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## Introduction

A vital part of geodynamics research is to quantitatively determine the tectonic stress that caused the past and present deformation in the crust and lithospheric mantle1,2,3,4. The presence of experimentally calibrated correlations between the stress and deformation microstructures (i.e., piezometers), suggests that tectonic stress conditions in the deformed rocks can be inferred from the latter. Hitherto, various piezometers have been proposed, among which three are most frequently used.

The first is the dislocation density piezometer. It is based on the relation between the density of a line defect (i.e., dislocation) and the stress in minerals such as olivine5,6,7,8 and quartz6. A higher dislocation density implies a larger stress. The second is the dynamically recrystallized grain size piezometer, which is based on an inverse empirical relation between stress and the grain size of dynamically recrystallized minerals in a mono-phase aggregate, such as olivine1,9,10,11, calcite12, hematite13, orthopyroxene14,15, and quartz16,17,18. The third is the subgrain-size piezometer, which is built on the empirical relation between the spacing (i.e., subgrain size) of adjacent low-angle boundaries (i.e., subgrain boundary (SGB) or dislocation wall) in the minerals and stress. This piezometer has also been calibrated for olivine6,10,11,19,20,21, quartz19,22 and calcite12,23. Similarly, there is also an inverse correlation between subgrain size and stress.

The deformation microstructures (i.e., dislocation density, dynamically recrystallized grain, and subgrain size) with relevance to the above piezometers can be measured by direct observations under optical microscope e.g.7,10,11,12, scanning electron microscope e.g.24, and transmission electron microscope e.g.7,9,25,26. Besides, they can also be calculated indirectly via high-angular resolution electron backscatter diffraction (EBSD)27 and/or traditional EBSD mapping13,14,16,19.

In this study, we investigated the deformation microstructures, especially the characteristics of low-angle misorientation boundaries (LAMBs) in olivine, of a natural harzburgite sample collected from the Yarlung Zangbo suture zone in the Tibetan Plateau (the sample location is shown in Fig. A1 and see Supplementary Text for more detailed geological background and sample information). Based on these results, we clarified the influence of EBSD mapping parameters, especially step size, on the observed LAMB features and its implications for stress estimations, and proposed other potential metrics to estimate stress and their cautions when applying to natural samples.

## Results

### Microstructures

The studied sample 13LQ117 shows a fresh harzburgitic composition characterized by mainly olivine (~ 78 vol.%) and orthopyroxene (~ 20 vol.%), minor clinopyroxene (~ 1.3 vol.%) and spinel (~ 0.2 vol.%), and a trace amount of amphibole (Fig. 1a, b; Supplementary Table S1). It displays a porphyroclastic texture, in which large olivine and orthopyroxene grains coexist with their fine-grained counterparts (Fig. 1a, b), with mean grain sizes of olivine and orthopyroxene at about 1.5 and 1 mm, respectively (Supplementary Fig. S1a, b; Table S1). The olivine long axes are aligned predominantly subparallel to the foliation and lineation (Supplementary Fig. S2a). Olivine shows distinct intracrystalline plasticity features characterized by widespread undulose extinction and subgrain boundaries (Fig. 1a). The abrupt color variations in the mis-to-mean (M2M, i.e., misorientation angle between each pixel and their mean orientation of a grain) map (Fig. 1c) and the bright streaks in the kernel average misorientation (KAM, i.e., average misorientation angle between each pixel and its nearest neighbors within the grain, depending on the choice of kernel size) map (Fig. 1d) match well with the locations of low-angle misorientation boundaries (LAMBs, see Results Section for its definition) (Fig. 1e). The unevenly distributed M2M, grain orientation spread (GOS, i.e., arithmetic mean of M2M in a grain), KAM and LAMBs reflect heterogeneous and extensive intracrystalline lattice distortions in olivines (Figs. 1c–f and Fig. A2 in the Supplementary Text). Olivine grains have straight to slightly curved boundaries and are weakly elongated (Fig. 1a, Supplementary Figs. S1c, e and Table S1). Notably, the orientations of LAMBs tend to be normal to the foliation (Fig. 1e and Supplementary Fig. S2d).

Similar to olivine, orthopyroxene grains have curvilinear boundaries and show some undulose extinction and LAMBs (Fig. 1a). Compared with olivine, orthopyroxene grains are smaller in size and have larger shape factor and aspect ratio (Supplementary Fig. S1b, d, f; Table S1). Besides, M2M, GOS, KAM and LAMB density of orthopyroxene are overall lower than those of olivine (Supplementary Fig. S3). Clinopyroxene lamellae can be observed in orthopyroxene, and vice versa. Clinopyroxene grains in the matrix have an irregular shape, a small mean grain size (~ 170 μm), and rarely develop LAMBs and undulose extinction. Orthopyroxene grains lack a discernable shape preferred orientation (Supplementary Fig. S2b), whereas spinel grains are well aligned, marking the foliation and lineation (Supplementary Fig. S2c) that are consistent with those defined by the olivine grains.

### Crystallographic preferred orientations

In sample 13LQ117, olivine shows a relatively strong CPO (J-index ~ 2.46 and M-index ~ 0.12) characterized by a high concentration of [100] axes subparallel to the lineation, the girdle-like distribution of the [010] axes with a subordinate concentration sub-perpendicular to the foliation (Fig. 2a). This CPO can be classified as A/D-type, which is intermediate between A- and D-type CPOs, see also28,29, and indicated by a high BA-index of 0.64. In comparison, both orthopyroxene (J-index ~ 1.43 and M-index ~ 0.02) and clinopyroxene (J-index ~ 2.53 and M-index ~ 0.03) display weaker and more dispersed CPOs. For both orthopyroxene and clinopyroxene, the maxima of [001] and [100] axes tend to align subparallel to the lineation and subnormal to the foliation, respectively (Fig. 2b, c), which are overall correlated with the olivine CPO, i.e., correlations of $$\left[ {001} \right]_{{{\text{Opx}}}} \parallel \left[ {100} \right]_{{{\text{Ol}}}}$$ and $$\left[ {100} \right]_{{{\text{Opx}}}} \parallel \left[ {010} \right]_{{{\text{Ol}}}}$$ with some clear obliquities (Fig. 2).

### Characteristics of olivine low-angle misorientation boundaries

#### Variations of LAMBs with step size

Distinct from subgrain boundary (SGB), which specifically indicates that a sharp lattice curvature in a discrete plane occurs over a short-range wavelength (e.g., < 1 μm in olivine) caused by dense stacking of dislocation arrays30,31,32,33,34, a low-angle misorientation boundary (LAMB) is defined here as the low-angle boundary whose misorientation angle is greater than 1° and less than 10°, derived purely from EBSD data regardless of mapping step size. In other words, the descriptive term LAMB has no meaning on the physical origins of the observed low-angle boundaries. Otherwise, the unambiguous term SGB is used in the text where its physical meaning is specially indicated. In fact, the use of a larger EBSD mapping step size may actually result in artificial SGBs (see Discussion). The density of LAMB (ρLAMB) in a grain is defined as the total length of LAMBs divided by the analyzed area representing the olivine in the field of view, using the unit μm−1.

For olivine, LAMBs can be observed in both large and small grains in the studied sample and they are mostly straight and align sub-perpendicular to the foliation and lineation (Fig. 1e and Supplementary Fig. S2d). Based on the density of LAMBs (ρLAMB) at the step size of 1 μm and its slope with step size at 1–5 μm, the 22 analyzed olivine grains can be divided into low-ρLAMB (0.169–0.891 × 10−2 μm−1), medium-ρLAMB (1.27–1.91 × 10−2 μm−1), and high-ρLAMB (2.12–2.66 × 10−2 μm−1) groups (Supplementary Table S2). In general, low-ρLAMB grains are mainly the large grains, whereas the medium- and high-ρLAMB grains are much smaller in size (Fig. 1f). The features of LAMBs in three representative grains, grain-1, grain-4 and grain-2, from each group above are shown below.

