The Traveling Salesperson
Problem
Algorithms and Networks
Contents
TSP and its applications
Heuristics and approximation algorithms
Construction heuristics, a.o.: Christofides,
insertion heuristics
Improvement heuristics, a.o.: 2-opt, 3-opt, LinKernig
9.4 1
9.42
9.43
9.44
9.45
What is the number of turns of a coil for which a flux change o f 40 x 10 '
Wb in 0.4 ms induces 70 V in
the coil?
Ans. 700 turns
Find the flux linking a 500-turn, 0.1-H coil carrying a 2-mA current.
Ans. 0.4 pWb
Find the approxi
9 0 V and extinguishes at 1OV. The gas tube has a 50-0 resistance when
firing and a 10lO-R
resistance when extinguished.
1 Mfl
Fig. 8-1 8
When extinguished, the gas tube has such a large resistance (10' R)
compared to the 1-MR resistance
of the resistor t
i = 0 A.
and the capacitor current is
M1
Fig. 8-10 Fig. 8-11
8.21 Find the time constant of the circuit shown in Fig. 8-12.
30 kfl 9 kfl
Fig. 8-12
I64 CAPACITORS AND CAPACITANCE [CHAP. 8
The time constant is T = R,C, where R , , is the Thevenin resistance
A = l,is2 + y 2
6 + j 9 = v'/62 + 92 /tan- ' (9/6) = 10.8/56.3'
-21.4 +j33.3 =(', -21.4)2 + 33.32 /tan-' [33.3/( -21.4)l = 39.6/122.7"
Typically, a calculator will give tan-' (-33.3/21.4) = -57.3", which differs
from the correct angle by
180". For such a
2.17 Find the conductance of 100 ft of No. 14 AWG iron wire, which has a
diameter of 64 mils. The
temperature is 20 C.
sistikity of iron can be obtained from Table 2-1. Thus,
The conductance formula is G = aA,il, in which a = 1,p and A = n(d 2). Of
course
CHAP. 161 TRANSFORMERS 353
a
Circuit 1
Circuit 2
Fig. 16-5
Figure 16-5 shows two circuits coupled by an air-core transformer. Current
i , produces a mutual
and current i, produces a mutual flux 4m2 and leakage flux 44,. As
The coeficient of coupling, with
pumping system, which is the required output poucr of the electric motor.
Some nccdcd data arc that I p l
of water uv5gtis 8.33 Ib, and that 1 hp = 550 ( f t . Ib) s. Thus.
CHAP. I ] BASIC CONCEPTS 13
1.35
1.36
1.37
1.38
Two systems are in cascade. One op
12.79
CHAP. 12) BASIC AC CIRCUIT ANALYSIS, IMPEDANCE. AND
ADMITTANCE: 26 1
12.80
12.81
12.82
12.83
12.84
12.85
12.86
12.87
12.88
12.89
12.90
Find the simplest series circuit that has the same total impedance at 400 Hz
as the parallel arrangement of
a 6 2
Now consider PSpice circuit file statements in general. The first line in the
circuit file must be a title
statement. Any comments can be put in this line. For future reference,
though, it is a good idea to identify
the circuit being analyzed. No other su
Average power can be measured by an instrument called a 1 w t t i m > t w ,
a s sho\vn In Fig. 15-1. I t has
two pairs of terminals: a pair of voltage terminals on the left-hand side and a
pair of current terminals
on the right-hand side. The bottom termi
inductance and resistance of a coil in terms of a capacitance standard. Find L
, and R,y if the
bridge is in balance for R I = 500 kQ, R , = 6.2 kQ, R , = 5 kQ, and C, =
0.1 /IF.
Fig. 14-36
First, general formulas will be derived for R , and L , in terms
magnitudes (M) and phases (P) of the desired voltages and currents:
VM(C1) specifies the magnitude of
the voltage across capacitor C1, and VP(C1) specifies its phase; IM(L1)
specifies the magnitude of the
current flowing through inductor L1, and IP(L1) sp
great as 30 kV, and of the same
for example,
VAA' VBB' VCC'
(b)
Fig. 17-2
frequency (60 Hz), but phase-shifted by 120 . These voltages might be,
i',.,., = 25 000 sin 377t V
z'BB, = 25 000 sin (377t - 120') V
and
If the voltages shown in Fig
zero. This zer
-41.1 V. Alternatively, the angle can be converted to degrees, 1 . 2 ~x 180 x
= 216 , and a calculator
operated in the more popular decimal degrees mode: ~ (m3s) = 70 sin 216
= -41.1 V.
