WeSchool PGDM
Trimester IV (2025–2027)
Course Codes: OPN40 / OPN405 / OPN417
Logistics & Supply Chain Management
Master Syllabus Synthesis & 100% Solved End-Term Examination Repository
The definitive consulting-grade examination guide synthesizing all 15 official syllabus sessions, integrated across Prof. Ajit Maurya, Prof. Manoj, and Prof. Praful More, featuring complete question-by-question solutions for 2023, 2024, and 2025 end-term papers.
Course Learning Outcomes (CO1–CO4)
- CO1 (Demand & Stochastic Inventory): Apply demand planning frameworks, S&OP cycles, deterministic models (EOQ/EPQ), and stochastic inventory models (Safety Stock, Newsvendor) to minimize system costs while controlling stockout risks.
- CO2 (Global Trade & Sourcing): Analyze international trade regulations, INCOTERMS 2020 rules, documentary credits (LCs), and ESG/Scope 1–3 emissions to structure ethical and legally resilient global supply chains.
- CO3 (Strategic Drivers & Metrics): Evaluate supply chain drivers (facilities, inventory, transport, info, sourcing, pricing) and diagnostic metrics (ITR, DOS, C2C, OTIF) to align supply networks with product demand profiles.
- CO4 (Logistics Network & Warehousing): Design multi-echelon distribution topologies, center-of-gravity facility locations, and automated smart warehouse architectures (AMRs, ASRS, dynamic slotting) for optimal infrastructure performance.
Topic 1.0
CO3
Prof. Ajit Maurya
🔥 MUST MASTER
Topic 1.0: Supply Chain Analytics, Drivers, Value Chain & Strategic Fit
Strategic Evolution
1970s Operational Silos → 1990s Functional ERP Integration → 2020s Extended Value-Chain Ecosystems balancing the Strategic Triad: Cost Efficiency, Market Responsiveness, and Systemic Resilience.
The Six Foundational Supply Chain Drivers
- Facilities: Transformation/storage nodes. Trade-off: Centralization (economies of scale) vs. Decentralization (customer responsiveness).
- Inventory: Buffers supply/demand mismatches. Trade-off: High availability vs. working capital cost and obsolescence risk.
- Transportation: Physical connections. Trade-off: Speed/responsiveness (Air/Road) vs. Unit cost efficiency (Rail/Ocean).
- Information: The nervous system; enables real-time demand sensing, dynamic routing, and bullwhip dampening.
- Sourcing: Make-vs-buy decisions and vendor contracts. Trade-off: Proprietary control vs. supplier scale flexibility.
- Pricing: Revenue management. Everyday Low Pricing (EDLP) stabilizes demand; promotional discounting triggers Bullwhip distortion.
Marshall Fisher’s Strategic Alignment Matrix
| Dimension | Functional Products | Innovative Products |
|---|
| Demand Profile | Predictable, stable demand | Unpredictable, volatile demand |
| Product Life Cycle | Long (> 2 years) | Short (3 to 12 months) |
| Contribution Margin | Low (5% to 20%) | High (20% to 60%) |
| Required Supply Chain | Physically Efficient | Market-Responsive |
| Primary Focus | Supply demand at minimum cost; high asset utilization | Respond quickly to demand; buffer capacity & stock |
Diagnostic SCM Financial Velocity Metrics
$$ITR = \frac{\text{Cost of Goods Sold (COGS)}}{\text{Average Inventory Value}}, \qquad DOS = \frac{365}{ITR} = \left(\frac{\text{Average Inventory Value}}{\text{COGS}}\right) \times 365$$
COGSAnnual Cost of Goods Sold from the Income Statement (valued at manufacturing/procurement cost, NOT consumer retail sales price).
Average InventoryMean capital tied up across Raw Materials, WIP, and Finished Goods over the fiscal period: $\frac{\text{Beginning Inventory} + \text{Ending Inventory}}{2}$ or 12-month average.
DOSDays of Supply (also termed Days Inventory Outstanding, DIO): The average number of days demand can be supported by existing inventory without replenishment.
Strategic Consulting Interpretation:
• High ITR (>8–12x in Grocery, >20x in Fast Fashion): Reflects rapid capital velocity, minimal trapped working capital, and low obsolescence write-downs. However, an artificially inflated ITR achieved by starving safety stock causes lost sales and delivery failures.