Accompanied with the decrease of step size from 30 to 1 μm, at least five variants of LAMBs can be observed (Fig. 3). Case-1 LAMBs are persistent, as they can exist at all step sizes. Case-2 LAMBs exist only at larger step sizes but fade away gradually with decreasing step size. Oppositely, case-3 LAMBs emerge only at smaller step sizes but disappear gradually with increasing step size. Case-4 LAMBs are more complicated as they occur only at medium step sizes and vanish at both larger and smaller step sizes. The four LAMB variants above have relative stable locations on the map when step size changes. In contrast, some LAMBs (case-5) disappear and seem to “split” into two or more LAMBs with decreasing step size. It appears that the LAMBs in the high- and medium-ρLAMB grains are dominated by case-1 and -3 (Fig. 3b, c), while more case-2 LAMBs appear in the low-ρLAMB grains (Fig. 3a). The variation of LAMB density with step size also provides clues on how these LAMBs vary. The low-ρLAMB grains show a nearly constant LAMB density (0.0015–0.0018 μm−1, Fig. 3a), which implies a nearly balanced disappearance and appearance rate of different LAMB variants with step size. Conversely, LAMB densities in the medium-ρLAMB (0.0046–0.0164 μm−1, Fig. 3b) and high-ρLAMB (0.0017–0.0246 μm−1, Fig. 3c) grains vary remarkably with step size, suggesting imbalanced evolutions of different LAMB variants.

### Dislocation slip systems inferred from SGBs

As shown in the Discussion below, LAMB is equivalent to SGB when the step size is small enough. Therefore, we assume that the LAMBs recovered at the minimum step size of 1 μm in this study are predominantly SGBs, from which different dislocation slip systems can be inferred. According to the geometrical relation between the SGB trace and the axis of rotation, SGB can be divided into tilt (rotation axis aligning subparallel to SGB trace) and twist (rotation axis aligning subperpendicular to the SGB trace) boundaries. The analyses of SGBs retrieved at the step size of 1 μm in all 22 olivine grains show that SGBs are dominantly tilt boundaries, as their percentages are 60.3–99.9%, 49.8–95.9% and 41.1–88.6% for the low-, medium- and high-ρLAMB grains, respectively. Overall, the percentages of tilt boundary are over 80% for most grains (Fig. 4a; Supplementary Table S2). The SGB length proportions of edge dislocation slip systems that form the tilt boundaries vary greatly among the all 22 grains, i.e., (010)[100] ~ 0–55.7%, (010)[001] ~ 0–6.7%, (100)[001] ~ 0–10.2%, {0kl}[100] ~ 5.1–98.3%, (001)[100] ~ 0.9–91.6% (Fig. 5a; Supplementary Table S2). When taken together, {0kl}[100], (001)[100] and (010)[100] are the majorities, as their average percentages are 52–56%, 28–31% and 12–14%, respectively (Fig. 5b, c; Supplementary Table S2). Likewise, no obvious correlation between the dislocation slip systems and LAMB density is observed.

### Relations between KAM, LAMB density, intercept length, and step size

Overall the median KAM increases with step size and varies in the ranges of 0.007°–0.37°, 0.009°–0.80° and 0.014°–0.65° in low-, medium- and high-ρLAMB grains, respectively (Fig. 6a; Supplementary Table S2). The KAMs in low-ρLAMB grains are markedly lower than those in the medium- and high-ρLAMB grains, and the latter two show largely overlapped KAM ranges, together implying a moderate correlation between KAM and LAMB density. Besides, KAM fluctuates considerably at step sizes larger than 10 μm in the medium- and high-ρLAMB grains (Fig. 6a).

The LAMB density fluctuates weakly and increases slightly with step size in the low-ρLAMB grains. However, in the medium- and high-ρLAMB grains, LAMB density shows striking fluctuation and increment with decreasing step size, which is especially prominent in high-ρLAMB grains (Figs. 3 and 6b). For comparison, the LAMB density varies in a large range of from 0 to 9.9 × 10−3 μm−1, with the area-weighted mean of 4.37 × 10−3 μm−1 in large-area mapping data (grey area and green star in Fig. 6b). This range of LAMB is mostly overlapped with that of the 22 olivine grains produced at the step size of 15 μm. The area-weighted mean LAMB density is similar to the LAMB density of low-ρLAMB grains, owing to the large area of such grains.

At the step size of 1 μm, the mean intercept length ranges from 88 to 692 μm in the low-ρLAMB grains, whereas smaller mean intercept lengths of 37–79 μm and 27–63 μm are found in the medium- and high-ρLAMB grains, respectively (Fig. 6c; Supplementary Table S2). This correlation is readily comprehensible, because higher LAMB density is equivalent to a narrower spacing between neighboring LAMBs. The intercept length generally decreases with the reducing step size, and it fluctuates dramatically at large step sizes in some grains, especially the high- and medium-ρLAMB grains, due to their small grain sizes (Fig. 6c).

### Stress

At the same step sizes, the stresses calculated by T1979 are larger than those estimated by G2020wHK2010 and G2020w/oHK2010 (Fig. 6d–f; Supplementary Table S2). For the 22 olivine grains and at the step size of 1 μm, the average stresses (mean ± one standard deviation) by T1979, G2020wHK2010, and G2020w/oHK2010 are 22 ± 15.4, 12.8 ± 7.7 and 9.3 ± 6.5 MPa, respectively. Stress displays a trend of decrease with increasing step size, which is especially striking at the small step sizes (Fig. 6d–f). The fluctuations of stress with step size are also opposite to those of intercept length (cf. Fig. 6c–f). Because of the higher LAMB density and smaller intercept length, stresses of the high-ρLAMB grains are considerably larger than those of the medium- and low-ρLAMB grains (Fig. 6d–f). This phenomenon is more pronounced at the step size of 1 μm, where the stresses estimated by T1979 are 7.4 ± 4.5, 25.7 ± 6.1 and 42.5 ± 11 MPa for the low-, medium- and high-ρLAMB grains, respectively. This difference weakens at the step size of 15 μm (Fig. 6d). The same patterns are found in the stresses estimated by G2020wHK2010 and G2020w/oHK2010 (Fig. 6e, f).

## Discussion

### Deformation mechanism

The porphyroclastic microstructure, undulose extinction, well-developed SGBs, and distinct CPOs clearly suggest that deformation is mainly accommodated by dislocation slip. The presence of an olivine A/D-type CPO is indicative of the activation of dominant (010)[100] and {0kl}[100] dislocation slip systems35, which is overall consistent with the dislocation slip systems inferred from SGBs (Figs. 4 and 5). However, the prevalence of the (001)[100] slip system (28–31%) over (010)[100] (12–14%), which was also recently reported by Lopez-Sanchez et al.34, is contradictory to the observation of an A-type CPO component. The reason for this discrepancy may be related to faster or more extensive recovery of the SGBs consisting mainly of the (010)[100] slip system compared to the SGBs composed majorly of the (001)[100] slip system on the geological time scale, thus more former SGBs than the latter ones evolve into grain boundaries28. Alternatively, the proportion of the different dislocation types in the SGBs needs not be in direct proportion to the activity of the different slip systems that accommodate the overall deformation. The reason is that (010)[100] dislocations can move unimpededly through the crystal without leaving any (or hardly any) misorientation gradients or dislocation walls (i.e., SGBs)34,36, or that grain-to-grain interaction can activate the hard-slip dislocations and thus vary the proportion of different dislocation types in SGBs locally34. In these cases, the presence of the (001)[100] slip system related SGBs does not contradict with the dominant (010)[100] and {0kl}[100] slip systems in the deformation process. This discrepancy has also been recently observed in the experimentally deformed samples34,37.

The alignments of [100]Ol, [001]Opx and [001]Cpx crystallographic axes subparallel to the lineation indicate coeval deformation of olivine and two pyroxenes. However, both orthopyroxene and clinopyroxene are characterized by relatively dispersed CPOs and weaker strength (J-index ~ 2.69 and 2.47) and an oblique angle between their [001]-axis maxima and the olivine [100]-axis maximum (Fig. 2). This oblique angle and dispersed CPO of pyroxenes are commonly observed in naturally deformed peridotites e.g.28,38,39,40 and are usually attributed to the lower finite strains accommodated by stronger pyroxenes compared with weaker olivine crystals41,42. The orthopyroxene CPO shows maxima of the [001] and [100] axes aligning subparallel to the lineation and subnormal to the foliation, respectively (Fig. 2b, also named as Type-AC CPO by Jung, et al.43. This CPO type of orthopyroxene has also been reported in many previous natural peridotite studies e.g.28,40,44,45, indicating that the deformation of orthopyroxene is mainly accommodated by the (100)[001] dislocation slip system46.

