 second. What are the productions pieces per hour at 100% and 75%efficiency?
(b) The floor-to-floor time (meaning a complete productive cycle) is 31 seconds. Find the gross units per hour at100% and 90% efficiency.
Answer:(a) t = 0.5 sec/piece 1/0.5 x 60 x 60 = 7200 pieces/hour @ 100%
7200 x 0.75 = 5400 pieces/hour @ 75%(b) t = 31 sec/piece 1/31 x 60 x 60 = 116.1 pieces/hour @ 100%
116.1 x 0.90 = 104.5 pieces/hour @ 90%
2.2 (a) If the hourly production is 11 units, find the hours per 1000 pieces.(b) The production is 29.5 units per hour. Find hours per 100, 1000 and 10,000 units.(c) Find pieces per hour and standard minutes per unit for 15.325 hours per 100 units.
Answer:(a) 1000/11 = 90.9091 hrs for 1000 units(b) 100/29.5 = 3.390 hrs for 100 units
1000/29.5 = 33.898 hrs for 1000 units 10000/29.5 = 338.98 hrs for 10000 units
(c) 100/15.325 = 6.5 pieces/hour 60 min / 6.5 pieces = 9.23 min / piece
2.3 A time study of an assembly operation is summarized:Description Frequency Element
MinutesRating Factor
Part A to 5 in. dowel 1/1 0.037 1.10Part B to 5 in. subassembly 2/1 0.064 1.15Short piece to base 1/1 0.089 1.20Long piece to base 1/1 0.129 1.10Grab pieces from tote box 1/5 0.185 1.10Place on conveyor 1/1 0.087 1.10
Allowances for this work total 9.5%. Find the standard minutes per unit, pieces per hour, hours per 100 units, andlabor cost per unit, where the wage rate is $14.75 per hour.
2.4 A mail room prepares the company’s advertising for mailing. A time study has been done on the job ofenclosing material in envelops. Develop the elemental normal time, cycle standard time, pieces per hour, andhours per 100 units, providing 15% for allowances, for the continuous stop watch data.(Hint: The time readingsare in hundredths of a minute.)
Cycle
Description 1 2 3 4 5 6 7Rating
%Get envelops 11 55 105 151 105Get and fold premium 22 41 65 83 116 134 160 115Insert premium in envelopeand seal
Total 73.64Note: Normal times in hundredths of a minute
Std time 73.64/100 x 1.176 = 0.866 minutes / piece60 / 0.866 = 69.3 pieces per hour100 / 69.3 = 1.44 hours per 100 pieces
2.5 A method has been engineered and an operator has been time studied. The method has produced an averageoverall time of 2.32 minutes by actual watch timing. The overall pace rating applied to the job was 125%. Thiscompany, which uses an incentive system, adds extra time into the production rates to cover personal time,fatigue, and unavoidable delays. The total allowance is 15%.
(a) Determine the rated time or normal minutes, the total allowed time or standard minutes, the standardhour rate for the job per 100 units, and the pieces per hour.(b) The labor rate per hour is $15.30. What is the standard labor cost per unit?
Answer: (a) 2.32 x 1.25 = 2.9 rated minutes
2.9 x 1.176 = 3.41 standard minutes3.41 / 60 = 0.0568 hr per piece
0.0568 x 100 = 5.68 hr / 100 pieces1 / 0.568 = 17.6 pieces per hr
2.6 Element time study observations that use the snap-back method are given as
Element description 1 2 3 4 5Rating
(%)Allowance
(%)Place part in jig 0.03 0.08 0.05 0.04 0.06 117 11Drill hole 0.31 0.38 0.37 0.39 0.33 100 5Remove part, place onconveyor belt
0.07 0.08 0.09 0.12 0.11 93 13
(a) Find the average, normal and standard times for each element.(b) Calculate the production per hour and hours per 100 units.(c) By an older method, 0.834 minute of standard time was required to complete this same process.Calculate the increase in output in percent and savings in time in percent.
