0 9 hiệu suất của hệ thống bản lề

... Systems: A Model-Based Approach Boudewijn R Haverkort Copyright © 199 8 John Wiley & Sons Ltd ISBNs: 0- 471 -97 228-2 (Hardback); 0- 4 70- 84 192 -3 (Electronic) Chapter Introduct I ion N this chapter we discuss ... parallelism (n + 00 ) The tasks taking t seconds on a single processor can certainly allelised Indeed, it is reasonable to assume that only a fraction not completely be para (0 < a < 1) can be ... sequentially As an example, we consider Q = 90 % We then find that the best we can is to reduce the completion at most lim,,, 1.2 Model Regarding time with a factor 10, i.e., lim,,, solution the solution...

Ngày tải lên: 15/12/2013, 08:15

... Little’s law and the MIMI1 queue lOE[S] WSI WSI 7E[Sl 6Jwl 5-wSl 4EPl WSI 2Jwl EPI 0 0.1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 P Figure 2.4: The mean waiting E[S]) as a function 2.3 Further Little’s time ... queueing model reading law was originally presented in 196 1 [186] Other proofs of Little’s results can be found in [81, 1 50, 193 , 2 69, 2 701 The analysis of the MIMI1 queue as presented here is ... E[W] as a function of p Note that E[N] = for p = 50% (at 50% utilisation, on average, there is one job in the queueing station) and E[W] = E[S] (at 50% utilisation, the mean waiting time equals the...

Ngày tải lên: 15/12/2013, 08:15

... of 48.7 hours a year ((1 - 0. 99 444) x 3.6 Semi-Markov chains 51 1 .00 0 0. 99 5 0. 9 90 0 .98 5 P31@) 0. 9 80 0 .97 5 0. 9 70 0 .96 5 0. 9 60 Figure 3.6: Transient t probabilities 10 for the TMR system (1) We ... (3.17), and p (0) (l,O, 0) Clearly, (0. 50, 0.28 ,0. 22) on and calculate = = p(O)P = (0. 6 ,0. 2 ,0. 2) Then, p(2) = p (0) P2 = p(l)P = We could go ibe p roceed with p(3) = p(2)P = (0. 4 60~ 0.324 ,0> 16) many ... (07 1) (0, 0) 2. 893 6 x 1O-2 5.7813 x 1O-4 5.7755 x 1O-6 4 .97 51 x 1O-3 that the system is operational can thus be computed as 0. 99 444 Although this number looks very good (it is very close to 100 %)...

Ngày tải lên: 15/12/2013, 08:15

... 0. 0562 0. 0326 0. 12 60 0 .00 00 0 .00 00 0 .01 92 0. 0114 0. 101 9 0. 08 40 0 .00 68 0. 0 703 0. 000 0 10 0 .00 00 0 .00 41 0. 002 4 0. 0 595 0. 0 508 Table 4.1: Blocking probabilities Figure 4.8: State transition 4 .9 The ... 2 .00 0. 167 0. 084 5 .92 0. 8 80 0.4 40 2.33 2.76 0. 324 0. 4 70 0.162 0. 235 7.21 0. 9 30 0 .96 3 0. 465 0. 482 3. 30 0. 602 0. 301 10 0 .98 2 0. 492 3 .98 4.85 0. 715 0. 808 0. 358 0. 404 11 12 0. 99 2 0. 99 7 0. 496 0. 497 ... 4 .9 The MIMI m multi-server m queue without buffer m p = 0. 2 p = 0. 6 p = 0. 9 1 .00 00 oooo 1.oooo 0. 1667 0. 0322 0. 37 50 0.1837 0. 4737 0. 298 9 0. 006 4 0. 099 3 0. 001 3 0. 21 20 0.1 602 0. 000 3 0. 000 0 0. 0562...

Ngày tải lên: 15/12/2013, 08:15

... with server breakdowns 1 09 service time E[S] = 50/ 100 000 = 0. 5 msec, and p = XE[H]E[S] = Interesting 75(5 x lo-4)2 (1 - 0. 0 49) 5 x 1o-4 2(1 - 0. 7 69 )0. 0 49 + (1 - 0. 7 69 )0. 0 49 = 43 mSec’ to note is ... discussed in Section 4.3  100 M 1 G l-FCFS queueing models 25 20 15 E[Wl 10 0 Figure 5.1: Comparison to bottom): 0. 1 0. 2 of the MIHall, the average waiting Furthermore, 0. 3 0. 4 0. 6 0. 7 the MIMI1 and the ... the channel transmission speed ~1 = 10 kbps/lO bits/packet = 100 0 p/s In a similar way we obtain we find p = ~~=, Xaii/pi = 0. 648 Applying p2 = 104 / 200 = 50 p/s For th e utilisation the formula...

