D/Solutions To Exercises - Rob J. Hyndman
D/Solutions to exercisesChapter 1: The forecasting perspective1.1 Look for pragmatic applications in the real world. Note that there are no fixedanswers in this problem.(a) Dow theory: There is an element of belief that past patterns will continueinto the future. So first, look for the patterns (support and resistance levels)and then project them ahead for the market and individual stocks. This is aquantitative time series method.(b) Random walk theory: This is quantitative, and involves a time series ratherthan an explanatory approach. However, the forecasts are very simple becauseof the lack of any meaningful information. The best prediction of tomorrow’sclosing price is today’s closing price. In other words, if we look at first differencesof closing prices (i.e., today’s closing price minus yesterday’s closing price) therewill be no pattern to discover.(c) Prices and earnings: Here instead of dealing with only one time series (i.e., thestock price series) we look at the relation between stock price and earnings pershare to see if there is a relationship—maybe with a lag, maybe not. Therefore this is an explanatory approach to forecasting and would typically involveregression analysis.1.2 Step 1: Problem definition This would involve understanding the nature of the individual product lines to be forecast. For example, are they high-demand products or specialty biscuits produced for individual clients? It is also importantto learn who requires the forecasts and how they will be used. Are the forecaststo be used in scheduling production, or in inventory management, or for budgetary planning? Will the forecasts be studied by senior management, or bythe production manager, or someone else? Have there been stock shortages sothat demand has gone unsatisfied in the recent past? If so, would it be betterto try to forecast demand rather than sales so that we can try to prevent this76
Chapter 1: The forecasting perspective77happening again in the future? The forecaster will also need to learn whetherthe company requires one-off forecasts or whether the company is planning onintroducing a new forecasting system. If the latter, are they intending it tobe managed by their own employees and, if so, what software facilities do theyhave available and what forecasting expertise do they have in-house?Step 2: Gathering information It will be necessary to collect historical data on eachof the product lines we wish to forecast. The company may be interested inforecasting each of the product lines for individual selling points. If so, it isimportant to check that there are sufficient data to allow reasonable forecaststo be obtained. For each variable the company wishes to forecast, at least afew years of data will be needed.There may be other variables which impact the biscuit sales, such as economicfluctuations, advertising campaigns, introduction of new product lines by acompetitor, advertising campaigns of competitors, production difficulties. Thisinformation is best obtained by key personnel within the company. It will benecessary to conduct a range of discussions with relevant people to try to buildan understanding of the market forces.If there are any relevant explanatory variables, these will need to be collected.Step 3: Preliminary (exploratory) analysis Each series of interest should be graphedand its features studied. Try to identify consistent patterns such as trendand seasonality. Check for outliers. Can they be explained? Do any of theexplanatory variables appear to be strongly related to biscuit sales?Step 4: Choosing and fitting models A range of models will be fitted. These modelswill be chosen on the basis of the analysis in Step 3.Step 5: Using and evaluating a forecasting model Forecasts of each product line willbe made using the best forecasting model identified in Step 4. These forecastswill be compared with expert in-house opinion and monitored over the periodfor which forecasts have been made.There will be work to be done in explaining how the forecasting models workto company personnel. There may even be substantial resistance to the introduction of a mathematical approach to forecasting. Some people may feelthreatened. A period of education will probably be necessary.A review of the forecasting models should be planned.
78Part D. Solutions to exercisesChapter 2: Basic forecasting tools2.1 (a) One simple answer: choose the mean temperature in June 1994 as the forecastfor June 1995. That is, 17.2 C.(b) The time plot below shows clear seasonality with average temperature higherin cise 2.1(b): Time plot of average monthly temperature in Paris (January 1994–May1995).2.2 (a) Rapidly increasing trend, little or no seasonality.(b) Seasonal pattern of period 24 (low when asleep); occasional peaks due to strenuous activity.(c) Seasonal pattern of period 7 with peaks at weekends; possibly also peaks duringholiday periods such as Easter or Christmas.(d) Strong seasonality with a weekly pattern (low on weekends) and a yearly pattern. Peaks in either summer (air-conditioning) or winter (heating) or bothdepending on climate. Probably increasing trend with variation increasing withtrend.2.3 (a) Smooth series with several large jumps or direction changes; very large rangeof values; logs help stabilize variance.(b) Downward trend (or early level shift); cycles of about 15 days; outlier at day8; no transformation necessary.(c) Cycles of about 9–10 years; large range and little variation at low points indicating transformation will help; logs help stabilize variance.
