315 The table below shows production data for five batches o

3.15 The table below shows production data for five batches of parts processed through a batclh production plant. Setup times (T) are given in hours. Operation cycle times (T) are given in min per cycle. Utilization fractions ( are the fractions of time during the 40-hour week that the machine is devoted to the production of these parts. The parts do not proceed through the machines in the same order. Determine (a) weekly production rate, (b) orkload, and (c) average utilization of this set of equipment. A spreadsheet calculator is recommended for this problem. Machine1 T. Machine 3 Machine 2 T. Part T. T. A 2.50 1.50 0.125 2.25 2.0 0.140 5.00 56 0.190 3.00 0.78 0.140 2.90 53 0.200 .25 2.39 0.230 1.50 1.50 0.150 1.80 4.000.345 2.30 150 0.170 D 4.00 7.20 0.250 .75 5.10 0.150 60 2.88 0.100 2.00 3.20 0.150 2.55 1.80 0.120 2.55 .80 0.120 3.20 (A) The break-even point is to be determined for two production methods, one manual and the other automated. The manual method requires two workers at S16.50 per hour each. Together, their production rate is 30 units per hour. The automated method has an initial cost of $125.000, a 4-year service life, no salvage value, and annual maintenance costs = $3000 No labor (except for maintenance) is required for the machine, but the power to operate it is 50 kW (when running). Cost of electric power is $0.05 per kWh. The production rate for the automated machine is 55 units per hour. (a) Determine the break-even point for the two methods, using a rate of return-25%. (b) How many hours of operation per year would be required for each method to reach the breakeven point?

Solution

Tb=t su + Q tc

Tb = batch processing time expressed in minutes

T su = Time to set up machine to produce batch expressed in minutes

T c = Operational cycle time per part expressed in minutes

Q= Number of parts in batch expressed in parts

Machine 1

Part A

0.125  =2.50 +1.50

0.125= 4

=32

PART B

0.140 = 3.00+0.78

0.140 = 3.78

=27

PART C

0.150=1.50+1.50

=20

PART D

0.250 = 4+ 7.20

=44.80

PART E

0.150 = 2 +3.20

0.150 =5.20

=34.67

Machine  2

Tb=t su + Q tc

Part A

0.140= 2.25+2.01

0.140 =4.26

=30.43

Part B

0.200=2.90+1.53

0.200=4.43

=22.15

Part C

0.345= 1.80 + 4.00

0.345 =5.8

=16.81

Part D

0.150 = 1.75 + 5.10

0.150 =6.85

=45.67

PART E

0.120= 2.55+ 1.80

0.120 =4.35

=36.25

MACHINE 3

Tb=t su + Q tc

PART A

0.190=5+1.56

=34.53

PART  B

0.230=1.25 + 2.39

0.230=3.64

=15.83

PART C

0.170= 2.30 +1.50

0.170 = 3.80

=22.35

PART D

0.100=1.60 + 2.88

0.100=4.48

44.80

PART E

0.120 = 2.55 + 1.80

0.120= 4.35

=36.25

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B)

Work load

Machine 1 =32 + 27 + 20 +44.80 +34.67=158.47

out of 40 hours =part D having more loads

Machine 2=30.43  + 22.15  + 16.81 + 45.67  + 36.25=151.31

out of 40 hours =part D having more loads

Machine 3 =34.53 +  15.83  + 22.35  + 44.80  + 36.25=153.76

out of 40 hours =part D having more loads

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c) Machine 1

=32 + 27 + 20 +44.80 +34.67=158.47 / 5=31.69

Machine 2

30.43  + 22.15  + 16.81 + 45.67  + 36.25=151.31 /5=30.27

Machine 3

34.53 +  15.83  + 22.35  + 44.80  + 36.25=153.76 /5

=30.75

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