SCM RP ADVANCED SUPPLY CHAIN RISK POOLING:
COMPREHENSIVE MATHEMATICAL MODELS
WITH DETAILED SOLUTIONS 2026/2027
Q1.Threewarehouses serve independent markets with normally distributed
demands: μ₁=500, σ₁=100; μ₂=800, σ₂=150; μ₃=600, σ₃=120. Calculate the
aggregate standard deviation if demands are pooled.
Answer: For independent demands, pooled variance = Σσᵢ².
σ_pooled² = 100² + 150² + 120² = 10,000 + 22,500 + 14,400 = 46,900
σ_pooled = √46,900 = 216.56 units
Risk pooling benefit = (370 - 216.56)/370 = 41.5% reduction
Q2.Fiveretail locations each carry safety stock equal to z·σ where z=1.96 and
σ=80 units. If centralized, calculate the percentage reduction in total safety
stock.
Answer: Decentralized total SS = 5 × (1.96 × 80) = 784 units
Centralized σ = √(5 × 80²) = 178.89 units
Centralized SS = 1.96 × 178.89 = 350.62 units
Reduction = (784 - 350.62)/784 = 55.3%
,Q3.Two markets have correlated demands with ρ=0.6, μ₁=1000, σ₁=200,
μ₂=1500, σ₂=300. Derive the pooled standard deviation.
Answer: σ²_pooled = σ₁² + σ₂² + 2ρσ₁σ₂
= 200² + 300² + 2(0.6)(200)(300) = 40,000 + 90,000 + 72,000 = 202,000
σ_pooled = 449.44 units
Correlation increases pooled risk by 24.6% vs independence
Q4.Calculate the theoretical maximum safety stock reduction for n=10 identical
locations with CV=0.8.
Answer: Maximum reduction = 1 - 1/√n = 1 - 1/√10 = 68.4%
CV magnitude affects absolute savings but not percentage reduction
High CV means substantial absolute benefit despite fixed percentage
Q5.Givencorrelation matrix for markets A and B: ρ_AB=0.3, σ_A=150,
σ_B=200, derive portfolio variance for equal weights (0.5, 0.5).
Answer: σ²_p = 0.25(150²) + 0.25(200²) + 2(0.5)(0.5)(0.3)(150)(200)
= 5,625 + 10,000 + 4,500 = 20,125
σ_p = 141.86 units
Diversification benefit: 18.9% reduction vs weighted average
Q6.A firm operates 20 identical locations pooling to 4 regional DCs. Calculate
change in total safety stock at z=2.33.
Answer: Original: 20 × 2.33 × 15 = 699 units
Each DC serves 5 locations: σ_dc = √5 × 15 = 33.54
Per DC SS = 78.15, Total = 4 × 78.15 = 312.6 units
Reduction = 55.3% (verified by 1 - √(4/20) = 55.3%)
Q7.Newsvendor with pooling: c=$40, p=$100, v=$10, 3 markets (μ=500, σ=100
each). Find optimal Q*.
Answer: CR = (100-40)/(100-10) = 0.667 → z ≈ 0.43
Pooled: μ=1500, σ=173.2
Q* = 1500 + 0.43(173.2) = 1,574.5 units
Inventory savings: 3.3% vs decentralized
, Q8.For Poisson demands (λ₁=30, λ₂=45, λ₃=25), calculate pooled standard
deviation.
Answer: For Poisson: μ=λ, σ²=λ
Pooled λ = 100, σ = √100 = 10 units
Individual σ sum = 17.19 vs pooled 10
Benefit = 41.8% reduction
Q9.Gamma distribution (k=4, θ=50) across 2 independent locations. Derive
pooled parameters.
Answer: Each location: μ=200, σ²=10,000
Sum of Gamma: k_total = 8, θ = 50 (unchanged)
Pooled: μ=400, σ=141.42
Verified: √(σ₁² + σ₂²) = 141.42 ✓
Q10. Calculate square root law violation when ρ=0.8 for n=9 locations
(σ=60 each).
