WeSchool PGDM Trimester IV Faculty: Prof. Praful More

Logistics & Supply Chain Management

Quantitative Inventory Theory, Stochastic Models & Sourcing Strategy

Reconstructed consulting-grade master notebook covering EOQ/EPQ, quantity discount breakevens, safety stock uncertainty formulations, the Newsvendor critical ratio, Kraljic matrix, master quiz deck, and 100% solved end-term examination problems.

Prof. Praful More's Quantitative Doctrine

"Inventory is money sitting in another form. Every inventory model is an optimization balance between two competing costs: the cost of acquiring too little (stockouts, production halts) versus the cost of holding too much (capital lockup, obsolescence)."

EOQ & EPQ Models → Newsvendor Model → Jump to Solved PYQs →

Session 1 CO1 Alignment

Module 1: Inventory Fundamentals, Economics & Cost Topologies

The Three Operational Cost Buckets

1. Inventory Carrying (Holding) Cost ($H = I imes C$):
- Capital Cost: Opportunity cost of capital / WACC hurdle rate (10% to 18%).
- Storage Space Cost: Warehouse rent, utilities, material handling, racking (3% to 5%).
- Inventory Service Cost: Insurance premiums, local property taxes (1% to 2%).
- Inventory Risk Cost: Obsolescence, physical shrinkage, pilferage, product expiration (4% to 10%).
- Total Annual Holding Cost Rate ($I$): Typically ranges between 18% to 35% of unit purchasing price per year.
2. Ordering / Setup Cost ($S$): Administrative procurement expenses, electronic PO transmission, supplier invoice auditing, machine changeover, calibration, and cleaning.
3. Stockout / Shortage Cost ($C_s$): Lost gross margin, contractual SLA penalty fees, emergency air freight expediting costs, and long-term customer churn.

Session 7 CO1 Alignment

Module 2: Deterministic Inventory Lot-Sizing Models

$$Q^* = \sqrt{\frac{2 \cdot D \cdot S}{H}}, \qquad TC(Q^*) = \sqrt{2 \cdot D \cdot S \cdot H} + (D \cdot C)$$ $$N^* = \frac{D}{Q^*}, \qquad T^* = \left(\frac{Q^*}{D}\right) \times \text{Working Days}$$

D

Annual customer demand in units (must be annualized!).

S

Fixed ordering cost incurred per replenishment purchase order (₹/order or $/order).

H

Annual inventory carrying/holding cost per unit per year ($H = i \times C$).

i

Annual inventory carrying charge as a percentage of item purchase cost (typically 18%–25%).

C

Unit 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!

$$Q_p^* = \sqrt{\frac{2 \cdot D \cdot S}{H \cdot \left(1 - \frac{d}{p}\right)}}, \qquad I_{\max} = Q_p^* \cdot \left(1 - \frac{d}{p}\right)$$ $$t_1 = \frac{Q_p^*}{p} \quad (\text{Production Run}), \qquad t_2 = \frac{I_{\max}}{d} \quad (\text{Pure Consumption Phase})$$

p

Daily production/manufacturing rate (units/day).

d

Daily demand/consumption rate (units/day, where $p > d$).

1 - d/p

Net accumulation rate factor: fraction of produced goods entering inventory storage.

I_max

Maximum 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 $Q_p^* / 2$!

Quantity Discount Breakeven Optimization

Algorithmic Decision Rules

Compute standard EOQ for the lowest unit price tier ($C_{ ext{lowest}}$).
If feasible, it is optimal. If infeasible, step up to next price tier until a feasible EOQ is found.
Compute Total Annual Cost ($TC$) at feasible EOQ and at every quantity breakpoint above it: $$TC(Q) = (D \cdot C) + \left(\frac{D}{Q} ight)S + \left(\frac{Q}{2} ight)(I \cdot C)$$
Select the order quantity $Q$ minimizing total annual procurement plus holding and ordering costs.

