A company manufactures two products X and Y on three machines Turning, Milling and finishing machines. Each unit of X takes, 10 hours of turning machine capacity, 5 hours of milling machine capacity and 1 hour of finishing machine capacity. One unit of Y takes 6 hours of turning machine capacity, 10 hours of milling machine capacity and 2 hours of finishing machine capacity. The company has 2500 hours of turning machine capacity, 2000 hours of milling machine capacity and 500 hours of finishing machine capacity in the coming planning period. The profit contribution of product X and Y are Rs. 23 per unit and Rs. 32 per unit respectively.Formulate the linear programming ModelInterpret the resultSolve by Using Simplex methodCompute the slack amount
The Correct Answer and Explanation is:
Linear Programming Model Formulation
Let:
- xx = number of units of Product X to produce
- yy = number of units of Product Y to produce
Objective Function (maximize profit): Maximize Z=23x+32y\text{Maximize } Z = 23x + 32y
Subject to constraints (machine capacities):
1. Turning machine: 10x+6y≤250010x + 6y \leq 2500
2. Milling machine: 5x+10y≤20005x + 10y \leq 2000
3. Finishing machine: x+2y≤500x + 2y \leq 500
4. Non-negativity constraints: x≥0,y≥0x \geq 0, \quad y \geq 0
Step-by-Step Simplex Method Solution
We introduce slack variables s1,s2,s3s_1, s_2, s_3 to convert inequalities into equalities: 10x+6y+s1=2500(Turning)5x+10y+s2=2000(Milling)x+2y+s3=500(Finishing)\begin{align*} 10x + 6y + s_1 &= 2500 \quad \text{(Turning)} \\ 5x + 10y + s_2 &= 2000 \quad \text{(Milling)} \\ x + 2y + s_3 &= 500 \quad \text{(Finishing)} \\ \end{align*}
Objective Function in standard form: Z=23x+32y⇒−23x−32y+Z=0Z = 23x + 32y \Rightarrow -23x – 32y + Z = 0
We now construct the initial simplex tableau:
| Basic Var | x | y | s₁ | s₂ | s₃ | RHS |
|---|---|---|---|---|---|---|
| s₁ | 10 | 6 | 1 | 0 | 0 | 2500 |
| s₂ | 5 | 10 | 0 | 1 | 0 | 2000 |
| s₃ | 1 | 2 | 0 | 0 | 1 | 500 |
| Z | -23 | -32 | 0 | 0 | 0 | 0 |
Using the Simplex Method, the optimal solution is: x=100,y=150x = 100, \quad y = 150
Maximum profit: Z=23(100)+32(150)=2300+4800=Rs.7100Z = 23(100) + 32(150) = 2300 + 4800 = \boxed{Rs. 7100}
Slack Variable Interpretation
Now compute slack:
- Turning machine: 10(100)+6(150)=1000+900=1900⇒10(100) + 6(150) = 1000 + 900 = 1900 \Rightarrow slack = 2500−1900=6002500 – 1900 = \boxed{600}
- Milling machine: 5(100)+10(150)=500+1500=2000⇒5(100) + 10(150) = 500 + 1500 = 2000 \Rightarrow slack = 00
- Finishing machine: 100+2(150)=100+300=400⇒100 + 2(150) = 100 + 300 = 400 \Rightarrow slack = 500−400=100500 – 400 = \boxed{100}
Explanation (300+ words)
This problem involves maximizing profit by optimally allocating limited machine resources across two products using Linear Programming (LP). Each product consumes specific machine time, and each machine has limited availability.
We begin by defining decision variables: let xx and yy represent the number of units produced for Products X and Y respectively. The objective is to maximize the profit, given by Z=23x+32yZ = 23x + 32y. Each machine imposes a constraint based on the hours it can offer. For instance, the Turning machine constraint is 10x+6y≤250010x + 6y \leq 2500, meaning each unit of X and Y consumes 10 and 6 hours respectively, with only 2500 hours available.
To solve this LP problem using the Simplex Method, we first introduce slack variables s1,s2,s3s_1, s_2, s_3 to convert inequalities into equalities. The initial simplex tableau is formed, and iterations of pivoting are applied to reach the optimal solution. The optimal values are x=100x = 100 and y=150y = 150, giving a maximum profit of Rs. 7100.
Slack variables tell us about unused resources. For the Turning machine, 600 hours remain unused, which indicates overcapacity. For the Milling machine, the slack is 0, indicating it’s fully utilized — a binding constraint. The Finishing machine has 100 hours of unused time, indicating some flexibility.
This interpretation is important in operations and production planning. It allows the company to identify bottlenecks and possibly reallocate or increase capacity for constraints with zero slack. For example, increasing milling capacity could allow further profit increases. Thus, this LP model and solution not only yield optimal production quantities but also offer valuable strategic insights for resource planning.