Ngày tải lên: 24/12/2013, 15:15

... 0. 000 02 0. 0233 0. 15 40 0 .01 88 0. 0222 0. 308 0 0. 0378 0. 46 20 0 .05 70 0.61 60 0 .07 63 0. 0547 0. 105 9 0. 198 7 0. 000 04 0. 0567 0. 7 700 0. 095 8 0. 4164 0. 000 10 0.4215 0. 92 40 0.1154 1.5183 0. 000 12 1.5244 0. 000 06 ... 0. 008 0. 012 1.2 1.8 0. 30 0.45 0. 0378 0. 05 70 0 .05 47 0. 105 9 0. 0375 0. 0563 0. 61 60 0.7 700 0. 195 3 0. 407 6 1.6 2 .0 0 .01 6 0. 0 20 2.4 3 .0 0. 60 0.75 0. 0763 0. 095 8 0. 198 7 0. 4164 0. 07 50 0 . 09 38 0. 92 40 1.4 803 ... 1 20 MIG 11 queueing x P EWl Ai models pi with Pb Ab various E[Wi] scheduling E[Wb] disciplines E[Tp] 0. 15 40 0 .02 22 0. 4 0. 004 0. 6 0. 15 0. 0188 0. 0222 0. 0188 0. 308 0 0. 46 20 0 .05 42 0. 104 6 0. 8 1.2 0. 008 ...

Ngày tải lên: 24/12/2013, 15:15

... GIMIl-FCFS 1 40 and GIGIl-FCFS queueing models 1 .0 0 .9 0. 8 0. 7 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 P 0. 6 0. 7 0. 8 0. 9 Figure 7.2: Values of o as a function of p for k = 1,2 ,9 (top to bottom) ... 1O-2 10- l 10- l 10- l 10- l 10- l 10- l 2 .06 x 1. 29 x 2 .02 x 7.55 x 1. 79 x 3.83 x 6.48 x 1. 20 2.52 9. 17 lo-’ 1O-3 1O-2 1O-2 10- l 10- l 10- l 5.26 x 1.77 x 3.33 x 5. 39 x 8.18 x 1.22 1.86 3 .00 5.67 1. 90 x ... c+PFEn(t) n=O = (1 - n=O 0) g n=O on (I- epPtz $I.!) i=o * 7.1 The GlMll queue 137 P 0. 05 0. 15 0. 25 0. 35 0. 45 0. 55 0. 65 0. 75 0. 85 0. 95 2 .06 1.28 1 .98 7 .02 1.52 2.61 3 .93 5.46 7.16 9 .01 x x x x x x x...

Ngày tải lên: 24/12/2013, 15:15

... 29 P 'I 0. 044 0. 008 36 0. 088 0. 02882 0. 131 models 0. 05 705 0. 175 0. 0 90 5 7 0. 2 19 0. 12788 0. 263 0. 16 800 0. 306 0. 2 101 8 0. 3 50 0.25 406 0. 394 10 13 16 19 0. 436 0. 5 69 0. 700 0. 831 0. 345 60 0.481 0. 613 0. 744 ... 2x 10 12 14 16 18 20 22 P 0. 0875 0. 17 50 0.2625 0. 3 500 0. 4375 0. 52 50 0.6125 0. 700 0 0. 7875 0. 87 50 0 .95 81 Jw,I SS LR 0. 002 3 0. 01 60 0 .04 92 0. 1122 0. 2 200 0. 398 5 0. 696 3 1.2186 2.2432 4.8242 17.58 29 11 ... 0. 345 60 0.481 0. 613 0. 744 0. 38288 12 0. 525 0. 5 392 7 0. 691 17 15 0. 656 0. 6 400 0 11 14 17 0. 299 28 0. 4 4 09 7 0. 5 893 6 18 0. 786 0. 74282 0. 794 95 20 0.875 0. 84752 21 0. 91 9 0. 90 0 51 0. 4 897 9 Table 8.3: The values...

