As shown in the above formula.

Given x, we are able to solve nR and thereby the pi.

We check that nR increase in x using x ranging from [0:1:50] by plotting. **But how to check the concavity of pi with respect to x to show the existence and uniqueness of x*** Finally, check how the optimal x* changes with alpha.

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`v0 = 0; v = 10; m = 4; f = 1.5; pT = 8; beta = 20; n = 100; gamma = 10; w0 = 10; alpha=0.2; c0=1; c1=0.3; alpha=0.2; x_gd = [0:1:10000]; for nk = 1:length(x_gd) x = x_gd(nk); funCustomer = @(y) y - n.*exp (v-m*f+beta.*gamma.*x./(m.*y)-w0)./(exp(v0)+... exp(v-m*f+beta.*gamma.*x./(m.*y)-w0)+exp(v-pT)); nR(nk) = fzero(@(y) funCustomer(y), [-1000, 1000]); nS(nk) = n.*exp (v-pT)./(exp(v0)+exp(v-m.*f+beta.*gamma.*x./(m.*nR(nk))-w0)+exp(v-pT)); pi_x(nk)=(1-0.2)*f*m*nS(nk)-c1*x-c0; end figure(1) plot(x_gd,nR) xlabel('x'); ylabel('nR'); figure(2) plot(x_gd,nS) xlabel('x'); ylabel('nS'); figure(3) plot(x_gd,pi_x) xlabel('x'); ylabel('pi_x'); `