Multi-branch tree を用いた予算配分法において評価者の与える評価値の揺らぎが組織全体の予算配分に与える影響に関する考察 : Pythonによる揺らぎ抽出アルゴリズムの構築とシミュレーションによる分析評価

Multi-branch tree を用いた予算配分法において

評価者の与える評価値の揺らぎが組織全体の

予算配分に与える影響に関する考察

－ Python による揺らぎ抽出アルゴリズムの構築とシミュレーションによる分析評価－

田 島 博 之　

2

Multi-branch tree ŬÀĻŐ ÛĜ.®

řľĻŔĂũĂ"Ś‹ŧ

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3

2．T-method による Algorithm

本節ではT-method について説明する。正確な解法の定義は文献[4]の第2 節にあるが、ここでは 本論を理解するために、単純なSample を使って異なる説明を行う。 Figure1.では𝑒𝑒𝑖𝑖(𝑖𝑖 = 1,2, … ,9)が評価者を表し、𝑝𝑝𝑗𝑗(𝑗𝑗 = 1,2, … ,10)を評価対象物として表す。各評 価者は、直下にある１本のブランチに対して、そのブランチ以下に繋がる評価対象物の中から１つ を選択し、相対評価を行う。例えば、評価者_𝑒𝑒1は左のブランチによって、_𝑒𝑒2に従属している ｛_{𝑝𝑝1, 𝑝𝑝2, 𝑝𝑝3, 𝑝𝑝4}｝から１つ。さらに右のブランチからは、_𝑒𝑒3に従属している｛_{𝑝𝑝5, 𝑝𝑝6, 𝑝𝑝7, 𝑝𝑝8, 𝑝𝑝9, 𝑝𝑝10}｝から１ つの評価対象物を選択し、相対評価値を与える。同様に評価者_𝑒𝑒2は、左のブランチから_𝑒𝑒4に従属 する｛_{𝑝𝑝1, 𝑝𝑝2}｝から１つ、右のブランチから_𝑒𝑒5に従属する｛_{𝑝𝑝3, 𝑝𝑝4}｝から１つを選択し、それぞれに相対 評価値を与える。

Figure1. Sample Tree

Figure1.に示されるツリー構造における各評価者と評価対象物の選択肢の関係を以下に示す。 𝑒𝑒1: {{𝑝𝑝1, 𝑝𝑝2, 𝑝𝑝3, 𝑝𝑝4}, {𝑝𝑝5, 𝑝𝑝6, 𝑝𝑝7, 𝑝𝑝8, 𝑝𝑝9, 𝑝𝑝10}} 𝑒𝑒2: {{𝑝𝑝1, 𝑝𝑝2}, {𝑝𝑝3, 𝑝𝑝4}} 𝑒𝑒3: {{𝑝𝑝5}, {𝑝𝑝6, 𝑝𝑝7, 𝑝𝑝8, 𝑝𝑝9, 𝑝𝑝10}} 𝑒𝑒4: {{𝑝𝑝1}, {𝑝𝑝2}} 𝑒𝑒5: {{𝑝𝑝3}, {𝑝𝑝4}} 𝑒𝑒6: {{𝑝𝑝5}} 𝑒𝑒7: {{𝑝𝑝6, 𝑝𝑝7}, {𝑝𝑝8, 𝑝𝑝9, 𝑝𝑝10}} 𝑒𝑒8: {{𝑝𝑝6}, {𝑝𝑝7}} 𝑒𝑒9: {{𝑝𝑝8}, {𝑝𝑝9}, {𝑝𝑝10}} 評価者_{𝑒𝑒𝑖𝑖 が評価対象物𝑝𝑝𝑗𝑗}に対して与える相対評価値を_{𝑣𝑣(𝑖𝑖,𝑗𝑗)}とする。上の選択肢から各評価者 が選択した評価対象物をTable1.に示す。

4 Table 1．Evaluation table

Table1.では評価者が選択した評価物を表している。そこで、この評価物に対する評価値𝑣𝑣(𝑖𝑖,𝑗𝑗)に 従って評価対象物_{𝑝𝑝𝑖𝑖(𝑖𝑖 = 1,2, … ,10)に対する、予算配分𝑊𝑊 = (𝑤𝑤1, 𝑤𝑤2, … , 𝑤𝑤10)を求める演算を行う。} 初めに、_{𝑝𝑝𝑖𝑖(𝑖𝑖 = 1,2, … ,10)に対して T-method のルール[1]}に従い、それぞれの評価者が与える相 対評価値_{𝑣𝑣(𝑖𝑖,𝑗𝑗)}から以下のようにウェイト

