1. **State the problem:** Calculate the cumulative distance based on the overall movement of material between processes for the given layout. 2. **Given data:** - Flow matrix (material movement between processes): $$\begin{bmatrix}0 & 100 & 55 & 0 & 0 & 45 \\ 20 & 0 & 0 & 50 & 0 & 5 \\ 20 & 0 & 0 & 55 & 0 & 0 \\ 0 & 20 & 0 & 0 & 25 & 0 \\ 55 & 0 & 120 & 0 & 0 & 0 \\ 15 & 0 & 20 & 0 & 0 & 15 \end{bmatrix}$$ - Distance matrix (feet) between rooms 1 to 6: $$\begin{bmatrix}- & 20 & 40 & 20 & 40 & 60 \\ - & - & 20 & 40 & 20 & 40 \\ - & - & - & 60 & 40 & 20 \\ - & - & - & - & 20 & 40 \\ - & - & - & - & - & 20 \\ - & - & - & - & - & - \end{bmatrix}$$ 3. **Room assignments:** Room 1 = Benders, Room 2 = Materials, Room 3 = Welding, Room 4 = Drills, Room 5 = Grinders, Room 6 = Lathes. 4. **Calculate cumulative distance:** The cumulative distance is the sum over all pairs of processes of (flow between processes) $\times$ (distance between their rooms). 5. **Find distances for each pair:** Since the distance matrix is upper triangular, use symmetry: distance(i,j) = distance(j,i). 6. **Calculate each term:** - Materials (Room 2) to Welding (Room 3): flow = 100, distance = 20 - Materials to Drills (Room 4): flow = 55, distance = 40 - Materials to Benders (Room 1): flow = 45, distance = 20 - Welding (Room 3) to Materials (Room 2): flow = 20, distance = 20 - Welding to Drills (Room 4): flow = 50, distance = 40 - Welding to Benders (Room 1): flow = 5, distance = 40 - Drills (Room 4) to Materials (Room 2): flow = 20, distance = 40 - Drills to Welding (Room 3): flow = 55, distance = 40 - Drills to Grinders (Room 5): flow = 25, distance = 20 - Grinders (Room 5) to Materials (Room 2): flow = 55, distance = 20 - Grinders to Welding (Room 3): flow = 120, distance = 40 - Benders (Room 1) to Materials (Room 2): flow = 15, distance = 20 - Benders to Welding (Room 3): flow = 20, distance = 40 - Benders to Benders (Room 1): flow = 15, distance = 0 (same room) 7. **Sum all products:** $$ \begin{aligned} &100 \times 20 + 55 \times 40 + 45 \times 20 + 20 \times 20 + 50 \times 40 + 5 \times 40 + 20 \times 40 + 55 \times 40 + 25 \times 20 + 55 \times 20 + 120 \times 40 + 15 \times 20 + 20 \times 40 + 15 \times 0 \\ =& 2000 + 2200 + 900 + 400 + 2000 + 200 + 800 + 2200 + 500 + 1100 + 4800 + 300 + 800 + 0 \\ =& 20700 \end{aligned} $$ 8. **Final answer:** The cumulative distance based on the overall movement of material between processes for this layout is **20700** feet.