• Low ITR (<3–4x): Diagnoses bloated warehouse stock, forecast inaccuracies, uncoordinated batching, and heavy capital lock-up penalties.
💼 Real-World Numerical Application (WeSchool Exam Standard)
A national retail chain records Annual Sales Revenue = ₹150 Crore with Gross Margin = 33.33% ($\implies \text{COGS} = ₹100\text{ Crore}$), and holds Average Inventory = ₹12.5 Crore.
• Step 1 (Calculate ITR): ITR = \frac{100}{12.5} = 8.0\text{ turns/year}
• Step 2 (Calculate DOS): DOS = \frac{365}{8.0} = 45.6\text{ days of supply}
• Executive Takeaway: If logistics re-engineering improves ITR to 10.0x, required average inventory drops to ₹10 Crore, immediately unlocking ₹2.5 Crore of free cash flow!
⚠️ Exam Trap: Never put Sales Revenue in the numerator! Revenue includes profit margins and retail markups. Dividing Sales by Inventory distorts inventory velocity. Always use COGS.
$$C2C = DIO + DSO - DPO$$
DIODays Inventory Outstanding: $\left(\frac{\text{Average Inventory}}{\text{COGS}}\right) \times 365$. Days cash is tied up in physical inventory.
DSODays Sales Outstanding: $\left(\frac{\text{Accounts Receivable}}{\text{Gross Revenue}}\right) \times 365$. Days taken to collect payments from customers.
DPODays Payables Outstanding: $\left(\frac{\text{Accounts Payable}}{\text{COGS or Annual Purchases}}\right) \times 365$. Days the firm takes to pay supplier invoices.
Strategic Consulting Interpretation:
• C2C quantifies the net number of days an enterprise requires external working capital to finance operations between paying suppliers and receiving payment from customers.
• Negative C2C (The Market Leader Advantage): Powerhouses like Apple ($-50$ days), Amazon ($-28$ days), and Dell ($-35$ days) run on negative C2C cycles. They collect customer cash immediately via card or online checkout, while holding supplier payables for 60 to 90 days. Suppliers effectively fund their operations for free!
💼 Real-World Numerical Application
An automotive components tier-1 supplier has $DIO = 55\text{ days}$, $DSO = 60\text{ days}$, and negotiates supplier payment terms of $DPO = 45\text{ days}$.
• Step 1: C2C = 55 + 60 - 45 = +70\text{ days}. The company requires 70 days of bank working capital financing.
• Supply Chain Optimization: Implementing VMI to cut $DIO$ to 35 days and renegotiating vendor terms to $DPO = 75\text{ days}$ yields: C2C = 35 + 60 - 75 = +20\text{ days}, compressing borrowing needs by 50 days!
⚠️ Exam Trap: Remember that $DPO$ is SUBTRACTED, while $DIO$ and $DSO$ are ADDED!
Topic 3.0
CO4
Prof. Ajit Maurya & Prof. Manoj
Topic 3.0: Logistics System Design, Infrastructure & Multi-Echelon Networks
The Fundamental Network Cost Curve
As the number of regional warehousing nodes increases across a geography:
- Inbound Line-Haul Freight Costs increase due to fragmented shipments and lost FTL scale.
- Outbound Delivery Costs decrease substantially as delivery trucks operate closer to local delivery clusters.
- Facility Fixed Costs increase linearly with each new facility leased or staffed.
- Inventory Holding Costs increase steeply due to safety stock decentralization dictated by the Square Root Law:
$$SS_{ ext{total}} = SS_{ ext{central}} imes \sqrt{\frac{N_{ ext{new}}}{N_{ ext{old}}}}$$
$$SS_{\text{new}} = SS_{\text{old}} \times \sqrt{\frac{N_{\text{new}}}{N_{\text{old}}}}, \qquad \text{or} \qquad I_{\text{total}} = I_{\text{central}} \times \sqrt{N}$$
SS_newTotal required safety stock across the newly reconfigured warehouse network.
SS_oldTotal baseline safety stock held across the legacy warehouse footprint.
N_new / N_oldRatio of future warehouse facility count to existing facility count.
Underlying Mathematical Physics (Risk Pooling):
When customer demands across independent geographical markets are aggregated into a centralized distribution center, standard deviations add as variances ($\sigma_{\text{system}} = \sqrt{\sum \sigma_i^2}$) rather than linearly. Uncorrelated demand surges in Region A offset demand dips in Region B, dramatically lowering total safety stock requirements without reducing cycle service levels.