2.7 A time study summary is given.Element description Freq/
UnitAvg.Time
Rating(%)
Get part from conveyor 1/1 0.040 100Get subassembly from conveyor 1/1 0.030 105Connect parts to subassembly 2/1 0.055 110Install 3 brackets 3/1 0.153 100Assemble side 1 1/1 0.234 90Mate side 2 1/1 0.183 90Put on conveyor 1/1 0.032 100
(a) With an allowance factor of 15% to cover P (personal), F (fatigue), and D (delay), what are thenormal minutes per piece, the standard minutes per unit, and the number of hours per 1000 pieces?(b) If the performance against standard has averaged 120%, what is the incentive hourly if day work ispaid at $15.75 per hour?
2.8 The raw continuous-timing data for a two element punch press operation are provided. The times are inminutes.
Description 1 2 3 4 5Handle Part 0.01 0.04 0.08 0.11 0.14Punch Part 0.03 0.06 0.10 0.12 0.16
(a) What is the average time for each element? If the ratings are +8% for element 1 and –9% for element 2 whatis the normal time? The PFD allowance is 20%. What is the standard minute per piece for this operation? Howmany pieces per hour? How many hours per 100 units?(b) If the PFD allowance only applies to element 1, what are the standard minutes per unit? How many hours per100 units?
2.9 Engineering aides prepare job tickets to track their daily work. After a sufficient period of data collection andclassification, the data are summarized. A personal allowance of 15% is used for this work. (Hint: Forsimplification of calculation, ignore holiday and vacation days.)
(a) Determine the standard labor hours per engineering drawing.(b) A new product design is anticipated and a separate staff will be assembled. The new product design willprobably result in 2500 drawings over a 1-year period. How many aides will be required, and at $25 per hour,how much should be budgeted for their work on the drawings for the new design?
2.10 The results from a working sampling study of a work force is divided into 12 categories. This sample coversa span of 25, 8-hour days. What are the percentages and expected hours per category?
2.11 The pediatrics department in a hospital is work sampled over a total of 608 hours. The results are shown.There were a total of 1578 observations. Find the percent of occurrences for each category, cumulativepercentages, element hours and cumulative hours for each category.
Category Count Category CountRoutine nursing 496 Other 79Idle or wait 263 Feeding 52Unit servicing 183 Bathing 22Report 129 Elimination 11Personal time 128 Transporting 8Intervention 102 Housekeeping 7Unable to sample 91 Ambulation 7
2.12 A work sampling study is taken of a department with the following information obtained: Number ofsampling days, 25; number of trips per day, 16; number of workers observed per trip, 3; and number of itemssampled, 4. The observations counted for each of the items was, A – 80; B – 320; C – 1600; and D – 2800.
(a) How many labor-days and observations were sampled?(b) What are the percentages and equivalent hours for the activities?(c) For a confidence level of 90%, what is the relative accuracy for each item?
2.13 We want to determine the percentage of idle time of a machine shop by work sampling. A confidencelevel of 95% and a relative accuracy of ±5% is desired, where a rough estimate of 25% is suspected foridle time.(a) How many observations are necessary?(b) Assume that the relative accuracy is ±2 _%. How many observations are required now?(c) What happens to the number of the observations required as the relative accuracy becomes less?
2.14 (a) To get a 0.10 interval on work observed by work sampling that is estimated to require 70% of theworker’s time, how many random observations will be required at the 95% confidence level? Repeat for90%.(b) If the average handling activity during a 20-day study period is 85% and the number of dailyobservations is 45, then what is the interval allowed on each day’s percent activity? Use 90%. Repeat for99%.(c) Work sampling is to be used to measure the not-working time of a utility crew. A preliminary studyshows that not-working time is likely to be around 35%. For a 90% confidence level and a desiredrelative accuracy of 5%, what is the number of observations required for this study? Compare to 95%confidence level.
