Using graphs and Laplacian eigenvalues to evaluate block designs R. - PowerPoint PPT Presentation

Concurrence graph The concurrence graph G of a block design ∆ has ◮ one vertex for each treatment, ◮ one edge for each unordered pair α , ω , with α � = ω , g ( α ) = g ( ω ) (in the same block) and f ( α ) � = f ( ω ) : this edge joins vertices f ( α ) and f ( ω ) . There are no loops. 12/52

Concurrence graph The concurrence graph G of a block design ∆ has ◮ one vertex for each treatment, ◮ one edge for each unordered pair α , ω , with α � = ω , g ( α ) = g ( ω ) (in the same block) and f ( α ) � = f ( ω ) : this edge joins vertices f ( α ) and f ( ω ) . There are no loops. If i � = j then the number of edges between vertices i and j is b ∑ λ ij = n is n js ; s = 1 12/52

Concurrence graph The concurrence graph G of a block design ∆ has ◮ one vertex for each treatment, ◮ one edge for each unordered pair α , ω , with α � = ω , g ( α ) = g ( ω ) (in the same block) and f ( α ) � = f ( ω ) : this edge joins vertices f ( α ) and f ( ω ) . There are no loops. If i � = j then the number of edges between vertices i and j is b ∑ λ ij = n is n js ; s = 1 this is called the concurrence of i and j , and is the ( i , j ) -entry of Λ = NN ⊤ . 12/52

Example 1: v = 4 , b = k = 3 1 2 1 3 3 2 4 4 2 13/52

Example 1: v = 4 , b = k = 3 1 2 1 3 3 2 4 4 2 1 2 4 ✟ ① ❍❍❍❍❍ ① ① ✟ ❅ � ✟ ❅ � ✟ ✟ ✟ ❍ ❅ � ① ① � � � ❍ ✟ ❍ ✟✟✟✟✟ � ❅ ❍ 1 2 ❍ � ❅ ❍ ❍ � ❅ ① ① ① 3 3 4 Levi graph concurrence graph 13/52

Example 1: v = 4 , b = k = 3 1 2 1 3 3 2 4 4 2 1 2 4 ✟ ① ❍❍❍❍❍ ① ① ✟ ❅ � ✟ ❅ � ✟ ✟ ✟ ❍ ❅ � ① ① � � � ❍ ✟ ❍ ✟✟✟✟✟ � ❅ ❍ 1 2 ❍ � ❅ ❍ ❍ � ❅ ① ① ① 3 3 4 Levi graph concurrence graph can recover design may have more symmetry 13/52

Example 1: v = 4 , b = k = 3 1 2 1 3 3 2 4 4 2 1 2 4 ✟ ① ❍❍❍❍❍ ① ① ✟ ❅ � ✟ ❅ � ✟ ✟ ✟ ❍ ❅ � ① ① � � � ❍ ✟ ❍ ✟✟✟✟✟ � ❅ ❍ 1 2 ❍ � ❅ ❍ ❍ � ❅ ① ① ① 3 3 4 Levi graph concurrence graph can recover design may have more symmetry more vertices 13/52

Example 1: v = 4 , b = k = 3 1 2 1 3 3 2 4 4 2 1 2 4 ✟ ① ❍❍❍❍❍ ① ① ✟ ❅ � ✟ ❅ � ✟ ✟ ✟ ❍ ❅ � ① ① � � � ❍ ✟ ❍ ✟✟✟✟✟ � ❅ ❍ 1 2 ❍ � ❅ ❍ ❍ � ❅ ① ① ① 3 3 4 Levi graph concurrence graph can recover design may have more symmetry more vertices more edges if k = 2 more edges if k ≥ 4 13/52

Example 2: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 14/52

Example 2: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ① ① � ✁ ❆ ❅ � ❅ 1 2 ❅ � ✁ ❆ ❅ ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ① ① ❍❍ ✟ 8 6 ✟ ❍ ✟ ① ① ① 4 � ❆ ✁ ❅ � ❅ 4 3 ❆ ✁ ❅ � ❅ ① ① ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 14/52

