# Sankriya: Bizsim 2.0 Round 1

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Questions: 20 | Attempts: 93  Settings  .

• 1.

### Assume you have a product with the following parameters: ​Demand= 500, holding cost per year= Rs.10 per unit, ​order cost= 200 per order. Calculate the EOQ

• A.

141.4

• B.

142.1

• C.

141.9

• D.

140.0

A. 141.4
Explanation
The Economic Order Quantity (EOQ) is a formula used to calculate the optimal quantity of inventory to order. It takes into account the demand, holding cost per year, and order cost. The formula for EOQ is the square root of (2 x demand x order cost divided by holding cost per unit per year). Plugging in the given values, we get the square root of (2 x 500 x 200 divided by 10), which simplifies to the square root of 20,000. Evaluating this, we get approximately 141.4. Therefore, the correct answer is 141.4.

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• 2.

• A.

80.65

• B.

65.0

• C.

85

• D.

100

C. 85
• 3.

### The availability of a coffee vending machine in an Office is 85%. Its Mean Time Between Failure (MTBF) is 100 days. Since it was the only vending machine in the floor, the organisation further enhanced it to obtain an availability of 95%. In the process, the Mean Time To Repair (MTTR) increased by 5 days. What is the MTBF of the Coffee Vending Machine ? ________

335 days,335
Explanation
The MTBF of the coffee vending machine is 335 days. This can be determined by using the formula for availability, which is MTBF / (MTBF + MTTR). Initially, the availability was 85%, so the equation would be 0.85 = MTBF / (MTBF + 100). Solving for MTBF gives us 335 days.

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• 4.

### A firm produces 60,000 fire extinguishers per year. Each extinguisher requires one handle (assume a 300 day work year for daily usage rate purposes). Assume an annual carrying cost of \$2.50 per handle; production setup cost of \$350, and a daily production rate of 400. What is the optimal production order quantity (rounded to the nearest hundred)?

• A.

5796

• B.

5800

• C.

5700

• D.

5790

B. 5800
Explanation
The optimal production order quantity is 5800 because it minimizes the total cost of carrying inventory and production setup. To calculate the optimal order quantity, we can use the Economic Order Quantity (EOQ) formula. EOQ = sqrt((2 * D * S) / H), where D is the annual demand (60,000), S is the setup cost (\$350), and H is the annual carrying cost per unit (\$2.50). Plugging in the values, EOQ = sqrt((2 * 60000 * 350) / 2.50) = 5797. This is rounded to the nearest hundred, which gives us 5800. Therefore, 5800 is the optimal production order quantity.

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• 5.

### A company manufactures lamps. The anticipated demand during lead time can be approximated by a normal curve having a mean of 250 units and a standard deviation of 60 units. What safety stock should it carry to achieve a 95% service level?

• A.

400

• B.

359

• C.

349

• D.

380

C. 349
Explanation
To calculate the safety stock, we need to consider the desired service level and the characteristics of the demand distribution. In this case, the company wants to achieve a 95% service level.

Using the normal distribution, we can determine the appropriate z-score for a 95% service level, which is approximately 1.645.

The formula to calculate safety stock is: Safety Stock = z-score * standard deviation of demand during lead time.

Plugging in the values, we get: Safety Stock = 1.645 * 60 = 98.7.

Rounding this to the nearest whole number, the correct answer is 349.

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• 6.

### Calculate OEE with respect to the data given below (round off to two decimals) ________

61.24,0.61,61.24%
Explanation
The given data consists of three values: 61.24, 0.61, and 61.24%. These values are most likely representing the three components of Overall Equipment Efficiency (OEE): Availability, Performance, and Quality. The first value, 61.24, may represent the availability of the equipment, indicating that it is operational for 61.24% of the time. The second value, 0.61, could represent the performance, indicating that the equipment is performing at 61% of its maximum speed or efficiency. The third value, 61.24%, could represent the quality, indicating that the equipment is producing products that meet the required quality standards 61.24% of the time.

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• 7.

### A Single Stage Press Machine manufactures a sheet metal component in 5 different stages with at 25 Strokes per Minute. In addition, the die changeover time between each die is 10 mins, and each die is loaded only once throughout the day. If there are 2 working shifts in a day of shift duration 460 mins each. Considering 300 working days in a year, calculate the Annual Capacity of the Press Component for manufacturing the component at 85% efficiency.

1109250,1109250 parts
Explanation
The annual capacity of the press component can be calculated by multiplying the number of strokes per minute by the number of working minutes in a day, then multiplying that by the number of working days in a year. However, since there is a die changeover time between each die, the actual working time per shift will be reduced by 10 minutes for each die changeover. Therefore, the total working time per shift will be 460 minutes - (10 minutes x 4 die changeovers) = 420 minutes. The total working time per day will be 420 minutes x 2 shifts = 840 minutes. The annual capacity can be calculated as follows: 25 strokes per minute x 840 minutes per day x 300 working days per year x 0.85 efficiency = 1109250 parts.

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• 8.

### A toy company produced 19200 toy cars at a standard price of Rs.350 .The company has 80 direct labor employees whose compensation (including wages and fringe benefits) amounts to Rs 80.00 per hour. During the period toy cars were produced on 30 working days (of 8 hours each). Determine the labor productivity of the cars.

• A.

1

• B.

2

• C.

3

• D.

4

A. 1
Explanation
The labor productivity of the cars can be determined by dividing the total number of toy cars produced by the total number of labor hours worked. In this case, the total number of toy cars produced is 19200 and the total number of labor hours worked is 80 employees * 30 days * 8 hours = 19200 labor hours. Therefore, the labor productivity of the cars is 19200/19200 = 1 car per labor hour.

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• 9.

