Homework SourseFirst shift employees at General Motors work from 6am to 2pm
First shift employees at General Motors work from 6am to 2pm Monday through Friday. Everyone on the assembly line is allowed a 30 minute lunch and two 20 minute breaks, but other cross-trained associates cover lunches and breaks for the associates on the assembly line so the line doesn’t stop. The processes of the assembly line have the following dependencies:
Task Predecessor Average Time (sec)
A none 45
B none 90
C A 110
D B 120
E none 100
F C 15
G E 115
H F, D, G 35
The Plant Manager needs first shift to achieve an output of 100 cars.
Compute the Takt time for the line .
What is the theoretical minimum number of workstations?
Draw the precedence diagram. Then assign tasks to work stations and you must use the largest eligible task heuristic rule to break any ties. Show which tasks are assigned to each workstation.
Compute the efficiency and percent idle time for the system .
What is the maximum output that could be achieved by a single assembly line in 8 hours?
Solution
The First Shift time is from 6AM to 2PM means 8 hours shift and assembly line does not stop in between these 8hours, therefore total time available is 8*60*60 = 28,800 seconds per day
Manager wants to have 100 cars as output from first shift working, therefore the assembly line is supposed to produce or a car should turn out every 288 seconds (28800/100). This is ( as 288 sec.)called cycle time of assembly line.
Now theoritical minimum number of workstations = Total time for all the tasks / Cycle time = 630/288 = 3 because Total time for all tasks = 45+90+110+120+100+15+115+35 = 630 and Number has to be integer or to be more specific a +ve whole number >= Total tasks time/ cycle time
Precedence diagram is like given below
A----------C------------F
Start------------B-----------------D---------------------H
E-----------------G
To begin with we can start A, B and E for which no precedence, then after A we can start C, after B we can start D and G can follow after E. Task F requires completion of C and finally H after completion of D,F and G
Assignment of tasks to Workstations is as follows:
Workstation 1 Cycle time is 288 seconds Eligible candidates are A, B and E as E has the maximum time so first assignment is E to workstation, now G is added as eligible candidate alongwith A and B. Next G is allocated to workstation1 as its time is maximum, after allocation we are left with available time of 73 seconds therefore only A can be allocated to Workstation1 out of A and B.
Workstation 2 Cycle time 288 seconds, eligible candidates B and C, first C is assigned based on higher task time, after its assignment F also becomes eligible but B is assigned next on the same basis of task time, after B assignment D becomes eligible but its time is 120 as against 88 available Therefore F is next task allocated to workstation 2.
Workstation 3 : first D is allocated and then H with total time of 155 for tasks and 133 seconds as idle time for the workstation3.
Efficiency and percent idle time for the system is calculated as follows:
Total time required for completion of att tasks = 630 seconds
Total time actual spend = cycle time multiplied by number of workstations = 288*3 = 864 seconds
Therefore efficiency = 630/864 (standard/Actual) % = 72.92 % apprx
Total idle time = Sum of idle times at all workstations = 28+73+133 = 234 or Actual -Standard = 864-630
Therefore percentage idle time = (234/864) % = 27.08% approx which is also 100-efficiency%
Maximum output that can be achieved= 28,800 (8hours)/630(total task times) = 45.71 rounded to 45 only