Example 3: v = 15 , b = 7 , k = 3 1 1 2 3 4 5 6 1 1 1 1 1 1 1 2 4 5 6 10 11 12 2 4 6 8 10 12 14 3 7 8 9 13 14 15 3 5 7 9 11 13 15 8 9 13 7 8 11 7 ① ① 10 ❉ ☎☎ ① ① ① ① ① ❉ ① ❜ ✧ ❜ ✧ ✧✧✧✧ ❜ ✧✧✧✧ ❜ 4 5 � ❏ ✡ ❅ ❜ ❜ � ❏ ❉ ☎ ✡ ❅ ① ① ❜ ❜ ① ① 6 11 ◗ ✑ ◗ ❏ ❉ ✑✑✑✑ ☎ ✡ ❜ ❜ ① ① ① ① ◗ ❏ ❉ ☎ ✡ ❚ ✔ ❵ ✥ ❵ ✥✥✥✥✥ ❚ ✔ ◗ ① ❵ ① 5 12 ❵ 10 1 2 14 ❵ ◗ ❵ ❏ ❉ ☎ ✡ ✥ ❵❵❵❵❵ ❚ ✔ ✥ ① ✥ ① ✑ ◗◗◗◗ 3 ✥ ✥ ✡ ❏ ✥ ❵ ✔ ✔ ❚ ✑ ✔ ① ① 1 4 13 ✑ ✡ ❏ ✔ ❚ ✔ ✑ ① ① ① 12 9 ✑ ✡ ❏ ◗ ❚ ① ① ❚ ✔ 3 14 6 ❅ ✡ ❏ � ❚ ✔ ❅ ✡ ❏ � ① ① ① 15 2 15 15/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; ◮ L ii = valency of i = ∑ λ ij . j � = i 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; ◮ L ii = valency of i = ∑ λ ij . j � = i 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; ◮ L ii = valency of i = ∑ λ ij . j � = i The Laplacian matrix ˜ L of the Levi graph ˜ G is a ( v + b ) × ( v + b ) matrix with ( i , j ) -entry as follows: 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; ◮ L ii = valency of i = ∑ λ ij . j � = i The Laplacian matrix ˜ L of the Levi graph ˜ G is a ( v + b ) × ( v + b ) matrix with ( i , j ) -entry as follows: ◮ ˜ L ii = valency of i � k if i is a block = replication r i of i if i is a treatment 16/52

Laplacian matrices The Laplacian matrix L of the concurrence graph G is a v × v matrix with ( i , j ) -entry as follows: ◮ if i � = j then L ij = − (number of edges between i and j ) = − λ ij ; ◮ L ii = valency of i = ∑ λ ij . j � = i The Laplacian matrix ˜ L of the Levi graph ˜ G is a ( v + b ) × ( v + b ) matrix with ( i , j ) -entry as follows: ◮ ˜ L ii = valency of i � k if i is a block = replication r i of i if i is a treatment ◮ if i � = j then L ij = − (number of edges between i and j )  0 if i and j are both treatments   = 0 if i and j are both blocks  − n ij if i is a treatment and j is a block, or vice versa.  16/52

Connectivity All row-sums of L and of ˜ L are zero, so both matrices have 0 as eigenvalue on the appropriate all-1 vector. 17/52

Connectivity All row-sums of L and of ˜ L are zero, so both matrices have 0 as eigenvalue on the appropriate all-1 vector. Theorem The following are equivalent. 1. 0 is a simple eigenvalue of L; 2. G is a connected graph; 3. ˜ G is a connected graph; 4. 0 is a simple eigenvalue of ˜ L; 5. the design ∆ is connected in the sense that all differences between treatments can be estimated. 17/52

Connectivity All row-sums of L and of ˜ L are zero, so both matrices have 0 as eigenvalue on the appropriate all-1 vector. Theorem The following are equivalent. 1. 0 is a simple eigenvalue of L; 2. G is a connected graph; 3. ˜ G is a connected graph; 4. 0 is a simple eigenvalue of ˜ L; 5. the design ∆ is connected in the sense that all differences between treatments can be estimated. From now on, assume connectivity. 17/52

Connectivity All row-sums of L and of ˜ L are zero, so both matrices have 0 as eigenvalue on the appropriate all-1 vector. Theorem The following are equivalent. 1. 0 is a simple eigenvalue of L; 2. G is a connected graph; 3. ˜ G is a connected graph; 4. 0 is a simple eigenvalue of ˜ L; 5. the design ∆ is connected in the sense that all differences between treatments can be estimated. From now on, assume connectivity. Call the remaining eigenvalues non-trivial . They are all non-negative. 17/52