Ngày tải lên: 24/12/2013, 15:15

... 3 .07 0 2. 793 0. 25 0. 35 0. 8667 1 .06 2 0. 90 0 0 1.115 0. 93 10 1.1 79 0. 895 3 1.1 19 0. 75 0. 85 3. 400 5 .93 3 3. 700 6. 500 5.286 15 .00 4.6 90 12.27 0. 45 1.327 1. 4 09 1.535 1.428 0. 95 18. 60 20. 50 - - Table 9. 1: ... E[W,] 0. 3 0. 421 0. 347 0. 525 0. 3 70 0.4 20 0.322 0. 5 1.12 0. 747 1.51 0. 775 1.12 0. 6 70 0.8 10. 6 2.33 55.7 2.31 9. 67 2.23 0. 3 0. 442 0. 444 0. 317 0. 558 1. 20 0. 399 0. 98 9 0. 506 0. 5 1.38 1.16 0. 861 1.38 0. 8 ... ’ (9. 16) 182 Polling P E[WE] E[WG] E[WL] J?@%] 0. 05 0. 600 0 0. 605 3 0. 608 5 0. 603 1 p -@[WE] 0. 55 E[WG] 1.711 B[WL] 1.833 models E[b] 2 .08 9 1 .93 1 0. 15 0. 7176 0. 7353 0. 7485 0. 7 302 0. 65 2.314 2. 500 ...

Ngày tải lên: 24/12/2013, 15:15

... QNA 0. 533 0. 497 7.2 QNA 3 .97 2 E[W4] E[W~] E[l&] E[W7] E[W8] E[V&] 3 .00 8 1.377 1 . 09 1 1.8 30 0.8 20 20. 22 2 .93 1 1. 301 0. 98 5 1. 695 0. 758 18. 40 2.6 5.8 10. 8 7 .9 8.2 9. 9 9. 1 89 1. 596 1.372 5. 604 3.273 34.51 ... 6.111 21 .00 6 .00 5 5.257 12. 29 6.7 19 113.1 SIM RE 3.2 20 23.4 10. 97 -16.2 1.5 59 2.4 1.282 7 .0 6 .02 6 -7.5 2.685 21 .9 70. 25 - 50. 9 SIM RE 6.228 -1 .9 22 .92 -8.4 6. 591 -9. 7 5 .95 6 -11.7 13.31 -7.7 7. 597 -11.6 ... ( 10. 53) i=l 218 10 Open queueing networks +-El@ x +-El@ Figure 10. 6: The nine-node example queueing network c; = 0. 5, c; = 0. 5 Cl = 6 .0, C; = 0. 5 Cl = 6 .0, c: = 4 .0 SIM RE E[Wl;3] QNA 0. 533 0. 497 ...

Ngày tải lên: 24/12/2013, 15:15

... 1.2 201 0. 98 51 0. 8746 0. 8227 0. 798 3 0. 7868 0. 7812 30. 000 0 39. 4 192 46.6 798 51.4 892 54.37 89 56 .02 19 56 .9 304 57.4267 57. 696 7 57.8436 57 .92 37 57 .96 74 1. 203 8 1.2 500 0. 77 60 0. 800 0 58 .02 03 57 .9 500 177 .05 97 ... 2.2 500 E[N( 60) 1 63 .00 00 93 .00 00 121.2575 143 .03 93 157.4676 166.1366 171 .06 57 173. 791 3 175.2 802 176 . 09 02 176.5 3 09 176.7711 2. 204 1 173.8 500 Table 11.3: Bard-Schweitzer approximation 2 49 ~9[ N3( 60) 1 ... G 1 .00 0. 20 0. 30 1.13 0. 08 2. 40 0.33 0. 80 1 .00 0. 33 0. 12 10 3. 90 Table 11.6: Parameters Vi 0. 67 0. 05 00 1 .00 of the five-node simulation i Xi non-BCMP queueing network approximation Pi 0. 3 39 E[Ni]...

Ngày tải lên: 24/12/2013, 15:15

... X(K) 0. 09 50 X(K) 0. 094 5 queueing networks states 0. 4731 0. 4731 56 11 16 21 1 .03 43 1 .03 43 364 1. 494 8 1 .94 68 1. 494 8 1 .94 68 96 9 200 2 26 2.3875 2.3875 3157 31 2.8133 2.8133 4312 36 3.2 191 3. 59 80 3 .94 16 ... 2.3875 3157 31 2.8133 2.8133 4312 36 3.2 191 3. 59 80 3 .94 16 5467 51 3.2 191 3. 598 1 3 .94 19 4.2 4 09 4.2 397 893 2 56 4.4861 4.48 30 100 87 41 46 6622 7777 Table 12.1: Comparing the throughputs obtained via ... = -+ - cy = 0) Let us now start the computations following normalising constants: We find for the subnetwork to be aggregated the k WA 1 .0 0.8 0. 64 0. 512 I G(3,k) 1 .0 2.6 5.32 10. 088 The throughput...

