Exam 11: Dynamic Programming

Correct Answer:

Verified

The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state s_n be the vector (R₁,R₂),where R_i is the slack in the i-th constraint,so s₁ = (13,9)and s₂ = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f₂ (R₁,R₂, x₂)= x₂,0 $\le$ x₂ $\le$ min{R₁,R₂},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f₁(13,9, x₁)= = .For 0 $\le$ x₁ $\le$ 2, $\implies$ f₁(13,9, x₁)= .Since for 0 ≤ x₁ ≤ 2,we now have at x₁ = 2.For x₁ $\le$ 2, $\implies$ f₁(13,9, x₁)= Since = 5 - 4x₁ < 0 for x₁ ≥ 0,it follows that at x₁ = 2.Since f₁(13,9,x₁)is maximized at x₁ = 2 over both the range 0 ≤ x₁ ≤ 2 and the range x₁ ≥ 2,the maximum must be at x₁ = 2 over the entire range of x₁ ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x₁ = 2 leaves a slack of R₁ = 13 - 2 = 5 and R₂ = 9 - = 5 in the two constraints,so = min{R₁,R₂} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.

Correct Answer:

Verified

Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x_n be the advertising dollars (in millions)spent on product n.Let s_n be the amount of advertising budget remaining.Let be the increase in sales of product n when x_n million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3, For n = 2, For n = 1, Hence,since s₂ = 6 - and s₃ = s₂ - ,there are two optimal plans as given in the table below.

Correct Answer:

Verified

The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state s_n be the remaining slack in the constraint ,so s₁ = 4 and s₂ = 4 - .For n = 2, .For n = 1, .To maximize (2x₁ + 4 - )over 0 ≤ x₁ ≤ 2,we take the first and second derivatives. (2x₁ + 4 - )= 2 - 2x₁ = 0,so x₁ = 1 is a critical point. (2x₁ + 4 - )= -2,so x₁ = 1 is indeed the maximum over 0 ≤ x₁ ≤ 2.Since x₁ = 1 leaves a slack of in the constraint,the overall optimal solution is = 1, ,with Z* = 5.