Generalized inverse Under the assumption of connectivity, the Moore–Penrose generalized inverse L − of L is defined by � − 1 � L + 1 − 1 L − = vJ v vJ v , where J v is the v × v all-1 matrix. (The matrix 1 vJ v is the orthogonal projector onto the null space of L .) 18/52

Generalized inverse Under the assumption of connectivity, the Moore–Penrose generalized inverse L − of L is defined by � − 1 � L + 1 − 1 L − = vJ v vJ v , where J v is the v × v all-1 matrix. (The matrix 1 vJ v is the orthogonal projector onto the null space of L .) L − of ˜ The Moore–Penrose generalized inverse ˜ L is defined similarly. 18/52

Estimation We measure the response Y ω on each experimenal unit ω . 19/52

Estimation We measure the response Y ω on each experimenal unit ω . If experimental unit ω has treatment i and is in block m ( f ( ω ) = i and g ( ω ) = m ), then we assume that Y ω = τ i + β m + random noise. 19/52

Estimation We measure the response Y ω on each experimenal unit ω . If experimental unit ω has treatment i and is in block m ( f ( ω ) = i and g ( ω ) = m ), then we assume that Y ω = τ i + β m + random noise. We will do an experiment, collect data y ω on each experimental unit ω , then want to estimate certain functions of the treatment parameters using functions of the data. 19/52

Estimation We measure the response Y ω on each experimenal unit ω . If experimental unit ω has treatment i and is in block m ( f ( ω ) = i and g ( ω ) = m ), then we assume that Y ω = τ i + β m + random noise. We will do an experiment, collect data y ω on each experimental unit ω , then want to estimate certain functions of the treatment parameters using functions of the data. We want to estimate contrasts ∑ i x i τ i with ∑ i x i = 0. 19/52

Estimation We measure the response Y ω on each experimenal unit ω . If experimental unit ω has treatment i and is in block m ( f ( ω ) = i and g ( ω ) = m ), then we assume that Y ω = τ i + β m + random noise. We will do an experiment, collect data y ω on each experimental unit ω , then want to estimate certain functions of the treatment parameters using functions of the data. We want to estimate contrasts ∑ i x i τ i with ∑ i x i = 0. In particular, we want to estimate all the simple differences τ i − τ j . 19/52

Variance: why does it matter? We want to estimate all the simple differences τ i − τ j . Put V ij = variance of the best linear unbiased estimator for τ i − τ j . The length of the 95% confidence interval for τ i − τ j is proportional to � V ij . 20/52

Variance: why does it matter? We want to estimate all the simple differences τ i − τ j . Put V ij = variance of the best linear unbiased estimator for τ i − τ j . The length of the 95% confidence interval for τ i − τ j is proportional to � V ij . (If we always present results using a 95% confidence interval, then our interval will contain the true value in 19 cases out of 20.) 20/52

Variance: why does it matter? We want to estimate all the simple differences τ i − τ j . Put V ij = variance of the best linear unbiased estimator for τ i − τ j . The length of the 95% confidence interval for τ i − τ j is proportional to � V ij . (If we always present results using a 95% confidence interval, then our interval will contain the true value in 19 cases out of 20.) The smaller the value of V ij , the smaller is the confidence interval, the closer is the estimate to the true value (on average), and the more likely are we to detect correctly which of τ i and τ j is bigger. 20/52

Variance: why does it matter? We want to estimate all the simple differences τ i − τ j . Put V ij = variance of the best linear unbiased estimator for τ i − τ j . The length of the 95% confidence interval for τ i − τ j is proportional to � V ij . (If we always present results using a 95% confidence interval, then our interval will contain the true value in 19 cases out of 20.) The smaller the value of V ij , the smaller is the confidence interval, the closer is the estimate to the true value (on average), and the more likely are we to detect correctly which of τ i and τ j is bigger. We can make better decisions about new drugs, about new varieties of wheat, about new engineering materials . . . if we make all the V ij small. 20/52