Ngày tải lên: 24/12/2013, 15:15

... FCFS, PS, LCFSPR IS nodes nodes, (13. 19) 300 13 BCMP The throughput for class r customers (through queueing networks the node with I,$r = 1) is given as: (13. 20) and the expected number of class ... between various algorithms [1 701 as well as ways to deal with large normalising constants [1 69] The books by Bruell clusively with computational applicability and Balbo [ 30] and by Conway and Georganas ... computation of the normalising constant would involve Kj+M-1 M-l A 13.2 Computational algorithms 299 computational steps (one for each state) where each step consists of multiple multiplications...

Ngày tải lên: 24/12/2013, 15:15

... fo 0 0 01 00 0 00 100 0 0 pa Pa O pb 0 b o Pa O 00 00 O Pa 0 \j&pb 0 00 \ Xa Ab xa Xb Q’ = in which we have indicated the partitioning G” = (I - C)-ID = D so that 00 0 Petri nets o\ \ 0 0 Ab Xb 0 ... simultaneous resource possessing example are easy to compute from Cf = with / -1 0 0 0 1 -1 -1 0 0 -1 0 0 -1 0 c= \ 0 -1 0 -1 -1 1 The first group of three equations from the system Cf = yields fi ... Systems: A Model-Based Approach Boudewijn R Haverkort Copyright © 199 8 John Wiley & Sons Ltd ISBNs: 0- 471 -97 228-2 (Hardback); 0- 4 70- 84 192 -3 (Electronic) Chapter 14 Stochastic I N this chapter Petri...

Ngày tải lên: 24/12/2013, 15:15

... 0. 400 0, 0. 200 0 ) ( 0. 400 5, 0. 399 7, 0. 199 8 ) ( 0. 400 0, 0. 400 0, 0. 200 0 ) ( 0. 399 9, 0. 399 9, 0. 200 1 ) ( 0. 400 0, 0. 400 0, 0. 200 0 ) ( 0. 400 1, 0. 399 9, 0. 200 0 ) 10 ( 0. 400 0, 0. 400 0, 0. 200 0 ) Table 15.1: ... 0. 199 1 ) ( 0. 400 0, 0. 400 0, 0. 200 0 ) ( 0. 398 1, 0. 398 1, 0. 203 7 ) ( 0. 397 9, 0. 402 3, 0. 199 8 ) ( 0. 402 8, 0. 398 1, 0. 199 1 ) ( 0. 400 1, 0. 399 9, 0. 200 0 ) ( 0. 399 7, 0. 399 7, 0. 200 6 ) ( 0. 400 0, 0. 400 0, 0. 200 0 ... 0. 3333, 0. 1667 ) ( 0. 38 89, 0. 38 89, 0. 2222 ) ( 0. 5385, 0. 307 7, 0. 1538 ) ( 0. 4 90 2 , 0. 3137, 0. 196 1 ) ( 0. 5833, 0. 5833, 0. 291 7 ) ( 0. 400 0, 0. 400 0, 0. 200 0 ) ( 0. 4167, 0. 38 89, 0. 194 4 ) ( 0. 3 796 , 0. 4213, 0. 199 1...

Ngày tải lên: 21/01/2014, 20:20

... 0. 5 1 .0 2 .0 5 .0 10. 0 20. 0 50. 0 100 .0 00 Petri net applications w> Jw(t)l qLs(t>l kl k kss 0. 000 92 1 0. 001 788 0. 004 097 0. 007 141 0. 011154 899 9.475 899 8 .9 80 899 7.657 899 5. 891 899 3.5 30 0 .01 56 79 0. 016864 ... 0. 016864 0. 01 698 3 0. 01 698 5 0. 01 698 5 0. 01 698 5 89 90 . 796 89 90 . 056 898 9 .97 8 898 9 .97 8 898 9 .97 8 898 9 .97 8 0. 0265 0. 103 9 0. 61 40 0 2.2475 7.64 20 32.153 80. 700 1 80. 757 481.4 30 98 2.555 - 0 0 19 - 12 15 ... applications TM VM q 4 301 4 508 74 69 7843 34617 36363 4 700 8188 4875 8 500 503 1 8775 5166 90 0 9 5278 91 98 5365 93 38 5425 94 25 5456 99 20 3 79 80 394 50 407 55 41877 42 798 43 500 4 396 5 446 40 Table 16.1: The...

