# Minimization and Manipulate[Plot] not working as expected

Quite new to Mathematica, and I have been trying to fix the code using online resources for the past few days now. I am trying to plot and minimise this equation to learn more about its properties:

(1) $$t=a+b$$
(2) $$p=u+(1-u)(1-e^{-\lambda b})$$
(3) $$u = \frac{(1-e^{-\lambda a})\sigma}{(1-e^{-\lambda a})\sigma + (e^{-\lambda a})(1-\sigma)}$$

$$\sigma, \lambda \text{ and } p$$ are constant terms and $$b$$ is determined uniquely by choice variable $$a$$ in equation (2)/(3). There are two other variables, $$t$$ and $$a$$ - I would like to explore the relationship between these.

I have tried to Plot this relationship for varying and specified levels of $$\sigma, \lambda \text{ and } p$$, but I get empty axes each time. This is what I have used.

t = a + b
p = u + (1 - u) (1 - E^{-λ b})
u = ((1 -
E^{-λ a}) (σ))/((1 -
E^{-λ a}) (σ) + (E^{-λ a}) (1 - σ))
Manipulate[Plot[t, {a, 0, 1}], {λ, 0, 1}, {σ, 0.5, 1}, {p, 0, 1}]


Being unable to find a plot, I have also tried to minimize $$t$$ with respect to $$k$$, with $$\sigma, \lambda \text{ and } p$$ as constants. Essentially the minimisation problem is:

$$\underset{k}\min t$$ subject to $$p=u+(1-u)(1-e^{-\lambda b})$$ where $$u = \frac{(1-e^{-\lambda a})\sigma}{(1-e^{-\lambda a})\sigma + (e^{-\lambda a})(1-\sigma)}$$

I have tried different variations of this:

Assuming[a > 0 && t > 0 && b > 0 && λ > 0 && 0 < p < 1 && 0.5 < σ < 1, Minimize[{t, p == u + (1 - u) (1 - E^(-λ b))}, a]]


And I have also tried substituting the value of $$u$$ as per equation (3). However I am unable to output any meaningful results.

Does anyone have any ideas on how I can reformulate the plot or minimisation expressions so I can output meaningful results?

Thank you

• First suggestions: use == (not =) in equations inside Minimize and simple () (not {}) for the exponents. – b.gates.you.know.what May 23 '18 at 11:47
• Thank you – b.gatessucks I have included those suggestions and now I am getting an indeterminate result - definitely not something that I expected. This is what I used now: Assuming[a > 0 && t > 0 && b > 0 && \[Lambda] > 0 && 0 < p < 1 && 0.5 < \[Sigma] < 1, Minimize[{t, p == u + (1 - u) (1 - E^(-\[Lambda] b))}, a]] And this is the output {-\[Infinity], {a -> Indeterminate}} – Elke May 23 '18 at 11:59
• You can try to solve (2) for u, substitute into (3) and solve for b; this way you will have an explicit expression to NMinimize. – b.gates.you.know.what May 23 '18 at 15:02

You can solve it for b, for given value of p.

       a = .1; σ = .2; λ = .2;
p == (1 - Exp[-λ a]*σ)/(1 -
Exp[-λ a]*σ +
Exp[-λ a] (1 - σ)) + (1 - (1 -
Exp[-λ a]*σ)/(1 - Exp[-λ a]*σ +
Exp[-λ a] (1 - σ))) (1 - Exp[-λ *b])

Solve[p == 0.506234 + 0.493766 (1 - E^(-0.2 b)), b]
(*out put*)
(*{{b -> -5. Log[-2.02525 (-1. + p)]}}*)
or
(*{{b -> -5. Log[-2.02525 (-1. + p)]}} /. Rule -> Equal*)

• Thank you for your answer. I have tried to keep $p$ a constant - perhaps it would be better the express that equation in terms of $a$? If I combine (2) and (3), then I get $p=\frac{(1-e^{-\lambda a})\sigma}{(1-e^{-\lambda a})\sigma + (e^{-\lambda a})(1-\sigma)}+(1-\frac{(1-e^{-\lambda a})\sigma}{(1-e^{-\lambda a})\sigma + (e^{-\lambda a})(1-\sigma)})(1-e^{-\lambda b})$ This leaves one degree of freedom, so when $a$ is chosen, $b$ should be pinned down. I have tried (unsuccessfully) to get Mathematica to form an expression for the above with a as the subject for my constraint – Elke May 23 '18 at 14:56
• I have updated the answes. Is it helpful? – Gopal Verma May 23 '18 at 16:13