### Genesis of various LAMBs and their relations to SGB

In sum, an LAMB is equivalent to an SGB when the EBSD mapping step size is small enough (case-1 and -3), while it develops into various spurious SGBs at larger step sizes (case-2, -4, -5 and more). Besides, it is noted that the manner that lattice curvature manifested in EBSD maps is essentially a continuum effect, in that there is an endless range of possible distributions and magnitudes of lattice curvature, which can be sampled by a continuous range of step size. This fact will generate a wide range of observational effects including and likely going beyond the five variants of LAMB observed in this study. The LAMB patterns documented here are explicit and their categorization could be helpful to illustrate the key differences among the various LAMBs discovered in this work. However, they should not be regarded simply as a discretization of the full continuum effects as mentioned above.

The LAMB variants observed in this study appear to correlate with the LAMB density in the grains (see Section of Variations of LAMBs with step size). The LAMBs are mainly case-1 and case-3 (real SGBs) in the high- and medium-ρLAMB grains, whereas case-2 LAMBs (artificial SGBs) are more prevalent in the low-ρLAMB grains (Fig. 3). At the early stage of deformation or lower strain and stress, dislocation density is low as indicated by a sparse distribution of dislocations, the number or density of LAMBs is therefore limited. The dislocations are loosely stacked in the vicinity of an LAMB, resulting in a low misorientation gradient and case-2 LAMB represented deformation bands. As deformation (i.e., strain) advances and stress increases, the dislocation density increases correspondingly, and dislocations slip and climb closer to form a narrow dislocation wall. This process stacks the previous loose dislocations on the LAMBs more densely and produces a larger misorientation gradient, forming case-1 and -3 LAMBs represented SGBs and a high LAMB or SGB density.

### Implications for stress estimation

#### Can KAM be used to estimate stress?

Kernel average misorientation (KAM), which is usually derived from the EBSD mapping data, is an important metric to measure the local lattice misorientation48. It currently has been widely employed to analyze the intracrystalline deformation microstructures in minerals e.g.49. KAM is thought to result mainly from the geometrically necessary dislocations (GNDs) that are caused by stress or plastic deformation, although inherent crystal defects such as solute atom, second phase, and grain boundary can also contribute partially to the local lattice curvature or distortion48. This argument is supported by the close relation between KAM and dislocation density50,51, suggesting that KAM, like dislocation density, can be used theoretically as a proxy to constrain stress.

In our sample, high KAM values correspond well with the locations of LAMBs (Fig. 1d, e), implying that KAM may also correlate closely with the density of LAMB or SGB, as well as with stress. For step sizes of 1 and 15 μm, the correlations between KAM and LAMB or SGB density (R2 = 0.49 and 0.66, Fig. 8a) are moderate and slightly higher correlations between KAM and stress are observed (R2 = 0.60 and 0.75, Fig. 9a, b). These fair correlations may partly be attributed to the arbitrary choice of kernel size over which KAM is calculated. In contrast, KAM and grain size are weakly correlated (R2 = 0.33 and 0.33, Fig. 8b). Compared to grain orientation spread (GOS), which is not a viable metric to estimate stress (see Supplementary Text for more explanations), KAM has several benefits. First, KAM avoids the problem caused by averaging misorientations over a grain, which can result in a low GOS but a high LAMB or SGB density. Second, KAM is estimated over the entire area of the grain considered and there is no need to weight the result by grain area to obtain a meaningful average value. Third, KAM is also straightforward to calculate using the built-in functions in MTEX. In this case, KAM is expected to be a viable metric to estimate stress in future.

### Can SGB density be used to estimate stress?

In a grain with a fixed size, a smaller distance between the neighboring LAMBs indicates that there exist more LAMBs (i.e., higher LAMB density). Our results support this argument, as intercept length has strong negative correlations with LAMB density (R2 = 0.98 and 0.93, Fig. 8c). Besides, based on the stress calculated from the intercept length, we found that there are also strong positive correlations between the LAMB density and the stress (R2 = 0.92 and 0.87 for step sizes of 1 and 15 μm, respectively, Fig. 9c, d).

The subgrain-size piezometer implies that the smaller size of the subgrain, the greater stress it represents. In this study, subgrain size is represented by the linear intercept length between neighboring LAMBs. However, this assumption is valid only when LAMBs are SGBs (i.e., case-1 and -3 LAMBs), which occur when the step size is sufficiently small. In other words, the estimated stress is meaningful unless the density of SGBs is measured. The minimum step size of 1 μm produces predominantly case-1 and -3 LAMBs (i.e., SGBs) in the EBSD maps (Fig. 3a9, b9 and c9), implying that the intercept length approximates subgrain size, that LAMB density approaches SGB density, and that the calculated stress is most likely sound in this situation, although an even smaller step size (< 0.5 μm) is suggested to be required to capture nearly all SGBs and authentic subgrain size in the experimentally deformed olivines19. It is noteworthy that the value of this critical step size apparently depends on the spacing of adjacent SGBs in a grain, which needs to be at least twice the step size19.

The appearance of artificial SGBs (e.g., case-2, -4, -5 and more LAMBs) indicates that the subgrain size and its deduced SGB density will yield erroneous stress estimates when using a large EBSD mapping step size (e.g., 15 μm). This case is clearly exemplified by the remarkably different stresses estimated from 1 μm and 15 μm step sizes (Figs. 6 and 9; Supplementary Table S2). The reason we raised this issue is that a large step size is very often used, especially when one maps natural coarse-grained samples in a large domain. To deduce a meaningful stress value using subgrain size and its derived piezometers, we must be cautious about the step size selected and examine whether the LAMBs it produces are real SGBs or not.

A clear negative correlation between LAMB or SGB density and grain size for the step size of 1 μm (R2 = 0.67, Fig. 8d) indicates that the stress inferred from an LAMB or SGB density may be consistent with the stress inferred from a dynamically recrystallized grain size e.g.9. The observation of olivines with a smaller grain size tending to develop a higher LAMB or SGB density and smaller subgrains (Fig. 8d), implies that smaller sized olivines may be affected by greater stress. This finding may be interpreted by the fact that the smaller grains have smaller contact areas with their neighboring grains, thus concentrating a considerably high stress in the small grains52,53. The widespread distributions of olivine subgrain size, LAMB or SGB density, and grain size may reflect that the distribution of stress is also spatially heterogenous in the sample, which is consistent with the inhomogenous distribution of GNDs in deformed polycrystalline aggregates52,54. Furthermore, it is noted that the orientation of a grain to the applied stress can determine the resolved shear stress on a dislocation slip system, and thus also affect the development of LAMBs or SGBs in individual grains. In other words, the density of LAMBs or SGBs in a single grain is a combined effect of grain orientation and stress intensity. However, their relative importances are still unknown but can be quantified using deformation experiments on single crystals in future.

Compared to measuring subgrain size using intercept lengths, it is easier to obtain LAMB or SGB density, because the density of LAMB or SGB can be calculated directly using the built-in functions of “subBoundaryLength” and “area” in the MTEX toolbox version 5.5 and above. This fact suggests that SGB density could be used as a convenient metric to estimate stress in future and it might also avoid the problem caused by direction of profiles to measure intercept length (see below), although their underlying physics is essentially the same.

### Cautions of employing experimentally calibrated piezometers

Besides the effect of the step size as mentioned above, many other parameters can also play important roles in deriving stress using subgrain-related or EBSD-based piezometers. First, the estimation of intercept length and SGB density is significantly affected by the choice of misorientation threshold that defines LAMB or SGB. A lower threshold results in a higher LAMB or SGB density and vice versa (Supplementary Fig. S4). Second, the choices of denoising filters and boundary smoothing also affect the appearance of LAMBs or SGBs55,56. Third, the direction of profiles (e.g., parallel to either X- or Y-axes or random in the specimen reference frame19,57,58) can also yield contrasting intercept lengths of neighboring LAMBs or SGBs, especially when LAMBs or SGBs tend to have a preferred direction like our sample (Fig. 1e and Supplementary Fig. S2d). In addition, more factors can affect the measurement of microstructures and the estimation of stress by employing other piezometers, including segmentation of recrystallized grains, calculation of dislocation density, and selection of analytical parameters (e.g., kernal size of KAM), as well as the effects of structural factors (e.g., grain pinning, grain-to-grain interaction, distribution heterogeneity in substructures, and sampling size)34,37,59,60. Because of the appreciable influences of various factors listed above, to derive a reliable stress for natural samples, it is optimal to apply the same parameters and similar structures that are originally used to calibrate piezometers.

