Linear Programming
Introduction
Supply chain management is a significant process of the business process. It is also known as a combination of the procurement, production, storage and delivery of goods. The concept of supply chain management can be seen in each sector of management. In the business process, it can be also seen that companies or management face the problem in the regard of the different processes such as procurement, production, storage and delivery of the goods. In order to solve these issues, management takes the help of linear programming. It is a method to achieve the best outcome in the business process by maximizing the profit profitability and minimizing the cost (Higle & Sen, 2013). Along with this, linear programming is a special case of mathematic programming. It is one of the simplest ways to perform optimization in business activity. It is helpful to solve some very complex problems in the operation and provide possible solutions.
Answer 1
In the provided case study of Mecha Wire Works, it is found that the main problem of the company is to identify which order should procedure first and which order can be held. At the same time, it is found that the company is growing and its market demand is increase. But, at the same time, the company has not had enough employees or labour so that company can complete its production process effectively
As the principle of lineal programming that a company uses linear programming for two purposes such as profit maximisation and cost reduction (Vanderbei, 2015). In the context of Mecha Wire Works, it has been assumed that the objective of the utilisation of linear programming is profit maximisation. In order to solve the problem of the given case study, the following assumptions are developed:
In order to profit maximisation, the production units of the Mecha Wire Works are:
Product W0075C = X1
Product W0033C = X2
Product W0005X = X3
Product W0007X = X4
At the same time, the given information also depict that the demand or order to the following products:
Order for Product W0075C = X1 = >= 150 Units
Order of units for Product W0007X = X4 = >= 600 Units
In the context of the plant capacity, the following information is provided in the case study of the Mecha Wire Works:
| Plant name | Capacity |
| Drawing | <= 4000 hours |
| Extrusion | <= 4200 hours |
| Winding | <= 2000 hours |
| Packaging | <= 2300 hours |
Each product provides some profit to the company after the deduction of the cost. The profit from each product is calculated in the below manner.
| Product | material | Labor | Overhead | Total expense | Selling price | Profit |
| W0075C | $33 | $9.90 | $23.10 | $66 | $100 | $34 |
| W0033C | $25 | $7.50 | $17.50 | $50 | $80 | $30 |
| W0005X | $35 | $10.50 | $24.50 | $70 | $130 | $60 |
| W0007X | $75 | $11.25 | $64.75 | $150 | $175 | $25 |
| Mecha Wire Works | ||||||||
| W0075C | W0033C | W0005X | WOOO7X | LHS | OPERATOR | RHS | SLACK | |
| Object function | 34 | 30 | 60 | 25 | 59900 | |||
| (Max. profit)drawing | 1 | 2 | 0 | 1 | 2200 | <= | 4000 | 1800 |
| Extrusion | 1 | 1 | 4 | 1 | 1950 | <= | 4200 | 2250 |
| Winding | 1 | 3 | 0 | 0 | 1850 | <= | 2000 | 150 |
| Packaging | 1 | 0 | 3 | 2 | 2300 | <= | 2300 | 0 |
| Solution value | 1100 | 250 | 0 | 600 | ||||
| W0075C | W0033C | W0005X | WOOO7X | ||||
| Number of units | 1,100.00 | 250.00 | 0.00 | 600.00 | |||
| Profit | $34 | $30 | $60 | $25 | $59,900.0 | ||
| Constraints | |||||||
| Drawing time | 1 | 2 | 1 | $2,200.0 | <= | 4000 | |
| Extrusion time | 1 | 1 | 4 | 1 | $1,950.0 | <= | 4200 |
| Winding time | 1 | 3 | $1,850.0 | <= | 2000 | ||
| Packaging time | 1 | 3 | 2 | $2,300.0 | <= | 2300 | |
| W75C orders | 1 | $1,100.0 | <= | 1400 | |||
| W33C orders | 1 | $250.0 | <= | 250 | |||
| W5X orders | 1 | $0.0 | <= | 1510 | |||
| W7X orders | 1 | $600.0 | <= | 1116 | |||
| Minimum W75C | 1 | $1,100.0 | >= | 150 | |||
| Minimum W7X | 1 | $600.0 | >= | 600 | |||
| LHS | Sign | RHS |
| Total Maximum Profit | |||||
| 34X1 + 30X2 + 60X3 + 25X4 | Profit for product X1 per unit = 34 | ||||
| Equation Developed Using the Constraints | Profit for product X2 per unit = 30 | ||||
