Question

Task	Time	Predecessor
1	0.2	-
2	0.4	1
3	0.1	1
4	0.4	2,3
5	0.1	2
6	0.3	3,4
7	0.2	5
8	0.2	5,6
9	0.4	7,8

Consider the operations given in Table 1. 47 feet of the oor space along the conveyor is available for installing assembly stations around a closed loop conveyor system. Eeach station is made 6 feet long in order for the operators to work comfortably, and at least 1 feet clearance between stations (except between input and output stations).

• Find the maximum number of stations on conveyor when the target production rate is 90 units per hr.

• Find the maximum number of stations on conveyor when the target production rate is 120 units per hr.

• Find the speed of conveyor when target production rate is 90 units per hr.

• Find the speed of conveyor when target production rate is 120 units per hr.

• Find the task to workstation assignments with Largest Candidate Rule when the target production rate is 90 units per hr.

• Find the total space occupied by the workstations and clearances when the target production rate is 90 units per hr.

• Find the task to workstation assignments with Largest Candidate Rule when the target production rate is 120 units per hr. Adjust the station-sizes using the method we covered in the class.

• Find the allocated space occupied to each workstation, and total space occupied including clearances when the target production rate is 120 units per hr.

Answer 1

The Total Task time, T is 2.3 mins, as given.

When the target production rate is 90 units per hr (1.5 units per min), the Cycle time, C (Production Time/ Output) is 0.67 mins.

The theoretical minimum number of stations on conveyor, Nt is T/C rounded up to the next highest integer.

Hence, Nt = 2.3/0.67 = 3.45 = 4 stations

The corresponding Efficiency, Et = T/ (Nt x C) = 0.86

When the target production rate is 120 units per hr (2 units per min), the Cycle time, C is 0.5 mins.

Hence, the theoretical minimum number of stations on conveyor, Nt = 2.3/0.5 = 4.6 = 5 stations

The corresponding Efficiency, Et = 0.92

The task to workstation assignments with Largest Candidate Rule when the target production rate is 90 units per hr, is as tabulated below. Based on this, there are 4 stations (Na) for which the space needed is 27 feet (given that each station is 6 feet long, and there is 1 foot clearance between stations - except between input and output stations.

The corresponding actual Efficiency, E = T/ (Na x C) = 0.86

	Task	Task time	Remaining unassigned time	Feasible remaining tasks
WS1	1	0.2	0.466666667	2
	2	0.4	0.066666667	None
WS2	3	0.1	0.566666667	4
	4	0.4	0.166666667	5
	5	0.1	0.066666667	None
WS3	6	0.3	0.366666667	7, 8
	7	0.2	0.166666667	None
WS4	8	0.2	0.466666667	9
	9	0.4	0.066666667

The task to workstation assignments with Largest Candidate Rule when the target production rate is 120 units per hr, is as tabulated below. Based on this, there are 6 stations for which the space needed is 41 feet (given that each station is 6 feet long, and there is 1 foot clearance between stations - except between input and output stations.

The corresponding actual Efficiency, E = 0.77

	Task	Task time	Remaining unassigned time	Feasible remaining tasks
WS1	1	0.2	0.3	3
	3	0.1	0.2	None
WS2	2	0.4	0.1	5
	5	0.1	0	None
WS3	4	0.4	0.1	None
WS4	6	0.3	0.2	7, 8
	7	0.2	0	None
WS5	8	0.2	0.3	None
WS6	9	0.4	0.1