Session 8 CO1 Alignment

Module 3: Managing Uncertainty, Safety Stock & Stochastic Inventory

$$\text{Case A (Demand Uncertain, Lead Time Fixed): } SS = Z \cdot \sigma_d \cdot \sqrt{L}, \qquad ROP = (\bar{d} \cdot L) + SS$$ $$\text{Case B (Lead Time Uncertain, Demand Fixed): } SS = Z \cdot d \cdot \sigma_L, \qquad ROP = (d \cdot \bar{L}) + SS$$ $$\text{Case C (Dual Uncertainty - Both Variable): } SS = Z \cdot \sqrt{\bar{L} \cdot \sigma_d^2 + \bar{d}^2 \cdot \sigma_L^2}, \qquad ROP = (\bar{d} \cdot \bar{L}) + SS$$

Z

Standard 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_d

Standard deviation of daily consumer demand.

\sigma_L

Standard 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.

$$TC(Q) = (D \cdot C_j) + \left(\frac{D}{Q}\right)S + \left(\frac{Q}{2}\right)(i \cdot C_j)$$

C_j

Unit acquisition price at discount tier $j$ (where price drops as order quantity crosses breakpoint $q_j$).

D \cdot C_j

Annual purchase cost (this is now a variable decision cost because price changes with quantity!).

i \cdot C_j

Unit 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.

Normal Distribution $z$-Factor Lookup Reference

Cycle Service Level (CSL)	Normal $z$-Factor	Managerial Stockout Risk
90.0%	$z = 1.282$	10.0% probability of stockout during lead time
95.0%	$z = 1.645$	5.0% probability of stockout during lead time
97.5%	$z = 1.960$	2.5% probability of stockout during lead time
99.0%	$z = 2.326$	1.0% probability of stockout during lead time
99.9%	$z = 3.090$	0.1% probability of stockout (extreme critical items)

Session 8 CO1 Alignment

Module 4: The Single-Period Stochastic (Newsvendor) Model

$$C_u = P - C, \qquad C_o = C - V$$ $$\text{Critical Fractile } (CR) = \frac{C_u}{C_u + C_o}, \qquad Q^* = \mu + Z^* \cdot \sigma \quad [\Phi(Z^*) = CR]$$

C_u

Cost of Underage: Gross profit margin forfeited per unit of unmet customer demand ($P - C$).

C_o

Cost of Overage: Net financial loss incurred per unsold unit salvaged or discarded ($C - V$).

P, C, V

Selling Price ($P$), Procurement Cost ($C$), Salvage/Liquidation Value ($V$).

CR

Critical Ratio: The exact cumulative probability fractile balancing marginal expected profit with marginal risk.

\mu, \sigma

Mean ($\mu$) 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 $C_o$!

The Double Marginalization Trap

In a decentralized supply chain, the retailer's critical fractile $CR_{ ext{retailer}} = \frac{P - W}{(P - W) + (W - S)}$ is strictly lower than the integrated supply chain's fractile $CR_{ ext{chain}} = \frac{P - C}{(P - C) + (C - S)}$. Consequently, individual store managers systematically under-order relative to the system optimum, destroying total channel profitability.

Session 9 & 10 CO2 Alignment

Module 5: Strategic Sourcing, Kraljic Matrix & Total Cost of Ownership

The Kraljic Portfolio Purchasing Matrix

Quadrant	Characteristics	Recommended Sourcing Strategy	Supplier Preferencing Quadrant
Strategic Items	High Profit Impact, High Supply Risk; single-source critical components (microchips, custom engines).	Long-term partnerships, joint R&D, strategic alliances, transparent cost-plus contracts.	Core: High spend, high supplier attractiveness; supplier prioritizes buyer.
Leverage Items	High Profit Impact, Low Supply Risk; standardized commodity materials available from multiple suppliers.	Exploit purchasing scale, aggregate orders, dynamic reverse auctions, short-term contracts.	Exploitable: High buyer spend, low attractiveness; supplier extracts premium margin.
Bottleneck Items	Low Profit Impact, High Supply Risk; specialized chemical additives or spare parts with proprietary specs.	Secure volume commitments, maintain safety stock buffers, design out dependencies.	Nuisance: Low spend, low attractiveness; supplier provides bare minimum service.
Routine (Non-Critical)	Low Profit Impact, Low Supply Risk; office supplies, MRO items, standard packaging materials.	Automate purchase-to-pay, corporate credit cards (P-Cards), standardized vendor catalogs.	Development: Low current spend, high potential; supplier invests to grow account.