Answers(a) Pi = 0.70, I = 0.10, 95% -- Z = 1.645
N = (4)(1.645)2(0.70)((0.3)/(0.10)2 = 22790% -- Z = 1.960, N = 323
(b) 20 x 45 = 900 observations, Pi = 0.85For 90% -- Z = 1.645, I = (2)(1.645)[(0.85)(0.15)/(900)]1/2
I = ±0.39%For 99% -- Z = 1.960, I = ±0.61%
(c) 90% -- Z = 1.645, Pi = 0.35, R.A. = 0.05I = (2)(0.35)(0.05) = 0.035N = (4)(1.645)2((0.35)(0.65)/(0.035)2 = 201095% -- Z = 1.960, N = 2854
2.15 A shipping department that constructs wooden boxes for large switch gear has five direct-labor workers.A work-sampling study is undertaken, and the following observations of work elements are recordedover a 15-day, 8-hour period:
Description CountSet up and dismantle 312Construct crates 264Load switch gear in crates 204Move materials 324Idle 96
A rating factor of 90% is found. The number of switch gear shipped during this period is 26. Thisfirm uses an allowance value of 10% for work of this kind. Average labor costs $18.75 per hours.
(a) Find the elemental costs.(b) What is the standard labor cost per box?(c) Estimate the actual cost.
Answers(a) Total hours observed: 15 x 8 x 5 = 600 (a)
Totals 1200 1.00 600 394.13* Adjustment for performance rating and PFD factor: hrs/elem x 0.9 x 1.1** Idle time is not included in the standard calculationsb) $394.13c) (600 x 18.75) / 26 = $432.69
2.16 An eight-person CAD department concerned with size A, B, and C drawings is work-sampled by amanagement consultant over a standard 4-week period. A chart summarizes for the categories asfollows:
Item CountDrafting and tracing 778Calculating 458Checking prints 110Classroom 125Professional time off 172Personal time, idle 270
During this period, 55 drawings (A = 20, B = 25, C = 10) were produced with a total payoff of $26,400.Let relative size A = 1, B = 2A, C = 2B.
a) Given a performance rating of 1.0 and a PF&D factor calculated from the personal time, idle element,determine the standard time per drawing unit (A size) and standard times for the B and C sizes.b) Determine the hourly pay rate. A new order is estimated at A = 10, B = 30 and C = 25 drawings has beenplaced. What is the estimated cost of this order?
AnswersTotal hours worked 8 x 4 x 40 = 1280
Item Count Fraction Hrs/elemDrafting and tracing 778 0.41 525Calculating 458 0.24 307Checking prints 110 0.057 73Classroom 125 0.065 83Professional time off 172 0.09 *Personal time, idle 270 0.141 *
Total 1913 1.003 988* Not included in standard times
a) PFD = 100 / (100 – 10) = 1.11 For 55 dwgs: 988 x 1.11 = 1098 std hours
Drawing units: 20 + (25 x 2) + (10 x 4) = 110Std time per dwg unit: 1280 / 110 = 11.64 hr
A = 11.64 hrB = 2 x 11.64 = 23.28 hrC = 4 x 11.64 = 46.56 hr
b) 26,400 / 1280 = $20.62 / hr(10 x 11.64 + 30 x 23.28 + 25 x 46.56) x 20.62 = $40,803
2.17 Find the effective gross hourly cost for drill press operators paid an average wage of $22 per hour. Noovertime is planned. Company policy allows six paid holidays, and the average entitled vacation is 10 days.There is no non-chargeable time or performance subsidy. Sick leave is charged against vacation time. FICAis at current rate, and workmen’s compensation is at 2% of the first $20,000. The company pays $400 forunemployment insurance.
FICA (45760 x 0.076) 3,478Workers Comp (20000 x 0.02) 400Unemployment Ins 400
Excess cost total 7,094Excess hourly cost (7094 / 1952) 3.63Hourly wage 22.00Effective gross hourly cost 25.63Percent increase over wage 16.5%
2.18 A tool and die worker is paid $26.50 per hour. The work year consists of 52 40-hour weeks withovertime schedules for 26 Saturdays. The company allows 8 paid holidays and 10 days of vacation. Four days ofsick leave are budgeted and have historically been used. Expected non-chargeable hours are 2%. There is nosubsidy for performance. Non-hourly costs include FICA taxes at current governmental rate, workmen’scompensation at 2% of regular wages up to $20,000, accident insurance sum of $600, a major supplementalmedical plan for $300, and unemployment insurance of $800.(a) Find the effective gross hourly cost.(b) What is the job cost if a job will require 10 man-hours with a productivity of 90%?