How do we calculate variance? Theorem Assume that all the noise is independent, with variance σ 2 . If ∑ i x i = 0 , then the variance of the best linear unbiased estimator of ∑ i x i τ i is equal to ( x ⊤ L − x ) k σ 2 . In particular, the variance of the best linear unbiased estimator of the simple difference τ i − τ j is � � L − ii + L − jj − 2 L − k σ 2 . V ij = ij 21/52

How do we calculate variance? Theorem Assume that all the noise is independent, with variance σ 2 . If ∑ i x i = 0 , then the variance of the best linear unbiased estimator of ∑ i x i τ i is equal to ( x ⊤ L − x ) k σ 2 . In particular, the variance of the best linear unbiased estimator of the simple difference τ i − τ j is � � L − ii + L − jj − 2 L − k σ 2 . V ij = ij (This follows from assumption Y ω = τ i + β m + random noise. by using standard theory of linear models.) 21/52

. . . Or we can use the Levi graph Theorem The variance of the best linear unbiased estimator of the simple difference τ i − τ j is � � L − ˜ ii + ˜ L − jj − 2˜ L − σ 2 . V ij = ij 22/52

. . . Or we can use the Levi graph Theorem The variance of the best linear unbiased estimator of the simple difference τ i − τ j is � � L − ˜ ii + ˜ L − jj − 2˜ L − σ 2 . V ij = ij (Or β i − β j , appropriately labelled.) 22/52

. . . Or we can use the Levi graph Theorem The variance of the best linear unbiased estimator of the simple difference τ i − τ j is � � L − ˜ ii + ˜ L − jj − 2˜ L − σ 2 . V ij = ij (Or β i − β j , appropriately labelled.) (This follows from assumption Y ω = τ i − ˜ β m + random noise. by using standard theory of linear models.) 22/52

How do we calculate these generalized inverses? We need L − or ˜ L − . 23/52

How do we calculate these generalized inverses? We need L − or ˜ L − . ◮ Add a suitable multiple of J , use GAP to find the inverse with exact rational coefficients, subtract that multiple of J . 23/52

How do we calculate these generalized inverses? We need L − or ˜ L − . ◮ Add a suitable multiple of J , use GAP to find the inverse with exact rational coefficients, subtract that multiple of J . ◮ If the matrix is highly patterned, guess the eigenspaces, then invert each non-zero eigenvalue. 23/52

How do we calculate these generalized inverses? We need L − or ˜ L − . ◮ Add a suitable multiple of J , use GAP to find the inverse with exact rational coefficients, subtract that multiple of J . ◮ If the matrix is highly patterned, guess the eigenspaces, then invert each non-zero eigenvalue. ◮ Direct use of the graph: coming up. 23/52

How do we calculate these generalized inverses? We need L − or ˜ L − . ◮ Add a suitable multiple of J , use GAP to find the inverse with exact rational coefficients, subtract that multiple of J . ◮ If the matrix is highly patterned, guess the eigenspaces, then invert each non-zero eigenvalue. ◮ Direct use of the graph: coming up. 23/52

How do we calculate these generalized inverses? We need L − or ˜ L − . ◮ Add a suitable multiple of J , use GAP to find the inverse with exact rational coefficients, subtract that multiple of J . ◮ If the matrix is highly patterned, guess the eigenspaces, then invert each non-zero eigenvalue. ◮ Direct use of the graph: coming up. Not all of these methods are suitable for generic designs with a variable number of treatments. 23/52

Electrical networks We can consider the concurrence graph G as an electrical network with a 1-ohm resistance in each edge. Connect a 1-volt battery between vertices i and j . Current flows in the network, according to these rules. 1. Ohm’s Law: In every edge, voltage drop = current × resistance = current. 2. Kirchhoff’s Voltage Law: The total voltage drop from one vertex to any other vertex is the same no matter which path we take from one to the other. 3. Kirchhoff’s Current Law: At every vertex which is not connected to the battery, the total current coming in is equal to the total current going out. Find the total current I from i to j , then use Ohm’s Law to define the effective resistance R ij between i and j as 1/ I . 24/52

Electrical networks: variance Theorem The effective resistance R ij between vertices i and j in G is � � L − ii + L − jj − 2 L − R ij = . ij 25/52

Electrical networks: variance Theorem The effective resistance R ij between vertices i and j in G is � � L − ii + L − jj − 2 L − R ij = . ij So V ij = R ij × k σ 2 . 25/52