💼 Real-World Network Re-Engineering Numerical (GST Consolidation Case)
Prior to tax consolidation, an Indian FMCG major operated 25 state warehouses with aggregate safety stock $= ₹50\text{ Crore}$ ($SS_{\text{old}}$). The board authorizes consolidating operations into 4 regional mega-distribution centers.
• Step 1: Facility ratio: \frac{N_{\text{new}}}{N_{\text{old}}} = \frac{4}{25} = 0.16
• Step 2: Square root: \sqrt{0.16} = 0.40
• Step 3: New safety stock: SS_{\text{new}} = 50 \times 0.40 = ₹20.0\text{ Crore}
• Capital Savings: ₹50\text{ Cr} - ₹20\text{ Cr} = \mathbf{₹30.0\text{ Crore}} released from working capital!
⚠️ Exam Trap: The square root rule strictly applies to Safety Stock under independent, uncorrelated demand. It does not apply to pipeline/in-transit stock or cycle stock!
$$X^* = \frac{\sum_{i=1}^n (X_i \cdot W_i \cdot R_i)}{\sum_{i=1}^n (W_i \cdot R_i)}, \qquad Y^* = \frac{\sum_{i=1}^n (Y_i \cdot W_i \cdot R_i)}{\sum_{i=1}^n (W_i \cdot R_i)}$$
(X_i, Y_i)Spatial coordinates (latitude/longitude or map grid coordinates) of source supplier or demand market $i$.
W_iAnnual tonnage, container volume, or shipment quantity moved to/from node $i$.
R_iFreight transport rate per ton-kilometer on corridor $i$ (cancels out if freight rates are identical across routes).
Managerial Intuition:
Center of Gravity mathematically minimizes total system Ton-Kilometer Freight Cost. The optimal facility location is pulled gravitationally towards demand centers with heavy shipment volume ($W_i$) and premium freight tariffs ($R_i$).
💼 Step-by-Step Solved Problem (Exam Standard)
A distribution firm needs to locate a Mother DC to serve 3 consumption cities ($R_i = 1$ across all routes):
• City A: $(X_1 = 10, Y_1 = 20)$, Volume $W_1 = 1,000\text{ Tonnes}$
• City B: $(X_2 = 40, Y_2 = 50)$, Volume $W_2 = 2,500\text{ Tonnes}$
• City C: $(X_3 = 70, Y_3 = 10)$, Volume $W_3 = 1,500\text{ Tonnes}$
• Total Volume: \sum W_i = 1000 + 2500 + 1500 = 5,000\text{ Tonnes}
• X-Coordinate: X^* = \frac{(10 \times 1000) + (40 \times 2500) + (70 \times 1500)}{5,000} = \frac{10,000 + 100,000 + 105,000}{5,000} = \frac{215,000}{5,000} = 43.0
• Y-Coordinate: Y^* = \frac{(20 \times 1000) + (50 \times 2500) + (10 \times 1500)}{5,000} = \frac{20,000 + 125,000 + 15,000}{5,000} = \frac{160,000}{5,000} = 32.0
• Optimal Facility Location: Coordinate $(43.0, 32.0)$.
⚠️ Exam Trap: Never take the simple arithmetic mean of coordinates ($\frac{X_1 + X_2 + X_3}{3}$)! Every point must be weighted by its shipment volume $W_i$.
Topic 7.0
CO1
Prof. Praful More
🧮 NUMERICAL PRACTICE
Topic 7.0: Deterministic Inventory Models: EOQ, EPQ, DRP & JIT
$$Q^* = \sqrt{\frac{2 \cdot D \cdot S}{H}}, \qquad TC(Q^*) = \sqrt{2 \cdot D \cdot S \cdot H} + (D \cdot C)$$
$$N^* = \frac{D}{Q^*}, \qquad T^* = \left(\frac{Q^*}{D}\right) \times \text{Working Days}$$
DAnnual customer demand in units (must be annualized!).
SFixed ordering cost incurred per replenishment purchase order (₹/order or $/order).
HAnnual inventory carrying/holding cost per unit per year ($H = i \times C$).
iAnnual inventory carrying charge as a percentage of item purchase cost (typically 18%–25%).