Despite the differences of parameters in calculating stress between those used in our study and those by Goddard, et al.19, their underlying principles are basically the same and our results still support the close relations between LAMB or SGB density, KAM, and stress. These correlations are built based on one natural sample and available subgrain-size piezometers, which may incur uncertainties and inaccuracies in the stress estimation. To derive more reliable piezometers based on SGB density and KAM, further calibrations using the experimentally deformed samples are needed.

## Methods

### Sample preparation

Owing to the prevalence of large grains, the sample 13LQ117 was cut into a large (12 × 6 cm2), thick (~ 150 μm) thin section in the XZ-plane, i.e., normal to the foliation (Z-direction) and parallel to the lineation (X-direction). The foliation and lineation were determined by the alignments of trails and shape preferred orientation of spinel in three nearly perpendicular sections. To obtain high quality electron backscattered patterns (EBSPs), the thin section was finally polished with an aqueous suspension containing 0.05 μm colloidal silica for > 2 h using a Buehler vibration polisher.

### EBSD data acquisition and treatment

Mineral phase and crystallographic orientation data are measured using an Oxford Symmetry EBSD detector attached to a Zeiss Sigma 300 field emission scanning electron microscope (FE-SEM) housed at the China University of Geosciences (Wuhan), China. An accelerating voltage of 15 kV, a spot size of 6, a beam current of 6 nA, and working distance of 16–23 mm were used. We mapped a large area (~ 4.3 × 1.8 cm2) with EBSD with a step size of 15 μm and small areas for 22 single olivine grains (without obvious dynamic recrystallization) with a step size of 1 μm (Supplementary Fig. S5 and Supplementary raw EBSD data). To examine the effect of step size on the derived microstructures at the same domains, the step size was then undersampled from 1 to 30 μm by selecting the points at regular row and column intervals in the 1 μm step-sized gridified EBSD data. The microstructures produced by artificially enlarged step sizes are very similar to those measured using various actual step sizes. The raw indexation rates for all mapping are larger than 95%. The raw EBSD data are cleaned by removing wild spikes (isolated pixels) and points with MAD > 0.8°. The denoising process is performed by applying Kuwahara filter with number of neighbors of 5, because this filter does not smooth out low-angle boundaries47,55.

The strength of the crystallographic preferred orientation (CPO) is quantified using M- and J-indices61,62. The M- and J-indices range from 0 to 1 and from 1 to infinity for random and single crystal fabrics, respectively. The BA-index is used to describe the olivine CPO symmetry63. This index classifies olivine CPOs into three types: (1) fiber-[010] ([010]-axis point plus [100]- and [001]-axes girdles, BA-index < 0.33), (2) orthorhombic ([100]-, [010]- and [001]-axes points, 0.33 < BA-index < 0.66), and (3) fiber-[100] ([100]-axis point plus [010] and [001]-axes girdles, BA-index > 0.66). The CPOs of olivine (Ol), orthopyroxene (Opx) and clinopyroxene (Cpx) are presented using one point per grain and shown in the pole figures.

A grain boundary is a high-angle boundary having misorientation angles larger than 10°–15° when defined using EBSD data64,65. To minimize the mistake of taking grain boundaries in the misorientation angle range of 10°–15° as low-angle boundaries, we selected the lower bound definition for a high-angle boundary, namely 10°, as the threshold to construct grain boundaries. In this case, a low-angle boundary can be defined as a boundary whose misorientation angle is greater than 1° and less than 10°. The choice of a lower limits of 1° (larger than the angular resolution of ~ 0.5° for our Symmetry EBSD system) for low-angle boundary and 10° for high-angle boundary is also adopted by many previous olivine and pyroxene studies e.g.19,45,57,66. To avoid the staircase effect generated during the grain reconstruction routine, grain boundaries and LAMBs are smoothed iteratively by 8 times.

Grains consisting of less than 10 pixels are omitted in the grain construction only for large-area EBSD map in which undersampling is not applied. The apparent grain size is reported as equivalent circular diameter and area-weighted mean grain size is calculated. The sinuosity of grain boundaries is quantified by the shape factor—the ratio between the actual perimeter of the grain and the perimeter of a circle with the same area. The ellipticity of grain is measured by aspect ratios—the length ratio of long to short axes of a fitted ellipse. The discrimination between twist and tilt LAMBs and inference of dislocation slip systems in olivine are based on the geometrical relations between misorientation axes, direction of LAMB trace, and olivine axis orientations at the LAMBs67. To quantify the degree of intracrystalline lattice distortion, M2M, GOS and KAM are also calculated.

Linear intercept length represents the distance between two neighboring LAMBs. Because of the low precision and large EBSD data volume, we did not calculate the linear-intercept length for the large-area mapping data. Instead, linear-intercept lengths are only calculated for the high spatial resolution ESBD maps of 22 olivine grains. For the olivine grains that show an obvious oblique alignment of LAMBs with respect to the horizontal direction, we rotated their EBSD data to make most of their LAMBs align vertically. Since most LAMBs are (sub)perpendicular to the horizontal X-axis after the rotation of EBSD data and they are parallel to each other, only X-axis parallel profiles are chosen. Along each profile, the points of intersection with the LAMB are detected and the intercept length between neighboring intersections is measured and recorded for further calculations (e.g., arithmetic mean of intercept length). Different from the method used by Goddard, et al.19, all X-axis parallel profile lines, which correspond to all rows in the gridified EBSD data, are used to calculate the intercept length. The complete number of intercept lines used in our study averts the unstable mean intercept length which varies significantly if the number of intercept lines is insufficient19. All above-mentioned EBSD data treatments were performed using the MTEX toolbox (ver. 5.5) in MATLAB (http://mtex-toolbox.github.io/)68,69,70.

### Stress estimation

In this study, three olivine subgrain-size piezometers are adopted to calculate stress, including an old version proposed by Toriumi20 and two new versions developed by Goddard, et al.19. The mean intercept length of LAMBs is used as a substitute for subgrain size in the formulae of piezometers. The new subgrain-size piezometers proposed by Goddard, et al.19 consider the cases with and without the correction of Holyoke and Kronenberg71. This correction accounts for the increased friction resulting from the Poisson effect on load column during specimen loading. For convenience, the three piezometers hereafter are abbreviated as T1979, G2020wHK2010, and G2020w/oHK2010.

It is noteworthy that differential stress is used by T1979 and von Mises equivalent stress is employed by G2020wHK2010 and G2020w/oHK2010. Because the geometry of the deformation experiment that T1979 adopted is axial compression7, differential stress is thus equal to von Mises equivalent stress72. Therefore, von Mises equivalent stress (in short, stress) is consistently applied in the three piezometers.


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# What is the circumference of the circle above with center O?

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What is the circumference of the circle above with center O?

(1) The perimeter of triangle OXZ is $$20 + 10\sqrt{2}$$.
(2) The length of arc XYZ is $$5\pi$$.
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26 Dec 2012, 05:05
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What is the circumference of the circle above with center O?

(1) The perimeter of triangle OXZ is $$20 + 10\sqrt{2}$$. Since <O=90 degrees and OX=OZ=r, then $$XZ=r\sqrt{2}$$. Thus we have that $$r+r+r\sqrt{2}=20 + 10\sqrt{2}$$ --> we can find r, thus we can find the circumference. Sufficient.

(2) The length of arc XYZ is $$5\pi$$. Again, since <O=90 degrees, then the length of arc XYZ is 1/4 of the circumference (360 degrees), thus the circumference = $$20\pi$$. Sufficient.

Answer: D.
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Re: What is the circumference of the circle above with center O?  [#permalink]

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12 Jan 2013, 08:33
Bunuel wrote:

What is the circumference of the circle above with center O?