| Drawing constraint | Profit for product X3 per unit = 60 | ||||
| 1X1 + 2X2+ 0X3+ 1X4 <= 4000 | Profit for product X4 per unit = 25 | ||||
| Extrusion constraint | |||||
| 1X1 +1X2 + 4X3 + 1X4 <= 4200 | Plant capacity (Hours) | ||||
| Winding Constraint | Drawing | Extrusion | Winding | Packaging | |
| 1X1 + 0X2 + 3X3 + 2X4 <= 2300 | 4000 | 4200 | 2000 | 2300 | |
| Packaging Constraint | Required hours/unit | ||||
| 1X1 + 0X2 + 3X3 + 2X3 + 2X4 <= 2300 | Product | Drawing | Packaging | Extrusion | Winding |
| X1 >= 150; X4 >= 600 | |||||
| X1 <= 1400 | W0075C | 1 | 1 | 1 | 1 |
| X2 <= 250 | W0033C | 2 | 0 | 1 | 3 |
| X3 <= 1510 | W0005X | 0 | 3 | 4 | 0 |
| X4 <= 1116 | W0007X | 1 | 1 | 0 | 2 |
On the basis of the results, it can be recommended to Jon smith that it should produce the following quantity in order to achieve the optimum profit. It will be helpful to increase the profit that the company wants to achieve (Plan & Vershynin, 2013). It will be also effective to satisfy all limitations:
W0075C = 1100 (in units) (It is >= 150)
W0033C = 250 (in units)
W0005X = 0 (in units)
W0007X = 600 (in units) (It is >= 600)
Along with this, it is also found by the production of such units, the company will be able to earn a total net profit of $59900.
In this, it is found that packaging hours will be used fully. But at the same time, the remaining hours will be unused including Drawing, extrusion.
Answer 2
Sensitivity analysis is a significant technique that is useful to demonstrate the impact of an independent variable on the dependent variables in the given data set. This analysis technique is used in some specific situations that depend on one or more input variables where the requirement is regarding measure the effect of change on other variables (Dantzig, 2016). For example, Roy is a sales manager in a company and they want to evaluate the impact of customer traffic on the sales of the company. Roy states that sale is an activity that contains that transaction of the price and product. In this, the price of the product is $1000 and Roy team sole 100 units. Hence, the total sales are $100000. Roy also states that a 10 per cent increase leads to 5 percent in the total sales that allows Roy to develop a financial model and sensitivity analysis. In this, the question of the sensitivity analysis can be developed as that what happen sales if customer traffic increases by 20% 40% and 60%. Based on the assumption, it can be calculated and found that a 20%, 40% and 60% increase in customer traffic equates to an increase in transactions by 10%, 20% and 30%. The findings of the sensitivity analysis determine that sales are highly sensitive to changes in customer traffic.
Answer 3
On the basis of the above discussion and finding, it is identified that there is no need of hiring a temporary in the drawing department. It is because hours for drawing are unemployed due to a lack of winding and packaging hours. Along with this, Mecha Wire Works also has not had sufficient hours of packaging in order to precede the product after extrusion and drawing. Hence, the final goods will not become carrying out for delivery (Agrawal, et. al. 2014).
Conclusion
From the above discussion, it is found the really linear programming is an effective technique that provides the solution of the operation process. It also enables the company to maximise profit and as well as also allows to reduce the cost. It brings effectiveness in the process at the plant.
References
Agrawal, S., Wang, Z., & Ye, Y. (2014). A dynamic near-optimal algorithm for online linear programming. Operations Research, 62(4), 876-890.
Dantzig, G. (2016). Linear programming and extensions. UK: Princeton university press.
Higle, J. L., & Sen, S. (2013). Stochastic decomposition: a statistical method for large scale stochastic linear programming. Germany: Springer Science & Business Media.
Plan, Y., & Vershynin, R. (2013). One‐Bit Compressed Sensing by Linear Programming. Communications on Pure and Applied Mathematics, 66(8), 1275-1297.
Vanderbei, R. J. (2015). Linear programming. Heidelberg: Springer.
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