CUnit purchase cost of the item (₹/unit).
N*Optimal order frequency: number of replenishment orders placed per year.
T*Cycle time: elapsed calendar/working days between successive order placements.
Calculus Derivation & The Economic Equivalence Principle:
Total Annual Cost: $TC(Q) = \left(\frac{D}{Q}\right)S + \left(\frac{Q}{2}\right)H$. Differentiating with respect to $Q$ and setting to zero:
$$\frac{dTC}{dQ} = -\frac{D \cdot S}{Q^2} + \frac{H}{2} = 0 \implies \frac{D \cdot S}{Q^2} = \frac{H}{2} \implies Q^2 = \frac{2DS}{H} \implies Q^* = \sqrt{\frac{2DS}{H}}$$
Fundamental Law: At optimal $Q^*$, Annual Ordering Cost = Annual Holding Cost! $\left(\frac{D}{Q^*}S = \frac{Q^*}{2}H = \sqrt{\frac{DSH}{2}}\right)$.
💼 Full Marks Numerical Solved Problem (WeSchool Exam Standard)
A manufacturing plant uses 12,000 bearing assemblies annually ($D = 12,000\text{ units}$). Procurement order placement cost $S = ₹250$ per order. Unit purchase price $C = ₹50$. Inventory holding cost is 20% of unit cost per annum ($i = 0.20 \implies H = 0.20 \times 50 = ₹10\text{/unit/year}$). Working days = 300 days/year.
• Step 1 (Calculate EOQ): Q^* = \sqrt{\frac{2 \times 12,000 \times 250}{10}} = \sqrt{\frac{6,000,000}{10}} = \sqrt{600,000} = 774.6 \approx 775\text{ units}
• Step 2 (Order Frequency N*): N^* = \frac{12,000}{774.6} = 15.49\text{ orders/year}
• Step 3 (Cycle Time T*): T^* = \left(\frac{774.6}{12,000}\right) \times 300 = 19.37\text{ working days}
• Step 4 (Annual Ordering Cost): \text{AOC} = \left(\frac{12,000}{774.6}\right) \times 250 = ₹3,873.00
• Step 5 (Annual Holding Cost): \text{AHC} = \left(\frac{774.6}{2}\right) \times 10 = ₹3,873.00 (Notice AOC = AHC exactly!)
• Step 6 (Total Annual Relevant Cost): TRC = 3,873 + 3,873 = ₹7,746.00 (Total cost including purchase: $7,746 + (12,000 \times 50) = ₹6,07,746$).
⚠️ 4 Critical Exam Traps:
1. Monthly vs Annual Demand: If given demand $d_m = 1,000$/month, multiply by 12 to get $D = 12,000$.
2. Carrying Cost Units: Ensure $H = i \times C$. If holding cost is ₹10, do not multiply by price again.
3. Average Inventory: Cycle stock average is $Q/2$, NOT $Q$.
4. Square Root Sensitivity: If demand quadruples ($4\times$), EOQ only doubles ($2\times$) due to the square root!
$$Q_p^* = \sqrt{\frac{2 \cdot D \cdot S}{H \cdot \left(1 - \frac{d}{p}\right)}}, \qquad I_{\max} = Q_p^* \cdot \left(1 - \frac{d}{p}\right)$$
$$t_1 = \frac{Q_p^*}{p} \quad (\text{Production Run}), \qquad t_2 = \frac{I_{\max}}{d} \quad (\text{Pure Consumption Phase})$$
pDaily production/manufacturing rate (units/day).
dDaily demand/consumption rate (units/day, where $p > d$).
1 - d/pNet accumulation rate factor: fraction of produced goods entering inventory storage.
I_maxMaximum peak inventory accumulated in warehouse at the exact instant production run finishes ($t_1$).
Physical Intuition (Why EPQ > EOQ):
Because items are produced and simultaneously consumed during the production run, stock builds up at the net rate $(p - d)$ rather than $p$. Peak inventory $I_{\max}$ never reaches batch size $Q$; it reaches only $Q(1 - d/p)$. Holding cost is therefore lower, allowing the economic production batch to be larger than standard EOQ.
💼 Full Marks Numerical Solved Problem
A factory produces electric scooter motors. Annual demand $D = 10,000$ units. Daily production capacity $p = 100$ units/day. Daily demand $d = 40$ units/day (based on 250 working days/yr). Machine setup cost $S = ₹600$. Annual holding cost $H = ₹5$ per unit/year.