(1) The perimeter of triangle OXZ is $$20 + 10\sqrt{2}$$. Since <O=90 degrees and OX=OZ=r, then $$XZ=r\sqrt{2}$$. Thus we have that $$r+r+r\sqrt{2}=20 + 10\sqrt{2}$$ --> we can find r, thus we can find the circumference. Sufficient.

(2) The length of arc XYZ is $$5\pi$$. Again, since <O=90 degrees, then the length of arc XYZ is 1/4 of the circumference (360 degrees), thus the circumference = $$20\pi$$. Sufficient.

Answer: D.

Is there a formula for this explanation?
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Re: What is the circumference of the circle above with center O?  [#permalink]

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13 Jan 2013, 03:29
fozzzy wrote:
Bunuel wrote:

What is the circumference of the circle above with center O?

(1) The perimeter of triangle OXZ is $$20 + 10\sqrt{2}$$. Since <O=90 degrees and OX=OZ=r, then $$XZ=r\sqrt{2}$$. Thus we have that $$r+r+r\sqrt{2}=20 + 10\sqrt{2}$$ --> we can find r, thus we can find the circumference. Sufficient.

(2) The length of arc XYZ is $$5\pi$$. Again, since <O=90 degrees, then the length of arc XYZ is 1/4 of the circumference (360 degrees), thus the circumference = $$20\pi$$. Sufficient.

Answer: D.

Is there a formula for this explanation?

Yes, check here: math-circles-87957.html (see Arc Length under Arcs and Sectors chapter)
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Re: What is the circumference of the circle above with center O?  [#permalink]

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02 Jan 2015, 14:06
2
Hi All,

Even though this DS question is about Geometry rules, it involves a great Algebra shortcut that you'll see at least once on Test Day.

We're given a circle with an ISOSCELES right triangle within it (with the 3 vertices at the center of the circle and on the circumference), so we have a 45/45/90 right triangle. We're asked for the circumference of the circle.

Fact 1: The perimeter of triangle OXZ = 20 + 10\sqrt{2}

Since the triangle is a 45/45/90, we know the sides are A, A and A\sqrt{2}, thus...

A + A + A\sqrt{2} = 20 + 10\sqrt{2}

We have ONE VARIABLE and ONE equation, so we CAN solve it (and since there are no additional possible answers, re. due to exponents, absolute values, etc. - there will be JUST ONE ANSWER.)
Fact 1 is SUFFICIENT.

Fact 2: Arc XYZ = 5pi

Since this arc is based on a 90 degree angle, we CAN figure out what fraction of the circumference we're dealing with (it's 1/4, but that math is not necessary). In this way we have the SAME shortcut that we had in Fact 1: ONE VARIABLE and ONE equation, so we CAN solve it.
Fact 2 is SUFFICIENT.

Final Answer:

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Manager Joined: 22 Nov 2016 Posts: 205 Location: United States Concentration: Leadership, Strategy GPA: 3.4 Re: What is the circumference of the circle above with center O? [#permalink] ### Show Tags 06 Jul 2017, 17:48 We need to find the value of $$2 \pi R$$ From the figure we can tell that OX=OZ = radius or R. Since these two sides are equal and the other angle is given as $$90^ {\circ}$$, triangle XOZ is an isosceles right triangle with sides $$x:x:x\sqrt{2}$$ Armed with this info, lets look at the statements: Statement 1: Perimeter = $$20+10\sqrt{2}$$ This straight away gives us the radius as 10..........................Sufficient Statement 2: Length of arc XYZ = $$5\pi$$ We know that Length of an arc = $$x^{\circ}$$ * Circumference/360 Therefore, $$5\pi$$ = 90 * 2 * $$\pi$$ * R/360 We can find the circumference..........................Sufficient Answer is D _________________ Kudosity killed the cat but your kudos can save it. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: What is the circumference of the circle above with center O? [#permalink] ### Show Tags 07 Jul 2017, 22:39 Walkabout wrote: Attachment: Circle.png What is the circumference of the circle above with center O? (1) The perimeter of triangle OXZ is $$20 + 10\sqrt{2}$$. (2) The length of arc XYZ is $$5\pi$$. Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. By Variable Approach method, we can identify a circle with just 1 varable, which is a radius. Thus D is the answer most likely. Let $$r$$ be the radius of the above circle. Condition 1) $$2 \cdot r + r\sqrt{2} = r \cdot (2 + \sqrt{2} ) = 20 + 10\sqrt{2}$$ Thus $$r = 10$$. This condition is sufficient. Condition 2) The arc XYZ is $$\frac{1}{4} \cdot 2 \pi r = 5 \pi$$. Thus $$r = 10$$. This condition is also sufficient. Therefore D is the answer as expected. -> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: What is the circumference of the circle above with center O?  [#permalink]

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07 Jul 2017, 22:48
Answer is D
1. Traingle is 45,45,90 degree triangle.
So two sides are same and length of radius is 10
Circumference of circle is 20pi
2. Is an arc is is known and angle subtending that arc is known than radius can be calculated.
So Both 1 and 2 are sufficient.

D

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1 Overview

While multi-state models can be used to estimate the parameters of a state transition model (STM) in a very flexible manner, data availability can make it difficult (or infeasible) to fit such a model. This is often the case when an evidence synthesis model based on summary level data is used to parameterize the STM. For example, in oncology, published articles of clinical trials often provide survival curves of progression-free survival (PFS) and overall survival (OS), but do not release information on time to event (and censoring) for each transition. In this setting partitioned survival analysis may consequently be a simpler approach.

We will use the same packages as in the “Semi-Markov Multi-state Model” tutorial.

library("rcea")
library("hesim")
library("data.table")
library("ggplot2")
library("flexsurv")

set.seed(101) # Make PSA reproducible

2 Theory

An 3-state partitioned survival model (PSM) simulates the probability that a patient is in each of 3 distinct health states at a given point of time when treated by a particular therapy. State membership is estimated from 2 survival curves (e.g., PFS and OS) using an “area under the curve” approach. Specifically, letting $$S_{PFS}(t)$$ be the PFS survival curve and $$S_{OS}(t)$$ be the OS curve, the probabilities of being in the stable, progression, and death states at time $$t$$ are $$S_{PFS}(t)$$, $$S_{OS}(t) - S_{PFS}(t)$$, and $$1 - S_{OS}(t)$$, respectively. For a more general N-state case, refer to the hesim vignette on partitioned survival models (PSMs) and the NICE Decision Support report on partitioned survival analysis.

Partitioned survival analysis can also be used to parameterize a time-inhomogeneous cohort discrete time state transition model (cDTSTM). This approach is identical to the PSM, except that discrete model cycles must be used. In a 3-state model, the transition probability matrix at time $$t$$ is given by,

$\begin{pmatrix} C & p_1 - p_2 & p_2\\ 0 & C & p_2 \\ 0 & 0 & 1, \end{pmatrix}$ where $$C$$ is refers to complement of the sum of the remaining elements in a row and $$p_1$$ and $$p_2$$ are transition probabilities derived from survival curves. In the PFS and OS example with model cycles of length $$u$$, $$p_1 = 1 - s_{PFS}(t)/s_{PFS}(t-u)$$ and $$p_2 = 1 - s_{OS}(t)/s_{OS}(t-u)$$. From this matrix, one can immediately see that a PSM—in contrast to a multi-state model—make the implausible assumption that the probability of a transition from stable to death and from progression to death are equal. See the heemod vignette on survival models for an example cDTSTM based on a partitioned survival analysis.

