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Chapter Six Facility Layout

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Chapter Six Facility Layout Formulas Cycle Time (CT) = Available Time / Desired Output N Min (Min. Number of Workstations) =

Max Output =

Total Available Time Longest Task Time

t CycleTime

t Efficiency =

N * CycleTime

* 100

or

100 – Balance Delay

Actual

 idleTime Balance Delay =

N * CycleTime

* 100 (to get a %)

Actual

1. A manufacturing company is designing an assembly line to produce its main product. The line should be able to produce 60 units per hour. The following data give the necessary information. Task Immediate Predecessor Task Time (sec) A None 35 B A 50 C A 21 D B 38 E C 25 F D,E 58 G F 15 (a) Which task is the bottleneck? (b) Draw a precedence diagram for the above information. (c) Compute the cycle time with a desired output of 60 units per hour.

Chapter Six Facility Layout

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2. An assembly line must be designed to produce 50 packages per hour. The following data give the necessary information. Task Immediate Predecessor Task Time (sec) A None 25 B A 60 C B 35 D B 45 E B 10 F C,D,E 50 (a) (b) (c) (d) (e)

Draw a precedence diagram. Compute the cycle time (in seconds) to achieve the desired output rate. What is the theoretical minimum number of stations? Which work element should be assigned to which workstation? What the resulting efficiency and balance delay percentages?

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3. An assembly line must be designed to produce 40 containers per hour. The following data gives the necessary information. Task Immediate Predecessor Task Time (sec) A None 60 B A 12 C B 35 D A 55 E D 10 F E 50 G F,C 5 (a) (b) (c) (d)

Draw a precedence diagram. Compute the cycle time (in seconds) to achieve the desired output rate. What is the theoretical minimum number of stations? Which work element should be assigned to which workstation?

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4.The ABC Corporation is designing its new assembly line. The line will produce 50 units per hour. The tasks, their times, and their immediate predecessors are shown in Table 10-15: Table 10-15 Task Information for ABC Corporation Task A B C D E F G H (a) (b) (c) (d)

Immediate Predecessor None A A B B,C C F G

Task Time (sec) 55 30 22 35 50 15 5 10

Which task is the bottleneck? Compute the cycle time with a desired output of 50 units per hour. Use a cycle time of 72 second/unit to assign tasks to workstations. Compute the theoretical minimum number of stations. Did you end up using more stations than the theoretical minimum? (e) Compute the efficiency of the line.

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5. Kiko Teddy Bear is a manufacturer of stuffed teddy bears. Kiko would like to be able to produce 40 teddy bears per hour on its assembly line. Use the information provided in Table 10-16 to answer following: Table 10-16 Task Information for Kiko Teddy Bear Work Element Task Description Immediate Predecessor Task Time (sec) A Cut teddy bear pattern None 90 B Sew teddy bear cloth A 75 C Stuff teddy bear B 50 D Glue eyes on C 20 E Glue nose on C 15 F Sew on mouth C 35 G Attach manufacturers label B 15 H Inspect and pack teddy bear D,E,F,G 40 (a) (b) (c) (d) (e) (f)

Draw a precedence diagram. What is the cycle time? What is the theoretical minimum number of stations? Assign tasks to specific workstations using the cycle time you computed in part (b). What are the efficiency of the line? Which task is the bottleneck?

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6. Use the longest task time rule to balance the assembly line described in Table 10-17 as the line can produce 30 units per hour. Table 10-17 Assembly Line Task Information Work Element Task Time (sec) Immediate Predecessor A 25 None B 30 A C 15 A D 30 A E 40 C,D F 20 D G 10 B H 15 G I 35 E,F,H J 25 I K 25 J (a) (b) (c) (d)

What is the cycle time? What is the theoretical minimum number of stations? Which work elements are assigned to which workstation? What is the resulting efficiency percentage?

Chapter Six Facility Layout

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7. A dress-making operation is being designed on an assembly line. Table 10-18 shows the task that need to be performed, their task times, and preceding tasks. If the goal is to produce 30 dresses per hour, answer the following questions: Table 10-18 Dress-Making Tasks Information Work Element Task Time (sec) Immediate Predecessor A. Cut dress body 30 None B. Cut sleeves 40 None C. Cut collar 20 None D. Sew dress body 100 A E. Sew sleeves to dress 25 B.D F. Sew collar to dress 50 C,D G. Hem dress 50 D,E,F H. Package dress 90 G (a) (b) (c) (d) (e) (f)

Compute the cycle time. Which task is the bottleneck? What is the maximum output for this line? Compute the theoretical minimum number of stations. Assign work elements to stations, using the longest task time rule. Compute the efficiency and balance delay of your assignment.

Chapter Six Facility Layout

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8.Twelve tasks, with times and precedence requirements as shown below are to be assigned to workstations using a cycle time of 1.5 minutes. Two heuristic rules will be tried: a. greatest positional weight b. greatest number of following tasks In each case, the tiebreaker will be the shortest task time: Task A B C D E F G H I J K L

Length 0.1 0.2 0.9 0.6 0.1 0.2 0.4 0.1 0.2 0.7 0.3 0.2

Follows Task -A B C -D, E F G H I J K

a. Draw the precedence diagram for this line b. Assign tasks to stations under each of the two rules. c. Compute the percentage of idle time for each rule. (Balance delay) Greatest Positional Weight Station What Tasks fits

Tasks Assigned

Time

Cycle time = 1.5/ Time remaining

Idle Time

Total Idle Time =

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9 Using the information below, balance the line and calculate delay and efficiency using the longest task time rule.

