Question

For an assembly line, the inventory was calculated by considering the production rate as 4 pieces per hour and the average processing time as 60 minutes. Now, if the production rate is kept the same and the average processing time is brought down by 30%, then the change in inventory is

• A. Decreased by 25%
• B. Increased by 25%
• C. Decreased by 30%
• D. Increased by 30%

Hint:

Since $I = R \times T$, if the production rate ($R$) is kept entirely constant, the inventory ($I$) becomes directly proportional to the processing time ($T$): \[ I \propto T \] Therefore, whatever percentage change happens to the processing time will directly map onto the inventory level! A 30

Answer 1

The Correct Option is C

Concept: The fundamental relationship governing inventory, production/throughput rate, and processing time (or cycle time) in a manufacturing system or stable queuing process is described by Little's Law. Little's Law states that the average inventory level ($I$) is equal to the product of the long-term average production/throughput rate ($R$) and the average processing/lead time ($T$). Mathematically, this is expressed as: \[ I = R \times T \]

Step 1: Calculate the initial inventory configuration ($I_1$).
We are given the initial processing characteristics of the assembly line: * Initial Production Rate, $R_1 = 4\text{ pieces per hour}$ * Initial Average Processing Time, $T_1 = 60\text{ minutes} = 1\text{ hour}$ Using Little's Law, we find the baseline initial inventory value: \[ I_1 = R_1 \times T_1 \] \[ I_1 = 4\text{ pieces/hour} \times 1\text{ hour} = 4\text{ pieces} \]

Step 2: Determine the modified processing properties ($R_2$ and $T_2$).
The problem presents a modification to the line parameters: 1. The production rate remains completely constant: \[ R_2 = R_1 = 4\text{ pieces per hour} \] 2. The average processing time is decreased by $30\%$: \[ T_2 = T_1 \times \left(1 - \frac{30}{100}\right) = 1\text{ hour} \times 0.70 = 0.70\text{ hours} \] Alternatively, in minutes: $T_2 = 60\text{ minutes} - (0.30 \times 60\text{ minutes}) = 42\text{ minutes} = \frac{42}{60}\text{ hours} = 0.70\text{ hours}$.

Step 3: Calculate the new inventory ($I_2$) and percentage change.
Re-applying Little's Law with our modified conditions gives: \[ I_2 = R_2 \times T_2 = 4\text{ pieces/hour} \times 0.70\text{ hours} = 2.8\text{ pieces} \] Now, compute the exact percentage variation in the inventory level ($\Delta I\%$): \[ \Delta I\% = \left( \frac{I_2 - I_1}{I_1} \right) \times 100\% \] \[ \Delta I\% = \left( \frac{2.8 - 4.0}{4.0} \right) \times 100\% = \left( \frac{-1.2}{4.0} \right) \times 100\% \] \[ \Delta I\% = -0.30 \times 100\% = -30\% \] The negative sign signifies a clear reduction. Hence, the inventory decreases by exactly $30\%$. This matches Option (3).