ｓ

𝑖𝑖(𝑖𝑖 = 1,2, … ,10)を算出する。 𝑝𝑝1: 𝑠𝑠1= 𝑣𝑣(1,1) 𝑝𝑝2: 𝑠𝑠2=𝑣𝑣(1,1)_𝑣𝑣 ∙ 𝑣𝑣(4,2) (4,1) 𝑝𝑝3: 𝑠𝑠3=𝑣𝑣(1,1)_𝑣𝑣 ∙ 𝑣𝑣(2,3) (2,1) 𝑝𝑝4: 𝑠𝑠4=𝑣𝑣(1,1)_𝑣𝑣 ∙ 𝑣𝑣(2,3)∙ 𝑣𝑣(5,4) (2,1)∙ 𝑣𝑣(5,3) 𝑝𝑝5: 𝑠𝑠5=𝑣𝑣(1,9)_𝑣𝑣 ∙ 𝑣𝑣(9,8)∙ 𝑣𝑣(7,6)∙ 𝑣𝑣(8,7)∙ 𝑣𝑣(3,5) (9,9)∙ 𝑣𝑣(7,8)∙ 𝑣𝑣(8,6)∙ 𝑣𝑣(3,7) 𝑝𝑝6: 𝑠𝑠6=𝑣𝑣(1,9)_𝑣𝑣 ∙ 𝑣𝑣(9,8)∙ 𝑣𝑣(7,6) (9,9)∙ 𝑣𝑣(7,8) 𝑝𝑝7: 𝑠𝑠7=𝑣𝑣(1,9)_𝑣𝑣 ∙ 𝑣𝑣(9,8)∙ 𝑣𝑣(7,6)∙ 𝑣𝑣(8,7) (9,9)∙ 𝑣𝑣(7,8)∙ 𝑣𝑣(8,6) 𝑝𝑝8: 𝑠𝑠8=𝑣𝑣(1,9)_𝑣𝑣 ∙ 𝑣𝑣(9,8) (9,9) 𝑝𝑝9: 𝑠𝑠9= 𝑣𝑣(1,9) 𝑝𝑝10: 𝑠𝑠10=𝑣𝑣(1,9)_𝑣𝑣∙ 𝑣𝑣(9,10) (9,9) 次にS=∑10_{𝑗𝑗=1𝑠𝑠𝑗𝑗}による正規化を行い以下に予算配分を決定する。 𝑤𝑤𝑗𝑗 =𝑠𝑠_𝑆𝑆𝑗𝑗(𝑗𝑗 = 1,2, … ,10) 本章ではT-method により、多階層からなる 9 人の評価者が相対評価を 9 回行うことで予算配分 を行う例を紹介した。次章では新たなツリー構造を設定し、T-method における１人の評価者の評価 値の揺らぎが、全体の予算配分の変化に与える影響についてシミュレーションを行う。

5 3.
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7 ŕĺũşŗĵWĞ_{!ĿŧYŃŚuĦŬBń!Ś.nŀĭŃŘŒŔĻũĶ} ņŚŦļřģhŀřĺũĂîŚW<Ğ_{!Ś"śW<Ğ!řsĻuĦŬĽĵ)Ś Û} Ĝ.ŚW<ĞŬŅŔĻũņŖŀ“ŧĿřŘŒŐĶ 4. )ŚćlĞŬ—dřŋũſƓƘ¤ėŚ‰` 3Öŕś
ŚĂîŚÍcĂ"ŬW<ňōŐŖŁĵĂîŚģhŀřĺũşŗ <sub>!, !</sub> ŚÍġśsŃŘŨĵW<Ğ<sub>!Ś"ŀV7ŋũņŖŀ.ĿŒŐĶņņŕśģhŽŬ2 ≤ ℎ ≤ 6ĵšŐƉƒ</sub> ƗŽŚŽŬ<sub>2 ≤ ! ≤ 6ŖÝOŬĢ`ʼnŐ! ℎ, ! řcʼnŔĵStepwise method ŬÀĻŔYĝMoƏƁ</sub> ƔŬ¤ÞŋũĶņŪřŦŒŔG`ňŪũƏƁƔśĵĂcĎ¹ŀ₂!_{= 4 Ŀŧ6}!_{= 46656 šŕ} ŬcĎŖŋũņŖŀŕŁũĶ_{! ℎ, ! řľĻŔ—h
ÆkŚĂîŚÍcĂ"ŬW<ňōĵ} WĞ_{!Ś"ŬþÛŋũĶ *ÈŘ ℎ, ! ŚâŢFūōś 25 ĖŨĺũŚŕĵ ŎŪŏŪ ! = 1,2, …,} 40000 ŚŶƌƐƕƘŶƑƗŬõĻĵ40,000 MIJ25 ĖŨ=1,000,000 MŚăõŬõļĶ