10.11 A current sine wave has a peak of 58 mA and a radian frequency
0.4 j0.5 -j0.8
- + _ + -~_= 2.5 - j0.75 S
for node 1, and
11
- + ~
0.5 -j0.8
= 2 +j1.25S
for node 2. The mutual admittance is 1/( -j0.8) = j1.25 S.
the current into node 2 from the dependent current source.
The controlling current I in terms of V , is
Fro
7.22 Without using PSpice, determine the output corresponding to the
following circuit file.
CIRCUIT
152 PSPICE DC CIRCUIT ANALYSIS [CHAP. 7
7.24 Without using PSpice, determine the output corresponding to the
following circuit file. (Hint: Consider an
op
saturated, however, the output voltage becomes constant and so the voltage
fed back cannot increase
in magnitude as the input voltage does.
In every op-amp circuit in this chapter, each op amp has a feedback resistor
connected between the
output terminal
Multiplying by 4 produces
8 V - 96 + 5V - 60 + 2 V - 32 = 0 or 1 5 V = 188 and
188
15
V = - ~ =~ 1-2.5 33 V
Consequently, the current into the 12-V battery with 0.5-0 internal resistance
is (12.533 - 12),10.5 = 1.07
A, and the current into the other 12-V
5.44 Find the Thevenin equivalent of the transistor circuit shown in Fig. 547. Reference V, positive toward
terminal a.
Ans. 5.88 kR, -29.4 V
CHAP. 51 DC EQUIVALENT CIRCUITS, NETWORK THEOREMS 107
5.45 Find f in the circuit shown in Fig. 5-48, which contai
The substitution of I , = - 4 A into the mesh 1 equation results in
22
11
1111 - 6( -4) = 46 and I , = - = 2 A
Fig. 4-12
64 DC CIRCUIT ANALYSIS [CHAP. 4
4.10 Determine the mesh currents in the circuit shown in Fig. 4-13.
The self-resistance of mesh 1 is 6
-I 10 R I5 R + + +
v2 j20 R VI -j30 R
-Fig. 12-16
The current can be found by dividing the voltage by the total impedance, and
this imped nce can b
found by combining impedances starting- at the end of-the circuit opposite
the source. There, the series
re
Convert each of the following to polar form:
(a) 8.1 + j l l (c) -33.4 +j14.7 (e) 16.2 +j16.2
(b) 16.3 -j12.2 (d) -12.7 -j17.3 (f) -19.1 +j19.1
Ans. (a) 13.7/53.6", (b) 20.4/- 36.8", (c) 36.5/156", (d) 21.51- 126", (e)
22.9/45", ( f ) 27/135"
Convert each
leads the input current, provided that this angle is positive. If i t is negative,
then the current leads the
voltage. A circuit with a positive impedance angle is sometimes called an
inhrcfii7e circuit because the
inductive reactances dominate the capaci
Since the transformer has an iron core, the turns ratio can be used to find the
secondary rms
C ' z = ( 1 L O C ; = (100 400)(240) = 60 V rms. Because the koltages Lary
sinusoidallj, they are induced
whcre 4, is the peak value of
(id, d t =
it follows
= .
of henryper meter and a unit symbol of H/m. (The henry, with unit symbol
H, is the SI unit of inductance.)
The permeability of vacuum, designated by p o , is 0 . 4 p~H /m.
Permeabilities of other materials are related
174
CHAP. 91 INDUCTORS, INDUCTANCE, A
17.15 A balanced three-phase A load has one phase current of I,f = 10m A.
The phase sequence is
ACB. Find the other phasor phase currents and also the phasor line currents.
The two other desired phase currents are those having angles that differ by
120 fr
Each line current is the difference of two phase currents, and each phase
current is the ratio of a phase
V,., = 480/40 V. And from the given ABC phase
Vnc = 480/-80- V
voltage and impedance. One phase voltage is the given
sequence, the other phase voltag
the 2 R of the series resistor plus the total impedance of the three branches to
the right of this resistor. Since
these branches extend between the same two nodes, they are in parallel and
have a total admittance Y that
is the sum of the individual admit
induced after the conductor has rotatcd through an iiiigle of 43 from its
hori/orital position.
Ans. k29.8 V
If the conductor of the alternator in Fig. 10-2 is rotating :it 400 H/, and i f the
induced taltage ha\ ;I 23-V
peak, find the induced voltage 0 2
5.21 Use superposition to find the power absorbed by the 12-52 resistor in
the circuit shown in Fig. 5-29.
1
I lFig. 5-29
Superposition cannot be used to find power in a dc circuit because the
method applies only to linear
quantities, and power has a squa
17 = 42 R.
What is the total resistance of thirty 6-R resistors connected in series?
The total resistance is the number of resistors times the common resistance
of 6 R: R,. = 30 x 6 = 180 R.
What is the total conductance of 4-, 10-, 16-, 20-, and 24-S res