Stochastic Models of Manufacturing Systems

Transcription

1 Stochastic Models of Manufacturing Systems Ivo Adan

2 Exponential closed networks 2/55 Workstations 1,..., M Workstation m has c m parallel identical machines N circulating jobs (N is the population size) Processing times in workstation m are exponential with rate Processing order is FCFS Buffers are unlimited Markovian routing: job moves from workstation m to n with probability p mn This network is also called Closed Jackson network

3 Exponential closed networks 3/55! #"%$&$('*),+-./0 4/8# : /0*4

4 Example: Robotic barn 4/55

5 Example: Robotic barn 5/55 How to design a robotic barn? How many robots?

6 Example: Robotic barn 6/55 Closed network with K circulating cows and 6 workstations: 1. Milking robot, 2. Concentrate feeder, 3. Forage lane, 4. Water trough, 5. Cubicle and 6. (artifical one) Walking.

7 Example: Robotic barn 7/55 Closed network with K circulating cows and 6 workstations: Milking Concentrate Forage Walking Water Cublices

8 Example: Zone-Picking 8/55 Main conveyor Segment entrance Segment conveyor Segment System entrance/exit

9 Example: Zone-Picking 9/55 Issues in design: What should be the layout of the network? Size of zones? Where to locate items? What number of pickers and zones? Required CONWIP level?

10 Example: Single Zone 10/55 Storage Zone Weight check Buffer System entrance/exit Recirculation Tote Order picker

11 Example: Single Zone 11/55 System entrance/exit e Conveyor c 1 Zone z 1. Conveyor c 2 Conveyor c 3 Zone z 2.. Closed network with K totes and 6 workstations

12 Example: KIVA robots 12/55

13 Example: KIVA robots 13/55 How to design a KIVA system? How many robots?

14 Example: KIVA robots 14/55 Buffer area in front of aisle T ravel time of robot within aisle T ravel time of robot in departure path of cross aisle 2 1 T ravel time of robot on arrival path of cross aisle N + 1 N + 2 N T ime spent in order picking by the order picker N + 3 Buffer area in front of LU P oint Closed queueing network model with K circulating robots

15 Example: Container terminal 15/55 How many AGVs needed for unloading ship?

16 Example: Container terminal 16/55 Figure 1: Overview Abstract of view the of model load/unload of a sea process container terminal. 1

17 Example: Container terminal 17/55 Figure 2: Closed Queuing queueing network. network The servers model withtwo K circulating circles represent AGVs the twophase servers. The servers with an arrow through the circle represent the load dependent servers.

18 Exponential single server network 18/55 States of network (k 1,..., k M ) where k m is number of jobs in workstation m Note that M k m = N m=1 so there are ( N+M 1) M 1 states! State probabilities p(k 1, k 2,..., k M ) satisfy balance equations (c m = 1) p(k) M ɛ(k m ) = m=1 M M n=1 m=1 p(k + e n e m )µ n p nm ɛ(k m ) where e m = (0,..., 1,..., 0) with 1 at place m and ɛ(k) = { 1 if k > 0 0 else

19 Exponential single server network 19/55 Product form solution Jackson s miracle p(k) = Cp 1 (k 1 )p 2 (k 2 ) p M (k M ), where C is normalizing constant and p m (k m ) = ( vm ) km, k m = 0, 1,... with v m the arrival rate to workstation m This is again the product of M/M/1 solutions, where the number in station m follows the M/M/1 solution with arrival rate v m and service rate!

20 Exponential single server network 20/55 v m is the relative arrival rate or visiting frequency to m, satisfying v m = M v n p nm, n=1 m = 1,..., M Remarks: Equations above determine v m s up to a multiplicative constant Set v 1 = 1, then v m is the expected number of visits to m in between two successive visits to station 1 Product form result also valid for fixed routing Although p(k) is again a product, the queues at stations are dependent!

