The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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The sequence of operation is A - E - D - C - B - A - F
The given parameters;
- <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
- <em>initial unit load specified in the first machine, L₁ = 50</em>
- <em>final unit load, L₂ = 100 </em>
- <em>the capacity of the vehicle = 1 unit load</em>
<em />
The given sequence of operation of the vehicle;
A - E - D - C - B - A - F
<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>
The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.
100 unit loads ----------------- 6 trips
2000 unit loads --------------- ?

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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Answer:
as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.
Explanation:
4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum
Explanation:
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