To determine the earliest time all 120 parts can be finished, we must analyze the production rates, setup times, and the constraint that partial parts do not count (integer completions). ### 1. Analyze Machine Parameters **Machine A:** * Production Rate ($R_A$): 8 parts/hour. * Setup Time ($S_A$): 30 minutes = $0.5$ hours. * Start Time: $0.5$ hours. * Production Time per part ($T_{A,part}$): $1/8$ hours = $7.5$ minutes. **Machine B:** * Production Rate ($R_B$): 5 parts/hour. * Setup Time ($S_B$): 12 minutes = $0.2$ hours. * Start Time: $0.2$ hours. * Production Time per part ($T_{B,part}$): $1/5$ hours = $12$ minutes. ### 2. Determine Total Parts Produced by Time $t$ Let $t$ be the total time elapsed in hours from time zero. The number of parts produced by a machine is determined by the time available for production ($t - S$) divided by the time per part, rounded down to the nearest integer (since partial parts don't count). * **Parts from A ($N_A$):** $$ N_A(t) = \left\lfloor \frac{t - 0.5}{1/8} \right\rfloor = \lfloor 8(t - 0.5) \rfloor $$ *(Note: If $t < 0.5$, $N_A = 0$)* * **Parts from B ($N_B$):** $$ N_B(t) = \left\lfloor \frac{t - 0.2}{1/5} \right\rfloor = \lfloor 5(t - 0.2) \rfloor $$ *(Note: If $t < 0.2$, $N_B = 0$)* We need to find the minimum $t$ such that: $$ N_A(t) + N_B(t) \ge 120 $$ ### 3. Estimate the Completion Time First, let's estimate the time assuming continuous production (ignoring integer constraints) to get a baseline. Total Rate = $8 + 5 = 13$ parts/hour. However, machines don't start at $t=0$. Let's approximate the average start time. Machine A starts later. Let's try to find an approximate $t$. If both ran continuously from $t=0$, time would be $120/13 \approx 9.23$ hours. Since Machine A loses 0.5 hours and Machine B loses 0.2 hours, the effective working time is reduced. Let's test values around $t = 10$ hours. **Test $t = 10$ hours:** * **Machine A:** * Available time: $10 - 0.5 = 9.5$ hours. * Parts: $\lfloor 8 \times 9.5 \rfloor = \lfloor 76 \rfloor = 76$ parts. * **Machine B:** * Available time: $10 - 0.2 = 9.8$ hours. * Parts: $\lfloor 5 \times 9.8 \rfloor = \lfloor 49 \rfloor = 49$ parts. * **Total:** $76 + 49 = 125$ parts. * This exceeds 120. The answer is likely less than 10 hours. **Test $t = 9$ hours:** * **Machine A:** * Available time: $9 - 0.5 = 8.5$ hours. * Parts: $\lfloor 8 \times 8.5 \rfloor = \lfloor 68 \rfloor = 68$ parts. * **Machine B:** * Available time: $9 - 0.2 = 8.8$ hours. * Parts: $\lfloor 5 \times 8.8 \rfloor = \lfloor 44 \rfloor = 44$ parts. * **Total:** $68 + 44 = 112$ parts. * This is less than 120. The answer is between 9 and 10 hours. ### 4. Refine the Search We need 8 more parts to reach 120 from the 112 produced at $t=9$. Let's check specific integer completion times. Since parts are produced in discrete steps, we should look at the times when either machine finishes a part. Let's calculate the number of parts produced by each machine at specific intervals or solve for the required counts. Let $n_A$ be parts from A and $n_B$ be parts from B. $n_A + n_B = 120$. Time required for Machine A to produce $n_A$ parts: $T_A = 0.5 + \frac{n_A}{8}$ hours. Time required for Machine B to produce $n_B$ parts: $T_B = 0.2 + \frac{n_B}{5}$ hours. The total time $t$ must be at least $\max(T_A, T_B)$. We want to minimize this maximum time. Since Machine A is faster (8/hr vs 5/hr), it will likely produce more parts. Let's test distributions of parts. **Scenario 1: Machine A produces 70 parts, Machine B produces 50 parts.