21 Normalizing constant 21/55 Let C(m, n) = k 1,..., k m 0 mi=1 k i = n ( v1 ) k1 ( ) k2 v2 µ 1 µ 2 ( vm ) km. So C(m, n) is sum of products in network with stations 1,..., m and population n. Clearly C = 1/C(M, N) Recursion (Buzen s algorithm): C(m, n) = C(m 1, n) + v m C(m, n 1) with initial conditions C(0, n) = 0, n = 1,..., N, C(m, 0) = 1, m = 1,..., M,

22 Normalizing constant 22/55 Recursion (Buzen s algorithm): C(m, n) = C(m 1, n) + v m C(m, n 1) with initial conditions C(0, n) = 0, n = 1,..., N, C(m, 0) = 1, m = 1,..., M, m M n N

23 Mean values 23/55 What is the real arrival rate λ m? Note that and λ M = v m C(M, N 1) C(M, N) λ m = v m v M λ M What is mean number E(L M ) in station M? E(L M ) = 1 C(M, N) N k M =0 ( ) km vm k M C(M 1, N k M ) µ M What is expected cycle time E(C) between two visits to station 1? E(C) = N λ 1 (Little s law)

24 Exponential multi server network 24/55 Product form solution p(k) = Cp 1 (k 1 )p 2 (k 2 ) p M (k M ), where C is normalizing constant and p m (k m ) = k m k=1 v m (k) where (k) = min(k, c m ) and v m the visiting frequency to workstation m This is product of M/M/c m solutions with arrival rate v m and service rate! Normalizing constant C can again be calculated via recursion (verify!)

25 Arrival theorem 25/55 Question: What is the state seen by job moving from one station to another? Total number of jumps per time unit that see the (single server) network in state k S(N 1) = {k 0 M i=1 k i = N 1} M m=1 p(k + e m ) = 1 C(M, N) p 1(k 1 ) p M (k M ) M v m, m=1 where p m (k m ) = ( vm ) km Total number of all jumps per time unit in the (single server) network M l S(N 1) m=1 p(l + e m ) = 1 C(M, N) l S(N 1) M p 1 (l 1 ) p M (l M ) v m, m=1

26 Arrival theorem 26/55 Fraction of jumps per time unit that see the network in state k S(N 1) 1 C(M,N) 1 C(M,N) p 1(k 1 ) p M (k M ) M m=1 v m l S(N 1) p 1(l 1 ) p M (l M ) = M m=1 v m 1 C(M, N 1) p 1(k 1 ) p M (k M ) which is probability that network with N 1 circulating jobs is in state k Conclusion: Arbitrary job moving from one station to another sees the network in equilibrium with a population with one job less (job does not see himself) Remarks: Also valid in multi-server networks (verify!) Also valid for jobs moving to a specific station (verify!) What is the impact of this result?

27 Mean value analysis 27/55 Define for network with population k E(S m (k)) = mean production lead time at station m m (k) = throughput of station m E(L m (k)) = mean number of jobs in station m For population k = 1, 2,..., N in single server network E(S m (k)) = E(L m (k 1)) m (k) = kv m Mn=1 v n E(S n (k)) (Arrival theorem) (Little) E(L m (k)) = m (k)e(s m (k)) with initially E(L m (0)) = 0 (Little)

28 Mean value analysis 28/55 In multi server network E(S m (k)) = m (k 1) ( 1 + E(L m (k 1)) ) m(k 1) c m where m (k 1) is probability that all servers are busy c m Approximate m (k 1) by probability of waiting in corresponding M/M/c m m (k 1) ( 1 m(k 1) c m 1 c m! ) cm 1 i=0 ( ) cm m (k 1) 1 i! ( m (k 1) ) i + 1 cm! ( m (k 1) ) cm If c m = (no waiting) E(S m (k)) = 1

29 General closed networks 29/55 In multi server station E(S m (k)) = m (k 1) E(R m) + (E(L m (k 1)) m (k 1)E(B m )) E(B m) c m c m +E(B m ) where m (k 1) is approximated by probability of waiting in M/M/c In single server station this reduces to E(S m (k)) = ρ m (k 1)E(R m ) + (L m (k 1) ρ m (k 1)) E(B m ) + E(B m ) where ρ m (k 1) = m (k 1) E(B m )