79Chapter 2: Basic forecasting tools(d) No clear trend; seasonality of period 12; high in July; no transformation necessary.(e) Initial trend; level shift end of 1982; seasonal period 4 (high in Q2 and Q3, lowin Q1); no transformation necessary.2.4 1-B, 2-A, 3-D, 4-C. The easiest approach to this question is to first identify D.Because it has a peak at lag 12, the time series must have a pattern of period 12.Therefore it is likely to be monthly. The slow decay in plot D shows the series hastrend. The only series with both trend and seasonality of period 12 is Series 3. Nextconsider plot C which has a peak at lag 10. Obviously this cannot reflect a seasonalpattern since the only series remaining which is seasonal is series 2 and that hasperiod 12. Series 4 is strongly cyclic with period approximately 10 and series 1 hasno seasonal or strong cyclic patterns. Therefore C must correspond to series 4. PlotA shows a peak at lag 12 indicating seasonality of period 12. Therefore, it mustcorrespond with series 2. That leaves plot B aligned with series 1.2.5 Y43.7044.422.478.022.94(b) Mean and median give a measure of center; MAD, MSE and St.dev. are measures of spread.(c) r 0.660. See plot on next page.(d) It is inappropriate to compute autocorrelations since there is no time componentto these data. The data are from 14 different runners. (Autocorrelation wouldbe appropriate if they were data from the same runner at 14 different times.)2.6 (a) See plot on following page.(b) and (c)Notation:Error 1 Error 2 (actual demand) (method 1 forecast)(actual demand) (method 2 forecast)
8046444240Y: maximal aerobic capacity48Part D. Solutions to exercises48505254565860X: running times200180160ActualForecast Method 1Forecast Method 2140Demand220240Exercise 2.5(c): Plot of running times versus maximal aerobic capacity.51015MonthExercise 2.6(a): Time plots of data and forecasts.20
81Chapter 2: Basic forecasting 23242239Analysis of errors(periods 1–20)Method 11221222235Error 1 18 834 20 81824 843701416162 20162204MEMAEMSEMPEMAPETheil’s U6.2514.45307.252.557.870.94Method 25232233243Error 2 31 2517 31 23414 15 626 6578 7 312 99 4 4.8014.29294.00 3.618.240.85On MAE and MSE, Method 2 is better than Method 1. On MAPE, Method 1is better than Method 2. Note that this is different from the conclusion drawnin Section 4/2/3 where these two methods are compared. The difference is thatwe have used a different time period over which to compare the results. Holt’smethod (Method 2) performs quite poorly at the start of the series. In Chapter4, this period is excluded from the analysis of errors.2.7 (a) Changes: 0.25, 0.26, 0.13, . . . , 0.09, 0.77. There are 78 observations inthe DOWJONES.DAT file. Therefore there are 77 changes.(b) Average change: 0.1336. So the next 20 changes are each forecast to be 0.1336.(c) The last value of the series is 121.23. So the next 20 are forecast to be:X̂79 121.23 0.1336 121.36X̂80 121.36 0.1336 121.50X̂81 121.50 0.1336 121.63etc.
82Part D. Solutions to exercisesIn general, X̂79 h 121.23 h(0.1336).120115110Dow Jones index(d) See the plot below.020406080100dayExercise 2.7(d): Plot of Dow Jones index (DOWJONES.DAT)(e) The average change is c Xn hc. Therefore,1n 1Pnt 2 (Xt Xt 1 ) and the forecasts are X̂n h nX̂n h1 X Xn h(Xt Xt 1 )n 1 t 2 Xn h(Xn X1 ).n 1This is a straight line with slope equal to (X n X1 )/(n 1). When h 0,X̂n h Xn and when h (n 1), X̂n h X1 . Therefore, the line is drawnbetween the first and last observations.2.8 (a) See the plot on the next page. The variation when the production is low ismuch less than the variation in the series when the production is high. Thisindicates a transformation is required.(b) See the plot on the next page.(c) See the table on page 84.
