Chapter 6: Line Balancing Homework
Problem 1, a,b,c,d,e
Problem 4 (a), 1,2,3,4
Problem 5 a, b, c, d (balance on Min. CT)
1. ∑t=18 minutes
OT = 450 minutes/day
tasks = 17
longest task time = 2.4 minutes
a. minimum cycle time = 2. 4 minutes
maximum cycle time = 18 minutes
b. what range of output is possible?
Output Capacitymin cycle time = OT
cycle time
= 450 minutes
2.4 minutes
= 187.5 units
Output Capacitymax. cycle time = 450 = 25 units
18
c. minimum # of workstations for maximum output?
Nmin = sum of all times
CT
= 18 minutes
2.4 minutes
= 7.5 stations (8 stations)
d. what cycle time will provide output of 125 units/day?
Output Capacity = OT
cycle time
125 = 450 minutes
CT
125 CT = 450
125 CT = 450
125 125
CT = 450 = 3.6 minutes
125
e. What output potential will result if cycle time is (1) 9 minutes (2) 15 minutes?
Output Capacity = OT
cycle time
= 450 minutes = 50 units
9 minutes
Output Capacity = OT
cycle time
= 450 minutes = 30 units
15 minutes
Problem 4
a(1) draw the precedence diagram
.2 .4
a ➙ b ➘
.3 1.3 .3 1.2
c ➙ d ➙ g ➙ h
.1 .8
e ➙ f ➚
a(2) Assign tasks according to > # of following tasks. Assume the cycle time is to be the minimum.
Min. Cycle time = 1.3 minutes
Nmin = sum of task times
CT
= 4.6 minutes
1.3 minutes
= 3.538 stations (4 stations)
Work Station #1
CT = 1.3 minutes
a -.2
1.1 minutes
b -.4
.7 minutes
c -.3
.4 minutes
e -.1
.3 minutes Idle Time
Work Station #2
CT = 1.3 minutes
d -1.3
0 minutes Idle Time
Work Station #3
CT = 1.3 minutes
f -.8
.5 minutes
g -.3
.2 minutes Idle Time
Work Station #4
CT = 1.3 minutes
h -1.2
.1 minutes Idle Time
a(3) determine % of Idle Time
Total Idle Time ______
.6 minutes
% Idle time = Idle time per cycle * 100
Nactual * Cycle time
= .6 minutes * 100
4 WS * 1.3 minutes
= .6 * 100
5.2
= .1153856 * 100 = 11.54%
a(4) output for 420 minute work day?
Output Capacity = OT
cycle time
= 420 minutes = 323.07 units
1.3 minutes
Problem 5
(ignore desired output of 240 units per day, balance on min. CT)
a. draw the precedence diagram
a ➙ b ➙ c ➘
f ➘
d ➙ e ➙ g
b. max. cycle time = ∑t
= 4.6 minutes
c. min. # of stations needed.
Min. cycle time = 1.2 minutes
Nmin = ∑ t
CT
= 4.6 minutes
1.2 minutes
= 3.83 stations (4 stations)
d. Assign tasks on ># of following tasks. Use > time as tiebreaker.
Work Station #1
CT = 1.2 minutes
a -.2
1.0 minutes
b -.4
.6 minutes
d -.4
.2 minutes
c -.2
0 minutes Idle Time
Work Station #2
CT = 1.2 minutes
f -1.2
0 minutes Idle Time
Work Station #3
CT = 1.2 minutes
e -1.2
0 minutes Idle Time
Work Station #4
CT = 1.2 minutes
g -1.0
.2 minutes Idle Time
e. compute % Idle time
Total Idle time = .2 minutes
% Idle time = Idle time per cycle * 100
Nactual * Cycle time
= .2 minutes * 100
4 WS * 1.2 minutes
= .2 * 100
4.8
= .041666 * 100 = 4.16%