2.19 Determine a company’s base annual cost for a top-grade worker. The base hourly rate is $15.75. Use thecurrent FICA costs of the base. State unemployment compensation runs 2%. Health insurance premiums cost $40per month. The company carries term life insurance that costs $45,000 per year for all employees (there are 60employees). In addition, the company profit-sharing plan usually pays 5% of the base wage.
Answer:Base wage: 15.75 x 2080 = $32,760FICA: 32760 x 0.76 = 2,490Unemployment: 32760 x 0.02 = 655Health Ins: 40 x 12 = 480Life Ins: 45000 / 60 = 750Profit sharing: 32760 x 0.05 = 1,638
Total $38,773Hourly wage 38773 / 2080 = $18,64
2.20 Find a typical hourly cost for direct labor. In this instance, the average workweek is 48 hours, of whichthe final 8 hours are premium at one and a half. The wage is $30.30 per hour. Each day a total of 20 minutes ispermitted as a coffee break, the final 15 minutes is cleanup. Annually there are 2 weeks of paid vacation on thebasis of 40 hours worked and five paid holidays. Taxes for FICA are at the current rate and there is a 1.2%workers’ compensation tax for the amount of $15,000. There is a requirement for $250 per month for company-paid medical and life insurance, and a $25 Christmas bonus.
2.21 A man worked 8 hours on incentive and non-incentive jobs. While on incentive, he completed 8 tasks, eachwith a 1-hour standard time. The non-incentive jobs took 2 actual hours. The incentive plan has a base rate of $13per hour, though the man’s wage rate for non-incentive tasks is $16 per hour. How much did the man earn?
Answer:Ha = 8 hrs, Hs = 1x8 = 8 hrs, Other = 2 hrs(8 x 13) + (16 x 2) = $136136 / 8 = $17 / hr
2.22. (a) A worker produces 56 units on an incentive plan during the 40-hour week. Her base hourly rate is $15per hour, and the standard for one unit of product is 1.10. What are her weekly earnings?(b) Find the full employee cost for the product if the employee has a FICA rate of 7.60%, unemploymentinsurance costs the company 2.1%, the workers’ compensation rate is 2.5%, 10 days of yearly vacation, and $400per month for medical insurance. Approximate the actual wage and fringe cost per hour. Per unit. Assume aweekly wage for a 52-week year. Yearly vacation is at base rate of $15 per hour.
Answer:a) Na = 56, Hs = 1.10 x 56 = 61.6 hrs, Ha = 40 hrs, base R = $15
2.23 A worker is paid the day-work rate if she earns less than 100% incentive premium. The standard is 75 unitsper hour, and the operator completes 140 good parts in 1.4 hours. The rate is $25 per hour.(a) Find the piece rate.(b) What are the earnings and labor efficiency?
Answer:a) 140 / 1.4 = 100 units/hr
$25 / 75 units = $0.33 / unit0.33 x 100 = $33 / hr
b) Cdl = HaRh + Rh(HsNp – Ha)(1.4 x 25) + 25[(1/75)(140) – 1.4] = $46.67
2.24 (a) A sheet metal operator is told that her standard is 1.875 hours per 100 units. If her wage is $16 per hourand she makes 800 units in 12 hours, what are her earnings?(b) If the actual time is 16 hours, what is her guaranteed pay and efficiency for the 100% plan?
Answer:a) 1.875 x (800/100) = 15 standard hours
15 x 16 = $240 earnings for the 12 hours, or $20 / hourb) Ha = 16, Rh = $16, Np = 800, Hs = 1.875/100 = 0.01875 hr / piece
Cdl = (16 x 16) + 16 x [(0.01875 x 800) – 16] = $240240 / 16 = $15 / hour
But, $16 / hour is guaranteed, so Cdl = 16 x 16 = $256Efficiency: 800 x (0.01875 / 16) x 100 = 93.75%
2.25 Efficiency of 125% for an operation is calculated. The standard is 10 hours per unit and the wage is$23.50 per hour. What is the expected cost for 210 units?
Answer:Need to find Ha:
Ha = (100NpHs) / E = (100 x 210 x 10) / 125 = 1680 hoursCdl = (1680)(23.50) + (23.50) x [(10x210) – 1680] = $49,350
2.26 A molding operator tends two machines and plastic products A and B are produced. Machines A and Bhave a production rate of 400 and 250 strokes per hour. Molds A and B produce three or four buttons perstroke respectively. Operator wage is $11 per hour. Dejoint these labor costs when allocation is based on(a) number of machines, (b) product output of machines A and B and (c) marketing believes that A to Bvalue is 5:4.