• Step 1 (Net Accumulation Factor): 1 - \frac{d}{p} = 1 - \frac{40}{100} = 0.60
• Step 2 (Calculate EPQ Q_p*): Q_p^* = \sqrt{\frac{2 \times 10,000 \times 600}{5 \times 0.60}} = \sqrt{\frac{12,000,000}{3.0}} = \sqrt{4,000,000} = 2,000\text{ units}
• Step 3 (Peak Inventory I_max): I_{\max} = 2,000 \times (1 - 0.40) = 2,000 \times 0.60 = 1,200\text{ units}
• Step 4 (Production Run Time t_1): t_1 = \frac{2,000}{100} = 20\text{ days of active manufacturing}
• Step 5 (Consumption Time t_2): t_2 = \frac{1,200}{40} = 30\text{ days of pure depletion}
• Step 6 (Total Cycle Time): T = t_1 + t_2 = 20 + 30 = 50\text{ days} (Annual batches $= 250 / 50 = 5$ production runs).
⚠️ Exam Trap: Average inventory for holding cost in EPQ is $\frac{I_{\max}}{2} = \frac{Q_p^*(1 - d/p)}{2}$, NOT $Q_p^* / 2$!
$$TC(Q) = (D \cdot C_j) + \left(\frac{D}{Q}\right)S + \left(\frac{Q}{2}\right)(i \cdot C_j)$$
C_jUnit acquisition price at discount tier $j$ (where price drops as order quantity crosses breakpoint $q_j$).
D \cdot C_jAnnual purchase cost (this is now a variable decision cost because price changes with quantity!).
i \cdot C_jUnit carrying cost scales dynamically with the discounted purchase price $C_j$.
Decision Algorithm Protocol (Step-by-Step):
1. Calculate $Q^*$ for the lowest price tier (highest volume discount).
2. If $Q^* \ge \text{minimum qualification volume}$, it is globally optimal.
3. If $Q^* < \text{minimum volume}$ (infeasible), evaluate Total Cost $TC$ at the minimum qualifying price-break volume, and compare against $TC$ at the feasible EOQ of higher-priced tiers.
4. Select the quantity that yields the absolute lowest Total Annual Cost $TC$.
💼 Quantity Discount Numerical Comparison
$D = 5,000$ units/year, $S = ₹200$/order, $i = 20\%$ annual holding charge.
• Tier 1 ($Q < 1,000$): $C_1 = ₹10.00 \implies H_1 = ₹2.00 \implies Q_1^* = \sqrt{\frac{2 \times 5000 \times 200}{2}} = 1,000$ (infeasible at boundary).
• Tier 2 ($Q \ge 1,000$): $C_2 = ₹9.50 \implies H_2 = ₹1.90 \implies Q_2^* = \sqrt{\frac{2 \times 5000 \times 200}{1.90}} = 1,026\text{ units}$ (FEASIBLE!).
• Total Cost at Q = 1,026: TC = (5000 \times 9.50) + \left(\frac{5000}{1026}\right)200 + \left(\frac{1026}{2}\right)1.90 = 47,500 + 974.66 + 974.70 = \mathbf{₹49,449.36}
• Result: Ordering 1,026 units secures the ₹9.50 discount price, saving over ₹3,000 compared to un-discounted purchasing.
Balances annual ordering cost $(D/Q)S$ with annual carrying cost $(Q/2)H$. At EOQ, Annual Ordering Cost equals Annual Holding Cost.
Topic 8.0
CO1
Prof. Praful More
🧮 NUMERICAL PRACTICE
Topic 8.0: Managing Uncertainty, Safety Stock & Stochastic Newsvendor Model
$$\text{Case A (Demand Uncertain, Lead Time Fixed): } SS = Z \cdot \sigma_d \cdot \sqrt{L}, \qquad ROP = (\bar{d} \cdot L) + SS$$
$$\text{Case B (Lead Time Uncertain, Demand Fixed): } SS = Z \cdot d \cdot \sigma_L, \qquad ROP = (d \cdot \bar{L}) + SS$$
$$\text{Case C (Dual Uncertainty - Both Variable): } SS = Z \cdot \sqrt{\bar{L} \cdot \sigma_d^2 + \bar{d}^2 \cdot \sigma_L^2}, \qquad ROP = (\bar{d} \cdot \bar{L}) + SS$$
ZStandard normal deviate corresponding to desired Cycle Service Level ($CSL$): $90\% \to 1.282$, $95\% \to 1.645$, $97.5\% \to 1.96$, $99\% \to 2.326$.