30 Example 30/55 Closed system with 4 single server stations and 10 circulating pallets:! #" $ "&% '(#)* '+, Processing characteristics: Station E(B m ) c 2 B m

31 Example 31/55 Mean value analysis: 1 (10) = parts per time unit Simulation: 1 (10) = ± parts per time unit Station E(S m (10)) amva sim ± ± ± ± 0.118

32 Example 32/55 Production system: C machines N pallets M operations to be performed each operation requires a specific tool set r m copies of tool set m v m E(B m ) is work load to be handled by tool set m

33 Example 33/55 Optimization problem: max T H(c 1, c 2,..., c m ) subject to M c m C, m=1 1 c m r m, m = 1, 2,..., M.

34 Example 34/55 Optimization problem: max T H(c 1, c 2,..., c m ) subject to M c m C, m=1 1 c m r m, m = 1, 2,..., M. Heuristic solution: Subsequently allocate tool sets to machines allocate tool set with maximum increase in throughput

35 Example: Robotic barn 35/55 Closed network with K circulating cows and 6 workstations: 1. Milking robot, 2. Concentrate feeder, 3. Forage lane, 4. Water trough, 5. Cubicle and 6. (artifical one) Walking.

36 Example: Robotic barn 36/ Histogram of the processing time (in min.) in the milking robot:

37 Example: Robotic barn 37/55 Processing times in the facilities of the barn: Processing time (in min.) Facility Routing probability Mean Standard deviation Milking robot Concentrate feeder Forage lane Water trough Cubicle

38 Example: Robotic barn 38/55 Milking Concentrate Forage Walking Water Cublices General closed network model of robotic dairy barn.

39 Type Cow 39/55 type cow = tuple (real arr; int stat); type cow_walk = tuple(cow x; timer t);

40 Buffer 40/55 proc B(chan? cow a; chan! cow b): list cow xs; cow x; end while true: select a?x: xs = xs + [x] alt size(xs) > 0, b!xs[0]: xs = xs[1:] end end

41 Machine 41/55 proc M(chan? cow a; chan! cow b, c; dist real u): cow x; end while true: a?x; b!x; delay sample u; c!x; end

42 Workstation 42/55 proc W(chan? cow a; chan! cow b, c; dist real u; int m): chan cow d; run B(a,d), unwind j in range(m): M(d, b, c, u) end end

43 Workstation Walking 43/55 proc WLK(chan? cow a; list chan! cow b; real walk): list cow_walk xst; cow x; list(1000) int dest; for i in range(1000): if i < 164: dest[i] = 0; elif i < 319: dest[i] = 1;... while true: select a?x: x.arr = time + walk; x.stat = dest[sample uniform(0, 1000)]; xst = xst + [(x, timer(walk))] alt not empty(xst) and ready(xst[0].t), b[xst[0].x.stat]!xst[0].x: xst = xst[1:]

44 Model Dairy Barn 44/55 model DairyBarn(): chan cow a, c; list(5) chan cow b; run G(a, 10), WLK(a, b, 5.0), W(b[0], c, a, exponential(8.41), 1), W(b[1], c, a, exponential(6.38), 1), W(b[2], c, a, exponential(15.0), 1), W(b[3], c, a, exponential(3.18), 1), W(b[4], c, a, exponential(38.9), 1), E(c, ) end

45 Multiple visits to work stations 45/55 n m distinct types of operations at (single server) work station m v mr visits to work station m for type r operation mean processing time for type r operation at work station m is E(B mr ) mean residual processing time is E(R mr )

46 Multiple visits to work stations 46/55 Define E(S mr (k)) = mean production lead time at station m for job of type r operation mr (k) = arrival rate at station m of jobs for type r operation E(L mr (k)) = mean number of jobs at station m for type r operation Then E(S mr (k)) = n m s=1 ρ ms (k 1)E(R ms ) + +E(B mr ) n m s=1 where ρ ms (k 1) = ms (k 1)E(B ms ) and kv mr mr (k) = Mn=1 nm s=1 v ns E(S ns (k)) E(L mr (k)) = mr (k)e(s mr (k)) (E(L ms (k 1)) ρ ms (k 1))E(B ms )

