Questions & explanations
1. Use Lawler's algorithm to schedule jobs: A (p=3, d=5), B (p=2, d=4), C (p=1, d=3). What is the optimal order?
Total processing time = 6. Jobs that can be last: those with d >= 6? None. So we need to consider jobs that can be last in a partial schedule. Actually, Lawler's algorithm: at each step, consider jobs whose due date is at least the sum of remaining processing times. Initially sum=6, no job has d>=6. So we look at jobs with d >= sum of remaining minus something? Standard algorithm: let T be total remaining processing time. A job j can be last if d_j >= T. Here none, so we pick the job with largest p among all? That would be A (3). But that's not correct. Let me recalc: Actually, Lawler's algorithm for total tardiness: at each step, find the job that minimizes something. The correct algorithm: for the set of unscheduled jobs, compute for each job j the cost if it were last. The job with smallest cost is placed last. The cost is max(0, T - d_j) plus the optimal cost for the remaining jobs. This is complex. For this example, optimal order is C, B, A: completion times 1,3,6; tardiness 0,0,1; total 1.
2. A product has 20 failures in 100 hours of testing. After fixing, it has 10 failures in the next 100 hours. Use the Duane model to estimate the cumulative failure rate at 200 hours.
Cumulative failures at 100h: 20; at 200h: 30. Cumulative failure rate = failures / time. At 100h: 20/100=0.2 failures/hour; at 200h: 30/200=0.15 failures/hour. The Duane model says cumulative failure rate = a * time^(b-1). Using two points: 0.2 = a*100^(b-1); 0.15 = a*200^(b-1). Divide: 0.2/0.15 = (100/200)^(b-1) => 1.333 = 0.5^(b-1). Take log: log(1.333) = (b-1)*log(0.5) => b-1 = log(1.333)/log(0.5) = 0.1249/(-0.3010) = -0.415. So b=0.585. Then a = 0.2/100^(-0.415)=0.2/0.146=1.37. At 200h, cumulative failure rate = 1.37*200^(-0.415)=1.37*0.146=0.2? Wait recalc: Actually at 200h, cumulative failure rate = 30/200=0.15, so the model fits. To estimate at 300h, use formula: 1.37*300^(-0.415)=1.37*0.119=0.163 failures/hour? But cumulative failures would be 0.163*300=48.9, so about 49 failures. This shows the rate is decreasing.
3. A newsvendor buys newspapers for $1 and sells for $2. Demand is equally likely to be 50, 60, or 70. How many should he buy?
The critical ratio is (selling price - cost) / (selling price) = (2-1)/2 = 0.5. The optimal order quantity is the smallest demand such that the cumulative probability is at least 0.5. Cumulative probabilities: 50 (0.33), 60 (0.67). So order 60. If he orders 60, expected profit is (0.33*50 + 0.67*60)*1 - 60*1 = (16.5+40.2)-60 = -3.3? Wait recalc: Actually profit per paper sold is $1, unsold loss $1. Expected profit = 0.33*(50*1 -10*1) + 0.33*(60*1) + 0.33*(60*1) = 0.33*40 + 0.33*60 + 0.33*60 = 13.2+19.8+19.8=52.8? Better to compute: Order 60, if demand 50, sell 50, profit 50-10=40; if demand 60, profit 60; if demand 70, profit 60. Expected profit = (40+60+60)/3 = 53.33. Order 50 gives profit 50 always? Actually if demand 50, profit 50; if 60, profit 50 (lost sales); if 70, profit 50. Expected = 50. So order 60 is better.
4. A factory has 5 machines (M1-M5) and 6 parts (P1-P6). The routes are: P1: M1-M2-M3; P2: M1-M2-M4; P3: M1-M2-M3; P4: M5-M4; P5: M5-M4; P6: M5-M4. Use production flow analysis to suggest part families and machine cells.
First, list which machines each part visits. Parts P1, P2, P3 all visit M1 and M2. P1 and P3 also visit M3; P2 visits M4. So P1, P2, P3 share M1 and M2, but they diverge. Parts P4, P5, P6 all visit M5 and M4. So two natural families: Family A: P1, P2, P3 (use M1, M2, plus M3 or M4). Family B: P4, P5, P6 (use M5 and M4). For cell 1, you could put M1, M2, M3, M4 to make all of family A. But M4 is also used by family B. To avoid sharing, you might duplicate M4 or assign P2 to a different cell. A simple solution: Cell 1: M1, M2, M3 for P1 and P3; Cell 2: M5, M4 for P4-P6; and move P2 to Cell 2 (but P2 needs M1 and M2). Better: keep M4 in both cells? That is costly. So PFA shows that P2 causes conflict; you might redesign part P2 or accept some inter-cell movement.
5. Compare 'part family formation' in FMS with 'group technology cell formation' in cellular manufacturing. How are they similar and different?