83Chapter 2: Basic forecasting tools10 6819504Logarithms of vehicles0Vehicles (thousands) Exercise 2.8 (a) and (b): Time plots of Japanese automobile production and the logarithmsof Japanese automobile production.
84Part D. Solutions to 29.359.419.419.419.459.47ForecastErrorError2 Error/Log 490.337 0.0970.1980.3190.2610.1350.1230.0940.0800.118 0.0780.0580.1220.0820.0850.0390.1360.012 0.0410.0350.0310.068 0.001 0720.00010.00010.00380.0027Exercise 2.8 (c) and (d).
Chapter 2: Basic forecasting tools85(d) MSE 0.059 (average of column headed Error 2 )MAPE 3.21% (average of values in last column multiplied by 100).(e) See graph. Forecast is e9.47 13026.(f ) There are a large number of possible methods. One method, which is discussedin Chapter 5, is to consider only data after 1970 and use a straight line fittedthrough the original data (i.e. without taking logarithms).(g) The data for 1974 is lower than would be expected. If this information could beincluded in the forecasts, the MSE and MAPE would both be smaller becausethe forecast error in 1974 would be smaller.
86Part D. Solutions to exercisesChapter 3: Time series 170.00187.40206.00230.00261.40298.80316.50339.673 71236.86268.29283.00298.80316.505 A3x3 MA5-MA5x5 MA7-MA1234567891011121314Exercise 3.1: Smoothers fitted to the shipments data.151617
87Chapter 3: Time series decompositionThe graph on the previous page shows the five smoothers. Because moving averagesmoothers are “flat” at the ends, the best smoother in this case is the one with thesmallest number of terms, namely the 3-MA.3.2T̂t 13 15 (Yt 3 Yt 2 Yt 1 Yt Yt 1 ) 51 (Yt 2 Yt 1 Yt Yt 1 Yt 2 ) 51 (Yt 1 Yt Yt 1 Yt 2 Yt 3 )1211115 Yt 3 15 Yt 2 5 Yt 1 5 Yt 5 Yt 1 215 Yt 2 115 Yt 3 .3.3 (a) The 4 MA is designed to eliminate seasonal variation because each quarterreceives equal weight. The 2 MA is designed to center the estimated trend atthe data points. The combination 2 4 MA also gives equal weight to eachquarter.(b) T̂t 18 Yt 2 14 Yt 1 41 Yt 14 Yt 1 81 Yt 2 .3.4 (a) Use 2 4 MA to get trend. If the end-points are ignored, we obtain the followingresults.Data:Trend:Y1 Y2 Y3 Y4Y1Y2Y3Y4Q199 120 139 160Q1110.250 129.250 150.125Q288 108 127 148Q2114.875 134.500 154.750Q393 111 131 150Q3 100.375 119.635 138.875Q4 111 130 152 170Q4 105.500 124.375 145.125Data – trend:Y1Y2Q19.750Q2–6.875Q3 –7.375 9.875–6.500Ave9.792–7.042–8.2926.000(b) Hence, the seasonal indices are:Ŝ1 9.8, Ŝ2 7.0, Ŝ3 8.3 and Ŝ4 6.0.The seasonal component consists of replications of these indices.(c) End points ignored. Other approaches are possible.3.5 (a) See the top plot on the next page. There is clear trend which appears close tolinear, and strong seasonality with a peak in August–October and a trough inJanuary–March.(b) Calculations are given at the bottom of the next page. The decomposition plotis shown at the top of the next page.
88Part D. Solutions to exercises120011010470 80 1600Plastic sales2YearJFData17426972741700389679349518615 103010322 12 MA Trend12 1000.5 1011.23 1117.4 1121.54 1208.7 1221.35 1374.8 easonal 0114.0111.4123.077.991.3104.8116.1977.0 977.0 977.1 978.4 982.7 990.41054.4 1065.8 1076.1 1084.6 1094.4 1103.9 1112.51163.0 1170.4 1175.5 1180.5 1185.0 1190.2 1197.11276.6 1287.6 1298.0 1313.0 1328.2 1343.6 3.6Exercise 3.5(a) and (b): Multiplicative classical decomposition of plastic sales data.