A

B

E

C

F

H G

A = 24 sec B = 12 sec C = 8 sec D = 20 sec

E = 23 sec F = 16 sec G = 20 sec H = 7 sec

Cycle time = 45 sec D Station

Available task

G B Assignment

Task Time

Remaining time

Idle time

Chapter Six Facility Layout

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10. Work the same problem above but base it on greatest positional weight rule (Task time for A and all that follows) Cycle time = 45 sec Station

Available task (Positional weight)

Solutions: problem 1

Assignment

Task Tme

Remaining time

Idle time

Chapter Six Facility Layout

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a. Task F is the bottleneck b. B

D

A

F C

G

E

c. C = (3600 seconds/hour)/(60 units/hour) = 60 seconds/unit Problem 2: a. C A

B

F D

b. C = (3600 seconds/hour)/(50 units/hour) = 72 seconds/unit E c. TM =

t =

d. Station 1 2 3 4 5

C

225 = 3.125 or 4 stations 72

Task A B, E D C F

e. Efficiency =

Problem 3:

t NC

(100%) =

225 (100%) = 62.5% (5)(72)

Balance Delay (%) = 37.5%

Chapter Six Facility Layout

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a. B

C G

A b. C = (3600 seconds/ho D (40 units E r) = 90 sec F s/unit c. TM =

t = C

227 = 2.52 or 3 stations 90

d. Station 1 2 3 Problem 4:

Task A, B D, C E, F, G

a. Task A is the bottleneck. b. C = (3600 seconds/hour)/(50 units/hour) = 72 seconds/unit c. Work Station 1 2 3 4

d. TM =

Task Selected A B D C E F G H

t = C

Task Time 55 30 35 22 50 15 5 10

Time Remainiing 17 42 7 50 0 57 52 42

Idle Time 17 7 0

42

222 = 3.08 or 4 stations. Same number of stations as the theoretical minimum 72

stations was used. e. Efficiency =

Problem 5: a. A

t NC

(100%) =

B

222 (100%) = 77.08% ( 4)(72)

C

D

F

H

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b. C = (3600 seconds/hour)/(40 units per hour) = 90 seconds/unit. c. TM =

t = C

340 = 3.78 or 4 stations. 90

d. Work Station 1 2 3 4

Task Selected A B G C F D E H

e. Efficiency =

Task Time 90 75 15 50 35 20 15 40

t NC

Time remaining 0 15 0 40 5 70 55 15

(100%) =

Idle Time 0 0 5

15

340 (100%) = 94.44% ( 4)(90)

Balance Delay = 5.56% f. Task A Problem 6: a. C = (3600 seconds/hour)/(30 units/hour) = 120 seconds/unit b. TM =

t = C

270 = 2.25 or 3 stations 120

c. Work Station 1

Task Selected A B D F

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C E G H I J K

2

3

d. Efficiency =

t NC

(100%) =

3270 (100%) = 75% (3)(120)

Balance Delay = 25%

Problem 7: a. C = (3600 seconds/hour)/(30 units/hour) = 120 seconds/dress b. Task D c. Maximum Output = (Available Time)/(Minimum Cycle Time) = (3600 seconds/hour)/(100 seconds/unit) = 36 dresses per hour. d. TM =

t = C

405 = 3.38 or 4 stations. 120

e. Work Station 1

2 3 4 5

Task Selected B A C D E F G H

f. Efficiency =

t NC

(100%) =

405 (100%) = 67.5% (5)(120)

Balance Delay = 32.5%

Problem 8

CT = 1.5, Total Time = 3.9 Greatest Positional Weight Station What Tasks fits

Tasks Assigned

Time

Cycle time = 1.5/ Time

Idle Time

Chapter Six Facility Layout

1

2

3

a,b,c,e

d,f,g,h,i

J,k,l

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a b c e d f g h i J K l

remaining 1.4 1.2 .3 .2 .9 .7 .3 .2 0 .8 .5 .3

.1 .2 .9 .1 .6 .2 .4 .1 .2 .7 .3 .2

.2

0

.3 Total Idle time = .5

9. Answer above is the same for the greatest number of tasks following

Positional Weight and Greatest # Will be the same Idle Rate (Balance delay)= =

10 Solution CT = 45 Station 1 2

3 4

.5 * 100 = 11.11 % 3*1.5

Available task A B,C,D B,C,G B,C C,E E,F F H

Assignment

Task Tme

A D (longest task time) G (longest task time) B C E F H

24 20 20 12 8 23 16 7

Remaining time 21 (45 – 24) 1 (21-20) 25 ( 45- 20) 13 (25 – 12) 5 (13 – 8) 22 (45-23) 6 (22 – 16) 38 (45 – 7 )

Idle time

1

5 6 38

Total Idle Time 50 Balance delay = Summation of idle time / N (CT)

* 100 so 50 / 4 (45)

* 100 = 28

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Eff

One way

100 – balance delay

Eff

2 nd way

Summation of time / N (CT)

11 solution Station Available task (Positional weight) 1 A (130) B (42), C (31), D (47) 2 B (42), C (31), G (27) C (31), G (27), E (30) G (27), E (30), F (23) 3 G (27), F (23), F (23), H (7)

so 100 – 28 = 72% so 130 / 4 (45) = .72 or 72 %

Assignment

Task Tme

A D B C E G F H

24 20 12 8 23 20 16 7

Remaining time 21 1 33 25 2 25 9 2

Idle time 1

2

2

Total Idle Time 5

Balance delay =

5 100 = 3.7 3 45

Eff = 100 - 3.7

= 96.3 %...