EIN 6336 HW Solution V12 (9-16-2016)
Chapter 2 HW Solution:
3.Week / Output / Number of
Workers / Material
(lbs.) / Labor Cost / Overhead Cost / Material Cost / Total Cost / MFP
1 / 30,000 / 6 / 450 / 2,880 / 4,320 / 2,700 / 9,900 / 3.03
2 / 33,600 / 7 / 470 / 3,360 / 5,040 / 2,820 / 11,220 / 2.99
3 / 32,200 / 7 / 460 / 3,360 / 5,040 / 2,760 / 11,160 / 2.89
4 / 35,400 / 8 / 480 / 3,840 / 5,760 / 2,880 / 12,480 / 2.84
Notes:
Labor Cost = Number of Workers x 40 hours x $12/hour
Overhead Cost = Labor Cost x 1.50
Material Cost = Material (lbs.) x $6/lb.
Total Cost = Labor Cost + Overhead Cost + Material Cost
Multifactor Productivity (MFP) = Output / Total Cost (rounded to two decimals)
Multifactor productivity dropped steadily from a high of 3.03 to a low of 2.84.
(The week one has the highest productivity)
5. Without scrap the output can be 80 pieces per hour
The increase in productivity would be 80 – 72 = 8 pieces per hour.
This would amount to an increase of (8 / 72) = 11.1%.
(The productivity would increase by 11.1%, by eliminating scrap)
6. Current period productivity =
Previous period productivity =
Thus, there was an increase of 4.44% in productivity.
Chapter 3 HW Solution:
4. Given:
Week / Requests1 / 20
2 / 22
3 / 18
4 / 21
5 / 22
a. Naïve approach forecast for Week 6 = Demand in Week 5 = 22
b. Four-period moving average forecast for Week 6:
(round to two decimals)
c. Exponential smoothing with alpha = 0.30 and a Week 2 Forecast = 20 (round to two decimals):
F3 = 20 + 0.30(22 – 20) = 20.60
F4 = 20.60 + 0.30(18 – 20.6) = 19.82
F5 = 19.82 + 0.30(21 – 19.82) = 20.17
F6 = 20.17 + 0.30(22 – 20.17) = 20.72
8. a. There appears to be a long-term upward increasing trend in the data. If we use an averaging technique, the forecast will underestimate when data values increase.
b.
t / Y / t*Y / t21 / 405 / 405 / 1
2 / 410 / 820 / 4
3 / 420 / 1,260 / 9
4 / 415 / 1,660 / 16
5 / 412 / 2,060 / 25
6 / 420 / 2,520 / 36
7 / 424 / 2,968 / 49
8 / 433 / 3,464 / 64
9 / 438 / 3,942 / 81
10 / 440 / 4,400 / 100
11 / 446 / 4,906 / 121
12 / 451 / 5,412 / 144
13 / 455 / 5,915 / 169
14 / 464 / 6,496 / 196
15 / 466 / 6,990 / 225
16 / 474 / 7,584 / 256
17 / 476 / 8,092 / 289
18 / 482 / 8,676 / 324
171 / 7,931 / 77,750 / 2,109
Round b & a to two decimals:
Forecasted demand for the next three week (round to two decimals):
Y19 = 397.01 + (4.59)(19) = 484.22
Y20 = 397.01 + (4.59)(20) = 488.81
Y21 = 397.01 + (4.59)(21) = 493.40
13. Given:
Quarter / Year 1 / Year 2 / Year 3 / Year 41 / 2 / 3 / 7 / 4
2 / 6 / 10 / 18 / 14
3 / 2 / 6 / 8 / 8
4 / 5 / 9 / 15 / 11
SA method (round season averages to three decimals and seasonal relatives to two decimals):
Quarter / Year 1 / Year 2 / Year 3 / Year 4 / Season Average / Seasonal Relative1 / 2 / 3 / 7 / 4 / 4.000 / 0.50 (4.000/8.000)
2 / 6 / 10 / 18 / 14 / 12.000 / 1.50 (12.000/8.000)
3 / 2 / 6 / 8 / 8 / 6.000 / 0.75 (6.000/8.000)
4 / 5 / 9 / 15 / 11 / 10.000 / 1.25 (10.000/8.000)
8.000
Overall Average
Sum of Seasonal Relatives = 0.50 + 1.50 + 0.75 + 1.25 = 4.00
Chapter 5 HW Solution:
1. a.
b.
c. This is not necessarily true. If the design capacity is relatively high, the utilization could be low even though the efficiency was high.