Both methods group parts that have similar manufacturing needs to improve efficiency. In cellular manufacturing, parts are grouped into families that are made in a dedicated cell of machines. In FMS, part families are formed for scheduling purposes, but the machines are not dedicated; any machine can make any part if it has the right tools. The similarity is that both aim to reduce setup and changeover time. The difference is that in cellular manufacturing, the grouping is fixed and machines are physically arranged into cells. In FMS, the grouping is flexible and can change over time as tooling is changed. FMS also considers tool availability and magazine capacity, while cellular manufacturing focuses on machine grouping.
6. How can contract theory address information asymmetry in supply chains?
Information asymmetry occurs when one party knows more than the other, like a supplier knowing their true cost but the buyer does not. Contract theory designs mechanisms to reveal private information or align incentives. For example, a menu of contracts can be offered: a high-fixed-fee, low-variable-fee contract for efficient suppliers, and a low-fixed-fee, high-variable-fee contract for inefficient ones. Suppliers self-select the contract that suits them, revealing their type. This is called screening. Another approach is signaling, where the informed party takes an action to credibly show their type, like offering a warranty. These contracts improve coordination even when information is hidden.
7. A car factory makes three models: sedan, SUV, and hatchback. Each model needs different parts and assembly time. If you schedule all sedans first, then all SUVs, then all hatchbacks, what problem might happen?
If you schedule all sedans first, then all SUVs, then all hatchbacks, the factory will have long changeover times between model groups. Changeover is the time to switch a machine or line from making one model to another. During changeover, the line stops, so no cars are made. Also, workers and machines may be very busy during sedan time and then idle during hatchback time. This uneven workload can cause delays and waste. Mixed-model assembly sequencing mixes the models throughout the day to level the workload and reduce changeover time. For example, you might make two sedans, then one SUV, then two hatchbacks, then repeat. This keeps the line running smoothly and uses resources evenly.
8. Compare a wholesale price contract with a revenue-sharing contract using game theory.
Under a wholesale price contract, the manufacturer charges a fixed price per unit, and the retailer keeps all revenue. The retailer bears all demand risk, so they order less than the chain-optimal quantity. In a revenue-sharing contract, the manufacturer charges a lower wholesale price but gets a percentage of the retailer's revenue. This shares the risk and incentive. Game theory shows that revenue-sharing can achieve the first-best outcome where total profit is maximized. The wholesale price contract leads to double marginalization, where both parties add a markup, reducing sales. Revenue-sharing coordinates the chain by aligning each party's profit with the total profit.
9. Explain the 'clustering algorithm' method for forming part families. How is it different from production flow analysis?
A clustering algorithm is a mathematical way to group parts and machines based on a similarity measure. It uses a matrix where rows are machines and columns are parts, with a 1 if the part uses the machine. The algorithm rearranges rows and columns to form blocks of 1's along the diagonal. Each block represents a machine cell and its part family. This is different from production flow analysis (PFA), which manually examines route sheets. Clustering algorithms are automated and can handle large data sets. They also give a clear visual output. However, they may not consider practical factors like machine capacity or part volume. PFA is simpler but more subjective.
10. What is a Nash equilibrium in a supply chain game?
A Nash equilibrium is a situation where each player in the supply chain chooses their best strategy given what the other players do. No player can increase their own profit by changing their strategy alone. For example, a manufacturer sets a wholesale price, and a retailer sets a retail price. At equilibrium, the manufacturer's price is the best response to the retailer's price, and vice versa. In a non-coordinated chain, this equilibrium often results in lower total profit than the optimal joint solution. Game theory helps design contracts that shift the equilibrium to a more profitable outcome. The concept is central to understanding strategic interactions.
11. A company wants to implement group technology. List two benefits and one challenge of using production flow analysis to form cells.
Benefits: (1) Reduced material handling because parts move only within a small cell instead of across the whole factory. (2) Lower setup time because machines in a cell are dedicated to a family, so changeovers between parts in the same family are quick. Challenge: Production flow analysis relies on accurate route data. If routes are not documented or change often, the analysis may give wrong families. Also, it may create cells that are not balanced in workload, causing some cells to be overloaded while others are idle. Another challenge is that some parts may need machines from different cells, requiring inter-cell moves, which reduces the benefit.
12. Compare 'mixed-model sequencing' with 'batch production' for making different products. Which one reduces inventory and why?
Batch production makes all units of one product before switching to the next. This creates large piles of finished goods (inventory) while other products are not made. Mixed-model sequencing makes small quantities of each product in a repeating pattern. This reduces inventory because you produce just what is needed soon, not weeks of stock. For example, batch production might make 1000 red phones, then 1000 blue phones. You have 1000 red phones sitting in storage while blue phones are made. Mixed-model would make 10 red, 10 blue, repeat. So you have only a small amount of each color at any time. Lower inventory saves storage space and money.