2. Step 2: Identifying the feasible area:
As per the direction of the constraints, the feasible area has been shown in the above graph
(above AED)
Step 3: Finding the optimum solution:
As per the Extreme Point Theorem, the optimum solution shall lie on one of the points A,
E or D.
𝐶𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝐴 = (0,4)
∴ 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐴:
𝑧 = 𝑥1 + 𝑥2
= 0 + 4
= 4
𝐼𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑖𝑑𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝐸, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑠𝑖𝑚𝑢𝑙𝑡𝑎𝑛𝑒𝑜𝑢𝑠𝑙𝑦 𝑠𝑜𝑙𝑣𝑒 𝑡ℎ𝑒
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 ( 𝑖) 𝑎𝑛𝑑 ( 𝑖𝑖):
5. 𝑨𝒏𝒔: 𝒙 𝟏 =
𝟐𝟏
𝟏𝟑
, 𝒙 𝟐 =
𝟏𝟎
𝟏𝟑
, 𝒛 =
𝟑𝟏
𝟏𝟑
Two-Phase Method
Here the solution of the LPP is completed in two phases. In the first phase of the method,
the sum of the artificial variables is minimised subject to the given constraints to get a basic
feasible solution. Th second phase minimises the original objective function starting with
the basic feasible solution obtained at the end of the first phase.
Steps of the Algorithm (Phase I)
1. Express the given LPP in the standard form.
2. Convert each of the constraints into equality by introducing slack, surplus or artificial
variables.
3. Solve the LPP by assigning a coefficient of ‘-1’ to each artificial variable in case of
maximisation problem and ‘+1’ in case of minimisation problem and zero to all other
variables in the objective function.
4. Apply the simplex algorithm to solve this LPP.
5. If Cj–Zj row indicates optimal solution and
a. the artificial variable appears as a basic variable, the given LPP has non-feasible
solution.
b. the artificial variable does not appear as a basic variable, the given LPP has a feasible
solution and we proceed to Phase II.
Phase II
6. Assign actual coefficients to the variables in the objective function and zero to the
artificial variables. That is, the last simplex table of phase I is used as the initial simplex
table for phase II. Now apply the usual simplex algorithm to the modified simplex table
to get the optimal solution to the original problem.
𝑀𝑖𝑛. 𝑧 = 𝑥1 + 𝑥2
𝑠/𝑡
2𝑥1 + 𝑥2 ≥ 4
𝑥1 + 7𝑥2 ≥ 7
𝑥1, 𝑥2 ≥ 0
𝐼𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑖𝑛𝑔 𝑠𝑢𝑟𝑝𝑙𝑢𝑠 𝑎𝑛𝑑 𝑎𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 (𝑃ℎ𝑎𝑠𝑒 𝐼 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 − 𝑝ℎ𝑎𝑠𝑒
𝑚𝑒𝑡ℎ𝑜𝑑)
𝑀𝑖𝑛. 𝑧 = 0𝑥1 + 0𝑥2 + 0𝑆1 + 0𝑆2 + 𝐴1 + 𝐴2
𝑠/𝑡
2𝑥1 + 𝑥2 − 𝑆1 + 𝐴1 = 4
𝑥1 + 7𝑥2 − 𝑆2 + 𝐴2 = 7
𝑥1, 𝑥2, 𝑆1, 𝑆2, 𝐴1, 𝐴2 ≥ 0
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = 6
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 = 2
∴ 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑛 − 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = 6 − 2 = 4
∴ 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = 6 − 4 = 2
𝑇ℎ𝑒 𝑡𝑤𝑜 𝑎𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝐴1 𝑎𝑛𝑑 𝐴2 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝐴1 = 4
𝐴2 = 7
7. Solve the following using two-phase method:
Maximise: z = 30x + 40y + 35z
Subject to the constraints:
3x + 4y +2z ≤ 90
2x + y +2z ≤ 54
X + 3y + 2z ≤ 93
x, y, z ≥ 0 (2013)