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experiment was conducted to determine if resistance to CDV was developing in these mice. Results indicated that resistance to CDV did not develop during the 30 days of CDV treatment and suggested that failure of CDV to protect against mortality in SCID mice was due to their inability to clear the virus because of their immunode cient state,26 and that once therapy was discontinued, virus replication immediately resumed and subsequently killed the animal. These studies con rm and expand upon previous reports of CDV ef cacy in murine models of systemic orthopoxvirus infections.22 25,40 42 Our results indicated that CDV given systemically as late as 96 h after CV or VV inoculation can protect BALB/c mice from death or delay the time to death in a dose- and timedependent manner. Since CDV is not effective when given orally, this delay in initiating therapy is necessary for attempting to plan postexposure intravenous treatments for potentially large numbers of people if an actual bioterrorist event occurred. Bray et al.22 reported similarly that a single treatment with 100 mg of CDV/kg i.p. on day 0, 2, or 4 after aerosol exposure to cowpox virus increased survival of BALB/c mice to 90 100%. Smee et al.40 reported that a single dose of 10 mg of CDV/kg administered intranasally (i.n.) at 24 h protected mice inoculated i.n. with CV. In our studies, a single dose of 3 100 mg/kg provided protection if given 5 days prior to infection or up to 3 days after infection. Bray et al.22 also reported that CDV increased the MDD of SCID mice but did not protect them from

multipath distortion By correlating the received midamble bits with the known value of the midamble, the receiver can get the timing of the burst and also estimate the multipath distortion on the radio channel It can then compensate for the multipath distortion and make an estimate of the received data bits This process is performed by the equalizer in the receive chain The stealing ags are always set to 1 except on traf c channels, where they are used to differentiate between TCH and FACCH bursts (0 = TCH, 1 = FACCH) In general (apart from FCCH, SCH and RACH), more than one burst is required to make a meaningful data block The data to be transmitted is rst passed through a forward error correction encoder (either one or two stages) then the bits are interleaved (reordered) and divided between a number of bursts (4, 8 or 19).

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Despite these challenges, the growing research into position-based routing and its associated challenges is re ective of its potential for providing scalable and ef cient data transport in sensor networks.

8.7 EXERCISES 1. In mechanisms that map void regions in advance, a path results. It is possible for multiple nodes along this path to identify themselves as stuck nodes, in which case each stuck node initiates the mapping protocol. This is a redundant exercise. How might mapping techniques be augmented to deal with this occurrence 2. Assume the unit-disk graph (UDG) is connected. Show that the Gabriel (GG) and relative neighborhood (RNG) graphs do not disconnect the unit-disk graph. In other words, show the intersections GG UDG and RNG UDG are connected. 3. In face-routing methods, a packet in recovery mode is routed greedily as soon as possible (i.e. some node is found that is closer to the destination than where recovery began). However, it is conceivable that only an edge sits closer to the destination (as shown in Figure 8.12). Modify face-routing so that it resolves this problem and maintains its delivery guarantees. 4. Nodes in CLDP probe incident links in order to nd and remove intersections. However, there are cases where concurrent probing may disconnect the network graph. Enumerate these cases (there are 3) and establish a mechanism to guarantee a disconnection does not occur. 5. In our discussion of BVR and other mechanisms relying on beacons, we left open any mechanism to replace a failed beacon. Assume that a minimum number r of beacons is required for correct operation. There are two operations at issue.

The encoding used is: Channel SCH Encoding Start with 25 bits of user data Append 10-bit CRC check code Pass through half-rate convolutional encoder This gives (25 + 10 + 4) 2 = 78 bits The extra 4 bits are needed to ush out the convolutional encoder Start with 8 bits of user data Append 6-bit CRC check code Pass through half-rate convolutional encoder This gives (8 + 6 + 4) 2 = 36 bits Start with 23 bytes = 184 bits of user data Pass through a FIRE block encoder that appends 40 parity check bits Then pass through a half-rate convolutional encoder This gives a total of (184 + 40 + 4) 2 = 456 bits These bits are divided between the transmitted bursts in four consecutive frames Start with 260 bits of compressed speech data.

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