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: 3/4

Due to overheating, the energy loss in the power line connecting G and Q has doubled. What is a feasible set of the energy loss values (in Rs. ’00) that have to be regulated at the capacitor banks P, Q, R and S respectively so that the energy loss from G to M across all paths is the same?

 A.

1, 3, 2, 3

 B.

0, 4, 2, 4

 C.

0, 1, 2, 1

 D.

1, 0, 2, 5

 E.

None of these

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Solution:
Option(C) 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 energy loss on one path has changed, we redraw our table to take it into account. The change to be made is that the path G – Q has energy loss = Rs. 600.

Table below can be scrolled horizontally

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

We now equate the energy losses for all paths, and solve for p, q, r and s.

Equating the first two expressions, we get,

16 + p = 15 + q + p

q = 1

Equating the third and fifth expressions, we get,

13 + q + r = 13 + s + r

s = q = 1

Equating the first and fourth expressions, we get,

16 + p = 15 + s

p = 0

Equating the second and third expressions, we get,

15 + q + p = 13 + q + r

r = 2

Hence, option C is the correct choice.


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