3 Model setup

We set up the model in the same way as in the multi-state model, except that since PSMs are cohort models, we no longer need to simulate a large number of patients to compute expected values. Instead, we will simulate separate cohorts of representative patients of varying ages and sexes. The overall estimates are (weighted) averages across patients, although in this example we will assume that each representative patient should receive equal weight. Note that we could have also used grp_id option in hesim::hesim_data() to assign patients to groups so that we could conduct subgroup analysis.

patients <- data.table(
patient_id = 1:4,
patient_wt = rep(1/4, 4), # Each patient has same weight
age = c(45, 45, 65, 65),
female = c(0, 1, 0, 1)
)

states <- data.table(
state_id = c(1, 2),
state_name = c("Stable", "Progression") # Non-death health states
)

strategies <- data.frame(
strategy_id = 1:2,
strategy_name = c("SOC", "New")
)

hesim_dat <- hesim_data(
patients = patients,
strategies = strategies,
states = states
)
print(hesim_dat)
## $strategies ## strategy_id strategy_name ## 1 1 SOC ## 2 2 New ## ##$patients
##    patient_id patient_wt age female
## 1:          1       0.25  45      0
## 2:          2       0.25  45      1
## 3:          3       0.25  65      0
## 4:          4       0.25  65      1
##
## $states ## state_id state_name ## 1: 1 Stable ## 2: 2 Progression ## ## attr(,"class") ## [1] "hesim_data" 4 Parameter estimation 4.1 Survival models Multi-state models estimate the hazards between all transitions in an integrated manner. PSMs are, in contrast, parameterized by estimating separate survival models for different endpoints. The data consists of one row per patient. In the oncology example, the two endpoints are PFS and OS. We created the function sim_pfs_os_data() to simulate such data. surv_est_data <- rcea::sim_pfs_os_data(n = 2000) surv_est_data[patient_id %in% c(1, 2)] ## patient_id strategy_id strategy_name female age pfs_time pfs_status ## 1: 1 2 New 0 66.40410 9.209143 1 ## 2: 2 2 New 0 57.64656 8.514136 1 ## os_status os_time ## 1: 1 9.209143 ## 2: 1 14.051688 To maintain consistency with the multi-state model, we will fit Weibull survival models for both PFS and OS and include an indicator for female as a covariate. The PFS and OS fits are stored in a hesim::flexsurvreg_list, which is just a list of models fit using flexsurv::flexsurvreg(). fit_pfs_wei <- flexsurv::flexsurvreg( Surv(pfs_time, pfs_status) ~ strategy_name + female, data = surv_est_data, dist = "weibull") fit_os_wei <- flexsurvreg( Surv(os_time, os_status) ~ strategy_name + female, data = surv_est_data, dist = "weibull") psfit_wei <- flexsurvreg_list(fit_pfs_wei, fit_os_wei) 4.2 Utility and costs Utility and medical costs are setup in the exact same way as in the multi-state model. # Utility utility_tbl <- stateval_tbl( data.table(state_id = states$state_id,
mean = c(.8, S = .6),
se = c(0.02, .05)
),
dist = "beta",
hesim_data = hesim_dat)
print(utility_tbl)
##    state_id mean   se
## 1:        1  0.8 0.02
## 2:        2  0.6 0.05
# Medical costs
medcost_tbl <- stateval_tbl(
data.table(state_id = states$state_id, mean = c(2000, 9500), se = c(2000, 9500) ), dist = "gamma", hesim_data = hesim_dat) print(medcost_tbl) ## state_id mean se ## 1: 1 2000 2000 ## 2: 2 9500 9500 Treatment costs are, on the other hand, treated slightly differently. Since the model is not semi-Markov, they cannot vary as a function of time since entering the progression state. One way to accommodate this is to use the survival models for PFS and OS fits to parameterize a cDTSTM as described above and create a number of tunnel states following progression. For instance, if we constructed a model with monthly cycles, then 4 progression states would be required (progression month 1, progression month 2, progression month 3, and progression month > 3). However, for simplicity, we will assume treatment costs are constant in the progression state and continue to use the area under the curve approach. drugcost_tbl <- stateval_tbl( data.table(strategy_id = strategies$strategy_id,
est = c(2000, 12000)),
dist = "fixed",
hesim_data = hesim_dat)
print(drugcost_tbl)
##    strategy_id   est
## 1:           1  2000
## 2:           2 12000

5 Simulation

5.1 Constructing the economic model

As in all models that perform probabilistic sensitivity analysis (PSA), we set the number of PSA iterations. Note that we use a smaller number of iterations because will will sample parameters from the survival models using bootstrapping, which takes longer to run than sampling directly from probability distributions like we have done previously. For instance, in the multi-state model, we sampled the parameters from their multivariate normal asymptotic distribution.

n_samples <- 100

5.2 Survival models

Survival predictions are made as a function of the fitted Weibull models and input data. The latter consists of each treatment strategy and patient combination.

# Input data for survival models
survmods_data <- expand(hesim_dat, by = c("strategies", "patients"))
print(survmods_data)
##    strategy_id patient_id strategy_name patient_wt age female
## 1:           1          1           SOC       0.25  45      0
## 2:           1          2           SOC       0.25  45      1
## 3:           1          3           SOC       0.25  65      0
## 4:           1          4           SOC       0.25  65      1
## 5:           2          1           New       0.25  45      0
## 6:           2          2           New       0.25  45      1
## 7:           2          3           New       0.25  65      0
## 8:           2          4           New       0.25  65      1

In addition, we must also specify arguments related to the PSA. As mentioned above, we sample the parameters via bootstrapping, whereby the survival models are refit repeatedly to resamples of the estimation dataset. The advantage of this approach is that it preserves the correlation between PFS and OS and ensures that the curves do not cross.

survmods <- create_PsmCurves(psfit_wei,
input_data = survmods_data,
n = n_samples,
bootstrap = TRUE, est_data = surv_est_data)

5.2.1 Utility and cost models

The same code from the multi-state model is used to setup the cost and utility models.

# Utility
utilitymod <- create_StateVals(utility_tbl, n = n_samples)

# Costs
drugcostmod <- create_StateVals(drugcost_tbl, n = n_samples)
medcostmod <- create_StateVals(medcost_tbl, n = n_samples)
costmods <- list(Drug = drugcostmod,
Medical = medcostmod)

5.2.2 Combining the disease progression, cost, and utility models

As in the multi-state model, we initialize the economic model by combining the disease, utility, and costs models using the $new() method. econmod <- Psm$new(survival_models = survmods,
utility_model = utilitymod,
cost_models = costmods)

5.3 Simulating outcomes

5.3.1 Survival curves

Once the PSM has been initialized, the $sim_survival() method can be used to generate survival curves for each endpoint. times <- seq(0, 20, by = .1) econmod$sim_survival(t = times)
head(econmod$survival_) ## sample strategy_id patient_id grp_id patient_wt curve t survival ## 1: 1 1 1 1 0.25 1 0.0 1.0000000 ## 2: 1 1 1 1 0.25 1 0.1 0.9998292 ## 3: 1 1 1 1 0.25 1 0.2 0.9992196 ## 4: 1 1 1 1 0.25 1 0.3 0.9981030 ## 5: 1 1 1 1 0.25 1 0.4 0.9964389 ## 6: 1 1 1 1 0.25 1 0.5 0.9941987 We can, in theory, plot these curves for each treatment strategy and patient. To summarize, however, we will plot the average (across patients) survival curves for each treatment strategy. The probabilities of being in the stable, progression, and death states are the area under the PFS curves, the area between the OS and PFS curves, and the area above the OS curve, respectively. 5.3.2 Health state probabilities These health state probabilities are computed using the $sim_stateprobs() method. For example, let’s look at probabilities in the progression state at year 12 using the first PSA sample.

econmod$sim_stateprobs() econmod$stateprobs_[sample == 1 & state_id == 2 & t == 12]
##    sample strategy_id patient_id grp_id patient_wt state_id  t      prob
## 1:      1           1          1      1       0.25        2 12 0.3115353
## 2:      1           1          2      1       0.25        2 12 0.2206730
## 3:      1           1          3      1       0.25        2 12 0.3115353
## 4:      1           1          4      1       0.25        2 12 0.2206730
## 5:      1           2          1      1       0.25        2 12 0.4933336
## 6:      1           2          2      1       0.25        2 12 0.4193890
## 7:      1           2          3      1       0.25        2 12 0.4933336
## 8:      1           2          4      1       0.25        2 12 0.4193890

5.3.3 Utility and costs

Utility and costs are computed in the same manner as in the other hesim models.

econmod$sim_costs(dr = .03) econmod$sim_qalys(dr = .03)

6 Cost-effectiveness analysis

Since the cost and QALY output is the same regardless of the model type, the cost-effectiveness analysis (CEA) proceeds in the same way as when using the Markov cohort and multi-state modeling approaches.