89Chapter 3: Time series decomposition(c) The trend does appear almost linear except for a slight drop at the end. Theseasonal pattern is as expected. Note that it does not make much differencewhether these data are analyzed using a multiplicative decomposition or anadditive 19.2899.5383.59ForecastŶt Tt St .11859.11805.41515.31280.03.7 (a) See the top of the figure on the previous page.(b) The calculations are given 70766277254 2 MA12 399.3 413.33 478.3 499.64 557.9 580.65 654.8 670.66 689.4 708.1Ratios1295.799.0398.9 102.7497.5 100.2595.9 105.4691.0 102.4Seasonal indicesAve95.8 115.7112.1113.2114.587.985.188.388.787.4113.787.5
9060010590 00dataPart D. Solutions to exercises23456Exercise 3.7: Decomposition plot for exports from French company.(c) Multiplicative decomposition seems appropriate here because the variance isincreasing with the level of the series. The most interesting feature of thedecomposition is that the trend has levelled off in the last year or so. Anyforecast method should take this change in the trend into account.3.8 (a) The top plot shows the original data followed by trend-cycle, seasonal andirregular components. The bottom plot shows the seasonal sub-series.(b) The trend-cycle is almost linear and the small seasonal component is very smallcompared to the trend-cycle. The seasonal pattern is difficult to see in timeplot of original data. Values are high in March, September and December andlow in January and August. For the last six years, the December peak andMarch peak have been almost constant. Before that, the December peak wasgrowing and the March peak was dropping. There are several possible outliersin 1991.
91Chapter 3: Time series decomposition(c) The recession is seen by several negative outliers in the irregular component.This is also apparent in the data time plot. Note: the recession could be madepart of the trend-cycle component by reducing the span of the loess smoother.3.9 (a) and (b) Calculations are given below. Note that the seasonal indices arecomputed by averaging the de-trended values within each 0.962 2 Adjusted .974(c) With more data, we could take moving averages of the detrended values foreach half-year rather than a simple average. This would result in a seasonalcomponent which changed over time.
92Part D. Solutions to exercisesChapter 4: Exponential smoothing methods4.1Period DataMA(3)SES(α 0.7)tYtŶtEtŶtEt1974 115.4225.35.40-0.10335.35.33-0.03445.6 5.330.27 5.310.291975 156.9 5.401.50 5.511.39267.2 5.931.27 6.480.72377.2 6.570.63 6.990.21487.107.14Accuracy statistics from period 4 through l’s U1.401.14Theil’s U statistic suggests that the naı̈ve (or last value) method is better thaneither of these. If SES is used with an optimal value of α chosen, then α 1 isselected. This is equivalent to the naı̈ve method. Note different packages may giveslightly different results for SES depending on how they initialize the method. Somepackages will also allow α 1.4.2 (a) Forecasts for May 2MA(7)55.6757.5α 0.341.91211.80α 0.533.11193.98MA(9)51.7860.8MA(11)53.11313.8(b) Forecasts for May 1992MethodForecastMSEα 0.145.51421.35α 0.729.11225.40α 0.928.71298.49(c) Of these forecasting methods, the best MA(k) method has k 7 and the bestSES method has α 0.5. However, it should be noted that the MSE values forthe MA methods are taken over different periods. For example, the MSE for theMA(7) method is computed only over 9 observations because it is not possibleto compute an MA(7) forecast for the first seven observations. So the MSE
93Chapter 4: Exponential smoothing methodsvalues are not strictly comparable for the MA forecasts. It would be better touse a holdout sample but there are too few data.4.3 Optimizing α for SES over the period 3 through 910.822.086.170.919.804.921.017.864.00The optimal value is α 1.With Holt’s method, any combination of α and β will give MAPE 0. This is sobecause the differences between successive values of (4.13) are always going to bezero with this errorless series. Using α 1 for SES and α 0.5 and β 0.5 forHolt’s method gives the following results.DataYt2468101214161820SESŶt EtHolt’sŶt a) Clearly Holt’s method is better as it allows for the trend in the data.(b) For SES, α 1. Because of the trend, the forecasts will always lag behind theactual values so that the forecast errors will always be at least 2. Choosingα 1 makes the forecast errors as small as possible for SES.(c) See above.4.4 (a) (b) and (c) See the table on the following page.(d) There’s not much to choose between these methods. They are both bad! Lookat Theil’s U values for instance. The last value method over the same period(13–28) gives MSE 6.0, MAPE 2.05 and Theil’s U 1.0.