ņŪŧŚ1,000,000 âŚƁƘŻŬŤŖřĵStepwise method ŬÀĻŔYĝMoƏƁƔŬ¤ÞŋũĶ ›ƏƁƔŕś_{ℎ, !ĵÀħŕĺũℎ ∙ !ĿŧŘũ 3 œŚN]ŬĈ“WŽŖʼnŔÀ€ʼnŐĶ} šŐĵËÈWŽŬW<Ğ_{!ŖʼnŐĶņŚŚŤŖřĵStepwise method ŬÀĻŔYĝMoƏƁƔ} Ŭ¬ţũĶþÛŚä Ŭ_{Table2.řÑŋĶEN]Ś!"śĵĻŌŪŤ 0 ŖŘŨ 3 œŚN]ŀĵĻŌŪ} Ť˜9ŕĺũņŖŀÑňŪŐĶ÷¨R2 Ś"ś 0.358ĵĝÍġ R Ś"ś 0.641 ŖĭĻĘF¼ŀÐĿ ţŧŪŐĶ

Table 2ƛStepwise method řŦũYĝMoƏƁƔ

Stepwise method řŦŒŔyŧŪŐYĝMoƏƁƔŬ§rřĽũĶ

_{= −0.0379ℎ − 0.0286! + 0.00194ℎ ∙ ! + 0.315} ℎ, ! Ś"ŀ’ÎŕĺũŖŁĵņŚĝMorŬļņŖřŦŒŔ—hřĻũ
ŚĂîŚ Ă"ƁƘŻŚ‹ŧłřcŋũ_{!ŚW<ĞŬ ³ŋũņŖŀCðŖŘũĶ} §řĵĂcĎ¹ŚçŽŬ_{!ĵſƓƘŚĭňŬℎŖʼnŐŖŁĵƉƒƗŽŚŽ! = !}!!ŖöōũĶņŪŬ rř(ŋũņŖŕĂcĎ¹Ś Ž_{!ŖĵģhŚĭňℎŬWŽŖʼnŐġŽ}   _{!, ℎ ŖŋũĶ}    _{!, ℎ = −0.0379ℎ − 0.0286!}!!_{+ 0.00194ℎ ∙ !}!!_{+ 0.315} ņŪřŦŒŔĵĂcĎ¹Ś Žŀ<sub>!Ŗ­`ʼnŔĻũŖŁĵ2 ≤ ℎ ≤ 6ŚÝOřľĻŔ</sub>   <sub>!, ℎ Ŭ</sub> —dŖŋũℎŚ"Ŭę…ŋũņŖřŦŒŔĵ—hŚĂŧł<sub>!ŀ)Ś ÛĜ.řcʼn</sub> Ŕ—ŤuĦŬ@ŠňŘĻſƓƘ¤ėŬ­`ŋũņŖŀCðŖŘũĶ

