Math Problem Statement

Dave's sandwich ship has an average demand of 140 sandwiches (units) per hour.

Station1 takes the order and prepares the bread.

Station2 applies the meat, cheese and sauce.

Station3 applies the vegies and gets payment from the customer.

The table below shows the processing time for each of the Stations.

Required:

Resource

Station1

Statioin2

Station3

Processing Time

33

43

35

[sec/unit]

What is the capacity per sec for each Station?

   1/33 

   1/43 

   1/35 

[units/sec]

What is the capacity per hour for each Station?

109.09

83.72

102.86

[units/h]

Which Station is the bottleneck?

X

Demand

140

[units/h]

What is the process capacity?

83.7

[units/h]

What is the flow rate of the process?

83.7

[units/h]

What is the cycle time of the process?

[sec/unit]

What is the utilization rate for each Station?

[%]

Solution

Let's go step-by-step to solve the given problem, as it's a typical production process or operations management scenario.

1. Capacity per second for each Station

The capacity per second is given as:

  • Station 1: 133\frac{1}{33} units/second
  • Station 2: 143\frac{1}{43} units/second
  • Station 3: 135\frac{1}{35} units/second

2. Capacity per hour for each Station

To find the capacity per hour, we multiply the capacity per second by 3600 (since there are 3600 seconds in an hour):

  • Station 1: 133×3600=109.09\frac{1}{33} \times 3600 = 109.09 units/hour
  • Station 2: 143×3600=83.72\frac{1}{43} \times 3600 = 83.72 units/hour
  • Station 3: 135×3600=102.86\frac{1}{35} \times 3600 = 102.86 units/hour

3. Which Station is the bottleneck?

The bottleneck is the station with the lowest capacity per hour. From the values calculated:

  • Station 1: 109.09 units/hour
  • Station 2: 83.72 units/hour
  • Station 3: 102.86 units/hour

Station 2 is the bottleneck, as it has the lowest capacity of 83.72 units/hour.

4. What is the process capacity?

The process capacity is determined by the capacity of the bottleneck. Since Station 2 is the bottleneck:

  • Process Capacity = 83.72 units/hour

5. What is the flow rate of the process?

The flow rate is limited by the process capacity, which is governed by the bottleneck. Therefore, the flow rate is:

  • Flow Rate = 83.72 units/hour

6. What is the cycle time of the process?

The cycle time is the inverse of the process capacity, measured in seconds per unit. Since the process capacity is 83.72 units/hour: Cycle Time=360083.72=43 seconds/unit\text{Cycle Time} = \frac{3600}{83.72} = 43 \text{ seconds/unit} Thus, the cycle time is 43 seconds per unit.

7. What is the utilization rate for each Station?

Utilization rate is the ratio of demand (flow rate) to capacity. Given that the demand is 140 units/hour and the actual flow rate is 83.72 units/hour, the utilization rate for each station is calculated as follows:

  • Station 1 Utilization: 83.72109.09×100=76.75%\frac{83.72}{109.09} \times 100 = 76.75\%
  • Station 2 Utilization (Bottleneck): 83.7283.72×100=100%\frac{83.72}{83.72} \times 100 = 100\%
  • Station 3 Utilization: 83.72102.86×100=81.38%\frac{83.72}{102.86} \times 100 = 81.38\%

Summary:

  1. Cycle time: 43 seconds/unit
  2. Utilization rate:
    • Station 1: 76.75%
    • Station 2: 100%
    • Station 3: 81.38%

Let me know if you'd like more details or clarifications on any of the steps!

Follow-up Questions:

  1. How does the bottleneck impact the overall system's efficiency?
  2. What would happen to the utilization rates if the demand decreased to 100 units/hour?
  3. How can Station 2's capacity be improved to remove the bottleneck?
  4. If Station 1's processing time is reduced to 30 seconds/unit, how would the process capacity change?
  5. How does the flow rate compare to the average demand, and what strategies can be used to address any discrepancies?

Tip: Reducing the cycle time of the bottleneck station can significantly improve the overall process capacity.

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Math Problem Analysis

Mathematical Concepts

Operations Management
Rate Calculations
Capacity Analysis
Cycle Time

Formulas

Capacity per second = 1 / Processing Time
Capacity per hour = Capacity per second × 3600
Cycle Time = 3600 / Process Capacity
Utilization Rate = (Flow Rate / Station Capacity) × 100

Theorems

-

Suitable Grade Level

Undergraduate/Operations Management