E1 = 0	L7 = 26
E2 = 0 + 5 = 5	L6 = 26 - 5 = 21
E3 = max of { 5 + 3, 0 + 10} = 10	L5 = 21 - 5 = 16
E4 = 10 + 4  = 14	L4 = 21 - 6 = 15
E5 = 10 + 6 = 16	L3 = min of {16 - 6, 15  - 4} = 10
E6 = max of { 14 + 6, 16 + 5 } = 21	L2 = 10 - 3 = 7
E7 = 21 + 5 = 26	L1 = min of {7 - 5, 10  - 10} = 0

Activity	Duration tij	EST	EFT=EST+tij	LST=LFT-tij	LFT
1-2	5	0	5	7-7 = 2	7
1-3	10	0	10	10-10 = 0	10
2-3	3	5	8	10-3 = 7	10
3-4	4	10	14	15-4 = 11	15
3-5	6	10	16	16-6 = 10	16
4-6	6	14	20	21-6 = 15	21
5-6	5	16	21	21-5 = 16	21
6-7	5	21	26	26-5 = 21	26

Since EFT and LFT values are the same in 1-3, 3-5, 5-6 and 6-7, the critical path is 1-3-5-6-7 and duration time is 26 days.

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 \(x_1+x_2 \leq 50; \ 3 x_1+x_2 \leq 90\)
\({ x_{ 1 } }+{ x }_{ 2 }=50\) is a line passing through the points (0,50) and (50,0).
[(0,50) is obtained by taking x₁ = 0 in \({ x_{ 1 } }+{ x }_{ 2 }=50\), (50,0) is obtained by taking \(x_2=0 \ in \ \left.x_1+x_2=50\right]\)
Any point lying on or below the line \({ x_{ 1 } }+{ x }_{ 2 }=50\). satisfies the constraint \(x_1+x_2 \leq 50\)
We follow the same steps for the following
\({ x_{ 1 } }+{ x }_{ 2 }=50\)

x1	0	50
x2	50	0

\(3{ x_{ 1 } }+{ x }_{ 2 }=90\)

x1	0	30
x2	90	0

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(30, 0), B(20,30), C(0,50).

Corner points	Z = 60x1 + 15x2
O(0,0)	0
A(30,0)	1800
B(20,30)	1200 + 450 = 1650
C(0,50)	7500

The optimal solution occurs at C(0,50)
x₁ = 0, x₂ = 50, Z_max = 7500.
Verification
\(x_1+x_2 =50 \)
\(3 x_1+x_2 =90 \)
\(-2 x_1 =-40 \)
\(x_1 =20 \)
\(x_2 =30 \)
B(20,30)

Since both the decision variables x₁ and x₂ are non-negative the solution lies in the quadrant of the plane.
consider the equations \(x_1+2 x_2=10;\ 3 x_1+4 x_2=24.\)
\({ x_{ 1 } }+{ 2x }_{ 2 }=10\) is a line passing through the points (0,5) and (10,0). Any point lying on or above the line \({ x_{ 1 } }+{ 2x }_{ 2 }=10\) satisfies the constraint \(x_1+2 x_2 \geq 10\).
We follow the same steps for the following
\({ x_{ 1 } }+{ 2x }_{ 2 }=10\)

x1	0	10
x2	5	0

\(3{ x_{ 1 } }+{ 4x }_{ 2 }=24\quad \quad \)

x1	0	8
x2	6	0

Draw the graph using the given constraints

The feasible regions ABC (since the problem is of minimization type we are moving towards a.origin).

Corner points	Z = 200x1 + 500x2
A(0,5)	2500
B(4,3)	800 + 1500 = 2300
C(0,6)	3000

Optimal solution is occurs at B(4,3)
x₁ = 4, x₂ = 3, Z_min = 2300.

Activity	Duration (in week)	EST	EFT	LST	LFT
1-2	8	0	8	0	8
1-3	7	0	7	8	15
1-5	12	0	12	9	21
2-3	4	8	12	11	15
2-4	10	8	18	8	18
3-4	3	12	15	15	18
3-5	5	12	17	16	21
3-6	10	12	22	15	25
4-6	7	18	25	18	25
5-6	4	17	21	21	25

Here the critical path is 1–2–4–6
The project completion time is 25 weeks.

 

E1  = 0	L8 = 35
E2 = 0 + 7 + 7	L7 = 35 - 18 - 17
E3 = 7 + 14 = 21	L6 = 18 - 11 = 7
E4 = 7 + 5 = 12	L5 = 36 - 4 = 32
E5 = max of {21 + 11, 12 + 7} = 32	L4 = 32 - 7 = 25
E6 = 0 + 6 = 6	L3 = 32 - 11 = 21
E7 = 6 + 11 = 17	L2 = min of {21 - 14, 25 - 5} = 7
E8 = max of {32 + 4, 17 + 18) = 36	L1 = 7 - 7 = 0

Activity	Duration tij	EST	EFT = EST + Tij	LST = LFT - tij	LFT
1-2	7	0	7	0	7
1-6	6	0	6	6	7
2-3	14	7	21	7	21
2-4	5	7	12	20	25
3 -5	11	21	32	21	32
4 - 5	7	12	19	7	14
6 - 7	11	6	17	7	18
5- 8	4	32	36	32	36
7 - 8	19	17	36	17	36

Since EFT and LFT are same in 1 - 2, 2 - 3, 3 - 5 and 5 -8, The critical path is 1 - 2 - 3 - 5 - 8 and the duration of time is 36 days.