Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The standard form LP problem is

$$\min -3x_1-7x_2-10x_3 \text{ s.t. }$$ $$x_3\leq 2$$ $$40x_3+40x_2+20x_1\leq 180$$ $$x_1,x_2,x_3\geq 0$$

My last lecture covered the Bellman equation with stationary discounted case getting $J^*(x)=\min_u\left\{g(x,u)+\alpha J^*(f(x,u))\right\}$ and the general form of the DP algorithm: $J_N(x_N)=g_N(x_N)$ and $J_k(x_k)=\min_{u_k}\left\{g_k(x_k,u_k)+J_{k+1}(f_k(x_k,u_k))\right\}$ where $k=0,1,...,N-1$.

I am trying to solve this LP as Dynamic Programming problem but I cannot see where to start. Is the $g(x)=-3x_1-7x_2-10x_3$? What is the $J_k(x_k)$ here? Does this hold $J_k(x_k)=J(x)$?

Solution to the problem solved as LP

By simplex, the optimum is $x_B=(x_2,x_3)=(2.5,2)$ with the $c_B'B^{-1}b=-37.5$.

share|cite|improve this question
up vote 1 down vote accepted

The basic idea is to use Backward Induction. The marginal benefits are $b=(b_1,b_2,b_3)=(0.15,0.175,0.25)$. So we will first look at $x_3$ with the largest marginal gain, then $x_2$ and lastly $x_1$. The optimality is based on the optimality principle of Dynamic programming in the following examples: suppose a-b-c is optimal from the state A to the state C, then b-c must be optimal from the state B to the state C.

Suppose 1-2-3 is optimal, then 1-2 and 2-3 must be optimal. 2-3 means the choice of $x_3$ when $x_3\leq 2$ so $40*2\leq 180$ so this is optimal choice. Now with 100 resources, 1-2 is optimal because $2*40\leq 100$ and $3*40\not \leq 100$. Now because all resources are used at end when $x_1=1$, we have found the optimal solution, the example with the integers -- similarly with the real numbers.

Integer problem

We use everything of $x_3$ so $x_3=2$ because of the constraint $x_3\leq 2$. Then we will look at $x_2$ and use that 2 units because otherwise $40*2+40*2=200\leq 180$, breaking a consraint. So the last unit is $x_1=1$ hence $x=(1,2,2)$ when $x\in\mathbb Z^3$.

Real problem

Similarly as above but the step 2 is different where we can use 2.5 units instead of just 2 units. If $x\in\mathbb R^3$ then $x=(0,2.5,2)$, the same result as you got with the Simplex.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.