An appliance repair shop has 5 vacuum cleaners, 12 TV sets, and 18 VCRs to be repaired. The store employs two part time repairmen. One repairman can repair at least one vacuum cleaner, three TV sets and three VCRs in 1 week, while the second repairman can repair at least one vacuum cleaner, two TV sets and six VCRs in 1 week. The first employee is paid $250 a week and the second is paid $220 a week. To minimise the cost (in terms of wages paid), how many weeks should each of the two repairmen be employed? Clearly outline the objective function and the constraints.
Objective function: Minimize the cost of wages paid.
Let:
- x = number of weeks the first repairman is employed
- y = number of weeks the second repairman is employed
The cost of wages paid can be calculated using the following equation:
Cost = (x * $250) + (y * $220)
Constraints:
1. Number of vacuum cleaners repaired:
- First repairman can repair at least one vacuum cleaner per week.
- Therefore, the number of vacuum cleaners repaired by the first repairman in x weeks should be at least x.
- Second repairman can also repair at least one vacuum cleaner per week.
- Therefore, the number of vacuum cleaners repaired by the second repairman in y weeks should be at least y.
- Total number of vacuum cleaners repaired should be equal to or less than 5.
- Therefore, x + y ≤ 5.
2. Number of TV sets repaired:
- First repairman can repair at least three TV sets per week.
- Therefore, the number of TV sets repaired by the first repairman in x weeks should be at least 3x.
- Second repairman can repair at least two TV sets per week.
- Therefore, the number of TV sets repaired by the second repairman in y weeks should be at least 2y.
- Total number of TV sets repaired should be equal to or less than 12.
- Therefore, 3x + 2y ≤ 12.
3. Number of VCRs repaired:
- First repairman can repair at least three VCRs per week.
- Therefore, the number of VCRs repaired by the first repairman in x weeks should be at least 3x.
- Second repairman can repair at least six VCRs per week.
- Therefore, the number of VCRs repaired by the second repairman in y weeks should be at least 6y.
- Total number of VCRs repaired should be equal to or less than 18.
- Therefore, 3x + 6y ≤ 18.
To summarize, the optimization problem can be formulated as follows:
Minimize: Cost = (x * $250) + (y * $220)
Subject to:
- x + y ≤ 5
- 3x + 2y ≤ 12
- 3x + 6y ≤ 18
Objective: Minimize the cost of wages paid to the repairmen.
Let:
x = number of weeks the first repairman should be employed
y = number of weeks the second repairman should be employed
The cost of wages can be calculated as follows:
Cost = (x * 250) + (y * 220)
Constraints:
1. The repairmen can only work a whole number of weeks, so x and y must be integers.
2. The repairmen have limited capacity to fix appliances.
- Repairman 1 can repair at least 1 vacuum cleaner, 3 TV sets, and 3 VCRs per week:
x <= 5 (for vacuum cleaners), x * 3 <= 12 (for TV sets), x * 3 <= 18 (for VCRs)
- Repairman 2 can repair at least 1 vacuum cleaner, 2 TV sets, and 6 VCRs per week:
y <= 5 (for vacuum cleaners), y * 2 <= 12 (for TV sets), y * 6 <= 18 (for VCRs)
Therefore, the mathematical formulation is as follows:
Minimize: Z = (x * 250) + (y * 220)
Subject to:
x <= 5
x * 3 <= 12
x * 3 <= 18
y <= 5
y * 2 <= 12
y * 6 <= 18
Note: Since the question asks for the minimum cost, there are no maximum limits given for the number of weeks each repairman can be employed.
Objective function: Minimize the cost (in terms of wages paid)
Let's define the decision variables:
Let x be the number of weeks the first repairman is employed
Let y be the number of weeks the second repairman is employed
Now, we need to formulate the constraints based on the repair capacity and the number of appliances to be repaired.
For the first repairman:
1 vacuum cleaner requires 1 week (x)
3 TV sets require 3 weeks (3x)
3 VCRs require 3 weeks (3x)
So the repair capacity of the first repairman can be represented as:
x + 3x + 3x <= 5 (vacuum cleaners)
This can be simplified to:
7x <= 5
For the second repairman:
1 vacuum cleaner requires 1 week (y)
2 TV sets require 2 weeks (2y)
6 VCRs requires 6 weeks (6y)
So the repair capacity of the second repairman can be represented as:
y + 2y + 6y <= 12 + 18
This can be simplified to:
9y <= 30
Additional constraint:
Both x and y should be non-negative (x >= 0, y >= 0)
Now, we can formulate the mathematical model as follows:
Minimize: 250x + 220y (cost function)
Subject to:
7x <= 5
9y <= 30
x >= 0
y >= 0
To solve the problem and find the optimal solution, we can use linear programming techniques such as the simplex method or graphical methods.