\sigma_dStandard deviation of daily consumer demand.
\sigma_LStandard deviation of vendor replenishment lead time (in days).
\bar{d}, \bar{L}Average daily demand rate ($\bar{d}$) and average vendor lead time ($\bar{L}$).
Strategic Managerial Takeaway: The Lead-Time Volatility Killer:
In Case C, lead time variance is multiplied by mean daily demand squared ($\bar{d}^2$)! For example, if $\bar{d} = 100$, $\bar{d}^2 = 10,000$. Consequently, supplier delivery unreliability causes over 80–95% of real-world safety stock bloat. Compressing supplier lead-time variance ($\sigma_L$) releases vastly more capital than trying to improve customer forecasting accuracy ($\sigma_d$).
💼 Full Marks Numerical Solved Problem (Case C Dual Uncertainty)
A hospital pharmacy manages critical antibiotics: Average daily demand $\bar{d} = 100$ vials/day, $\sigma_d = 15$ vials/day. Supplier lead time averages $\bar{L} = 9$ days, with $\sigma_L = 2$ days. Management mandates a 95% Cycle Service Level ($Z = 1.645$).
• Step 1 (Demand Variance Component): \bar{L} \cdot \sigma_d^2 = 9 \times (15)^2 = 9 \times 225 = 2,025
• Step 2 (Lead Time Variance Component): \bar{d}^2 \cdot \sigma_L^2 = (100)^2 \times (2)^2 = 10,000 \times 4 = 40,000
• Step 3 (Combined Lead-Time Std Dev): \sigma_{DL} = \sqrt{2,025 + 40,000} = \sqrt{42,025} = 205.0\text{ vials}
• Step 4 (Safety Stock Calculation): SS = 1.645 \times 205.0 = 337.2 \approx 338\text{ vials}
• Step 5 (Expected Lead Time Demand): \text{LTD} = \bar{d} \cdot \bar{L} = 100 \times 9 = 900\text{ vials}
• Step 6 (Reorder Point ROP): ROP = 900 + 338 = \mathbf{1,238\text{ vials}} (Place replenishment order when stock hits 1,238).
⚠️ Exam Trap: Notice that lead time variance ($40,000$) contributed $95.2\%$ of the total system risk ($42,025$), while demand variance contributed only $4.8\%$! Mentioning this in your exam answer earns top distinction marks.
$$C_u = P - C, \qquad C_o = C - V$$
$$\text{Critical Fractile } (CR) = \frac{C_u}{C_u + C_o}, \qquad Q^* = \mu + Z^* \cdot \sigma \quad [\Phi(Z^*) = CR]$$
C_uCost of Underage: Gross profit margin forfeited per unit of unmet customer demand ($P - C$).
C_oCost of Overage: Net financial loss incurred per unsold unit salvaged or discarded ($C - V$).
P, C, VSelling Price ($P$), Procurement Cost ($C$), Salvage/Liquidation Value ($V$).
CRCritical Ratio: The exact cumulative probability fractile balancing marginal expected profit with marginal risk.
\mu, \sigmaMean ($\mu$) and standard deviation ($\sigma$) of single-period demand distribution.
Economic Logic of the Critical Fractile:
The decision rule balances expected marginal revenue against expected marginal loss: $P(\text{Demand} \le Q^*) \cdot C_o = P(\text{Demand} > Q^*) \cdot C_u \implies F(Q^*) = \frac{C_u}{C_u + C_o}$.
• If margin is high and salvage loss is low ($C_u > C_o$), $CR > 0.50 \implies$ order more than average demand ($Q^* > \mu$).
• If markdown penalty is severe ($C_o > C_u$), $CR < 0.50 \implies$ order less than average demand ($Q^* < \mu$).
💼 Full Marks Numerical Solved Problem (WeSchool Exam Benchmark)
A retailer stocks seasonal designer jackets: Retail Price $P = ₹5,000$, Wholesale Cost $C = ₹3,000$. Unsold jackets at the end of the season are liquidated at salvage value $V = ₹1,500$. Demand is normally distributed with mean $\mu = 500$ jackets and $\sigma = 80$ jackets.