Answer: Square root law (ρ=0): σ=60√9=180
With correlation: σ²=9(60²)+72(0.8)(60²)=239,760
σ_pooled = 489.65
Violation: 2.72× higher than predicted
COMPREHENSIVE MATHEMATICAL MODELS
WITH DETAILED SOLUTIONS 2026/2027
Q1.Threewarehouses serve independent markets with normally distributed
demands: μ₁=500, σ₁=100; μ₂=800, σ₂=150; μ₃=600, σ₃=120. Calculate the
aggregate standard deviation if demands are pooled.
Answer: For independent demands, pooled variance = Σσᵢ².
σ_pooled² = 100² + 150² + 120² = 10,000 + 22,500 + 14,400 = 46,900
σ_pooled = √46,900 = 216.56 units
Risk pooling benefit = (370 - 216.56)/370 = 41.5% reduction
Q2.Fiveretail locations each carry safety stock equal to z·σ where z=1.96 and
σ=80 units. If centralized, calculate the percentage reduction in total safety
stock.
Answer: Decentralized total SS = 5 × (1.96 × 80) = 784 units
Centralized σ = √(5 × 80²) = 178.89 units
Centralized SS = 1.96 × 178.89 = 350.62 units
Reduction = (784 - 350.62)/784 = 55.3%
,Q3.Two markets have correlated demands with ρ=0.6, μ₁=1000, σ₁=200,
μ₂=1500, σ₂=300. Derive the pooled standard deviation.
Answer: σ²_pooled = σ₁² + σ₂² + 2ρσ₁σ₂
= 200² + 300² + 2(0.6)(200)(300) = 40,000 + 90,000 + 72,000 = 202,000
σ_pooled = 449.44 units
Correlation increases pooled risk by 24.6% vs independence
Q4.Calculate the theoretical maximum safety stock reduction for n=10 identical
locations with CV=0.8.
Answer: Maximum reduction = 1 - 1/√n = 1 - 1/√10 = 68.4%
CV magnitude affects absolute savings but not percentage reduction
High CV means substantial absolute benefit despite fixed percentage
Q5.Givencorrelation matrix for markets A and B: ρ_AB=0.3, σ_A=150,
σ_B=200, derive portfolio variance for equal weights (0.5, 0.5).
Answer: σ²_p = 0.25(150²) + 0.25(200²) + 2(0.5)(0.5)(0.3)(150)(200)
= 5,625 + 10,000 + 4,500 = 20,125
σ_p = 141.86 units
Diversification benefit: 18.9% reduction vs weighted average
Q6.A firm operates 20 identical locations pooling to 4 regional DCs. Calculate
change in total safety stock at z=2.33.
Answer: Original: 20 × 2.33 × 15 = 699 units
Each DC serves 5 locations: σ_dc = √5 × 15 = 33.54
Per DC SS = 78.15, Total = 4 × 78.15 = 312.6 units
Reduction = 55.3% (verified by 1 - √(4/20) = 55.3%)
Q7.Newsvendor with pooling: c=$40, p=$100, v=$10, 3 markets (μ=500, σ=100
each). Find optimal Q*.
Answer: CR = (100-40)/(100-10) = 0.667 → z ≈ 0.43
Pooled: μ=1500, σ=173.2
Q* = 1500 + 0.43(173.2) = 1,574.5 units
Inventory savings: 3.3% vs decentralized
, Q8.For Poisson demands (λ₁=30, λ₂=45, λ₃=25), calculate pooled standard
deviation.
Answer: For Poisson: μ=λ, σ²=λ
Pooled λ = 100, σ = √100 = 10 units
Individual σ sum = 17.19 vs pooled 10
Benefit = 41.8% reduction
Q9.Gamma distribution (k=4, θ=50) across 2 independent locations. Derive
pooled parameters.
Answer: Each location: μ=200, σ²=10,000
Sum of Gamma: k_total = 8, θ = 50 (unchanged)
Pooled: μ=400, σ=141.42
Verified: √(σ₁² + σ₂²) = 141.42 ✓
Q10. Calculate square root law violation when ρ=0.8 for n=9 locations
(σ=60 each).
Answer: Square root law (ρ=0): σ=60√9=180
With correlation: σ²=9(60²)+72(0.8)(60²)=239,760
σ_pooled = 489.65
Violation: 2.72× higher than predicted
