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).
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 |
Consider the following data: x 1 2 3 4 5 P(X=x) 0.2 0.1 0.1 0.2 0.4 Step 5 of 5 : Find the value of P(X≤1). Round your answer to one decimal place
For question 3, assume the following about an economy: 1. y = f(k)=k", where a = 0.25 2. S=0.3 3. 8=0.2 4. n=0.05 5. g=0.02 3. Answer the following: [10 Points] a) [4 Points) Find the steady state capital per effective worker, output per effective worker, investment per effective worker, and consumption per effective worker. b) [4 Points) Find the steady-state growth rate of capital per worker, output per worker, investment per worker, and consumption per worker. c) (2 Points)...
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