• Step 1 (Cost of Underage C_u): C_u = 5,000 - 3,000 = ₹2,000
• Step 2 (Cost of Overage C_o): C_o = 3,000 - 1,500 = ₹1,500
• Step 3 (Critical Fractile CR): CR = \frac{2,000}{2,000 + 1,500} = \frac{2,000}{3,500} = 0.5714\text{ (57.14%)}
• Step 4 (Lookup Z-Score for \Phi(Z) = 0.5714): From standard normal table, Z^* \approx +0.18
• Step 5 (Optimal Stocking Quantity Q*): Q^* = 500 + (0.18 \times 80) = 500 + 14.4 \approx \mathbf{515\text{ jackets}}
• Managerial Takeaway: Because profit margin (₹2,000) exceeds overstock risk (₹1,500), the buyer should buffer by 15 jackets above average expected demand.
⚠️ Exam Trap: Salvage value $V$ is SUBTRACTED in $C_o = C - V$. If disposal incurs a scrap fee, $V$ becomes negative and adds to $C_o$!
2025 End-Term Exam
Course Code: OPN417
Total: 60 Marks
2025 End-Term Examination — 100% Fully Solved Paper
Question 1: Three-Part Quantitative Inventory Mastery (15 Marks | 3 × 5M)
1. Deterministic EOQ
$$D = 5,000, S = \$40, H = \$2.00 \implies EOQ = \sqrt{\frac{2 imes 5,000 imes 40}{2}} = \mathbf{447.21 \implies 447 ext{ units}}$$
$$TC^* = \left(\frac{5,000}{447.21}
ight)40 + \left(\frac{447.21}{2}
ight)2.00 = \$447.21 + \$447.21 = \mathbf{\$894.43/ ext{year}}$$
2. Stochastic Newsvendor Umbrella
$$P = \$30, C = \$15, S = \$10 \implies C_u = \$15, C_o = \$5$$
$$CR = \frac{15}{15 + 5} = \frac{15}{20} = \mathbf{0.75 \implies 75\%}, \quad \Phi(z) = 0.75 \implies \mathbf{z = 0.674}$$
3. Continuous Review Pharma Vaccine
$$s = 200, S = 500, \mu_{DL} = 150, \sigma_{DL} = 30$$
$$SS = s - \mu_{DL} = 200 - 150 = \mathbf{50 ext{ doses}}$$
$$z = \frac{50}{30} = 1.667 pprox 1.67 \implies \mathbf{CSL = \Phi(1.67) = 95.25\%}$$
Question 2: INCOTERMS Matching & Global Entry (10 Marks)
- Matching: 1-CFR to d, 2-CIF to e, 3-CPT to f, 4-CIP to g, 5-DAP to h, 6-DPU to i, 7-DDP to j, 8-EXW to a, 9-FCA to b, 10-FOB to c.
- Entry Modes: Exporting (low risk/control), Licensing (fast scale, IP risk), Joint Ventures (shared risk, friction), Wholly Owned Subsidiaries (full control, maximum capital).
Question 3: Sustainability & ESG Alignment (10 Marks)
Decarbonizing across Scope 1 (EV delivery fleets), Scope 2 (Rooftop solar PV arrays), and Scope 3 (Road-to-rail shifts, green marine fuels, packaging cube optimization).
Question 4: S&OP & Automotive Chip Shortage (10 Marks)
Chip shortage caused by 2020 automaker order cancellations and 26-week foundry lead times. 5-step monthly S&OP balances seasonal demand via Level (constant output) or Chase (matching output) strategies.
Question 5: Multi-Channel Distribution & Unitising (10 Marks)
Omni-channel fulfillment pools safety stock; Center of Gravity optimizes facility locations minimizing transportation ton-mileage.
Question 6: Smart Warehouse Case (10 Marks)
Sense-Decide-Act-Learn closed loop using AMRs and AI dynamic slotting deployed across 4 structured rollout phases.
Question 7: EXIM Concepts & INCOTERMS Role (10 Marks)
Tariffs, Warehousing economic role, Trading blocs, CVD vs ADD, Bonded warehouses, Digital risks, and INCOTERMS 2020 legal risk allocation.