Ngày tải lên: 21/01/2014, 20:20

... by Figure 17 .9 Since the 17.5 Application studies 399 0. 0 70 I set - 0. 065 0. 0 60 0 .05 5 0. 0 50 0 .04 5 WI 0. 0 40 0 .03 5 0. 0 30 0 .02 5 I 0. 0 20 Figure I 50 17 .9: The expected connection 100 x (P/S> setup ... these quantities can be 398 17 Infinite-state 0. 30 I I SPNs I set set 0. 28 0. 26 0. 24 0. 22 Wwl 0. 20 0.18 0. 16 0. 14 0. 12 50 Figure 17.8: The expected bandwidth 100 x (p/s) 1 50 200 E[Bw] (in Mbp s ) ... matrices 402 17 Infinite-state 2 X 5 .0 6.5 8 .0 ; 0. 13 10 0.13 10 0.25 10 a 21 .0 p 11 .0 21 .0 11 .0 SPNs 21 .0 11 .0 Table 17.2: Parameters used in the three scenarios 3 .0 E[buf f er] 2.5 1 .0 l Figure...

Ngày tải lên: 21/01/2014, 20:20

... = 0. 9 a = 0. 95 a = 0. 99 10 15 20 25 30 40 50 60 70 80 90 100 k > 100 4. 605 6.253 7.7 79 9.236 10. 645 12 .01 7 13.362 14.684 15 .98 7 22. 307 28.412 34.382 40. 256 51. 805 63.17 74. 40 85.53 96 .58 107 .6 ... 1.28)2 5 .99 1 7.817 9. 488 11 .07 1 12. 592 14 .06 7 15. 507 16 .91 9 18. 307 24 .99 6 31.4 10 37.653 43.773 55.7 59 67. 505 79 .08 2 90 . 531 101 .8 79 113.145 124.342 9. 2 10 11.356 13.277 15 .08 6 16.812 18.475 20. 0 90 21.666 ... 2 .04 8 2.763 1.833 2.262 3.2 50 30 2.7 50 10 1.812 3.1 69 60 1. 697 1.671 2 .04 2 2.228 2 .00 0 2.6 60 12 14 16 1.782 2.1 79 2.145 2.1 20 3 .05 5 90 1.662 2.632 2 .97 7 1 20 1.658 2 .92 1 00 1.645 1 .98 7 1 .9 80 1 .96 0...

Ngày tải lên: 21/01/2014, 20:20

... côn: 0. 92 - 0. 95 ηch = ηcon ηcon = 0. 92 x0 .92 = 0. 8464 (2 – 6) Vậy công suất cần thiết động cơ: Pdc = - Pct ηch = 0. 0718 = 0. 085(W) 0. 8464 (2 – 7) Tỉ số truyền chung: ich = nDC ntrucPIN = 200 = 300 ... = m.g.R = 21x9.8x0.1 = 20. 58 (N.m) ω: tốc độ góc số vòng quay đó: ω = 3.49x 10- 3(rad/s) Công suất trục PIN là: P = M x ω = 20. 58x3.49x 10- 3 =0. 0718 (W) (2 – 5)  Hiệu suất chung hệ thống truyền ... bị điện sử dụng Mỗi pin mặt trời có công suất khác như: 30Wp, 40Wp, 45Wp, 50Wp, 75Wp, 100 Wp, 125Wp, 150Wp Điện áp pin thường 12VDC Công suất điện áp hệ thống tuỳ thuộc vào cách ghép nối pin lại...

Ngày tải lên: 05/09/2015, 22:09

... 200 5 286. 206 .02 9. 158 Năm 200 6 305 .514. 894 . 394 Năm 200 7 3 19. 494 . 406 .8 40 Năm 200 8 443.652 .00 0 .00 0 Năm 200 9 497 .02 8.7 10. 553 ( Nguồn: Công ty bảo hiểm Bưu Điện-BCKT) 68. 200 .00 00. 000 107 . 800 .00 0 .00 0 ... vụ 200 3 Tài sản-KT 101 .64 Hàng hải 15 ,92 Phi hàng 38,18 200 4 117,22 22 ,08 3 67 ,01 3 200 5 132,58 21, 591 1 10, 83 200 6 121,86 27 131,8 200 7 1 09 , 09 5 25,5 69 157, 602 200 8 200 9 242,874 285 ,93 37, 608 46,28 ... 107 . 800 .00 0 .00 0 144. 400 .00 0 .00 0 166 .99 4. 408 .176 178.714 .07 1.152 Bảng 3: Quỹ dự phòng nghiệp vụ Đơn vị: Việt nam đồng STT Năm Quỹ dự phòng nghiệp vụ 284. 897 .284.175 Năm 200 6 3 10. 386 .95 9.123 Năm 200 7 306 .775.247.156...

Ngày tải lên: 17/07/2013, 11:11