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zhannawk [14.2K]
2 years ago
15

Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.

A unit load size of 50 is initially specified at the first machine. However, due to lot sizing decisions, the unit load size isdoubled after processing on machine D. If one vehicle (e.g., lift truck) is used to transport the unit loads from machine to machine, determine the total number of trips that the vehicle has to make, assuming that the vehicle capacity is one unit load.
Engineering
1 answer:
Vlad1618 [11]2 years ago
6 0

The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>

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 --------------- ?

= \frac{2000}{100} \times 6\\\\= 120 \ trips

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

Learn more here:brainly.com/question/21468592

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when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

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