** * $T_A = 0.5 + 70/8 = 0.5 + 8.75 = 9.25$ hours. * $T_B = 0.2 + 50/5 = 0.2 + 10 = 10.2$ hours. * Max Time = 10.2 hours. (Too slow, we saw 10 hours was enough). **Scenario 2: Machine A produces 76 parts, Machine B produces 44 parts.** * $T_A = 0.5 + 76/8 = 0.5 + 9.5 = 10.0$ hours. * $T_B = 0.2 + 44/5 = 0.2 + 8.8 = 9.0$ hours. * Max Time = 10.0 hours. Let's check if we can finish earlier by balancing the load better or checking the integer constraints near $t=9.5$. **Check $t = 9.5$ hours:** * **Machine A:** * Available time: $9.5 - 0.5 = 9.0$ hours. * Parts: $\lfloor 8 \times 9.0 \rfloor = 72$ parts. * **Machine B:** * Available time: $9.5 - 0.2 = 9.3$ hours. * Parts: $\lfloor 5 \times 9.3 \rfloor = \lfloor 46.5 \rfloor = 46$ parts. * **Total:** $72 + 46 = 118$ parts. * Still short by 2 parts. **Check $t = 9.6$ hours:** * **Machine A:** * Available time: $9.6 - 0.5 = 9.1$ hours. * Parts: $\lfloor 8 \times 9.1 \rfloor = \lfloor 72.8 \rfloor = 72$ parts. * **Machine B:** * Available time: $9.6 - 0.2 = 9.4$ hours. * Parts: $\lfloor 5 \times 9.4 \rfloor = \lfloor 47 \rfloor = 47$ parts. * **Total:** $72 + 47 = 119$ parts. * Still short by 1 part. **Check $t = 9.7$ hours:** * **Machine A:** * Available time: $9.7 - 0.5 = 9.2$ hours. * Parts: $\lfloor 8 \times 9.2 \rfloor = \lfloor 73.6 \rfloor = 73$ parts. * **Machine B:** * Available time: $9.7 - 0.2 = 9.5$ hours. * Parts: $\lfloor 5 \times 9.5 \rfloor = \lfloor 47.5 \rfloor = 47$ parts. * **Total:** $73 + 47 = 120$ parts. * **Target Reached.** ### 5. Verify the "Immediately Preceding Completion Time" The question asks to check the immediately preceding completion time. This implies we should verify if the count was lower just before this moment or if there is a gap between integer completions. At $t = 9.7$ hours: * Machine A finished its 73rd part at exactly $T_{A,73} = 0.5 + \frac{73}{8} = 0.5 + 9.125 = 9.625$ hours. * Machine B finished its 47th part at exactly $T_{B,47} = 0.2 + \frac{47}{5} = 0.2 + 9.4 = 9.6$ hours. Wait, let's re-evaluate the exact completion times of the specific parts that push the total to 120. At $t=9.7$, we have 73 from A and 47 from B. Let's look at the completion times of the **last** parts produced by each machine to reach these counts. 1. **Machine A:** * To produce 73 parts, it needs $73/8 = 9.125$ hours of running time. * Start time: 0.5 hours. * Completion time of 73rd part: $0.5 + 9.125 = 9.625$ hours. * At any time $t \ge 9.625$, Machine A has produced 73 parts. 