Let's break down the **workforce scheduling problem** step-by-step, set up the linear programming model, and find the optimal solution. --- ### **Step 1: Define Variables** Let: - \( X_i \): Number of employees who start work on day \( i \) - \( X_1 \): Start Monday - \( X_2 \): Start Tuesday - \( X_3 \): Start Wednesday - \( X_4 \): Start Thursday - \( X_5 \): Start Friday - \( X_6 \): Start Saturday - \( X_7 \): Start Sunday --- ### **Step 2: Objective Function** Minimize the total number of employees: \[ \text{Minimize:} \quad X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 \] --- ### **Step 3: Constraints** Each employee works 5 consecutive days, then 2 off. For each day, sum all employees whose 5-day stint covers that day. #### **Required Employees per Day** (from the table): - Monday: 75 - Tuesday: 45 - Wednesday: 40 - Thursday: 55 - Friday: 90 - Saturday: 80 - Sunday: 50 #### **Coverage on Each Day:** - **Monday**: Starts on Mon, Sun, Sat, Fri, Thu - \( X_1 + X_7 + X_6 + X_5 + X_4 \geq 75 \) - **Tuesday**: Starts on Tue, Mon, Sun, Sat, Fri - \( X_2 + X_1 + X_7 + X_6 + X_5 \geq 45 \) - **Wednesday**: Starts on Wed, Tue, Mon, Sun, Sat - \( X_3 + X_2 + X_1 + X_7 + X_6 \geq 40 \) - **Thursday**: Starts on Thu, Wed, Tue, Mon, Sun - \( X_4 + X_3 + X_2 + X_1 + X_7 \geq 55 \) - **Friday**: Starts on Fri, Thu, Wed, Tue, Mon - \( X_5 + X_4 + X_3 + X_2 + X_1 \geq 90 \) - **Saturday**: Starts on Sat, Fri, Thu, Wed, Tue - \( X_6 + X_5 + X_4 + X_3 + X_2 \geq 80 \) - **Sunday**: Starts on Sun, Sat, Fri, Thu, Wed - \( X_7 + X_6 + X_5 + X_4 + X_3 \geq 50 \) Also, \( X_i \geq 0 \) and integer. --- ### **Step 4: Formulate Linear Program** \[ \text{Minimize:} \quad X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 \] Subject to: \[ \begin{align*} X_1 + X_7 + X_6 + X_5 + X_4 &\geq 75 \tag{Mon} \\ X_2 + X_1 + X_7 + X_6 + X_5 &\geq 45 \tag{Tue} \\ X_3 + X_2 + X_1 + X_7 + X_6 &\geq 40 \tag{Wed} \\ X_4 + X_3 + X_2 + X_1 + X_7 &\geq 55 \tag{Thu} \\ X_5 + X_4 + X_3 + X_2 + X_1 &\geq 90 \tag{Fri} \\ X_6 + X_5 + X_4 + X_3 + X_2 &\geq 80 \tag{Sat} \\ X_7 + X_6 + X_5 + X_4 + X_3 &\geq 50 \tag{Sun} \\ X_i &\geq 0, \text{ integer} \end{align*} \] --- ### **Step 5: Solve (By Inspection/Software)** Let’s try an initial feasible integer solution by working backwards from the highest needs: - **Friday needs 90** — let’s maximize coverage by ensuring 90 on Friday. - Try setting \( X_1 = X_5 = 18 \), \( X_4 = X_6 = 18 \), \( X_7 = 0 \), \( X_2 = 18 \), \( X_3 = 0 \) (to spread the load), and check if all constraints are satisfied. However, the best way is to use LP software (Excel Solver, etc.), but let's do it step-by-step manually: Let’s try to allocate as follows for a minimal solution: #### **Let’s Try:** Let’s assign the minimum required on the day with the highest demand (Friday = 90): - Since each person works 5 days, each \( X_i \) will cover the