47 Closed multi-class networks 47/55 Workstations 1,..., M Workstation m has c m parallel identical machines R job types N r circulating jobs of type r Processing times in workstation m are exponential with rate (so processing times are job-type independent!) Processing order is FCFS Buffers are unlimited Markovian routing: type r job moves from workstation m to n with probability p r mn (so each job type has its own Markovian routing) This network is also called Closed multi-class Jackson network

48 Closed multi-class networks 48/55 States of network (k 1,..., k M ) where k m = (k m1,..., k m R ) is the aggregate situation in station k k mr is the number of type r jobs in workstation m Note that for each r M k mr = N r m=1 v mr is the relative visiting frequency to station m of type r jobs satisfying v mr = M n=1 v nr pnm r, m = 1, 2,..., M.

49 Closed multi-class networks 49/55 Jackson s miracle p(k) = Cp 1 (k 1 )p 2 (k 2 ) p M (k M ), where C is normalizing constant If c m = 1 p m (k m ) = (k m1 + k m2 + + k m R )! k m1!k m2! k m R! If c m > 1 p m (k m ) = (k m1 + k m2 + + k m R )! k m1!k m2! k m R! where (k) = min(k, c m ) ( vm1 ) km1 ( ) km2 vm2 ( vm R ) km R v k m1 m1 vk m2 m2 vk m R m R (1) (2) (k m1 + k m2 + + k m R )

50 Arrival theorem 50/55 Arbitrary type r job moving from one station to another sees the network in equilibrium with a population with one job of his own type less (job does not see himself) N = (N 1, N 2,..., N R ) is the population vector So jumping type r job sees the network in equilibrium with population N e r

51 Mean value analysis Define for network with population N E(S mr (N)) = mean production lead time at work station m for type r job mr (N) = throughput of type r jobs of station m E(L mr (N)) = mean number of type r jobs in station m 51/55 In single-server network r E(S mr (N)) = E(L ms (N e r )) s=1 N r v mr mr (N) = Mn=1 v nr E(S nr (N)) E(L mr (N)) = mr (N)E(S mr (N)) with initially E(L ms (0)) Recursion over population vector N, starting from k = 0 to k = N!

52 Priority stations Define for network with population N E(W mr (N)) = mean waiting time at work station m for type r job mr (N) = throughput of type r jobs of station m E(Q mr (N)) = mean number of type r jobs waiting in station m 52/55 In non-preemptive (single server) priority station m (type 1 highest priority) R E(W mr (N)) = ρ ms (N e r ) 1 r + s=1 r 1 + s=1 where ρ mr (N) = mr (N) 1 s=1 ms (N e r )E(W mr (N)) 1 E(Q ms (N e r )) 1

53 Fixed-point modeling 53/55 Breaking the recursion: Assume jumping type r job sees the system in equilibrium with population N (instead of N e r ) In FCFS single server station m E(S mr (N)) = r s=1 E(L ms (N)) So mean number seen on arrival is mean number in system including himself

54 Fixed-point modeling 54/55 Breaking the recursion: Assume jumping type r job sees the system in equilibrium with population N (instead of N e r ) In FCFS single server station m E(S mr (N)) = r s=1 E(L ms (N)) So mean number seen on arrival is mean number in system including himself To avoid self queueing E(S mr (N)) = s =r E(L ms (N)) 1 + N r 1 N r E(L mr (N)) 1 + 1

55 Fixed-point modeling 55/55 3M R equations for 3M R unknowns E(S mr (N)), mr (N) and E(L mr (N)) E(S mr (N)) = s =r E(L ms (N)) 1 + N r 1 N r E(L mr (N)) mr (N) = N r v mr Mn=1 v nr E(S nr (N)) E(L mr (N)) = mr (N)E(S mr (N)) Solution by successive substitutions