This is a good clarification.
|Table 1 – Summary Table
|Day Full-Time Employee Starts Shift
||Minimum Number of Workers For Day
|1 = Monday
|2 = Tuesday
|3 = Wednesday
|4 = Thursday
|5 = Friday
|6 = Saturday
|7 = Sunday
These are the decision variables for this problem:
|Table 2 – Decision Variables
||Number of employees starting on Monday
||Number of employees starting on Tuesday
||Number of employees starting on Wednesday
||Number of employees starting on Thursday
||Number of employees starting on Friday
||Number of employees starting on Saturday
||Number of employees starting on Sunday
The results from each of these decision variables specifically show the number of workers beginning their 5-day shift on that particular day. For instance, the result for X1 represents the number of employees starting their 5-day long shift on Monday; the result for X2 represents the number of employees starting their 5-day long shift on Tuesday.
The goal of this problem is to minimize the number of employees to fulfill the Post Office’s daily workforce size demand. Since the decision variables quantify the number of employees starting on each day, there is no error of duplicity. Therefore, the following objective function has been made:
Min z =X1+X2+X3+X4+X5+X6+X7
The main constraint in this problem is the specific number of employees needed per day, as shown:
One can interpret the chart as this: “For Monday, X1, X4, X5, X6, and X7 employees will be working. And, the required number of employees working on Monday is 17 people. Therefore, the sum of X1, X4, X5, X6, and X7 must be at least 17.” It should be noted that X1 ≠ X1, because X1 means the number of employees beginning their 5-day long shift on Monday, and X1 refers to the group of employees beginning their 5-day long shift on Monday. In other words, X1 is the number of people in X1.
By using the information in Figure 1, the S.T. Equations have been created, as shown below:
Another constraint that should be taken into consideration is sign restriction, as shown below:
X1, X2, X3, X4, X5, X6, X7 ≥0
This constraint is necessary because the essence of the problem can never allow a negative number of people. With this restriction, it prohibits the optimum solution from having any negative values for the decision variables.