5. Demand = 30,000 = Q
FC = $25,000
VC = $.37/pen
a. Revenue = $1.00/pen
Breakeven Point:
b. Target profit = $15,000 @ demand of 30,000,
Solving for Rev:
Rev = $1.71 [rounded up]
11. a. Given: 10 hrs. or 600 min. of operating time per day per machine.
250 days x 600 min. = 150,000 min. per year operating time per machine.
Machine purchase costs: A = $40,000; B = $30,000; C = $80,000.
Total processing time by machineProduct / A / B / C
1 / 48,000 / 64,000 / 32,000
2 / 48,000 / 48,000 / 36,000
3 / 30,000 / 36,000 / 24,000
4 / 60,000 / 60,000 / 30,000
Total / 186,000 / 208,000 / 122,000
Options:
Buy two A machines at a total purchase cost of 2 x $40,000 per machine = $80,000.
Buy 2 B machines at a total purchase cost of 2 x $30,000 per machine = $60,000.
Buy 1 C machine at a total purchase cost of $80,000 for one machine.
Conclusion: We should buy 2 of the B machines at a total cost of $60,000.
b. Given: Operating Costs: A = $10/hour/machine; B = $11/hour/machine; C = $12/hour/machine.
Total cost for each type of machine:
A (2): 186,000 min. / 60 min./hour = 3,100.00 hrs. x $10 = $31,000 + $80,000 = $111,000
B (2): 208,000 min. / 60 min./hour = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133
C(1): 122,000 min. / 60 min./hour = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400
Conclusion: Buy 2 Bs—these have the lowest total cost.
Chapter 6 HW Solution
11. Determine if the department locations shown below are appropriate. If not, modify the assignments.
2 / 5 / 41 / 8 / 6
7 / 3
Check whether the following conditions have been satisfied:
A Links / X Links / E Links / I Links1-3 / 1-2 (no) / 1-6 (no) / 3-7
1-8 / 1-7 (no) / 2-6 (no)
2-7 (no) / 2-4 / 2-8
3-5 (no) / 7-8
3-6
3-8
4-5
4-8
5-7 (no)
5-8
6-8
Next, re-arrange departments to determine if all conditions can be satisfied. We will start with our table of links shown below.
A Links / X Links / E Links / I Links1-3 / 1-2 / 1-6 / 3-7
1-8 / 1-7 / 2-6
2-7 / 2-4 / 2-8
3-5 / 7-8
3-6
3-8
4-5
4-8
5-7
5-8
6-8
Rather than drawing clusters, we can start with a blank layout and then try to satisfy the A and the X conditions above. After that, we can shift departments around to satisfy E conditions followed by I conditions.
Looking at the A links, we can see that Department 8 appears most frequently, followed by Departments 3 & 5, Departments 1, 4, 6, & 7, and then Department. 2. Department 8 is a good candidate for a center position. We can try to satisfy those conditions in order in our layout, shifting departments around to meet all A and X conditions. After that, we can shift departments more if needed using trial and error to meet E conditions and then I conditions.
The layout below satisfies the A and the X conditions. It does not satisfy one E condition (1-6) and the single I condition (3-7).
1 / 5 / 43 / 8 / 7
6 / 2
13. Given: We must arrange departments in a 3 x 3 grid, and Department 5 must be located in the lower left corner (shown in the shaded area in the layout).
We start by placing the Muther grid information into a table.
A Links / X Links / E Links / I Links1-3 / 1-2 / 1-4 / 2-4
1-7 / 1-6 / 1-9
1-8 / 2-3 / 2-9
2-5 / 3-4 / 3-7
2-6 / 3-6 / 4-8
2-7 / 3-8 / 6-7
3-9 / 4-5
4-6 / 4-9
4-7 / 5-6
5-7 / 5-8
5-9 / 6-9
7-8 / 8-9
7-9
Looking at the A links, we can see that that Department 7 appears most frequently followed by Departments 1, 2, 5, & 9, and then Departments 3, 4, 6, & 8.
Department 7 is a good candidate for a central location. Then, we satisfy the remaining A conditions along with the X conditions. After that, we can use trial and error to move departments around to satisfy the E and I conditions.
The layout below satisfies all conditions.
3 / 1 / 89 / 7 / 4
5 / 2 / 6
Chapter 7 HW Solution
1. Given: Average time = 10.40 minutes. Standard deviation = 1.20 minutes for a worker with a performance rating of 125%. Assume an allowance of 16% of job time.
