Homework 6a Operations
Justin Prater
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?
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?