ce_sim <- econmod$summarize() cea_pw_out <- cea_pw(ce_sim, comparator = 1, dr_qalys = .03, dr_costs = .03) icer_tbl(cea_pw_out, colnames = strategies$strategy_name)
##                   SOC New
## Incremental QALYs "-" "0.99 (0.83, 1.14)"
## Incremental costs "-" "105,172 (99,504, 118,350)"
## Incremental NMB   "-" "-55,829 (-69,066, -48,682)"
## ICER              "-" "106,572"
## Conclusion        "-" "Not cost-effective"


text:
OpenStudy (anonymous):

find dy/dx given that y=x+5+3/√x

3 years ago
OpenStudy (anonymous):

you may use the power rule for this

3 years ago
OpenStudy (anonymous):

but first I would rewrite the term 3/(sqrt(x)

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

I'll check just a sec :3

3 years ago
OpenStudy (anonymous):

almost, there should be a negative before the first fraction y=x+5+3x^(-1/2) dy/dx = 1 - 3/2 x^(-3/2)

3 years ago
OpenStudy (anonymous):

oh ok thanx can you just explain how it becomes a negative

3 years ago
OpenStudy (anonymous):

sure. it starts off as $\frac{ 3 }{ \sqrt{x} }$but sqrt is the same thing as ^(1/2) $\frac{ 3 }{ x ^{1/2} }$which is the same thing as.... $3x ^{\frac{ -1 }{ 2 }}$ so then when we apply the power rule, it becomes $(-\frac{ 1 }{ 2 }*3)x ^\frac{ -3 }{ 2 }$ so the negative comes from the fact that the exponent on the x was negative

3 years ago
OpenStudy (anonymous):

Thank you for your support Appreciate it :)

3 years ago
OpenStudy (anonymous):

do you understand why $\frac{ 3 }{ x ^{\frac{ 1 }{ 2 }} } = 3x ^{\frac{ -1 }{ 2 }}$ ? that's the only step I dont think I fully explained

3 years ago
OpenStudy (anonymous):

Yes i do understand Thank you I'm an A level Student - just got mixed up with the signs Thankyou loads!

3 years ago


text:
# How can I constrain an imported netlist in Vivado?

I have a pre-compiled netlist (created by Xilinx ISE 14.7), which is imported into Vivado 2015.4 and used in synthesis to assemble my complete design. Vivado reports unconstrained paths for the imported netlist.

• According to the Vivado Constraint Guide, Vivado does not apply xdc constraints (Xilinx design constraints) to netlists. This feature will be implemented in the future.

• Setting an xcf constraint file (XST constraint file), while synthesizing the netlist with XST does not help. The xcf is read and processed, but is seems that Vivado does not apply annotated/embedded constraints from ngc files.

• Storing a ncf files (netlist constraint file) besides the ngc file does not help, because Vivado does not parse ucf style constraint files. (ISE reads ncf files and applies them to netlists, if the have the same name and location as the ngc files.)

• I cannot synthesize the netlist with Vivido because there are bugs in Vivado, which prevent me from changing the tool chain.

How can I apply the required timing constrains?

Notes:

• I can open the inner netlist in Vivado and see its hierarchy.
• I can rerun XST synthesis to recreate the ngc netlist.
• Just out of curiocity, if you have access to the HDL code, and can rerun the XST, why don't you just add the files into your Vivado project? – FarhadA Jan 18 '16 at 9:14
• The problem is that Vivado does not compile the source code due to a lack of VHDL-93 support and an intetnal memory leek when compiling this code ... – Paebbels Jan 18 '16 at 10:15
• I am surprised, because I thought Vivado 2015.x supports VHDL-2008 so VHDL-93 should not be any problem. – FarhadA Jan 18 '16 at 10:32
• It just supports a few features of VHDL-2008, while missing already implemented features of ISE. Moreover these features are not fully supported by xSim (the 'new' Vivado simulator ). Even if we rewrite the code to work around the usage of physical types, there is still the memory leak... – Paebbels Jan 18 '16 at 10:42
• Have you tried to ask this question on Xilin forums, this seems to be a problem highlighted in 2014 version of Vivado and was supposed to be fixed. – FarhadA Jan 18 '16 at 13:17

One solution is to write an XDC file including the required constraints and attach it to the entity name of the imported netlist using the XDC file property SCOPED_TO_REF. (Attaching it to a specific instance with SCOPED_TO_CELLS should also work.)
When using the SCOPED_TO_REF property, it must be set to the original (top-level) entity named stored within the netlist. Attaching it to the entity name of the black-box does not work (*). Moreover, the XDC file must be marked as to use "only for implementation", otherwise the synthesis step complains a critical warning. The setting "only for implementation" already includes the post-synthesis timing analysis.


text:
# Efficient principal pivots

It was suggested I should try posting this question here from Mathematics

Background

I'm working on a numerical linear algebra package in C#. I'm trying to implement a variety of "principal pivoting" methods to solve optimization problems (specifically linear complementarity problems, but I think similar schemes are used in linear and quadratic programming). These methods require the ability to multiply a column vector against the "principal pivotal transform" of a matrix.

The principal pivot transform (PPT) of a matrix $M$ is defined as:

$$PM^'P^T = \left (\begin{array}{cc} M_{\alpha\alpha}^{-1} &\ -M_{\alpha\alpha}^{-1} M_{\alpha\beta} \\ M_{\beta\alpha}M_{\alpha\alpha}^{-1} &\ M_{\beta\beta} - M_{\beta\alpha} M_{\alpha\alpha}^{-1}M_{\alpha\beta} \end{array} \right )$$

Where $\alpha$ is the set of pivoted row/column indices, and $\beta$ is the set of unpivoted row/column indices. $P$ is a permutation matrix that keeps the pivoted and unpivoted rows/columns together to make the block matrix form above cleaner. $M_{\alpha\beta}$ is a submatrix of $M$ with the rows in $\alpha$ and the columns in $\beta$, and the other submatrices of $M$ are defined similarly. Also, $M_{\alpha\alpha}^{-1} = (M_{\alpha\alpha})^{-1}$ and not $(M^{-1})_{\alpha\alpha}$

...

Each iteration of a principal pivot method swaps a single index from $\alpha$ to $\beta$ or vice-verse. This single pivoting operation can be expressed as a low rank update of the matrix. Say we're pivoting index $r$ and using $M^'$ to represent the updated matrix:

$$\begin{matrix} M_{rr}^' & = & & \frac{1}{M_{rr}} & \\ \\ \\ M_{ir}^' & = & & \frac{M_{ir}}{M_{rr}}, & (i \neq r) \\ \\ \\ M_{ri}^' & = & & \frac{-M_{ri}}{M_{rr}}, & (i \neq r) \\ \\ \\ M_{ij}^' & = & M_{ij} - & \frac{M_{ir} * M_{ri}}{M_{rr}}, & (i \neq r, j \neq r) \end{matrix}$$

This is what I'm currently doing. Unfortunately, this is basically calculating $M_{\alpha\alpha}^{-1}$ using something like Gaussian elimination. I have a few test cases where the results are suffering from numerical problems even on on small problems, even using double precision floating point. To help combat this, if I remove an index from $\alpha$ (that is, if I unpivot an index), I recalculate the matrix from scratch using one pivot at a time with the above scheme. But this makes the Principal Pivot algorithm I have quite slow.

Question

I know there are reasonable ways to calculate the product $M_{\alpha\alpha}^{-1} z$ for some column vector $z$ without calculating $M_{\alpha\alpha}^{-1}$ directly (such as LU decomposition or QR decomposition). I could calculate this decomposition fresh every iteration from the current pivot set ($\alpha$), and probably get a pretty robust method. But then each iteration of the pivoting algorithm would take $O(n^3)$; that's too slow. I could do a low rank update of the LU or QR decomposition, but since most of the algorithms pivot a single index at a time, I lose most of the benefit of these decomposition schemes after a few iterations. I could perform the full decomposition only every few iterations, but then the question becomes how often should I do so? And it still would be quite slow.

Is there something more clever that can be done? I think I basically need to find the decomposition of principal submatrices. Is there any literature on the subject? Any other ways to approach the problem?