8 ıİį šŖţ į į øŽŚcĎ¹řcʼnŔ ÛŬĜ.ŋũĂŚ‘®řœĻŔśĵņŪšŕś _{Analytic Hierarchy} ProcessƙAHPƚ[10]Ɵ[14]ŀ%Òؑ®ŖʼnŔ°ÀňŪŔŁŐĶšŐĵAHP ŕśøŽŚĂîřŦũ‘ ®ŤÏÕňŪŔĻũ_[15]ĶĻŌŪŚƏƁƔřľĻŔŤĵĂîŀĂcĎ¹Ŭ
cªĔŋũņŖŕ ĂcĎ¹řcŋũĂ"ŬÛ-ŋũņŖřŘũĶʼnĿʼnŘŀŧĵ_{AHP řśxžĿŧĂcĎ¹} Ś V 7 ř  Ļ ª Ĕ Ă  Ŭ õ ļ M Ž ŀ Ĭ Ē È ř V 7 ʼn Ŕ ʼn š ļ Ŗ Ļ ļ J Ī ŀ ĺ Œ Ő Ķ Ŏ Ś ¶ T-method śĵĂîŬſƓƘ¤ėřĜêʼnŔEĂîŀñēřgŋũƉƒƗŽŚŽőńĂŋ ŪŜŦŃĵĂřĿĿũŴŸƂŚ²řZĻřď»ʼnŔĻũĶʼnĿʼnŘŀŧĵ
ŚĽũÍcĂ Ś"ŀ)Ś ÛĜ.řŗŚÓqŚuĦŬĽũĿřœĻŔŚý@śňŪŔĻŘĿŒŐĶ

Ŏņŕ›ċŕśĵºñŚĂġŽŬ2 œ`ìʼnĵMonte Carlo method ŬÀĻŐŶƌƐƕƘŶƑƗ řŦũ£āŬõŒŐĶä ŖʼnŔĵĂîŚĂ"ŚW;ŀƏƁƔŚģhřĺũşŗĵ Û )řĽũW;ŀZŁĻņŖŬ“ŧĿřʼnŐĶņŪśĵT-method řľńũĂŋŞŁſƓƘ¤ė řľńũĵŎŪŏŪŚģhŕĂîŚĂ"ŀĘ/ř ÛĜ.ňŪũŖĻļņŖŕĺŨĵâéřľ ńũîŚĐŚĝňŚŽĞ<ř‚6ʼnŐŖýĽũĶ §řĵ_{2 ≤ ! ≤ 6
2 ≤ ℎ ≤ 6ŖŘũÝOřľńũ! ℎ, ! Ŀŧĵ—ģhĿŧ
ŚĂîŬ} †-ʼnĵŎŚ‹ŧłŬ¬ţũŶƌƐƕƘŶƑƗŬõŒŐĶņņŕyŧŪŐ_{1,000,000 ŚƁƘŻźžƂĿŧ} Stepwise method ŬÀĻŔĝÍġŚ R "ŀ 0.641 ŖŘũYĝMoƏƁƔŬyŐĶ ņŪřŦŒŔĵ T-method ŬÀĻũĤŚ—ŧłĿŧ¿Ŋũ)Ś ÛĜ.)řĽũuĦŬŽĞ< ʼnŔ ³ŋũņŖŬCðřʼnŐĶňŧřĵMoƏƁƔŬ¤ÞŋũņŖřŦŒŔĵĂcĎ¹Ś Žŀ VZʼnŐŖŁřĵT-method ŬÀĻũĤŚ¾ÈŘƉƒƗŽŽŖģhŽŬ ³CðřʼnŐĶ wśĵ›ÏÕŬSÊŖʼnŐ,ğƒÅŬíbŋũŖŖŤřĵT-method ŬŦŨ½aÈŘƏƁƔřŋ ũŐţřĵŶƌƐƕƘŶƑƗř2ÀŋũŽġŽŚę…ťĵřĘʼnŐƏƁƔřcŋũíbŀ{ ùřŘũŖíĽũĶ Č ĕ ›ċŬRÙŋũřĺŐŒŔĵ5Z_ãK_ěį iÁ:'¿řśĵŽYŃŚ˜ÉŘŇ8ýŬđ ŨšʼnŐĶņŚUŬľ!ŨʼnšʼnŔĵzŦŨČŚ€ŬöʼnšŋĶ

9

? í  »

[1] Melkers, J.; Willoughby, K. Models of Performance-Measurement Use in Local Governments: Understanding Budgeting, Communication, and Lasting Effects. Publ. Adm. Rev. 2005, 65, 180-190.

[2] Mousavi-Nasab, S.; Safari, J.; Hafezalkotob, A. Resource allocation based on overall equipment effectiveness using cooperative game. Int. J. Cybern. Syst. Manag. Sci. 2019, 49, 819-834.

