Scenario 9.14
Demand = 100,000 units/year
Number of days per year = 250 weeks
Standard deviation of weekly demand = 50 units
Average lead time = 5 days
Standard deviation of lead time: 2 days
Cycle-service level = 99%
A continuous review system is used to control inventory.

Use the information in Scenario 9.14. What is the standard deviation of demand during lead time?

Question

Scenario 9.14
Demand = 100,000 units/year
Number of days per year = 250 weeks
Standard deviation of weekly demand = 50 units
Average lead time = 5 days
Standard deviation of lead time: 2 days
Cycle-service level = 99%
A continuous review system is used to control inventory.

Use the information in Scenario 9.14. What is the standard deviation of demand during lead time?

BlackTom AI · Accepted Answer

We start by restating the problem to ensure clarity: Scenario 9.14 provides the annual demand of 100,000 units/year, 250 days per year, a weekly demand standard deviation of 50 units, an average lead time of 5 days with a lead-time standard deviation of 2 days, and a cycle-service level of 99%. The question asks for the standard deviation of demand during lead time (sigma_DL) in a continuous review system.

Option 1: 'greater than 0 but less than or equal to 500 units'—This would imply a very small variability during lead time. To judge this, we should use the standard formula for the standard deviation of demand during lead time: sigma_DL = sqrt( (sigma_d)^2 * L + (mu_d)^2 * (sigma_L)^2 ), where mu_d is the mean demand per period, sigma_d is the standard deviation of demand per period, L is the lead time in periods, and sigma_L is the standard deviation of lead time in periods. Here, mu_d is the weekly mean demand, which is the annual demand divided by the number of weeks (approximately 100,000 / 52 ≈ 1923 units/week). The term (mu_d)^2 * (sigma_L)^2 is driven by lead-time variability and, given mu_d ≈ 1923 and sigma_L ≈ 2 days converted to weeks (~0.286), contributes a substantial amount to sigma_DL. The resulting sigma_DL is well above 500 units, so this option is inconsistent with the calculations.

Option 2: 'greater than 500 units but less than or equal to 1,500 units'—This aligns with the computation using the standard deviation formula for demand during lead time. Plugging in mu_d ≈ 1923 units/week, sigma_d = 50 units/week, L ≈ 5 days ≈ 0.714 weeks, and sigma_L ≈ 2 days ≈ 0.286 weeks, we get sigma_DL ≈ sqrt( (50)^2 * 0.714 + (1923)^2 * (0.286)^2 ). The first term is about 2500 * 0.714 ≈ 1785. The second term is about (1923)^2 * 0.0816 ≈ 3.7e6 * 0.0816 ≈ 302,000. Summing gives roughly 303,800, and the square root is about 551 units. This falls within the range >500 and ≤1,500, making this option plausible and consistent with the math.

Option 3: 'greater than 2,500 units'—Given the calculated sigma_DL is around 550 units, this option overstates the variability. The combined effect of weekly demand variability and lead-time variability does not produce a standard deviation in the thousands here, so this option is incompatible with the computed value.

Option 4: 'greater than 1,500 units but less than or equal to 2,500 units'—While closer to the exact numeric result than Option 3, our rough calculation shows sigma_DL around 550 units, which is below 1,500. Therefore, this option would also be incorrect because the standard deviation does not reach the 1,500–2,500 range.

In summary, Option 2 is the only choice that matches the expected magnitude of the standard deviation of demand during lead time when applying the standard variance propagation formula with the given parameters, while the other choices misrepresent the scale of variability introduced by lead-time uncertainty and demand fluctuations.

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