Electrical networks: variance Theorem The effective resistance R ij between vertices i and j in G is � � L − ii + L − jj − 2 L − R ij = . ij So V ij = R ij × k σ 2 . Effective resistances are easy to calculate without matrix inversion if the graph is sparse. 25/52

Example 2 calculation: v = 8 , b = 4 , k = 3 5 ③ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✟ ③ ③ ❍ ✟✟✟✟✟ ❍ ❍ 1 2 ❍ ❍ ❍ ③ ❍❍❍❍❍ ✟ ③ 8 6 ✟ ✟ ✟ 4 3 ✟ ❍ ✟ ③ ③ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ③ 7 26/52

Example 2 calculation: v = 8 , b = 4 , k = 3 5 ③ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✟ ③ ③ ❍ ✟✟✟✟✟ ❍ ❍ 1 2 ❍ ❍ ❍ ❍❍❍❍❍ ③ ✟ ③ 8 6 ✟ ✟ ✟ 4 3 ✟ ❍ ✲ ✟ ③ ③ ❆ ✁ [ 0 ] 10 [ 10 ] ❆ ✁ ❆ ✁ ❯ 5 ✕ 5 ❆ ✁ ❆ ✁ ❆ ✁ ③ [ 5 ] 7 26/52

Example 2 calculation: v = 8 , b = 4 , k = 3 5 ③ [ 9 ] ✁ ❆ ✁ ❆ ❯ 3 ✁ ❆ 3 ✕ ✁ ❆ [ 6 ] ✁ ❆ [ 12 ] 6 ✁ ✲ ❆ ✟ ③ ③ ❍ ✟✟✟✟✟ 3 ❍ ❍ 1 2 ❍ ✯ ❍ ❍ ❍❍❍❍❍ ③ ✟ ③ 8 6 6 ✻ ✟ ✟ [ 3 ] ❨ ✟ 4 3 ✟ ❍ ✲ ✟ 3 ③ ③ ❆ ✁ [ 0 ] 10 [ 10 ] ❆ ✁ ❆ ✁ ❯ 5 ✕ 5 ❆ ✁ ❆ ✁ ❆ ✁ ③ [ 5 ] 7 26/52

Example 2 calculation: v = 8 , b = 4 , k = 3 5 ③ [ 9 ] ✁ ❆ ✁ ❆ ❯ 3 ✁ ❆ 3 ✕ ✁ ❆ [ 6 ] ✁ ❆ [ 12 ] 6 ✁ ✲ ❆ ✟ ③ ❍ ③ ✟✟✟✟✟ 3 ❍ 11 ❍ ❥ 1 2 ❍ ✯ ❍ ❍ ❍❍❍❍❍ ③ ✟ ③ 8 6 2 6 ✻ ✻ ✟ ✟ [ 3 ] ❨ ✯ [ 23 ] ✟ 4 3 ✟ ❍ ✲ ✟ 3 ③ ③ 13 ❆ ✁ [ 0 ] 10 [ 10 ] ❆ ✁ ❆ ✁ ❯ 5 ✕ 5 ❆ ✁ ❆ ✁ ❆ ✁ ③ [ 5 ] 7 26/52

Example 2 calculation: v = 8 , b = 4 , k = 3 R = 23 V = 23 I = 24 24 5 ③ [ 9 ] ✁ ❆ ✁ ❆ ❯ 3 ✁ ❆ 3 ✕ ✁ ❆ [ 6 ] ✁ ❆ [ 12 ] 6 ✁ ✲ ❆ ✟ ③ ③ ❍ ✟✟✟✟✟ 3 ❍ 11 ❍ ❥ 1 2 ❍ ✯ ❍ ❍ ③ ❍❍❍❍❍ ✟ ③ 8 6 2 6 ✻ ✻ ✟ ✟ [ 3 ] ❨ ✯ [ 23 ] ✟ 4 3 ✟ ❍ ✲ ✟ 3 ③ ③ 13 ❆ ✁ [ 0 ] 10 [ 10 ] ❆ ✁ ❆ ✁ ❯ 5 ✕ 5 ❆ ✁ ❆ ✁ ❆ ✁ ③ [ 5 ] 7 26/52