2. **Machine B:** * To produce 47 parts, it needs $47/5 = 9.4$ hours of running time. * Start time: 0.2 hours. * Completion time of 47th part: $0.2 + 9.4 = 9.6$ hours. * At any time $t \ge 9.6$, Machine B has produced 47 parts. **Combining them:** * At $t = 9.6$ hours: * Machine B has just finished its 47th part. * Machine A has been running for $9.6 - 0.5 = 9.1$ hours. * Parts from A: $\lfloor 8 \times 9.1 \rfloor = \lfloor 72.8 \rfloor = 72$ parts. * Total Parts: $72 + 47 = 119$ parts. * At $t = 9.625$ hours ($9$ hours and $37.5$ minutes): * Machine A has just finished its 73rd part. * Machine B has been running for $9.625 - 0.2 = 9.425$ hours. * Parts from B: $\lfloor 5 \times 9.425 \rfloor = \lfloor 47.125 \rfloor = 47$ parts. * Total Parts: $73 + 47 = 120$ parts. Therefore, the 120th part is completed exactly when Machine A finishes its 73rd part, because Machine B had already finished its 47th part earlier (at 9.6 hours). ### 6. Final Calculation and Formatting **Step-by-step Arithmetic:** 1. **Identify Critical Completion Times:** We need $N_A + N_B = 120$. Let's test the distribution where Machine A does the bulk of the work due to higher speed. If $N_A = 73$ and $N_B = 47$: * Time for A to make 73 parts: $T_A = 0.5 + \frac{73}{8} = 0.5 + 9.125 = 9.625$ hours. * Time for B to make 47 parts: $T_B = 0.2 + \frac{47}{5} = 0.2 + 9.4 = 9.6$ hours. * Since $T_A > T_B$, Machine A is the bottleneck for this specific distribution. The total time is $9.625$ hours. Let's check if a different distribution yields a faster time. If $N_A = 72$ and $N_B = 48$: * Time for A to make 72 parts: $T_A = 0.5 + \frac{72}{8} = 0.5 + 9 = 9.5$ hours. * Time for B to make 48 parts: $T_B = 0.2 + \frac{48}{5} = 0.2 + 9.6 = 9.8$ hours. * Since $T_B > T_A$, Machine B is the bottleneck. Total time is $9.8$ hours. * $9.8 > 9.625$, so this is slower. If $N_A = 74$ and $N_B = 46$: * Time for A to make 74 parts: $T_A = 0.5 + \frac{74}{8} = 0.5 + 9.25 = 9.75$ hours. * Time for B to make 46 parts: $T_B = 0.2 + \frac{46}{5} = 0.2 + 9.2 = 9.4$ hours. * Since $T_A > T_B$, Machine A is the bottleneck. Total time is $9.75$ hours. * $9.75 > 9.625$, so this is slower. The optimal distribution is $N_A=73, N_B=47$. 2. **Convert to Minutes:** $0.625$ hours $= 0.625 \times 60$ minutes $= 37.5$ minutes. Total Time $= 9$ hours and $37.5$ minutes. 3. **Check Immediately Preceding Completion Time:** The question asks to check the immediately preceding completion time. The last part to be completed is the 73rd part of Machine A. Before this part was completed (i.e., at any time $t < 9.625$), Machine A had only completed 72 parts. At $t = 9.625 - \epsilon$ (just before A finishes part 73): * Machine A has completed 72 parts. * Machine B has completed 47 parts (since B finished part 47 at $t=9.6$, which is before $9.625$). * Total parts $= 72 + 47 = 119$. Thus, the count jumps from 119 to 120 at $t = 9.625$ hours. ### Conclusion The earliest time all 120 parts can be finished is **9.625 hours** (or **9 hours and 37.5 minutes**). **Answer:** The earliest time all 120 parts can be finished is **9.625 hours** (or **9 hours and 37.5 minutes**). **Arithmetic Summary:** * **Optimal Allocation:** Machine A produces 73 parts; Machine B produces 47 parts. * **Machine A Time:** $0.5 \text{ h} + \frac{73}{8} \text{ h} = 0.5 + 9.125 = 9.625 \text{ h}$. * **Machine B Time:** $0.2 \text{ h} + \frac{47}{5} \text{ h} = 0.2 + 9.4 = 9.6 \text{ h}$. * **Bottleneck:** Machine A determines the completion time ($9.625 > 9.6$). * **Preceding Check:** At $t < 9.625$, Machine A has only produced 72 parts. With Machine B's 47 parts, the total is $72+47=119$. At $t=9.625$, Machine A completes its 73rd part, bringing the total to $73+47=120$.