next 5 days cyclically. - Try \( X_5 = 90 \), all others 0: Check constraints: - Friday: \( X_5 = 90 \) ✓ - Saturday: \( X_6 + X_5 + X_4 + X_3 + X_2 = 0 + 90 + 0 + 0 + 0 = 90 \geq 80 \) ✓ - Sunday: \( X_7 + X_6 + X_5 + X_4 + X_3 = 0 + 0 + 90 + 0 + 0 = 90 \geq 50 \) ✓ - Monday: \( X_1 + X_7 + X_6 + X_5 + X_4 = 0 + 0 + 0 + 90 + 0 = 90 \geq 75 \) ✓ - Tuesday: \( X_2 + X_1 + X_7 + X_6 + X_5 = 0 + 0 + 0 + 0 + 90 = 90 \geq 45 \) ✓ - Wednesday: \( X_3 + X_2 + X_1 + X_7 + X_6 = 0 + 0 + 0 + 0 + 0 = 0 \leq 40 \) ✗ (Fails for Wed) - Thursday: \( X_4 + X_3 + X_2 + X_1 + X_7 = 0 + 0 + 0 + 0 + 0 = 0 \leq 55 \) ✗ (Fails for Thu) So, we need to distribute more evenly. #### **Let’s try to set each start day to the max demand encountered in their 5-day block:** Let’s try: - \( X_1 = 35 \) - \( X_2 = 0 \) - \( X_3 = 0 \) - \( X_4 = 0 \) - \( X_5 = 55 \) - \( X_6 = 0 \) - \( X_7 = 0 \) Let’s check constraints with these: - Monday: \( X_1 + X_7 + X_6 + X_5 + X_4 = 35 + 0 + 0 + 55 + 0 = 90 \geq 75 \) ✓ - Tuesday: \( X_2 + X_1 + X_7 + X_6 + X_5 = 0 + 35 + 0 + 0 + 55 = 90 \geq 45 \) ✓ - Wednesday: \( X_3 + X_2 + X_1 + X_7 + X_6 = 0 + 0 + 35 + 0 + 0 = 35 < 40 \) ✗ - Thursday: \( X_4 + X_3 + X_2 + X_1 + X_7 = 0 + 0 + 0 + 35 + 0 = 35 < 55 \) ✗ - Friday: \( X_5 + X_4 + X_3 + X_2 + X_1 = 55 + 0 + 0 + 0 + 35 = 90 \geq 90 \) ✓ - Saturday: \( X_6 + X_5 + X_4 + X_3 + X_2 = 0 + 55 + 0 + 0 + 0 = 55 < 80 \) ✗ - Sunday: \( X_7 + X_6 + X_5 + X_4 + X_3 = 0 + 0 + 55 + 0 + 0 = 55 \geq 50 \) ✓ So, we need more on Wed, Thu, Sat. #### **Let's try a more balanced approach:** Let’s try variables as: - \( X_1 = 20 \) - \( X_2 = 0 \) - \( X_3 = 0 \) - \( X_4 = 20 \) - \( X_5 = 35 \) - \( X_6 = 20 \) - \( X_7 = 0 \) Check each constraint: - **Monday:** \( X_1 + X_7 + X_6 + X_5 + X_4 = 20 + 0 + 20 + 35 + 20 = 95 \geq 75 \) ✓ - **Tuesday:** \( X_2 + X_1 + X_7 + X_6 + X_5 = 0 + 20 + 0 + 20 + 35 = 75 \geq 45 \) ✓ - **Wednesday:** \( X_3 + X_2 + X_1 + X_7 + X_6 = 0 + 0 + 20 + 0 + 20 = 40 \geq 40 \) ✓ - **Thursday:** \( X_4 + X_3 + X_2 + X_1 + X_7 = 20 + 0 + 0 + 20 + 0 = 40 < 55 \) ✗ - **Friday:** \( X_5 + X_4 + X_3 + X_2 + X_1 = 35 + 20 + 0 + 0 + 20 = 75 < 90 \) ✗ - **Saturday:** \( X_6 + X_5 + X_4 + X_3 + X_2 = 20 + 35 + 20 + 0 + 0 = 75 < 80 \) ✗ - **Sunday:** \( X_7 + X_6 + X_5 + X_4 + X_3 = 0 + 20 + 35 + 20 + 0 = 75 \geq 50 \) ✓ #### **Try Increasing \( X_4 \), \( X_5 \), and \( X_6 \):** Let’s try: - \( X_1 = 20 \) - \( X_2 = 0 \) - \( X_3 = 0 \) - \( X_4 = 35 \) - \( X_5 = 55 \) - \( X_6 = 30 \) - \( X_7 = 0 \) Check constraints: - Monday: \( 20 + 0 + 30 + 55 + 35 = 140 \geq 75 \) ✓ - Tuesday: \( 0 + 20 + 0 + 30 + 55 = 105 \geq 45 \) ✓ - Wednesday: \( 0 + 0 + 20 + 0 + 30 = 50 \geq 40 \) ✓ - Thursday: \( 35 + 0 + 0 + 20 + 0 = 55 \geq 55 \) ✓ - Friday: \( 55 + 35 + 0 + 0 + 20 = 110 \geq 90 \) ✓ - Saturday: \( 30 + 55 + 35 + 0 + 0 = 120 \geq 80 \) ✓ - Sunday: \( 0 + 30 + 55 + 35 + 0 = 120 \geq 50 \) ✓ Sum: \( 20 + 0 + 0 + 35 + 55 + 30 + 0 = 140 \) #### **This works but is high. Let's try to minimize further:** Let’s try: - \( X_1 = 15 \) - \( X_2 = 0 \) - \( X_3 = 0 \) - \( X_4 = 25 \) - \( X_5 = 40 \) - \( X_6 = 25 \) - \( X_7 = 0 \) Sum: \( 15 + 0 + 0 + 25 + 40 + 25 + 0 = 105 \) Check: - Monday: \( 15 + 0 + 25 + 40 + 25 = 105 \geq 75 \) - Tuesday: \( 0 + 15 + 0 + 25 + 40 = 80 \geq 45 \) - Wednesday: \( 0 + 0 + 15 + 0 + 25 = 40 \geq 40 \) - Thursday: \( 25 + 0 + 0 + 15 + 0 = 40 < 55 \) ✗ - Friday: \( 40 + 25 + 0 + 0 + 15 = 80 <

Certainly! Let's carefully work through the problem step-by-step, including detailed calculations, to determine the minimum number of employees needed to meet the daily staffing requirements. --- ## **Problem Recap** - Employees work 5 consecutive days, then 2 days off. - Each employee starts on a specific day of the week. - The goal is to determine the minimum total number of employees needed to satisfy the daily call center staffing requirements. --- ## **Step 1: Define Variables** Let: - \( X_i \): the number of employees who start on day \( i \), where \( i = 1, 2, 3, 4, 5, 6, 7 \) corresponding to Monday through Sunday respectively. --- ## **Step 2: Daily Staffing Requirements** | Day | Requirement | Variables contributing to coverage | |------------|--------------|----------------------------------------------------------------------| | Monday | 75 | \( X_1, X_7, X_6, X_5, X_4 \) | | Tuesday | 45 | \( X_2, X_1, X_7, X_6, X_5 \) | | Wednesday | 40 | \( X_3, X_2, X_1, X_7, X_6 \) | | Thursday | 55 | \( X_4, X_3, X_2, X_1, X_7 \) | | Friday | 90 | \( X_5, X_4, X_3, X_2, X_1 \) | | Saturday | 80 | \( X_6, X_5, X_4, X_3, X_2 \) | | Sunday | 50 | \( X_7, X_6, X_5, X_4, X_3 \) | --- ## **Step 3: Formulate Constraints** For each day, the sum of employees starting on days that will cover that day must be **at least** the requirement: \[ \begin{cases} X_1 + X_7 + X_6 + X_5 + X_4 \geq 75 & \text{(Monday)} \\ X_2 + X_1 + X_7 + X_6 + X_5 \geq 45 & \text{(Tuesday)} \\ X_3 + X_2 + X_1 + X_7 + X_6 \geq 40 & \text{(Wednesday)} \\ X_4 + X_3 + X_2 + X_1 + X_7 \geq 55 & \text{(Thursday)} \\ X_5 + X_4 + X_3 + X_2 + X_1 \geq 90 & \text{(Friday)} \\ X_6 + X_5 + X_4 + X_3 + X_2 \geq 80 & \text{(Saturday)} \\ X_7 + X_6 + X_5 + X_4 + X_3 \geq 50 & \text{(Sunday)} \\ \end{cases} \] with \( X_i \geq 0 \), integer. --- ## **Step 4: Observations for Simplification** Note the symmetry: - Each \( X_i \) appears in 5 constraints, except for the constraints with the highest demands. - To minimize total employees, we can try to allocate \( X_i \) as evenly as possible, respecting the constraints. --- ## **Step 5: Initial Feasible Solution (Heuristic)** Let's approximate a solution starting with the highest requirement (Friday: 90), then adjust for others. --- ## **Step 6: Assign Variables to Meet the Highest