. . . Or we can use the Levi graph If i and j are treatment vertices in the Levi graph ˜ G and ˜ R ij is the effective resistance between them in ˜ G then V ij = ˜ R ij × σ 2 . 27/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ① ① � ✁ ❆ ❅ � ❅ 1 2 ❅ � ✁ ❆ ❅ ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ① ① ❍❍ ✟ 8 6 ✟ ❍ ✟ ① ① ① 4 � ❆ ✁ ❅ � ❅ 4 3 ❆ ✁ ❅ � ❅ ① ① ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ✲ ① ① � ✁ ❆ ❅ � ❘ ❅ 1 2 ❅ � ✁ ❆ ✒ ❅ ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ❄ ① ① ❍❍ ✟ 8 6 ✟ ✻ ❍ ✟ ① ① ① 4 � ❆ ✁ ❅ � ❅ 4 3 ❆ ✁ ❅ � ❅ ① ① ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ✲ ① ① � ✁ ❆ ❅ � ❘ 3 ❅ 1 2 ❅ � ✁ ❆ ✒ ❅ 3 3 ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ❄ ① ① ❍❍ ✟ 8 6 3 3 ✟ ✻ ❍ ✟ ① ① [ 0 ] ① [ 15 ] 4 � ❆ ✁ ❅ � ❅ 4 3 ❆ ✁ ❅ � ❅ ① ① ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ✲ ① ① � ✁ ❆ ❅ � ❘ 3 ❅ 1 2 ❅ � ✁ ❆ ✒ ❅ 3 3 ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ❄ ① ① ❍❍ ✟ 8 6 3 3 ✟ ✻ ❍ ✟ ① ① [ 0 ] ① [ 15 ] 4 � ❆ ✁ ❅ ❘ 5 5 � ❅ 4 3 ❆ ✁ ❅ 5 ✲ ✒ � ❅ ① ① ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 2 3 4 1 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ✲ ① ① � ✁ ❆ ❅ � ❘ 3 ❅ 1 2 ❅ � ✁ ❆ ✒ ❅ 3 3 ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ❄ ① ① ❍❍ ✟ 8 6 3 3 ✟ ✻ ❍ ✟ ① ① [ 0 ] ① [ 15 ] 4 � ❆ ✁ ❅ ❘ 5 5 ❘ � ❅ 4 3 ❆ ✁ ❅ 5 ✲ ✒ � ❅ 8 ① ① [ 23 ] ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Example 2 yet again: v = 8 , b = 4 , k = 3 1 2 3 4 R = 23 ˜ V = 23 I = 8 2 3 4 1 8 5 6 7 8 5 ① 5 � 8 1 ① � ✁ ❆ ✲ ① ① � ✁ ❆ ❅ � ❘ 3 ❅ 1 2 ❅ � ✁ ❆ ✒ ❅ 3 3 ① ① ✟ ❍ ① 2 � ✟✟ ❍ ❍ ❄ ① ① ❍❍ ✟ 8 6 3 3 ✟ ✻ ❍ ✟ ① ① [ 0 ] ① [ 15 ] 4 � ❆ ✁ ❅ ❘ 5 5 ❘ � ❅ 4 3 ❆ ✁ ❅ 5 ✲ � ✒ ❅ 8 ① ① [ 23 ] ❆ ✁ � ① � 3 6 � 7 ① 7 Levi graph concurrence graph 28/52

Optimality: Average pairwise variance The variance of the best linear unbiased estimator of the simple difference τ i − τ j is k σ 2 = R ij k σ 2 . � � L − ii + L − jj − 2 L − V ij = ij 29/52

Optimality: Average pairwise variance The variance of the best linear unbiased estimator of the simple difference τ i − τ j is k σ 2 = R ij k σ 2 . � � L − ii + L − jj − 2 L − V ij = ij We want all of the V ij to be small. 29/52

Optimality: Average pairwise variance The variance of the best linear unbiased estimator of the simple difference τ i − τ j is k σ 2 = R ij k σ 2 . � � L − ii + L − jj − 2 L − V ij = ij We want all of the V ij to be small. Put ¯ V = average value of the V ij . Then V = 2 k σ 2 Tr ( L − ) 1 = 2 k σ 2 × ¯ , v − 1 harmonic mean of θ 1 , . . . , θ v − 1 where θ 1 , . . . , θ v − 1 are the nontrivial eigenvalues of L . 29/52