Standard Time = Normal Time x Allowance Factor
ST = NT x AF
NT = Observed Time x Performance Rating
NT = OT x PR = 10.4 x 1.25 = 13.00 minutes (round to 2 decimals)
AFjob = 1 + A, where A = Allowance % based on job time
AFjob = 1 + .16 = 1.16
ST = NT x AFjob = 13.0 minutes x 1.16 = 15.08 minutes (round to 2 decimals)
1. / Average time per cycle (OT) = 10.4 minutesStandard deviation (s) = 1.2 minutes
Performance rating (PR) = 1.25
Job time allowance = 16%
AFjob = 1+A= 1.16
Normal time (NT):
NT = OT x PR = 10.4 x 1.25 = 13.0 minutes
Standard time (ST)
ST = NT x AFjob = 13.0 x 1.16 = 15.08 minutes
2. Given: Average time = 1.2 minutes per piece. Performance rating = 95%. Workday allowances = 10%.
a. Observed Time (OT) = 1.2 minutes
b. Normal Time (NT):
NT = Observed Time x Performance Rating
NT = OT x PR = 1.2 x .95 = 1.14 minutes (round to 2 decimals)
c. Standard Time:
Standard Time = Normal Time x Allowance Factor
ST = NT x AF
A = .10 of work time
(round to 3 decimals)
ST = NT x AFday = 1.14 minutes x 1.111 = 1.27 minutes (round to 2 decimals)
2. / OT = 1.2 minutesPerformance rating (PR) = .95
Workday allowance (A) = 10%
a. / OT = 1.2 minutes
b. / NT = OT x PR = 1.2 x .95 = 1.14 minutes
c. / ST = NT x AFday = 1.14 x 1.111 = 1.27 minutes
3. Given: A time study was conducted of a job with four elements. The observed times and performance ratings for six cycles are shown in the table below. Allowance factor = 15% of job time.
OBSERVATIONS (minutes per cycle)Element / PR / 1 / 2 / 3 / 4 / 5 / 6
1 / .90 / 0.44 / 0.50 / 0.43 / 0.45 / 0.48 / 0.46
2 / .85 / 1.50 / 1.54 / 1.47 / 1.51 / 1.49 / 1.52
3 / 1.10 / 0.84 / 0.89 / 0.77 / 0.83 / 0.85 / 0.80
4 / 1.00 / 1.10 / 1.14 / 1.08 / 1.20 / 1.16 / 1.26
a. Average cycle time (OT) for each element (round to 2 decimals):
OBSERVATIONS (minutes per cycle)Element / 1 / 2 / 3 / 4 / 5 / 6 / Average (OT) (min.)
1 / 0.44 / 0.50 / 0.43 / 0.45 / 0.48 / 0.46 / 0.46
2 / 1.50 / 1.54 / 1.47 / 1.51 / 1.49 / 1.52 / 1.51
3 / 0.84 / 0.89 / 0.77 / 0.83 / 0.85 / 0.80 / 0.83
4 / 1.10 / 1.14 / 1.08 / 1.20 / 1.16 / 1.26 / 1.16
b. Normal Time (NT) for each element (round to 2 decimals):
NT = Observed Time x Performance Rating
Element / Average (OT) / PR / NT(OT x PR)
1 / 0.46 / .90 / 0.41
2 / 1.51 / .85 / 1.28
3 / 0.83 / 1.10 / 0.91
4 / 1.16 / 1.00 / 1.16
c. Standard time for this job (round to 2 decimals):
Standard Time = Normal Time x Allowance Factor
ST = NT x AFjob
AFjob =1 + A = 1 + .15 = 1.15
Element / NT / AFjob / ST1 / 0.41 / 1.15 / 0.47
2 / 1.28 / 1.15 / 1.47
3 / 0.91 / 1.15 / 1.05
4 / 1.16 / 1.15 / 1.33
Standard Time for Job: 0.47 + 1.47 + 1.05 + 1.33 = 4.32 minutes
3. / Element / PR / x / OT / = / NT / x / AFjob / = / ST1 / .90 / .46 / .414 / 1.15 / .476
2 / .85 / 1.505 / 1.280 / 1.15 / 1.472
3 / 1.10 / .83 / .913 / 1.15 / 1.050
4 / 1.00 / 1.16 / 1.160 / 1.15 / 1.334
12. Given: An operation yielded times of 5.2, 5.5, 5.8, 5.3, 5.5, and 5.1. The goal is to estimate the mean time to within 2% of its true value with a confidence of 99%. The standard deviation is given already as 0.253 minutes per cycle.
Mean Time = (5.2 + 5.5 + 5.8 + 5.3 + 5.5 + 5.1) / 6 = 5.4 minutes per cycle