ie 419 1 work design: productivity and safety dr. andris freivalds class #5
DESCRIPTION
IE Worker-Machine Relationships (Ch. 2, pp ) Worker-machine relationships –Synchronous servicing – regular cycles –Random (asynchronous) servicing – random –Combination – real life N ≤ (l + m) (l + w)TRANSCRIPT
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IE 419 1
IE 419 Work Design:
Productivity and Safety
Dr. Andris FreivaldsClass #5
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IE 419 2
IE 419 Work Design:
Productivity and Safety
Dr. Andris FreivaldsClass #5
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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)
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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)
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Ex. 3 – No Loadl = 0, m = 1.2, w = 2.0, oper = $10/hr, mach = $15/hr
Time Oper Mach
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Ex. 3 – No Loadl = 0, m = 1.2, w = 2.0, oper = $10/hr, mach = $15/hr
Time Oper1 Oper2 Mach
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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! pm q(n-m) m! (n-m)!
p = prob of down time
q = prob of up time = 1-p
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Ex #1: n=3, p=0.1, q=0.9, % idle time?
Mach down (m)
Probability
n! pm 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
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Ex #1 con’t: oper = $10/hr, mach = $500/hr, prod = 120 units/hr
1 oper 2 oper 3 operidle
Prod(8 hr)Cost(8 hr)Unit cost
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Ex #2: n=3, p=0.4, q=0.6
m Probability Hrs lost(1 oper)
Hrs lost(2 oper)
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Ex #2 con’t: oper = $10/hr, mach = $60/hr, prod = 60/hr
1 oper 2 oper 3 operidle
Prod(8 hr)Cost(8 hr)Unit cost
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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)}
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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)}
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(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
Oper ST/op Delay time
ST ReqST
#Oper New ST/op
New Delay
1 0.522 0.483 0.654 0.415 0.55
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Straight Line Balancing (Ex #1 con’t)
Oper ST/op Delay time
ST ReqST
#Oper New ST/op
New Delay
1 0.52 0.132 0.48 0.173 0.65 0.004 0.41 0.245 0.55 0.10
• % efficiency (E) = 100x∑ST/∑AT• % idle = 100x∑DT/∑AT• Req. production (R) = 3,200/day → required
standard (cycle) time• # oper = Rx∑ST = ∑ST/ CT=
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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
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Positional Weight (PW) Matrix
Op ST DT 1 2 3 4 5 6 7 8 9 PW IP
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
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Fill Workstations (Ex #2 con’t)# workstations = Rx∑ST = ∑ST/ CT=
Oper IP PW ST Station T Delay T