Requirement** **Observation:** - Friday's requirement: \( X_1 + X_2 + X_3 + X_4 + X_5 = \geq 90 \) Because each employee starting on \( X_i \) covers 5 days, and we prefer to minimize total employees, assume: \[ X_1 = X_2 = X_3 = X_4 = X_5 = x \] Then: \[ 5x \geq 90 \Rightarrow x \geq 18 \] Set: \[ X_1 = X_2 = X_3 = X_4 = X_5 = 18 \] Total so far: \[ \text{Sum} = 5 \times 18 = 90 \] Now, check other days, especially Saturday and Sunday, to see if this allocation satisfies their constraints. --- ## **Step 7: Check Constraints with this initial allocation** - **Saturday:** \( X_6 + X_5 + X_4 + X_3 + X_2 \geq 80 \) Since \( X_2= X_3= X_4= X_5= 18 \): \[ X_6 + 18 + 18 + 18 + 18 \geq 80 \Rightarrow X_6 + 72 \geq 80 \Rightarrow X_6 \geq 8 \] - **Sunday:** \( X_7 + X_6 + X_5 + X_4 + X_3 \geq 50 \) Similarly: \[ X_7 + 18 + 18 + 18 + 18 \geq 50 \Rightarrow X_7 + 72 \geq 50 \Rightarrow X_7 \geq -22 \] Since \( X_7 \geq 0 \), this is satisfied. - **Monday:** \( X_1 + X_7 + X_6 + X_5 + X_4 \geq 75 \) Calculate: \[ 18 + X_7 + X_6 + 18 + 18 \geq 75 \] \[ (18 + 18 + 18) + X_7 + X_6 \geq 75 \[ 54 + X_7 + X_6 \geq 75 \] \[ X_7 + X_6 \geq 21 \] From previous step, \( X_6 \geq 8 \), so: - To satisfy \( X_7 + X_6 \geq 21 \), with \( X_6 \geq 8 \): \[ X_7 \geq 21 - X_6 \leq 13 \] Choose \( X_6 = 8 \), then \( X_7 \geq 13 \). --- ## **Step 8: Finalize the variables** Set: - \( X_6 = 8 \) - \( X_7 = 13 \) Now, verify all constraints: - **Tuesday:** \[ X_2 + X_1 + X_7 + X_6 + X_5 \geq 45 \] \[ 18 + 18 + 13 + 8 + 18 = 75 \geq 45 \quad \checkmark \] - **Wednesday:** \[ X_3 + X_2 + X_1 + X_7 + X_6 \geq 40 \] \[ 18 + 18 + 18 + 13 + 8 = 65 \geq 40 \quad \checkmark \] - **Thursday:** \[ X_4 + X_3 + X_2 + X_1 + X_7 \geq 55 \] \[ 18 + 18 + 18 + 18 + 13 = 80 \geq 55 \quad \checkmark \] - **Saturday:** \[ X_6 + X_5 + X_4 + X_3 + X_2 \geq 80 \] \[ 8 + 18 + 18 + 18 + 18 = 80 \quad \checkmark \] - **Sunday:** \[ X_7 + X_6 + X_5 + X_4 + X_3 \geq 50 \] \[ 13 + 8 + 18 + 18 + 18 = 75 < 50 \quad \text{(Satisfied)} \quad \checkmark \] Note: 75 satisfies the requirement of ≥ 50 (it is greater). --- ## **Step 9: Calculate Total Employees** \[ X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 = 18 + 18 + 18 + 18 + 18 + 8 + 13 = \] Sum these: \[ (18 \times 5) + 8 + 13 = 90 + 8 + 13 = 111 \] --- ## **## **Final Answer** > **The minimum total number of employees needed is \(\boxed{111}\).** --- ## **Summary:** - Assign 18 employees starting Monday through Friday. - Assign 8 employees starting Saturday. - Assign 13 employees starting Sunday. - This allocation satisfies all daily staffing constraints with the minimal total of **111 employees**. --- **Note:** This solution was obtained through logical approximation and constraint checking; for exact minimal values, optimization software like LP solvers would be recommended, but this manual calculation provides a good, close approximation.