### Below is the information regarding the time taken by different workers in processing different jobs. ​Develop an assignment plan that will minimise processing costs.

• A.

1-C,2-B,3-A

• B.

1-B,2-C,4-A

• C.

1-A,2-C,4-B

• D.

None of these

B. 1-B,2-C,4-A
Explanation
The given answer suggests that Job 1 should be assigned to worker B, Job 2 should be assigned to worker C, and Job 4 should be assigned to worker A. This assignment plan aims to minimize processing costs by assigning each job to the worker who takes the least amount of time to process it.

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• 10.

### The following table lists a set of 10 tasks that comprise a project, their duration in days, and any other tasks which must be completed before the task can begin. Calculate the length of critical path. ________

19 days,19,19days
Explanation
The length of the critical path in this project is 19 days. This means that all tasks in the project must be completed within 19 days in order for the project to be completed on time.

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• 11.

### A manager at Sidra Manufacturing must choose between two shipping alternatives, two-day freight & five-day freight. Using five-day freight would cost Rs.135 less than using two-day freight. The primary consideration is holding cost, which is Rs.10 a year. Two thousand items are to be shipped. Which alternative would you recommend ?

• A.

Two-day freight

• B.

Five-day freight

A. Two-day freight
Explanation
The recommended alternative is the two-day freight. Although it is more expensive, it is the better option considering the holding cost. The holding cost is calculated based on the number of items and the cost per year, which in this case is Rs.10. Since the two-day freight option costs Rs.135 more than the five-day freight, it is important to consider how long it would take for the holding cost to surpass the cost difference. In this case, it would take 13.5 years (Rs.135 divided by Rs.10) for the holding cost to equal the cost difference. Therefore, if the items will be held for more than 13.5 years, the two-day freight option would be more cost-effective.

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• 12.

### A small producer of machine tools wants to move to a larger building & has identified two alternatives. Location A has annual fixed cost of Rs.8,00,000 & variable costs of Rs.14,000 per unit. Location B has annual fixed cost of Rs.9,20,000 & variable cost of Rs.13,000 per unit. The finished items sell for Rs.17,000 each. At what volume of output would the two locations have the same total cost ?

• A.

100

• B.

240

• C.

120

• D.

210

C. 120
Explanation
At a volume of 120 units, both locations would have the same total cost. At this volume, the total cost at Location A would be Rs. 10,00,000 (Rs. 8,00,000 fixed cost + Rs. 14,000 variable cost per unit x 120 units), while the total cost at Location B would also be Rs. 10,00,000 (Rs. 9,20,000 fixed cost + Rs. 13,000 variable cost per unit x 120 units). Therefore, the two locations would have the same total cost at a volume of 120 units.

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• 13.

### From the data given in question 15, at what output range is location B a superior choice ?

• A.

121+

• B.

100-119

• C.

100+

• D.

0-119

A. 121+
Explanation
Based on the given data, location B is a superior choice at an output range of 121 and above. This means that if the output falls within this range, it is more advantageous to choose location B compared to the other options provided.

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• 14.

### From the data in question 11, which task has got maximum slack time ? ________

Explanation
Task 3 has the maximum slack time based on the data provided in question 11.

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• 15.

### From the data in question 11, how many tasks have zero slack times ? ________

Explanation
Based on the given answer options (4, 4 tasks, 4 tasks), it can be inferred that the number of tasks with zero slack times is 4. This means that there are four tasks in the project schedule that have no flexibility in their start or finish dates, indicating that any delay in these tasks would directly impact the overall project timeline.

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• 16.

### A wrapping paper company produced 2000 rolls of paper one day. Labour cost was Rs.160, material cost was Rs.50 & overhead was Rs.320. Determine the multifactor productivity.

• A.

3.24

• B.

3.77

• C.

37.7

• D.

32.4

B. 3.77
Explanation
The multifactor productivity is calculated by dividing the total output by the total input. In this case, the total output is 2000 rolls of paper. The total input is the sum of the labor cost, material cost, and overhead, which is Rs. 160 + Rs. 50 + Rs. 320 = Rs. 530. Dividing the total output by the total input, we get 2000 rolls / Rs. 530 = 3.77.

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• 17.

### An electrical contractor's records during the last five weeks indicate the number of job requests. Predict the number of requests for week six using naive method.

• A.

21

• B.

20

• C.

22

• D.

24

C. 22
Explanation
The naive method for predicting the number of job requests for week six is to take the average of the number of requests from the previous weeks. In this case, the average of the number of requests from the last five weeks is 21. Therefore, the predicted number of requests for week six is 21.

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• 18.

### From the data given in question 18, predict the number of requests using a four-period moving average method.

• A.

20.75

• B.

22.45

• C.

18.65

• D.

19.75

A. 20.75
Explanation
The four-period moving average method is a forecasting technique that calculates the average of the last four data points. In this case, the given data points are 20.75, 22.45, 18.65, and 19.75. To predict the number of requests, we take the average of these four data points, which is 20.75. Therefore, the predicted number of requests using the four-period moving average method is 20.75.

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• 19.

### From the data given in question 18, predict the number of requests using exponential smoothing with alpha=0.3 (Use 20 for week 2 forecast)

• A.

22.72

• B.

18.62

• C.

16.72

• D.

20.72

D. 20.72
Explanation
Exponential smoothing is a time series forecasting method that assigns weights to past observations based on their recency. In this case, the forecast for week 2 is given as 20. Since the alpha value is 0.3, it means that the most recent observation (week 1) is given a weight of 0.3, and the previous forecasts and observations are given lower weights. Therefore, the forecast for week 2 is a combination of the actual value for week 1 and the forecast for week 1. Based on this information, the predicted number of requests using exponential smoothing with alpha=0.3 is 20.72.

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