94Part D. Solutions to exercisesPeriod Data Forecast racy criteria: periods 13–28ME-1.51MAE3.12MSE14.40MAPE3.23Theil’s 0Calculations for Exercise 4.4-0.102.9612.643.031.45
95Chapter 4: Exponential smoothing methods4.5 (a) (b) and (c)Smoothing parametersForecast DayForecast DayForecast DayForecast DayMAEMSEMAPETheil’s U31323334PaperbacksSESHoltα 0.213 α 0.335β ESHoltα 0.347 β 0.437β .4327.328.61060.61273.013.514.30.810.92For both series, SES forecasting is performing better than Holt’s method.(d) SES forecasts are “flat” and Holt’s forecasts show a linear trend. Both seriesshow an upward linear trend and we would expect the forecasts to reflect thattrend. Perhaps an out-of-sample analysis would give a better indication of themerits of the two methods.(e) The autocorrelation functions of the forecast errors in each case are plotted onthe next page. In each case, there is no noticeable pattern. Only a few spikesare just outside the critical bounds which is expected.4.6 Here is a complete run for one set of values (β 0.1 and α 1 0.1). Note that in thisprogram we have chosen to make the first three values of α be equal to the startingvalue. This is not crucial, but it does make a 00.8200.4250.1970.4110.5550.4140.1660.165
96Part D. Solutions to exercises0.2-0.2-0.468101214246810LagLagSES hardbacksHolt 440.42ACF0.0ACF0.0-0.4-0.2ACF0.20.4Holt paperbacks0.4SES paperbacks246810121424Lag6810LagExercise 4.5 (e): Autocorrelation functions of forecast errors.For other combinations of values for β and starting values for α, here is what thefinal α value is:α 0.1α 0.2α 0.3β 0.10.1650.0580.140β 0.30.3270.4540.618β 0.50.7320.7830.797β 0.70.1430.1330.133The time series is not very long and therefore the results are somewhat fickle. Inany event, it is clear that the β value and the starting values for α have a profoundeffect on the final value of α.4.7 Holt-Winters’ method is best because the data are seasonal. The variation increaseswith the level, so we use Holt-Winters’ multiplicative method. The optimal smoothing parameters (giving smallest MSE) are a 0.479, b 0.00 and c 1.00. Thesegive the following forecasts (read left to right):
97Chapter 4: Exponential smoothing .1376.5324.4342.8361.2379.64.8 First choose any values for the three parameters. Here we have used α β γ 0.1.Different values will choose different initial values. Our program uses the methoddescribed in the textbook and gave the following eil’s U2.6Now compare with the optimal values: α 0.917, β 0.234 and γ 0.000. Usingthe same initialization, we obtain the results in Table 4-11, ��s U0.29
98Part D. Solutions to exercisesChapter 5: Simple regression5.1 (a) 0.7, almost 1, 0.2(b) False. The correlation is negative. So below-average values of one are associatedwith above-average values of the other variable.(c) Wages have been increasing over time due to inflation. At the same time,population has been increasing and consequently, new houses need to be built.So, because they are both increasing with time, they are positively correlated.(d) There are many factors affecting unemployment and it is simplistic to draw acausal connection with inflation on the basis of correlation. As in the previousquestion, both vary with time and the correlation could be induced by theirtime trends. Or they could both be related to some third variable such asbusiness confidence or government spending.(e) The older people in the survey had much less opportunity for education than theyounger people. This negative correlation is caused by the increase in educationlevels over time.PP5.2 (a) X̄ 5, Ȳ 25, (Xi X̄)2 20, (Xi X̄)(Yi Ȳ ) 78. So b 78/20 3.9and a 25 3.9(5) 5.5. Hence, the regression line is Ŷ 5.5 3.9X.pPP(b)(Ŷi Ȳ )2 304.20, (Yi Ŷi )2 25.80, σe 25.80/(5 2) 2.933. SoF 304.20/(2 1) 35.4.25.80/(5 2)This has (2 1) 1 df for the numerator and (5 2) 3 df for the denominator.From Table C in Appendix III, the P -value is slightly smaller than 0.010. (Usinga computer, it is 0.0095.) Standard errors:qs.e.(a) (2.93) 15 2520 3.53q1s.e.(b) (2.93) 20 0.656.On 3 df, t 3.18 for a 95% confidence interval. Hence 95% intervals areα:β:5.500 3.18(3.53) [ 5.7, 16.7]3.900 3.18(0.656) [1.8, 6.0]