• Is $M$ dense, positive-definite and/or symmetric in the general case? – Aron Ahmadia Apr 23 '12 at 7:53
• In my case it's symmetric and PSD. It's actually formed from the normal equations of some rank deficient dense matrices. I know there are some iterative schemes that might work well here, but I wanted to explore pivoting methods. – Jay Lemmon Apr 23 '12 at 16:26
• If it's SPD, then you can take the Cholesky decomposition, and there is a fair amount of literature on how to stably update the Cholesky decomposition using low-rank factors as you are suggesting. This is very commonly used in implementations of simplectical solvers, so I would not be surprised if it turns out to be the most efficient for you. – Aron Ahmadia Apr 24 '12 at 7:51
• Ah, that's actually an interesting idea. I'll look in to it. – Jay Lemmon Apr 24 '12 at 20:10


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# Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If

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Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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17 Oct 2017, 23:31
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Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If Jim’s son is now 5 years older than his daughter, how old is Jim’s daughter now?

(A) 5
(B) 10
(C) 13
(D) 16
(E) 18

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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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18 Oct 2017, 02:27
Let jim daughter be d
Jim son be x
X-8=2d-8
X=d+5
d+5-8=2d-8
d-3=2d-d
d=5
A is correct

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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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18 Oct 2017, 02:39
1
EQ 1 Jim's son = S
Jim's Daughter = D

(S-8)= 2 (D-8)
S-8 = 2D- 16
S-2D= -8
EQ2 S= D+5 or S- D = 5

Equating eq 1 and 2

D=13
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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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18 Oct 2017, 11:16
s=son's age now
d=daughter's age now

(s-8)/(d-8)=2
s-d=5

d=13 (C)

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Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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18 Oct 2017, 15:09
Bunuel wrote:
Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If Jim’s son is now 5 years older than his daughter, how old is Jim’s daughter now?

(A) 5
(B) 10
(C) 13
(D) 16
(E) 18

Exactly 8 years ago, Jim’s son, S, was twice as old as Jim’s daughter, D

S - 8 = 2(D - 8)

Son is now five years older than daughter:

S = D + 5
Substitute for S in first equation

(D + 5) - 8 = 2(D - 8)
D - 3 = 2D - 16
D = 13

(Check: D is 13 now; 8 years she ago was 5. S is 18 now (= D + 5), so 8 years ago he was 10. 10 = twice as old as 5, back then. Correct.)

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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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19 Oct 2017, 03:47
ewmike wrote:
Let jim daughter be d
Jim son be x
X-8=2d-8
X=d+5
d+5-8=2d-8
d-3=2d-d
d=5
A is correct

Hi,
There is an errot here. You did not mulyply 2 * 8 too. it should be

2 (d -8) = 2d - 16
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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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23 Oct 2017, 11:11
-I set the original equation up as (s-8)=2(d-16), which simplifies to s-8=2d-16.
-we are given s=d+5, so I substitute that in for s.
-d-3=2d-16, simplified d=13.

Initially, I made a small mistake and got d=5, quickly checking my math, I realized the daughter cannot be 5 because 8 years ago she wasn't born yet. So I redid it until I figured out my mistake. I wasted a lot of time on something pretty simple here.

Simple mistakes are my biggest problem, I know the material pretty well, but I just make a lot of dumb errors. Any guidance would be appreciated.
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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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23 Oct 2017, 11:55
If Son = X
Daughter=X-5

EQ: x-8=2(X-5-8)
X-8=2X-26
X=18

Daughter = 18-5= 13

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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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24 Oct 2017, 06:06
Bunuel wrote:
Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If Jim’s son is now 5 years older than his daughter, how old is Jim’s daughter now?

(A) 5
(B) 10
(C) 13
(D) 16
(E) 18

We can let Jim’s son’s current age = s and Jim’s daughter’s current age = d. We can also state that 8 years ago, his son was (s - 8) years old and his daughter was (d - 8) years old. We can create the following equation:

s - 8 = 2(d - 8)

s - 8 = 2d - 16

s = 2d - 8

and

s = 5 + d

Thus:

2d - 8 = 5 + d

d = 13

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Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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26 Oct 2017, 03:49
we can also check from answer choices if we are unable to form the equation.
Start from C
Let Jim's daughter's age be 13 today
Jim's son age today will be 13+5=18
Now 8 years ago the son was 2 times the age of the daughter.
find the age of both of them 8 years ago
Daughter=13-8=5
Son=18-8=10
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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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28 Aug 2018, 06:56
Top Contributor
Bunuel wrote:
Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If Jim’s son is now 5 years older than his daughter, how old is Jim’s daughter now?

(A) 5
(B) 10
(C) 13
(D) 16
(E) 18

Jim’s son is now 5 years older than his daughter
Let D = the daughter's age NOW
So D + 5 = the son's age NOW

This also means that D - 8 = the daughter's age EIGHT YEARS AGO
This also means that D + 5 - 8 = the son's age EIGHT YEARS AGO
Or we can say, D - 3 = the son's age EIGHT YEARS AGO

8 years ago, Jim’s son was twice as old as Jim’s daughter.
8 years ago, the daughter's age was D - 8 and the son's age was D - 3
So, we can write: D - 3 = 2(D - 8)
Expand right side to get: D - 3 = 2D - 16
Solve to get D = 13

Since D = the daughter's present age, the correct answer is C

Cheers,
Brent
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Re: Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If  [#permalink]

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28 Aug 2018, 07:11
Bunuel wrote:
Exactly 8 years ago, Jim’s son was twice as old as Jim’s daughter. If Jim’s son is now 5 years older than his daughter, how old is Jim’s daughter now?

(A) 5
(B) 10
(C) 13
(D) 16
(E) 18

$$? = D$$

$$\left\{ \begin{gathered} \left( {S - 8} \right) = 2\left( {D - 8} \right) \hfill \\ S = D + 5 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\, \sim \,\,\,\,\,\,\,\left\{ \begin{gathered} \left( {S - 8} \right) = 2\left( {D - 8} \right) \hfill \\ \left( {S - \underline 8 } \right) = \left( {D - \underline 8 } \right) + 5 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {\, - \,} \right)} \,\,\,\,\,\,\,\,\left( {D - 8} \right) - 5 = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = D = 13$$

The above follows the notations and rationale taught in the GMATH method.
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# Conformal Invariance and the Diffusion on Moduli Space for Radial SLE

 MATHEMATICAL PHYSICS SEMINAR Topic: Conformal Invariance and the Diffusion on Moduli Space for Radial SLE Speaker: Robert Bauer Affiliation: University of Illinois, at Urbana-Champaign Date: Thursday, November 4 Time/Room: 4:00pm - 5:00pm/S-101

We show that a random simple curve in a planar n-connected domain that is conformally invariant and satisfies a Markovian-type property, can be described by a diffusion on a moduli space of dimension 3n-2. Under a natural symmetry condition, satisfied for example by the boundaries of clusters in critical percolation or by loop-erased random walk, the diffusion is determined up to a single real parameter. The resulting one-parameter family of random simple curves (or, more generally, random growing compacts) is called SLE. We discuss the "radial" case, where the random curve grows from a boundary component to a point in the interior of the domain. We show that for well chosen values of the parameter the random simple curve has special properties. It satisfies locality, a property known from percolation, when the parameter is 6, and the restriction property, known from the self-avoiding random walk, when the parameter is 8/3.


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# How to implement modular exponentiation efficiently in Shor's algorithm?

I'm a noob in quantum computing and I'm trying to get Shor's algorithm working on Q# (the language is unrelated). However, I'm stuck on computing $$f(x)$$s in the quantum circuit. Let $$N\sim \log_2(n)$$ be the number of qubits needed for finding the period. As far as I know, we can initialize $$s=1$$, and multiply it with $$a,a^2,a^4 \cdots a^{2^{N-1}} \mod n$$ according to the $$N$$ control qubits. However, all methods I can think of needs around $$O(N^2)$$ modulus and thus $$O(N^2)$$ ancillary clean qubits (for example, compute $$a,a^2,a^4 \cdots a^{2^{N-1}}\mod n$$ beforehand, use additions, comparing numbers and subtracting accordingly). Is there a way to implement modulus as some reversible circuit? Or, am I all wrong?