E1 = 0	L10 = 22
E2 = 0 + 4 = 4	L9 = 22 - 7 = 15
E3 = 0 + 1 = 1	L8 = 22 - 5 = 17
E4 = Max of {4 + 1, 1 + 1} = 5	L7 = 17 - 2 = 15
E5 = 1 + 6 = 7	L6 = 17 - 1 = 16
	L5 = Min of {16-4, 15-8} = 7
E6 = 7 + 4 = 11	L4 = 15 - 5 = 10
E8 = Max of {15 + 2, 11 + 1}= 17	L3 = Min of {10-1, 7-6] = 1
E9 = 5 + 5 = 10	L2 = 10 - 1 = 9
E10 = Max of {10 + 7, 17 + 5} = 22	L1 = 0

Activity	Duration tij	EST	EFT = EST + Tij	LST = LFT - tij	LFT
1-2	4	0	4	9-4 = 5	9
1-3	1	0	1	1-1 = 0	1
2-4	1	4	5	10-1 = 9	10
3-4	1	1	2	10-1 = 9	10
3-5	6	1	7	7-6 = 1	7
4-9	5	5	10	15-5 = 10	15
5-6	4	7	11	16-4 = 12	16
5-7	8	7	15	15-8 = 7	15
6-8	1	11	12	17-1 = 16	17
7-8	2	15	17	17-2 = 15	17
8-10	5	17	22	22-5 = 17	22
9-10	7	10	17	22-7 = 15	22

Since EFT and LFT is same on 1 - 3, 3 - 5, 5 -7 and 7 - 8 and 8 -10 the critical path is 1- 3 - 5 -7 - 8 - 10 and the duration is 22 time units.

Earliest start time (EST) and latest finish time (LFT) of each activity are given in the following network.

E₁ = 0
E₂ = E₁ + t₁₂ = 0 + 8 = 8
E₃ = E₁ + t₁₃ = 0 + 4 = 4
E₄ = E₂ + t₂₄ or E₃ + t₃₄ = 8 + 10 = 18
(take E₂ + t₂₄ or E₃ + t₃₄ whichever is maximum)
E₅ = (E₂ + t₂₅ or E₄ + t₄₅) = 18 + 3 = 21
(take E₂ + t₂₅ or E₄ + t₄₅ whichever is maximum)
L₅ = 21
L₄ = L₅ – t₄₅ = 21 – 3 = 18
L₃ = L₄ – t₃₄ = 18 – 5 = 13
L₂ = L₅ – t₂₅ or L₄ – t₂₄ = 18 – 10 = 8
(take L₅ – t₂₅ or L₄ – t₂₄ whichever is minimum)
L₁ = L₂ – t₁₂ or L₃ - t₁₃ = 8 – 8 = 0
(take L₂ – t₁₂ or L₃ – t₁₃ whichever is minimum)
Here the critical path is 1-2-4-5, which is denoted by double lines.

Activity	Duration(tij)	EST	EFT=EST+tij	LST=LFT-tij	LFT
1-2	8	0	8	0	8
1-3	4	0	4	9	13
2-4	10	8	18	8	18
2-5	2	8	10	19	21
3-4	5	4	9	13	18
4-5	3	18	21	18	21

The longest duration to complete this project is 21 days.
The path connected by the critical activities is the critical path(the longest path).
Critical path is 1-2-4-5 and project completion time is 21 days.

(i) variables:
Let x₁ and x₂ denote the number of ordinary and auto-cut voltage stabilizers.
(ii) Objective function:
Profit on x₁ ordinary = 100x₁  
Profit on x₂ auto-cut = 150x₂
Total Profit = 100x₁ + 150x₂
Let Z = 100x₁ + 150x₂, which is the objective function.
Since the total profit is to be maximized, we have to maximize z = 100x₁ + 150x₂
(iii) Constraints:
Manufacturing capacity 720 hours at present is available per week.
0.8x₁ + 1.2x₂ ≤ 720
[\(\because \) 0.8 hour each for ordinary and 1.20 hours each for auto-cut]
Weekly sale of ordinary: x₁ ≤ 600
Weekly sale of auto-cut: x₂ ≤ 400
(iv) Non-negative restrictions:
Since the number of ordinary anad auto-cut voltage stabilizer, we have x₁, x₂ ≥ 0
Thus, the mathematical formulation of the LPP is
Max Z = 100x₁ + 150x₂
subject to the constraints
0.8x₁ + 1.2x₂ ≤ 720
x₁ ≤ 600
x₂ ≤ 400
x₁, x₂ ≥ 0  

We find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines
x₁ + x₂ = 30; x₂ = 12; x₁ = 20
x₁ + x₂ = 30 is a line passing through the points (0,30) and (30,0)
x₂ = 12 is a line parallel to x₁–axis
x₁ = 20 is a line parallel to x₂–axis.
The feasible region satisfying all the conditions x₁ + x₂ ≤ 30; x₂ ≤ 12 ; x₁ ≤ 20 and x₁, x₂ ≥ 0 is shown in the following graph.

The feasible region satisfying all the conditions is OABCD.
The co-ordinates of the points are O(0,0); A(20,0); B(20,10); C(18,12) and D(0,12).

Corner points	Z = 2x1 + 3x2
O(0,0)	0
A(20,0)	40
B(20,10)	70
C(18,12)	72
D(0,12)	36

Maximum value of Z occurs at C. Therefore the solution is x₁ = 18 , x₂ = 12, Z_max = 72.

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines x₁ + 4x₂ = 24 ; 3x₁ + x₂ = 21; x₁ + x₂= 9
x₁ + 4x₂ = 24 is a line passing through the points (0, 6) and (24, 0)
[(0, 6) is obtained by taking x₁ = 0 in x₁ + 4x₂ = 24 , (24, 0) is obtained by taking x₂ = 0 in x₁+ 4x₂ = 24].
Any point lying on or below the line x₁ + 4x₂ = 24 satisfies the constraint x₁ + 4x₂ ≤ 24 .
3 x₁ + x₂ = 21 is a line passing through the points (0, 21) and (7, 0). Any point lying on or below the line 3 x₁ + x₂ = 21 satisfies the constraint 3 x₁ + x₂ ≤ 21.
x₁ + x₂ = 9 is a line passing through the points (0, 9) and (9, 0). Any point lying on or below the line x₁ + x₂ = 9 satisfies the constraint x₁ + x₂ ≤ 9.
Now we draw the graph

The feasible region satisfying all the conditions is OABCD.The co-ordinates of the points are O(0,0) A(7,0);B(6,3) [the point B is the intersection of two lines x₁ + x₂ = 9 and 3 x₁ + x₂ = 21]; C(4,5) [the point C is the intersection of two lines x₁ + x₂ = 9 and x₁ + 4x₂ = 24] and D(0,6).
Corner points Z = 2x₁ + 5x₂ O(0,0) 0 A(7,0) 14 B(6,3) 27 C(4,5) 33 D(0,6) 30 Maximum value of Z occurs at C. Therefore the solution is x₁ = 4, x₂ = 5, Z_max = 33