99Chapter 5: Simple regressionOutput from Minitab for Exercise 5.2:Regression AnalysisThe regression equation isY 5.50 3.90XPredictorConstantXCoef5.5003.9000S 2.933StDev3.5310.6557R-Sq 92.2%T1.565.95P0.2170.010R-Sq(adj) 89.6%Analysis of 80330.00MS304.208.60F35.37P0.010(c) R2 0.922, rXY rY Ŷ 0.960.(d) The line through the middle of the graph is the line of best fit. The 95%prediction interval shown is the interval which would contain the Y value withprobability 0.95 if the X value was 17. The 80% prediction interval shown isthe interval which would contain the Y value with probability 0.80 if the Xvalue was 26. The dotted line at the boundary of the light shaded region givesthe ends of all the 95% prediction intervals. The dotted line at the boundaryof the dark shaded region gives the ends of all the 80% prediction intervals.5.3 (a) See the plot on the next page and the Minitab output on page 101. The straightline is Ŷ 0.46 0.22X.(b) See the plot on the next page. The residuals may show a slight curvature (Λshaped). However, the curvature is not strong and the fitted model appearsreasonable.(c) R2 90.2%. Therefore, 90.2% of the variation in melanoma rates is explainedby the linear regression.(d) From the Minitab output:Prediction: 9.286. Prediction interval: (6.749, 11.823)
100624melanoma810Part D. Solutions to 5Exercise 5.3(a): Scatterplot of melanoma rate against ozone depletion.10203040ozoneExercise 5.3(b): Scatterplot of residuals from the linear regression.
101Chapter 5: Simple regressionOutput using Minitab for Exercise 5.3:MTB Regress ’Melanoma’ 1 ’Ozone’;SUBC predict 40.The regression equation isMelanoma 0.460 0.221 OzonePredictorConstantOzoneCoef0.45980.22065S 0.9947StDev0.62580.02426R-Sq 90.2%T0.739.09P0.4810.000R-Sq(adj) 89.1%Analysis of Dev Fit0.517SS81.8228.90590.727(MS81.8220.98995.0% CI8.116, 10.456)F82.70(P0.00095.0% PI6.749, 11.823)Note that it is the prediction interval (PI) we want here. Minitab also givesthe confidence interval (CI) for the line at this point, something we have notcovered in the book.(e) This analysis has assumed that the susceptibility to melanoma among peopleliving in the various locations is constant. This is unlikely to be true due tothe diversity of racial mix and climate over the locations. Apart from ozonedepletion, melanoma will be affected by skin type, climate, culture (e.g. issun-baking encouraged?), diet, etc.5.4 (a) See plot on the next page and computer output on page 103.(b) Coefficients: a 4.184, b 0.9431. Only b is significant, showing the relationship is significant. (We could refit the model without the intercept term.)(c) If X 80, Ŷ 4.184 0.9431(80) 79.63. Standard error of forecast is 1.88(from computer output).
10230405060Production rating708090Part D. Solutions to exercises20406080Manual dexterity100Exercise 5.4(a): Scatterplot of production rating against manual dexterity te
2.1 (a) One simple answer: choose the mean temperature in June 1994 as the forecast for June 1995. That is, 17.2 C. (b) The time plot below shows clear seasonality with average temperature higher in summer. Month Celsius 1994 Jan 1994 Feb 1994 May 1994 Jul 1994 Sep
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Preface This manual contains solutions/answers to all exercises in the text Precalculus: Functions and Graphs, Thirteenth Edition, by Earl W. Swokowski and Jeffery A. Cole.A Student's Solutions Manualis also available; it contains solutions for the odd-numbered exercises in each section and for the Discussion Exercises, as well as solutions for all the exercises in the
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