8/2/2019 HW Chapter 15

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Due on September 5,11

Digit 1 L:D = 58/36 = 1.61 Code 1Digit 2 External Cone Code 7Digit 3 Internal to both and no sahpe Code 4Digit 4 No plane surface machining Code 0Digit 5 No auxiliary holes Code 0

9 8 7 6 5 4 3 2 1 0BinaryValue

512 256 128 64 32 16 8 4 2 1

Machine A B C D E F G H I J2 1 1 1 1 1 5583 1 1 1 1 5567 1 1 1 1 5506 1 1 1 1 4491 1 1 1 1 3375 1 1 1 1 2094 1 1 1 1 46

Machine E H A G I D J B C F BinaryValue

DecimalEquivalent

2 1 1 1 1 1 64 9923 1 1 1 1 32 9607 1 1 1 1 16 9284 1 1 1 1 1 8646 1 1 1 1 8 301 1 1 1 1 4 295 1 1 1 1 2 27

113 113 112 97 81 14 14 12 10 6

15.8 The following table lists the weekly quantities and routings of ten parts that are being considered for cellularmaunfacturing in a machine shop. Parts are identified by letters, and machines are indentified numerically. For the data givena.) Develop the part-machine incidence matrixb.) apply the rank order clustering technique to the part-machine incidence matrix to identify logical part families and mchainegroup

Chapter 15 Homework222523 Manufacturing Management

Pornpunsa Wuttisanwattana15.2 Develop the form code ( first five digits) in the Opitz System for the part illustrated in Figure P15.2

Decimal Equivalent

DecimalEquivalent

8/2/2019 HW Chapter 15

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8/2/2019 HW Chapter 15

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Hollier method applied to Group 1

Cell 1 for part E H A I GFrom/To 2 3 4 7

2 145 62 207

3 167 0

4 140 140

7 12 12 So we skip 0 and choose machine 7 to place at the lastTo Sum 167 0 157 202 So get __,__,__, 7

From/To 2 3 42 145 1453 167 1674 0

To Sum 167 0 145so machine 3 is chosen to be placed at the beginning and Machine 4 is chosen to be placed at the last

Bypass track = 62/526% = 11.8 %

Back track = 12/526% = 2.28 %

In Sequence = 452/526% = 85.9 %

Cell 2 for part D J B C FFrom/To 1 5 6

1 0 Choose Machine 1 to be placed at the last5 70 15 85 __, __, 16 35 85 120

To Sum 105 85 15

From/To 5 6

5 15 15

6 85 85

To Sum 85 15

Bypass track = 35/205% = 17.1 %Back track = 15/205% = 7.32 %

In Sequence = 155/205% = 75.6 %

Machine # 6 is chosen to be palced at the beginning and machine # 5 is chosen to be for the last position. Then the lineshould be ordered as 6 1 5 respectively

From Sum

From Sum

From Sum

If a minimum To sum is equal to a minimum From sum, then bothmachines are selected and placed at the beginning and end of thesequence, respectively.

From Sum If a minimum To sum is equal to a minimum From sum, then bothmachines are selected and placed at the beginning and end of the sequence,respectively.

If both To and From sums are equal for aselected machine, it is passed over and themachine with the next lowest sum is selected.

so get 3,__, 7, 4 and machine 2 is chosen for the beginning. Then the line should be ordered as 2,3,7,4

3 2 4 7

12+50 = 62

12

6 5 1

15+20

15

8/2/2019 HW Chapter 15

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a.) If 35hrs/wk are worked, deermine how many of each product will be made by the cell

b.) What is the Utilization of each machine in the cell

A B C D ENo. to produce 16 12 8 8 4 **Solved by Solver

Contraints A/E 4 = 4B/E 3 = 3C/E 2 = 2D/E 2 = 2

Total Time 2004

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cell.onoss