Critical Path
Logical Reasoning

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Common Information

The flow of energy from Generator (G) to Motor (M) in a power distribution network is as shown. Points P, Q, R and S are capacitor banks (denoted by CB) in the network. The arrows mark the direction of the energy flow. The energy loss (in Rs. ’00) in the power lines during the flow is indicated by the numbers adjacent to the arrows.

Common information image for Critical Path, Logical Reasoning:2296-1

An energy analyst has to make sure that when the energy flows from G to M, it flows through a path where the total energy loss (in Rs. ’00) is minimal. There is an additional energy loss (in Rs. ’00) at the capacitor banks. This loss can be regulated. For example, if the energy analyst selects a path G – P – M (using CB P) then the total energy loss would be Rs. 1,000 + Rs. 600 + the regulated energy loss at the CB P.

Q.

Common Information Question: 2/4

Suppose the power line between Q and R is being repaired and hence energy cannot flow between Q and R. If the energy analyst wants to ensure that the energy loss from G to M across all paths is still the same, then a feasible set of regulated energy losses (in Rs. ’00) at the capacitor banks P, Q, R and S respectively to achieve this goal is:

 A.

4, 4, 2, 3

 B.

3, 4, 2, 4

 C.

1, 4, 2, 0

 D.

1, 4, 2, 1

 E.

None of these

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Solution:
Option(B) is correct

Let us analyze the given information.

Let us assume that the energy losses (in Rs. ’00) at the capacitor banks P, Q, R and S are p, q, r and s respectively.

Draw a table showing the energy losses (in Rs. ’00) for all possible routes between G and M in terms of the variables above.

Table below can be scrolled horizontally

Route Energy loss(in Rs. ’00)
G – P – M 10 + p + 6 = 16 + p
G – Q – P – M 3 + q + 3 + p + 6 = 12 + q + p
G – Q – R – M 3 + q + 4 + r + 3 = 10 + q + r
G – S – M 8 + s + 7 = 15 + s
G – S – R – M 8 + s + 2 + r + 3 = 13 + s + r

Since the line Q – R is being repaired, the path G – Q – R – M will not be used. Hence, equate the energy losses on all other lines, and solve for the values of p, q, r and s that way.

Equating the first two expressions, we get,

16 + p = 12 + q + p

q = 4

Equating the fourth and fifth expressions, we get,

15 + s = 13 + s + r

r = 2

Equating the first and fourth expressions, we get,

16 + p = 15 + s

s - p = 1

Of all the options provided, only option 2 satisfies all the above conditions.

Hence, option B is the correct choice.


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