Master Quiz Deck CO2 & CO3 Alignment

Module 6: Smart Warehousing, ESG & Maritime Chokepoints

Global Strategic Maritime Chokepoints

Geopolitical Vulnerabilities

Bab el-Mandeb / Red Sea & Suez Canal: Disruptions forced container vessels to bypass the canal and circumnavigate Africa via the Cape of Good Hope, adding 10–14 sailing days, absorbing 10% of global container ship capacity, and tripling spot container rates.
Panama Canal: Severe climate-induced freshwater droughts restricted daily vessel transit slots, causing massive ship queues and shifting bulk traffic to longer maritime lanes.
Strait of Malacca & Hormuz: Critical narrow shipping bottlenecks connecting the Indian Ocean to East Asia and Middle Eastern petroleum hubs.

Decarbonization Levers across Emissions Scopes

Scope 1 (Direct Operations): Fleet transition to battery-electric delivery vehicles (EVs); warehouse electrification.
Scope 2 (Indirect Energy): On-site rooftop solar arrays; 100% renewable power purchase agreements (PPAs).
Scope 3 (Extended Value Chain — 80%+ of impact): Intermodal modal shifts from road to rail; green shipping fuels (bio-LNG, green methanol); packaging cube optimization.

Classroom Cases

Core Case Study Analyses

Case 1: Star Clinic Panadol / Clinical Triage

Dilemma: Chief Nursing Officer demands 100% availability for all medical supplies; Financial Director wants minimal inventory.

Resolution: Triage segmentation. Setting non-critical Panadol at 99% CSL ($z = 2.33$) requires holding only 28 more tablets than 97.5% CSL, adding just \$2.80 in annual holding cost. This minor expenditure satisfies clinical safety without working capital bloat.

Case 2: Just Baked Cupcake Stocking

Corporate cost is \$1.20, wholesale price to franchisee is \$1.50, retail price is \$3.00. Unsold cupcakes have zero salvage value.

Franchisee $CR = (3.00 - 1.50) / 3.00 = 50\%$. Integrated Chain $CR = (3.00 - 1.20) / 3.00 = 60\%$. Franchisee under-stocks relative to the supply chain optimum; resolved via centralized allocations or revenue-sharing contracts.

PYQ Repository

Solved End-Term Examination Papers (2023–2025)

End-Term 2025 · Question 1 (15 Marks)

Three-Part Quantitative Inventory Mastery

(a) Deterministic EOQ: $$EOQ = \sqrt{\frac{2 imes 5,000 imes 40}{2}} = \sqrt{200,000} pprox \mathbf{447.21 \implies 447 ext{ units}}$$ $$TC^* = \left(\frac{5,000}{447.21} ight)40 + \left(\frac{447.21}{2} ight)2.00 = \$447.21 + \$447.21 = \mathbf{\$894.43/ ext{year}}$$
(b) Stochastic Newsvendor: $$C_u = \$30 - \$15 = \$15, \quad C_o = \$15 - \$10 = \$5$$ $$CR = \frac{15}{15 + 5} = \frac{15}{20} = \mathbf{0.75 \implies 75\%}, \quad \Phi(z) = 0.75 \implies \mathbf{z = 0.674}$$
(c) Multi-Period Continuous Review: $$s = 200, \quad S = 500, \quad \mu_{DL} = 150, \quad \sigma_{DL} = 30$$ $$SS = s - \mu_{DL} = 200 - 150 = \mathbf{50 ext{ doses}}$$ $$z = \frac{SS}{\sigma_{DL}} = \frac{50}{30} = 1.667 pprox 1.67 \implies \mathbf{CSL = \Phi(1.67) = 95.25\%}$$

End-Term 2024 · Question 2 (15 Marks)

ElectroTech Solutions Category A Inventory Math

$$D = 50,000, \quad S = \$100, \quad H = \$2.00$$
$$EOQ = \sqrt{\frac{2 imes 50,000 imes 100}{2}} = \sqrt{5,000,000} pprox \mathbf{2,236.07 \implies 2,236 ext{ units}}$$
$$N^* = \frac{50,000}{2,236.07} = \mathbf{22.36 ext{ orders per year}}$$
$$ ext{Annual Holding Cost} = \left(\frac{2,236.07}{2} ight) imes 2.00 = \mathbf{\$2,236.07/ ext{year}}$$