Abstract This study copes with a class of principal-agent problems where information asymmetry represents an important characteristic. The paper examines the relationship between the principal and agents. The principal has to perform two agents’ screening and discipline tasks. To complete his duties, the principal lacks complete information concerning the agents’ behavior and rarely has partial information regarding the failure or success of launched tactics, alliances, rationalization, etc. We analyze the type of retention contracts (implicit) used by the principal to replace or retain agents. Consistent with literature findings, we demonstrated that agents could be extremely active in showing their competencies; the relationship between dismissal and bad performance is invalid; and occasionally, the principal dismisses qualified agents. Then we determined the rules under which electorates urge political parties to acquire information and choose optimal policies from the voter’s viewpoint.

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[16] Yahaya A., Mahat F., and Abdulkadir J., “Effect of Corporate Governance Practice and Bank Regulatory Capital on Performance: Evidence from Deposit Money Banks in Nigeria,” International Journal of Social, Political and Economic Research, vol. 7, no. 4, pp. 838-862, 2020. Zina Houhamdi received her Ph.D. in Software Engineering in 2004. She is a Professor at the Department of Cybersecurity, College of Engineering, Al Ain University, United Arab Emirates. Her research work has been published in several academic journals and presented at scientific conferences. Her main research interest is in the Internet of Things, Artificial Intelligence, particularly in Multi- Agent Systems Modelling, Testing and Applications. She is published several papers in journals and international peer-reviewed conferences. Belkacem Athamena holds a Ph.D. in System Analysis and Applications. He is a Professor at the Department of Business Administration, College of Business, Al Ain University, United Arab Emirates. His main research interest is in system modeling and analysis, multi-agent, fuzzy logic, software testing, Petri nets, formal methods, data quality, and fault diagnosis. He has published many refereed journal articles, contributed chapters, and presented papers at conferences. Ghaleb El Refae is a Professor in Financial Economics with expertise in higher education management, risk management in higher education institutions and university corporate governance. He endeavoured in researches that cover many themes including higher education quality, international higher education, diversity and education, risk management in higher education institutions, university corporate governance, asymmetric information in higher education as well as topics pertaining to financial economics and industrial organisation. Appendix This appendix determines the office's present discounted value. Assume that in iteration ������ the party � is the tenured party. � Represents the probability of the tenured reelection in iteration (I+1) and �������+1� is the equilibrium value for the elected party in iteration ������ and �������+1������ is the equilibrium value for the dismissed party in iteration I. The probability that party � is the tenures party in iteration ������ is: �������+1=��������+(1−�)(1−��)=(2�−1)�������+(1−�) The solution of equation (17) is �������=������(2�−1)������+(1−������) 1−(2������−1)=������(2�−1)������+12 Where ������ is a random constant. Remember that in iteration I=1, � is the tenured party which implies that for I=1, �1=1. Thus, ������=1 2(2������−1) and equation (18) solution is: �������=1 2(2�−1)������−1+1 2 We compute �������+1� �������+1�=∑�������(1 2(2�−1)������−1+1 2)(�−�′) ∞ ������=0 =(1−��)� (1−�)(1−2��+�)(�−�′) Where � defines the received utility by the tenured party and �′ defines the received utility by the opposition party. Now assume that in iteration I=0, L is dismissed. Then, in iteration I+1 party R will be tenured, then �1= 0. Since ������=−1 2(2������−1) and the equation solution is: �������=−12(2�−1)������−1+12 We compute �������+1������ in iteration I=0 as �������+1������=∑�������(−1 2(2�−1)������−1+1 2)(�−�′) ∞ ������=0 =(�−��)� (1−�)(1−2��+�)(�−�′) We deduce that �������+1�−�������+1������=�1−2�������+�(�−�′) (17) (18) (19) (20) (21) (22) (23)