[3] Nasseri, S.H.; Baghban1, A.; Mahdavi, I. A new approach for solving fuzzy multi-objective quadratic programming of water resource allocation problem. J. Ind. Eng. Manag. 2019, 6, 78-102.

[4] Oyamaguchi, N.; Tajima, H.; Okada, I. Model of Multi-branch Trees for Efficient Resource Allocation. Algorithms 2020, 13, 55.

[5] Oyamaguchi, N.; Tajima, H.; Okada, I. Tournament Method Using a Tree Structure to Resolve Budget Conicts. In Intelligent Decision Technologies 2019, Smart Innovation, Systems and Technologies, vol193; Springer, Singapore, 2020, 525-532.

[6] Cameron.B Edward P, Daniel W, Simon L, Peter I. C, Philipp R,Stephen Tavener, Diego Perez, Spyridon Samothrakis and Simon Colton A Survey of Monte Carlo Tree Search Methods IEEE TRANSACTIONS ON COMPUTATIONAL INTELLIGENCE AND AI IN GAMES, VOL. 4, NO. 1, MARCH, 2012,1-43

[7] Robert S.; Jian-Sheng W., Nonuniversal Critical Dynamics in Monte Carlo Simulation, Physical Review Letters 58(2),February 1987,86-88

[8] Bruce T. Stepwise Regression and Stepwise Discriminant Analysis Need Not Apply here: A Guidelines Editorial Educational and Psychological Measurement Volume: 55 issue: 4, 1995, 525-534

[9] Michael C.; A. Afifi, Comparison of Stopping Rules in Forward Stepwise Discriminant Analysis, Journal of the American Statistical Association, Vol. 74, No. 368. Dec., 1979, pp. 777-785.

[10] Thomas L. Saaty Decision making with the analytic hierarchy process Int. J. Services Sciences, Vol. 1, No. 1, 2008, 83-98

[11] JOSÉ A.,TERESA L. CONSISTENCY IN THE ANALYTIC HIERARCHY PROCESS:A NEW APPROACH, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems Vol. 14, No. 4, 2006, 445-459

[12] Thomas L. Saaty How to make a decision:The Analytic Hierarchy Process, European Journal of Operational Research 48, 1990, 9-26

[13] Jiří F.Aleš K. Judgment scales and consistency measure in AHP, ScienceDirect Procedia Economics and Finance 12 , 2014 , 164- 173

[14] Evangelos T.,Stuart H. M. USING THE ANALYTIC HIERARCHY PROCESS FOR DECISION MAKING IN ENGINEERING APPLICATIONS: SOME CHALLENGES,Inter’l, Journal of Industrial Engineering: Applications and Practice, Vol. 2, No. 1, 1995, 35-44 [15] María Teresa Escobar & José María Moreno-jiménez Aggregation of Individual Preference

Structures in Ahp-Group Decision Making Group Decision and Negotiation volume 16, 2007, 287-301

9 [1] Melkers, J.; Willoughby, K. Models of Performance-Measurement Use in Local Governments:

Understanding Budgeting, Communication, and Lasting Effects. Publ. Adm. Rev. 2005, 65, 180-190.

[2] Mousavi-Nasab, S.; Safari, J.; Hafezalkotob, A. Resource allocation based on overall equipment effectiveness using cooperative game. Int. J. Cybern. Syst. Manag. Sci. 2019, 49, 819-834.

[3] Nasseri, S.H.; Baghban1, A.; Mahdavi, I. A new approach for solving fuzzy multi-objective quadratic programming of water resource allocation problem. J. Ind. Eng. Manag. 2019, 6, 78-102.

[4] Oyamaguchi, N.; Tajima, H.; Okada, I. Model of Multi-branch Trees for Efficient Resource Allocation. Algorithms 2020, 13, 55.

[5] Oyamaguchi, N.; Tajima, H.; Okada, I. Tournament Method Using a Tree Structure to Resolve Budget Conicts. In Intelligent Decision Technologies 2019, Smart Innovation, Systems and Technologies, vol193; Springer, Singapore, 2020, 525-532.

