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IE 419 1 Work Design: Productivity and Safety Dr. Andris Freivalds Class #5.

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Presentation on theme: "IE 419 1 Work Design: Productivity and Safety Dr. Andris Freivalds Class #5."— Presentation transcript:

1 IE 419 1 Work Design: Productivity and Safety Dr. Andris Freivalds Class #5

2 IE 419 2 Work Design: Productivity and Safety Dr. Andris Freivalds Class #5

3 IE 419 3 Worker-Machine Relationships (Ch. 2, pp. 50-55) Worker-machine relationships –Synchronous servicing – regular cycles –Random (asynchronous) servicing – random –Combination – real life N ≤ (l + m) (l + w)

4 IE 419 4 Ex. #1 l = 1.0, m = 2.0, w = 0.1 Oper=$10/hr, Mach=$20/hr 3 machines ($1.28) 2 machines ($1.25)

5 IE 419 5 Ex. 3 – No Load l = 0, m = 1.2, w = 2.0, oper = $10/hr, mach = $15/hr TimeOperMach

6 IE 419 6 Ex. 3 – No Load l = 0, m = 1.2, w = 2.0, oper = $10/hr, mach = $15/hr TimeOper1Oper2Mach

7 IE 419 7 Random Servicing Machine servicing is not regular Most likely it is random (don’t know when) Use probability theory to estimate % idle time (binomial expansion) Probability of m (out of n) machines down = n! p m q (n-m) m! (n-m)! p = prob of down time q = prob of up time = 1-p

8 IE 419 8 Ex #1: n=3, p=0.1, q=0.9, % idle time? Mach down (m) Probability n! p m q (n-m) / m! (n-m)! Mach hrs lost (1 op) Mach hrs lost (2 op) Mach hrs lost (3 op) 0 1 2 3 Total

9 IE 419 9 Ex #1 con’t: oper = $10/hr, mach = $500/hr, prod = 120 units/hr 1 oper2 oper3 oper idle Prod (8 hr) Cost (8 hr) Unit cost

10 IE 419 10 Ex #2: n=3, p=0.4, q=0.6 mProbabilityHrs lost (1 oper) Hrs lost (2 oper)

11 IE 419 11 Ex #2 con’t: oper = $10/hr, mach = $60/hr, prod = 60/hr 1 oper2 oper3 oper idle Prod (8 hr) Cost (8 hr) Unit cost

12 IE 419 12 Complex Relationships (n↑) n > 6 Use Wright’s formula i = interference, % of l n = # machines x = m/l n ≤ 6 i = 50{[(1+x-n) 2 +2n] ½ - (1+x-n)}

13 IE 419 13 Ex #3: m = 7, l =1, n = 6 n > 6 Use Wright’s formula i = interference, % of l n = # machines x = m/l n ≤ 6 i = 50{[(1+x-n) 2 +2n] ½ - (1+x-n)}

14 IE 419 14 (Assembly) Line Balancing (Ch. 2, pp. 56-64) Worker-machine relationship determining ideal number of workers/workstations in production line Simple straight line (Ex #1): 1 2 3 4 5 OperST/opDelay time ST ReqST #OperNew ST/op New Delay 10.52 20.48 30.65 40.41 50.55

15 IE 419 15 Straight Line Balancing ( Ex #1 con’t) OperST/opDelay time ST ReqST #OperNew ST/op New Delay 10.520.13 20.480.17 30.650.00 40.410.24 50.550.10 % efficiency (E) = 100 x ∑ST/∑AT % idle = 100 x ∑DT/∑AT Req. production (R) = 3,200/day → required standard (cycle) time # oper = R x ∑ST = ∑ST/ CT=

16 IE 419 16 Complex Line Balancing ( Ex #2) Time set by slowest station (cycle time) Assign operators to a workstation: –Until cycle time is about to be exceeded –In order of decreasing positional weight –PW = ∑ST for all with “1” relationship –As allowed by precedence (i.e. immediate predecessors (IP) need to be assigned) 2 5 1 4 7 9 3 6 8

17 IE 419 17 Positional Weight (PW) Matrix OpSTDT123456789PWIP 1.05 2.03 3.04 4.05 5.01 6.04 7.05 8.04 9.06 2 5 1 4 7 9 3 6 8

18 IE 419 18 Fill Workstations (Ex #2 con’t) # workstations = Rx∑ST = ∑ST/ CT= OperIPPWSTStation TDelay T


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