2.27 A time study observes the operation of producing two parts, which are labeled 1 and 2. The elements andtheir standard minutes are given as follows:
Element Part Standard minutesLoad part in vise 1 0.17Load second part 2 0.08Balance and tighten vise 1,2 0.17Start machine 1,2 0.06Machine 1,2 1.17Stop, unload 2 0.13Start, retighten 1 0.15Fine machining 1 0.83Stop, unload 1 0.14Clean vise 1,2 0.16
The labor wage is $23 per hour. Dejoint the cost using allocation methods of (a) number of parts, (b) timerequired by part, and (c) potential sales price ratio of 3:2.
1.56 1.29 0.211.56 / 2 = 0.78 for 1 and 2Standard hours: Part 1 = 0.78 + 1.29 = 2.07
Part 2 = 0.78 + 0.21 = 0.99Total for parts 1 & 2 = 3.06
a) 2 parts / 3.06 min x 60 min / hr = 39.2 parts / hour (19.6 each 1 & 2)23 / 39.2 = $0.59 / part
b) Part 1: 2.07 / 3.06 = 67.6% Part 2: 0.99 / 3.06 = 32.4% 23 x .676 = $15.56 23 x .324 = $7.44 15.56 / 19.6 = $0.79 / part 7.44 / 19.6 = $38 / part
c) Part 1: 3 / 5 x 23 = $13.80 Part 2: 2 / 5 x 23 = $9.20 13.80 / 19.6 = $0.70 / part 9.20 / 19.6 = $0.47 / part
2.28 One operator controls four automatic machines. After those machines are set up they produce partsindependent of the operator except for occasional inspection. A cam controls the unit time to make one piece andis 4 seconds at 100% efficiency. Ignore the setup time because it is small.(a) If the actual efficiency is 85%, the operator controls four machines, and the labor wage rate is $15 per hour,
then what is the dejointed labor cost per unit?(b) Repeat for hours per 100 units.(c) Repeat for dollars per 100 units.
Answer:4 sec /piece = 4 / (60 x 60) = 0.00111 hr / piece at 100%
a) 0.00111 / 0.85 = 0.00131 hr / piece0.00131 x 15 / 4 = $0.0049 / piece or 0.49¢ / piece
b) 0.0031 x 100 / 4 = 0.033 hr / 100 piecesc) 0.0049 x 100 = $0.49 / 100 pieces
2.29 A technician tests and repairs printed circuit boards. On the average, printed circuit boards require 10 minutesfor testing and 12 minutes for testing and repairing, if necessary.(a) If, during a labor-hour study, 18 units tested OK and 5 failed requiring repair, then what proportion of the labor
wage is due to testing and repair?(b) If the gross labor rate is $21.75 per hour, then what is the labor charge per unit for testing and repair? Use
proportional methods to dejoint cost.
Answer:a)
Testing 18 x 10 = 180 75%Testing & Repair 5 x 12 = 60 25%
Total 24025% of time is due to testing and repair units
b) (240 min / 60 min/hr) x 0.25 x 21.75 / 5 = $ 4.35simply: 60 min / 60 min/hr x 21.75 / 5 = $4.35
2.30 The fire department is concerned about the speed of the crews assigned to the all-purpose trucks. The chiefwants to know the time required after receiving the alarm before starting to fight the blaze. To answer the chief’squestions, a time study of the activities of one crew on seven different alarms is conducted. On the observation sheetthat follows, continuous (no reset) electronic watch readings in hundredths of minutes are recorded, indicating fullminutes only when it changed.
*Watch stopped because of variable nature of distances.
(a) Determine the average time for the elements.(b) If the crew is rated at 110% for all elements, find normal time in minute per occurrence.(c) Determine the cycle-time standard for a 20% allowance.
