10 3. 7. 8. 2.

Consider the provided diagram,

In order to analyze the network given above by linear programming.

let x_i (>=0) represent the cost at which we start activity i.

This time is our choice and hence is a decision variable.

• the x_i can be constrained to take integer (whole number) values if we so wish using integer programming.
• the x_i are NOT necessarily the same as the earliest times.

Considering the precedence relationship between activity 1 and activity 2, it is clear that we must have,

$x_2⩾x_1+6$

That means,  the starting time for activity 2 must be at least as great as the cost for an immediate predecessor activity plus the cost for this predecessor activity. Note that we do not put $x_2=x_1+6$ here. This is because the inequality ($⩾$) gives us more flexibility than the equality (=).

we must also have

$$\begin{gathered} x_3⩾x_1+10 \\ {x}_3⩾x_2+4 \\ {x}_4⩾x_1+7 \\ {x}_4⩾x_3+5 \\ {x}_5⩾x_2+5 \\ {x}_5⩾x_3+2 \\ {x}_6⩾x_4+8 \\ {x}_6⩾x_3+4 \\ {x}_7⩾x_5+7 \\ {x}_7⩾x_6+5 \end{gathered}$$