[6] Cameron.B Edward P, Daniel W, Simon L, Peter I. C, Philipp R,Stephen Tavener, Diego Perez, Spyridon Samothrakis and Simon Colton A Survey of Monte Carlo Tree Search Methods IEEE TRANSACTIONS ON COMPUTATIONAL INTELLIGENCE AND AI IN GAMES, VOL. 4, NO. 1, MARCH, 2012,1-43

[7] Robert S.; Jian-Sheng W., Nonuniversal Critical Dynamics in Monte Carlo Simulation, Physical Review Letters 58(2),February 1987,86-88

[8] Bruce T. Stepwise Regression and Stepwise Discriminant Analysis Need Not Apply here: A Guidelines Editorial Educational and Psychological Measurement Volume: 55 issue: 4, 1995, 525-534

[9] Michael C.; A. Afifi, Comparison of Stopping Rules in Forward Stepwise Discriminant Analysis, Journal of the American Statistical Association, Vol. 74, No. 368. Dec., 1979, pp. 777-785.

[10] Thomas L. Saaty Decision making with the analytic hierarchy process Int. J. Services Sciences, Vol. 1, No. 1, 2008, 83-98

[11] JOSÉ A.,TERESA L. CONSISTENCY IN THE ANALYTIC HIERARCHY PROCESS:A NEW APPROACH, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems Vol. 14, No. 4, 2006, 445-459

[12] Thomas L. Saaty How to make a decision:The Analytic Hierarchy Process, European Journal of Operational Research 48, 1990, 9-26

[13] Jiří F.Aleš K. Judgment scales and consistency measure in AHP, ScienceDirect Procedia Economics and Finance 12 , 2014 , 164- 173

[14] Evangelos T.,Stuart H. M. USING THE ANALYTIC HIERARCHY PROCESS FOR DECISION MAKING IN ENGINEERING APPLICATIONS: SOME CHALLENGES,Inter’l, Journal of Industrial Engineering: Applications and Practice, Vol. 2, No. 1, 1995, 35-44 [15] María Teresa Escobar & José María Moreno-jiménez Aggregation of Individual Preference

Structures in Ahp-Group Decision Making Group Decision and Negotiation volume 16, 2007, 287-301

10

Appendix : Simulation code A using Python (

Python ver.3.8.5) import numpy as np import pandas as pd import random as r r.seed(2) import matplotlib.pyplot as plt import seaborn as sns class Global(): const = {

'level' : 3, # No. levels of a tree

'branch': 3, # No. branches in each assessor 'N' :10000, # No. trials

}

def calError(org,rev): n = float(len(org)) total = 0.

for ele1,ele2 in zip(org,rev): total += (ele2 - ele1) ** 2 tt=total**0.5 return tt def makeOnes(): branch = Global.const['branch'] tmp = [] for b in range(branch): tmp.append(r.uniform(1.,10.)) total = sum(tmp) vOne = [] for ele in tmp: vOne.append(ele/total) return vOne

11 def makeData(): level = Global.const['level'] branch = Global.const['branch'] vAll = [] for l in range(level): assessors = branch ** l vLevel = []

for assessor in range(assessors): vLevel.append(makeOnes()) vAll.append(vLevel) return(vAll) def makeRev(data,target): level = Global.const['level'] branch = Global.const['branch'] rev = []

for ele1 in data: rev2 = []

for ele2 in ele1: rev3 = []

for ele3 in ele2:

rev3.append(ele3) rev2.append(rev3) rev.append(rev2) rev[target][0] = makeOnes() alpha = calError(data[target][0],rev[target][0]) return rev,alpha def calWeight(value): level = Global.const['level'] branch = Global.const['branch'] num = branch ** level

weight = [0. for i in range(num)] for b in range(branch):

12 for l in range(1,level):

for n in range(branch ** l):

p = n * (branch ** (level - l)) q = branch ** (level - l - 1) for pos in range(1,branch):

weight[p+q*pos] = weight[p] * value[l][n][pos] / value[l][n][0] total = sum(weight)

ans = []

for ele in weight:

ans.append(ele / total) return ans ######################## if __name__ == "__main__": N = Global.const['N'] Level = Global.const['level'] Branch = Global.const['branch'] fig=plt.figure(figsize=(10,3), dpi=150) dataBP = []

for level in range(Level):

plt.subplot(1,Level,level+1) #plt.ylim(-.001,.5) #plt.xlim(-.001,1.0) X = [] Y = [] for i in range(N): orgData = makeData()

revData, alpha = makeRev(orgData,level) X.append(alpha)