2.31 We are interested in finding the estimated direct-labor cost for the job description of industrial electrician. Theyear consists of 52 40-hour weeks, and overtime is seasonal for 12 weeks consisting of Saturday work of 8 hours. Thecontract allows for nine paid holidays and two weeks of vacation at regular time. Four days of sick leave are paid.Expected non-chargeable hours are 5%. A subsidy for performance will be 15%. Non-hourly costs include FICA taxesat the current federal rate, workmen’s compensation at 1% of regular wages up to $15,000, accident insurance sum of$200, major medical plans for $500, and Christmas gift of $50. The hourly base is $27.10.(a) Find the effective gross hourly cost.(b) What is the job cost for this electrician if a job requires 25 man-hours?(c) What is the loss or gain if efficiency will be 85% or 115% for the job?
FICA (60270) x 0.076 4,581Workers Comp (15000 x 0.01) 150Accident Ins 200Major medical plan 500Christmas Gift 50
Excess cost total 22,210Excess hourly cost (22210 / 1888) 11.76Hourly wage 27.10Effective gross hourly cost 38.86Percent increase over wage 43.4%
b) 25 x 38.86 = $971.50c) 85% efficiency: 25 / 0.85 = 29.41, 29.41 x 38.86 = $1,142.87 115% efficiency: 25 / 1.15 = 21.74, 21.74 x 38.86 = $884.82
If a contract was bid at 100% ($971.50), then an 85% efficiency wouldresult in a loss of $171.37.An efficiency of 115% would result in a gain of $86.68
2.32 A preliminary work sampling study is made of an assembly operation. The work elements and the results areshown in the table. The desired interval for each of the elements is also shown.
ElementNumber of
ObservationsDesired
Interval (I)1. Assemble components 31 0.042. Setup 8 0.033. Wait for material 6 0.024. Painting 2 0.025. Inspection 3 0.02
(a) For a 90% confidence interval, find the number of observations required for the complete study.(b) At the completion of the whole work sampling study, the productive elements amounted to 83% of the time. Thestudy took 4 weeks and observed a crew of 6. Given a PF&D of 20% and a performance rating of 0.95 find the standardtime per part if 23 parts were completed during the study.
Answer:a) For a 90% confidence interval, Z = 1.645
For the study, use the maximum N, 2858, or approximately 2860 observations.b) (0.83)(4 x 40 x 6)(0.95)(1+0.20) x (1/23) = 39.5 hrs / unit
PRACTICAL APPLICATION
Perform a preliminary work sampling study to determine the amount of program and commercial time for prime-timetelevision. Determine times for 60 observations at random intervals between 8:00 and 11:00 PM. Carry out theobservations on one night for a major network (ABC, NBC, or CBS). (Hint: The selection of random numbers can befound using a telephone directory, where the last four numbers are considered “practically” random. A random numbertable from a statistics book can also be used. Spreadsheet programs will have a random number generator. It will benecessary to think up an algorithm for this conversion of a number called “random” to the observing time.)
Make snap observations of the following categories: Programming, commercial, promotion of other programsor local station and news update or news related promotions which will include weather. (Hint: Remember the “snap”notion of instantaneous recording.)
Calculate the amount of actual program time per hour. Prepare your report considering night, network,observation times, summary table of observations (number of observations for each category), and observations outsideof the specified categories along with the description. Talk about your coding of the observing times. Conclude with adiscussion of the belief of your work sampling conclusions.
Your instructor will provide additional instructions and reporting guidelines.
Answer: Times obtained may vary. Approximate times are:39 minutes of program per hour17 minutes of commercials per hour
“We can’t make any profit on that job. There’s too much labor cost in it,” said Dick, the foundry superintendent, toGeorge, engineer for the Endicott Iron Foundry.
The Endicott Iron Foundry, like its competitors, has always estimated costs on a per pound basis for thedelivered casting. Difficult castings are quoted at a higher price per pound than simple castings, but the difference inprice (often based on the estimated cost) did not seem to be great enough to warrant the extra labor costs.
Dick suggests that the company is making little profit and sometimes even a loss on jobs that take considerablelabor. What will happen if Endicott starts quoting higher prices for casting requiring extra labor? Should it recover thefull cost? What ideas can you suggest to improve the estimates? Should the engineer depend on the knowledge of Dickas final? What constitutes a loss or profit for the estimate?
In your consideration of the case study you may want to call foundries and inquire how they estimate futurework.
