# Homework 6a Operations

Chapter 6A

3, 4, 9, and 14

3) a.

b. Cycle time = production time per day/required output per day = [(8 hour/day)(3600 seconds/hour)]/240 units per day = 120 seconds per unit c.

Work station| Task| Task time| Idle time|

I| AD| 6050| 10|

II| BC| 8020| 20|

III| EF| 9030| 0|

IV| GH| 3060| 30|

d. Efficiency = Task time / [# stations X avg. cycle time) = = .875 or 87.5%

4) a.

A

B

F

C

D

E

G

30

35

15

65

H

30

35

40

25

A

B

F

C

D

E

G

30

35

15

65

H

30

35

40

25

b.

What is the workstation cycle time?

Cycle time = production time per day/required output per day Cycle time = 27000 / 360 = 75 seconds.

c. Calculate this balance line using the largest number of following tasks. Use the longest task time as a secondary criterion.

275 / 75 = 3.66

There should be a minimum of 4 work stations

Work station| Task| Task time| Idle time|

I| ACE| 303015| 0|

II| BD| 3535| 5|

III| F| 65| 10|

IV| GH| 4025| 10|

d. What is the efficiency of your balance line?

91.7%

9.

A

B

F

C

D

E

G

20

7

15

10

H

20

22

16

8

A

B

F

C

D

E

G

20

7

15

10

H

20

22

16

8

a.

Cycle time = production time per day/required output per day Cycle time = 25200 / 750 = 33.6 seconds.

b.

What is the theoretical number of workstations?

# workstations = sum of task times / cycle time

118 / 34 = 3.5

There is a minimum number of 4 workstations.

c.

Draw the precedence diagram

d.

Balance the line using sequential restrictions and the longest-operating time rule. Work station| Task| Task time| Idle time| Feasible Rem. T.| Tasks w/ most foll| Task w/ longest t.| | I| AB| 207| 147| B| B| B| |

II| DF| 2210| 122| F| F| F| |

III| C| 20| 14| | | | |

IV| EG| 1516| 193| G| G| G| |

V| H| 8| 26| | | | |

e.

What is the efficiency of the line balanced as in d?

Efficiency = Sum of task times / (# workstations X Cycle time) = 118 / (5 X 34) = .70

Efficiency = 70%

f.

Suppose that demand rose from 750 to 800 units per day. What would you do? Show any amounts or calculations. New cycle time = 25200/800 = 31.2 seconds

Reduce cycle time to 31 seconds and work 62-3 minute overtimes. g.

Suppose that demand rose from 750 to 1000 units per day. What would you do? Show any amounts or calculations. New cycle time = 25200/1000 = 25.2 seconds

1.89 hours overtime; may be better to rebalance.

14)

Develop two alternative layouts.

What is the efficiency of your layouts?

A

B

F

C

D

E

G

1

1

3

1

H

2

1

1

2

1

I

A

B

F

C

D

E

G

1

1

3

1

H

2

1

1

2

1

I

Cycle time = 4 minutes

What is the theoretical number of workstations?

# workstations = sum of task times / cycle time

13 / 4 = 3.25

Work station| Task| Task time| Idle time| Feasible Rem. T.| Tasks w/ most foll| Task w/ longest t.| | I| ABC| 11 2 | 32 0| B,F C,D| B C,D | B,F C| | IIIIIIV| DEFGHI| 131121| 303203| EGH| EGH| EGH| |

Work station| Task| Task time| Idle time| Feasible Rem. T.| Tasks w/ most foll| Task w/ longest t.| | I| AFG| 11 1 | 32 1| B,F G| B G | B,F G| | IIIIIIV| BCDEHI| 311321| 313021| C,DEI| C,DEI| C,DEI| |

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