Y.append(calError(calWeight(orgData),calWeight(revData))) dataBP.append(Y)

13

coef = np.corrcoef(X, Y) # Find the correlation coefficient print(coef) print("---") plt.scatter(X,Y,marker="x") plt.title('Level = ' + str(level+1),fontsize=10) plt.grid(True)

plt.savefig('Fig1.png', format='png', dpi=150) plt.show()

plt.boxplot(dataBP, labels=range(Level), whis="range") plt.savefig('Fig2a.png', format='png', dpi=150)

plt.show()

plt.violinplot(dataBP,showmeans=False,showmedians=True) plt.savefig('Fig2b.png', format='png', dpi=150)

plt.show() a=np.array(dataBP) print(a) print("---") l1=np.median(a[0]) l2=np.median(a[1]) l3=np.median(a[2]) print(l1,l2,l3) print("---") f = open('test.txt', 'w') for i in range(10000): strs=str(X[i])+'\t'+str(a[0,i])+'\t'+str(a[1,i])+'\t'+str(a[2][i])+'\n' f.writelines(strs) f.close()

14

Appendix : Simulation code B using Python (

Python ver.3.8.5) import pandas as pd import numpy as np import random as r import matplotlib.pyplot as plt r.seed(2) level_min,level_max=2,6 branch_min,branch_max=2,6 number_trials=40000 f = open('dataBP_6x6(40000(seed2)).txt', 'w') f.writelines('L_size'+'\t'+'B_size'+'\t'+'level'+'\t'+'dataBP[i]'+'\n') print('L_size','\t','B_size','\t','level')

for level_size in range(level_min,level_max+1): const = {}

const['N']=number_trials # No. trials

const['level']=level_size # No. levels of a tree for branch_size in range(branch_min,branch_max+1):

const['branch']=branch_size # No. branches in each assessor def calError(org,rev):

n = float(len(org)) total = 0.

for ele1,ele2 in zip(org,rev): total += (ele2 - ele1) ** 2 tt=(total)**0.5/n return tt def makeOnes(): branch = const['branch'] tmp = [] for b in range(branch): tmp.append(r.uniform(1.,10.)) total = sum(tmp) vOne = [] for ele in tmp:

15 vOne.append(ele/total) return vOne def makeData(): level = const['level'] branch = const['branch'] vAll = [] for l in range(level): assessors = branch ** l vLevel = []

for assessor in range(assessors): vLevel.append(makeOnes()) vAll.append(vLevel) return(vAll) def makeRev(data,target): level = const['level'] branch = const['branch'] rev = []

for ele1 in data: rev2 = []

for ele2 in ele1: rev3 = [] for ele3 in ele2:

rev3.append(ele3) rev2.append(rev3) rev.append(rev2) rev[target][0] = makeOnes() alpha = calError(data[target][0],rev[target][0]) return rev,alpha def calWeight(value): level = const['level'] branch = const['branch'] num = branch ** level

weight = [0. for i in range(num)] for b in range(branch):

16 weight[b * (branch ** (level - 1))] = value[0][0][b]

for l in range(1,level):

for n in range(branch ** l): p = n * (branch ** (level - l)) q = branch ** (level - l - 1) for pos in range(1,branch):

weight[p+q*pos] = weight[p] * value[l][n][pos] / value[l][n][0] total = sum(weight)

ans = []

for ele in weight:

ans.append(ele / total) return ans N = const['N'] Level = const['level'] Branch = const['branch'] dataBP = []

for level in range(1): X = []

Y = []

for i in range(N):

orgData = makeData()

revData, alpha = makeRev(orgData,level) X.append(alpha) Y.append(calError(calWeight(orgData),calWeight(revData))) dataBP.append(Y) for i in range(1): print(level_size,'\t',branch_size,'\t',i+1,'\n',dataBP[i],'\n') print(level_size,'\t',branch_size,'\t',i+1,'\n') for j in range(number_trials): strs=str(level_size)+'\t'+str(branch_size)+'\t'+str(i+1)+'\t'+str(dataBP[i][j])+'\n' f.writelines(strs) f.close() □