Under the influence of the control system will w3 from the state (t, 0) to state (m + 1, 1). Connect these points of the segment on which to put the costs equal to the cost of repair in the reporting year, plus the cost of operation: fo (r) + r (x, 0);
2. Suppose the system is in £ = (t, f), t> 0. Then, as before, we have H '{uc) == (t +1, £ +1) at a cost of r (r, /); g / (w3) = (t + l, 1), at a cost / (m , t) + g (t, 0).
Using Table. 3 and 4, mark the segments The corresponding possible state transitions from state g 'in Fig. 10. For example, over the segment connecting the points £ = (3, 2) and g / == (4, 1), is the number of 5.1, it is - the amount of repair costs, is 2.3 (see Table 3.) and operating costs for the 4th year, if after the repair was 0 years, is 2.8 (see Table. 4).
Constrained optimization draw on the resulting network.
The optimization of the 6-th step. The final states of the system are known - the point (6; t). Analyze how to get to each end state. For this, we consider all the possible states that occur after the 5th step.
State (5, 0). From it you can get to (6, 0), making the cost of 10 (only operation), and the status (6, 1) with costs 9.1 (repair and subsequent operation). It follows that if the penultimate step led to the point (5, 0), we should go to the point (6, 1) (we note this direction arrow), and the minimum (unavoidable) costs related to this transition are equal to 9.1 [put this value in the circle points (5, 0)].
State (5, 1). From it you can get to the point (6, 1) costs 3 4.1 = 7.1 and to the point (6, 2) the cost of 4.4. Choose the second administration, noted his arrow, and the minimal cost to put down in a circle point (5, 1).

Arguing in the same way for each of the penultimate step, we will find for any outcome of the 5th step of the conditional optimal control on the 6-th step, we note it in Fig. 10 arrows and, in addition, relatively minimal costs in the last step to put in the appropriate circle.
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