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... & Outlook 2 S ou rc e: w es tu m fa hr un g. ch (2 00 8) Target 3 fm: Street restructuring ls: Wollishofen ... Slide 1 “Westumfahrung Zurich”: Real World Study with MATSim May 22, 2008, IVTSeminar, ETH Zurich ... & Outlook 2 S ou rc e: w es tu m fa hr un g. ch (2 00 8) Target 3 fm: Street restructuring ls: Wollishofen ... Slide 1 ... Slide 1 ...
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... Space 0 1 2 3 4 5 j i 5 4 3 2 1 0 N = 4 j ≤ i i ≤ N = 4 0 ≤ j 0 ≤ i D = { (i,j) | 0 ≤ i ≤ N ∧ 0 ≤ j ≤ i ... spcl.inf.ethz.ch @spcl_eth 5 Mapping Computation to Device 0 1 2 0 1 2 0 1 2 3 0 1 2 3 0 0 1 1 Device Blocks ... Space 0 1 2 3 4 5 j i 5 4 3 2 1 0 N = 4 j ≤ i i ≤ N = 4 0 ≤ j 0 ≤ i D = { (i,j) | 0 ≤ i ≤ N ∧ 0 ≤ j ≤ i ... spcl.inf.ethz.ch @spcl_eth 5 Mapping Computation to Device 0 1 2 0 1 2 0 1 2 3 0 1 2 3 0 0 1 1 Device Blocks ... Slide 1 ...
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... acetone scan with 0.5 cm- 1 resolution 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 1210 1220 1230 1240 1250 1260 1270 ... p o w e r , m w 0 1 2 3 4 5 V o l t a g e 3.16 um 5.14 um 7.16 um 3.16 um 5.14 um 7.16 um SOLUTIONS ... acetone scan with 0.5 cm- 1 resolution 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 1210 1220 1230 1240 1250 1260 1270 ... p o w e r , m w 0 1 2 3 4 5 V o l t a g e 3.16 um 5.14 um 7.16 um 3.16 um 5.14 um 7.16 um SOLUTIONS ... Slide 1 ...
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... lasers Ikesue Cold press Yanagitani, Ueda 1.46 kW Year 95 96 97 98 99 00 01 02 03 04 0 5 06 A v e r a g e ... nm Confocal Raman Spectroscopy 2 5 0 2 7 5 3 0 0 3 2 5 3 5 0 3 7 5 4 0 0 4 2 5 # 1 # 2 # 3 E g A 1 g ... lasers Ikesue Cold press Yanagitani, Ueda 1.46 kW Year 95 96 97 98 99 00 01 02 03 04 0 5 06 A v e r a g e ... nm Confocal Raman Spectroscopy 2 5 0 2 7 5 3 0 0 3 2 5 3 5 0 3 7 5 4 0 0 4 2 5 # 1 # 2 # 3 E g A 1 g ... Slide 1 ...
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... the system, while the failure likelihood of the components is given qi = 0.01, i = 1, 2, 3, 4. Q2 ... qj = 0.01, j = 1, 2, 3, 4. ... the system, while the failure likelihood of the components is given qi = 0.01, i = 1, 2, 3, 4. Q2 ... qj = 0.01, j = 1, 2, 3, 4. ... 1 ...
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... 1 Reliability of Technical Systems Tutorial # 3 Solution 1. Consider the probability density ... = 1/1.28×10−4= 7812.5 hours MTTFs= MTTFc/4=1953.125 hours 3. A space vehicle requires three out four ... 1 Reliability of Technical Systems Tutorial # 3 Solution 1. Consider the probability density ... = 1/1.28×10−4= 7812.5 hours MTTFs= MTTFc/4=1953.125 hours 3. A space vehicle requires three out four ... 1 ...
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... 1 Reliability of Technical Systems Tutorial # 3 Due : October 12, 2010 1. Consider the probability ... component ? 3. A space vehicle requires three out four of its main engines to operate in order to achieve ... 1 Reliability of Technical Systems Tutorial # 3 Due : October 12, 2010 1. Consider the probability ... 1 ... 1 ...
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... 1 1 Reliability of Technical Systems Tutorial #5 (FTA Cont) Due: October 26th, 2010 The cooling ... operation situation and make further actions. 1. Your task is to help the maintenance personnel to draw a ... 1 ... 1 1 Reliability of Technical Systems Tutorial #5 (FTA Cont) Due: October 26th, 2010 The cooling ... 1 ...
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... [ 1/hour] • Failure rate of the spare tire: 𝜆𝑆= 10- 3 [ 1/hour] • Repair duration (time to install the ... tire (you may assume repair rate is about 6⋅10- 3 ( 1/hour) ). In this case, what is the probability that ... [ 1/hour] • Failure rate of the spare tire: 𝜆𝑆= 10- 3 [ 1/hour] • Repair duration (time to install the ... tire (you may assume repair rate is about 6⋅10- 3 ( 1/hour) ). In this case, what is the probability that ... 1 ...
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... 1 Ecological Systems Analysis Spring 2012 mandatory exercise I page 1 Ecological Systems Analysis ... do not hand in computer prints of the same solution for several people. Exercise 1 (Environmental ... 1 ... 1 Ecological Systems Analysis Spring 2012 mandatory exercise I page 1 Ecological Systems Analysis ... 1 ...
