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automatically if it remains in the money at expiration Options usually achieve a delta of 100 only if they are deep in the money and there is little time remaining until expiration This also occurs in periods with low interest rates because the value of the money leveraged with the option is hypothetically capable of gaining interest somewhere else in higher interest-rate conditions Let s look at an example of some options at parity In Figure 411 you can see a capture of March call options for Johnson & Johnson with only four days remaining until expiration You can also see that the 45 call option has a premium of 610 and a delta of 100 (100%) or at parity If we exercise the option today, we receive the stock at 4500 with a current market price of 5073 so we lose 37 cents of premium over selling the call outright If you are holding a bunch of these options, be careful in your analysis Now let s look at the gure assuming you were holding the 50 call The 50 call option has a premium of 110 and a delta of 678, not at parity at all, nor is it deep in the money If you look carefully though, the numbers are almost the same as the 45 call The 50 call exercised would give 110 in premium but be 73 in the money, leaving a 37 loss of premium with only four days to go until expiration The need to exercise what because of a deep in-the-money option is usually coupled with a liquidity problem in that the bid and offer are far too wide apart or there is insuf cient quantity for the size of a trade you are trying to execute Exercising the option cures the wide bid offer, and the investor can then hold or liquidate the underlying for the current value This situation is more common in derivative trading as the option bids and offers on deep in-the-money options can be wide at times especially for sizable positions in nonindex or agricultural markets.

Units sec GPS week number Seconds of GPS week sec (Group delay) (Clock data issue) sec sec/sec2 sec/sec sec (Orbit data issue) rad/sec rad m sec rad meters rad meters rad rad rad rad rad rad/sec rad/sec

Oracle Exadata Database Machine X2-2

C2 SATELLITE POSITION CALCULATIONS Variable A n0 tsv t tk n Mk Ek vk k uk rk ik uk rk ik Xk Yk k xk yk zk Value 26559 10 14587 10 4 32965 10 5 40327 105 71271 103 14587 10 4 12033 12073 12113 32735 10 1 20003 10 6 14310 102 54489 10 8 32735 10 1 26519 107 97477 10 1 25111 107 85267 106 27116 101 567841101 106 249239629 107 705651887 106

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Section C2 presented the standard equations use to compute the position pi (ti ) of the i-th satellite vehicle at the time of transmission ti as represented in the ECEF frame at time ti A subtlety in the equations presented in Table C2 is the fact that the computed position for each satellite is in a distinct frame-of-reference and are therefore not compatible The ECEF is not an inertial frame Due to the rotation of the frame, the ECEF frame a time t1 is distinct from the ECEF frame at time t2 Let [Pi ]e(t) denote the coordinates of the point Pi with respect to the ECEF frame at time t Even in the case that P1 is xed in inertial space, in general [P1 ]e(t1 ) = [P1 ]e(t2 ) for t1 = t2 Use of time-of-transit to measure range is premised on the fact that the speed-of-light through a vacuum is constant relative to inertial reference frames Therefore, if the coordinates of two points were known in an ineri i tial frame [P1 ] and [P2 ] the time for light to transit between the points through a vacuum would be [P1 ] [P2 ] c

Oracle Exadata Database Machine X2-8

When the coordinates of the two points are known in distinct reference a b frames a and b, [P1 ] and [P2 ] , neither of which is an inertial frame, then they could be transformed to an inertial frame: Ri [P1 ] Ri Ra [P2 ] a a b c

Because the norm of a vector is invariant under rotational transformation, this is equivalent to a b [P1 ] Ra [P2 ] b = c There are two important points First, to calculate the distance between these two points P1 and P2 , they must rst be represented in the same reference frame Second, there is not a unique best frame The following discussion will use the ECEF frame at an arbitrary time denoted as ta Often the selected time is tr the time at which the receiver measured the pseudoranges e(t ) e(t ) Given [P1 ] 1 and [P2 ] 2 the coordinates of P1 and P2 with respect to the ECEF frame at time ta are [P1 ] [P2 ]

.

Total Server Node 16 CPUs/ CPUs and Cores 64 cores Total Memory Total Flash InfiniBand Switches 256GB N/A 4

e(ta ) e(ta )

= Re(ta) [P1 ] 1 = Re(ta) [P2 ] 2

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