Type of Problem
The case problem is a maximization problem, given that the goal is to maximize the jewelry store's profit. The store makes and sells necklaces and bracelets. Its aim is to maximize the earnings it receives from the sale of the two products. The availability of few resources means that there are limits to the necklaces and bracelets the store can make and sell. The necklaces and bracelets are made from precious minerals such as gold and platinum, which are scarce in nature. The store holds limited amounts of the two minerals at any given time. Resultantly, the number of necklaces and bracelets it sells will typically be limited. Demand is also limited given the fact that a specific number of the products, especially the bracelets, can be purchased by the store's customers.
Variables and Criteria
In the case problem, there are two decision variables identified as X 1 and X 2 . These variables depict the final solution and decide the output of solver analysis. X 1 represents the number of necklaces, while X 2 depicts the number of bracelets sold by the store. The linear programming model has been set up in Excel and solved through the software’s solver tool. The objective function is the initial function formulated to show the store’s goal to maximize profits from the sale of necklaces and bracelets. For the profits to be maximized, an optimal number of necklaces and bracelets should be sold by the store. In Excel, the total profit amount is determined through the sum-product of the number of products sold and the per-unit price. In this respect, the total profit amount is the sum of the product of the number of bracelets and per-unit price and the product of the number of necklaces and per-unit price. After setting up the objective function, the constraints are then taken into account. The resources to be used in creating the bracelets and necklaces are expected to be within the limits of the resources available. The bracelets and necklaces to be sold should not exceed the amount of the gold and platinum resources available at the store. In this respect, the constraint functions will be set up while taking into account limits on the gold and platinum resources in addition to the expected demand. The total amount of resources used is also determined by the sum-product of the number of mineral resources required for each unit and the number of products to be sold. These totals should not exceed the established limits on the minerals available and expected demand.
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LP Model
The LP model for the problem is illustrated below.
Max Z=300X 1 +400X 2 |
ST |
3X 1 +2X 2 < 18 |
2X 1 +4X 2 < 20 |
X 2 < 4 |
X 1 , X 2 > 0 |
The objective function comprises two decision variables, namely the number of necklaces and bracelets sold by the store. The model has three constraints, with the first constraint showing the limitation on the number of ounces of gold held by the store. Currently, the store has only eighteen ounces of gold. The second constraint depicts the number of ounces of platinum held by the store. The store has only twenty ounces of platinum. The third constraint is the demand level. According to the store, the demand for the bracelets cannot exceed four. There are also non-negative restrictions on the number of bracelets and necklaces made, given that the model is linear in nature. Decision variables of linear models typically take non-negative values.
The problem’s optimal solution is indicated below.
Objective Cell (Max) | ||||||
Cell | Name | Original Value | Final Value | |||
$C$16 | Max Z= Profit | 0 | 2400 | |||
Variable Cells | ||||||
Cell | Name | Original Value | Final Value | Integer | ||
$C$20 | values X1 | 0 | 4 | Contin | ||
$D$20 | values X2 | 0 | 3 | Contin | ||
Constraints | ||||||
Cell | Name | Cell Value | Formula | Status | Slack | |
$E$23 | constraint 1 total | 18 | $E$23<=$G$23 | Binding | 0 | |
$E$24 | constraint 2 total | 20 | $E$24<=$G$24 | Binding | 0 | |
$E$25 | constraint 3 total | 3 | $E$25<=$G$25 | Not Binding | 1 |
The optimal objective function value, which is the optimal total profit from selling the optimal number of necklaces and bracelets, is $2400. The optimal number of bracelets and necklaces that should be sold to maximize profit are three and four, respectively. Both the first and second constraints, that is, the number of ounces of gold and platinum, are binding, meaning that at the optimum, their functions hold with equality. In this case, boundary solutions are given on the two constraints. Such constraints cause substantial changes in the optimal solution as a result of variations and, therefore, tighten the feasibility criteria. The third constraint, that is, demand level, is non-binding, meaning that it does not alter optimality under the model. The constraint does not lead to changes in the optimal solution due to its variation.
Sensitivity Analysis and Shadow Price
The sensitivity analysis results are depicted below.
Variable Cells | |||||||
Final | Reduced | Objective | Allowable | Allowable | |||
Cell | Name | Value | Cost | Coefficient | Increase | Decrease | |
$C$20 | values X1 | 4 | 0 | 300 | 300 | 100 | |
$D$20 | values X2 | 3 | 0 | 400 | 200 | 200 | |
Constraints | |||||||
Final | Shadow | Constraint | Allowable | Allowable | |||
Cell | Name | Value | Price | R.H. Side | Increase | Decrease | |
$E$23 | constraint 1 total | 18 | 50 | 18 | 12 | 4 | |
$E$24 | constraint 2 total | 20 | 75 | 20 | 2.666666667 | 8 | |
$E$25 | constraint 3 total | 3 | 0 | 4 | 1E+30 | 1 |
Based on the table, if the variation interval for objective coefficient 300 is , the current optimal solution is held. In addition, if the variation interval for the objective coefficient 400 is , the current optimal solution is also held.