Two new MIFIs, radically different speeds

I have two Verizon Jetpack MIFIs, both about two months old and both running on 3G through a Wilson booster in my home.  Each MIFI is on a separate contract with Verizon.  The speed test for one MIFI is 2.62Mbps up, .87Mbps down. The other is .12 up, and .18 down when I can get the full test to run because of its slowness.  The test software says that the faster MIFI is using a Towerstream server in Seattle, the other MIFI is connecting through a Wichita server by KS Fibernet.  I live near Spokane, Washington, which is about 200 miles from Seattle and a LONG way from Wichita.
Question: why would one MIFI be providing adequate speeds, while the other is essentially useless? 

AFter some telephone conversation and behind-the scenes troubleshooting
with Verizon experts, Verizon sent me a replacement MIFI and sim
card.  After swapping parts around between my MIFI and the
replacement unit, it because obvious that the replacement sim card was
dead on arrival, and installing my sim card in the replacement MIFI didn't
make any difference in the performance of the MIFI. 
I'll
be returning the replacement MIFI and sim card to Verizon.  On to
plan C.
Don
Verizon Wireless Customer Support
Verizon Wireless Customer Support
created the discussion
"Re: Two new
MIFIs, radically different speeds"
To view the
discussion, visit:
https://community.verizonwireless.com/message/1016140#1016140
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    SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'C' MIDDLE_NAME FROM DUAL UNION ALL 
    SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'W' MIDDLE_NAME FROM DUAL  UNION ALL 
    SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'X' MIDDLE_NAME FROM DUAL  UNION ALL 
      /* Scenario2 NULL can be equal to any value if there is only one value to equate with
        In the below case we have two values that can equate with null so assign three diffrent ids */
    SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , 'J' MIDDLE_NAME FROM DUAL UNION ALL 
      SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , NULL MIDDLE_NAME FROM DUAL  UNION ALL
    SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , 'L' MIDDLE_NAME FROM DUAL  UNION ALL
      /* Scenario3 NULL can be equal to any value if there is only one value to equate with
        In the below case we have ONE VALUE  that can equate with null so DONT ASSIGN ANY IDS*/
    SELECT 17540300 ID , 'AMARONE' LAST_NAME , 'JOSEPH' FIRST_NAME , 'J' MIDDLE_NAME FROM DUAL UNION ALL  
    SELECT 17540300 ID , 'AMARONE' LAST_NAME , 'JOSEPH' FIRST_NAME , NULL MIDDLE_NAME FROM DUAL
    Select * from names
    o/P Required
    ID    LAST_NAME    FIRST_NAME    MIDDLE_NAME
    47540310    WO      ROBERT          C
    99999990    WO      ROBERT          W                   -- New Sequence Number
    99999991     WO      ROBERT         X                   -- New Sequence Number
    47540300    AMATUZIO    ALBERT    J
    99999992    AMATUZIO    ALBERT                          -- New Sequence Number
    99999993    AMATUZIO    ALBERT    L                     -- New Sequence Number
    17540300    AMARONE    JOSEPH    J
    17540300    AMARONE    JOSEPH    Thanks
    Edited by: new learner on Jun 7, 2012 2:12 PM

    I don't understand what this line is doing:
    FIRST_VALUE (fn_get_person_id) -- sequence for the functionI have the impression that the only difference with my query is to select distinct values not to consider duplicates. Correct me if I'm wrong.
    To me this one should give the same result as your query:
    WITH NAMES AS ( /* Scenario1 -- Three dirrerent people so assign Three diffrrent ID, keeping 1 id as is and assign two
    new ids from sequence*/
                   SELECT   47540310 ID,
                            'WO' LAST_NAME,
                            'ROBERT' FIRST_NAME,
                            'C' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   47540310 ID,
                            'WO' LAST_NAME,
                            'ROBERT' FIRST_NAME,
                            'C' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   47540310 ID,
                            'WO' LAST_NAME,
                            'ROBERT' FIRST_NAME,
                            'W' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   47540310 ID,
                            'WO' LAST_NAME,
                            'ROBERT' FIRST_NAME,
                            'X' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   /* Scenario2 NULL can be equal to any value if there is only one value to equate with
                     In the below case we have two values that can equate with null so assign three diffrent ids */
                   SELECT   47540300 ID,
                            'AMATUZIO' LAST_NAME,
                            'ALBERT' FIRST_NAME,
                            'J' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   47540300 ID,
                            'AMATUZIO' LAST_NAME,
                            'ALBERT' FIRST_NAME,
                            NULL MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   47540300 ID,
                            'AMATUZIO' LAST_NAME,
                            'ALBERT' FIRST_NAME,
                            'L' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   /* Scenario3 NULL can be equal to any value if there is only one value to equate with
                     In the below case we have ONE VALUE  that can equate with null so DONT ASSIGN ANY IDS*/
                   SELECT   17540300 ID,
                            'AMARONE' LAST_NAME,
                            'JOSEPH' FIRST_NAME,
                            'K' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   17540300 ID,
                            'AMARONE' LAST_NAME,
                            'JOSEPH' FIRST_NAME,
                            'Y' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   17540300 ID,
                            'AMARONE' LAST_NAME,
                            'JOSEPH' FIRST_NAME,
                            NULL MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   17540300 ID,
                            'AMARONE' LAST_NAME,
                            'JOSEPH' FIRST_NAME,
                            'M' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   17540300 ID,
                            'AMARONE' LAST_NAME,
                            'JOSEPH' FIRST_NAME,
                            'M' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   99999999 ID,
                            'LO' LAST_NAME,
                            'CHRISTY' FIRST_NAME,
                            NULL MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   99999999 ID,
                            'LO' LAST_NAME,
                            'CHRISTY' FIRST_NAME,
                            'M' MIDDLE_NAME
                     FROM   DUAL
                   UNION ALL
                   SELECT   99999999 ID,
                            'LO' LAST_NAME,
                            'CHRISTY' FIRST_NAME,
                            'M' MIDDLE_NAME
                     FROM   DUAL),
    sel_names AS (  SELECT id
                              , last_name
                              , first_name
                              , middle_name
                              , COUNT (middle_name) OVER (PARTITION BY id ORDER BY middle_name) cnt
                              , ROW_NUMBER () OVER (PARTITION BY id  ORDER BY middle_name) rown
                           FROM (SELECT DISTINCT * FROM names) -- use SELECT DISTINCT here to remove duplicates
                       ORDER BY 1, 2, 3, 4
    SELECT 8888888 new_id, id, last_name, first_name
         , middle_name
      FROM sel_names
    WHERE cnt > 1 AND rown > 1;As you can see I have only replace the the table with an inline view using SELECT DISTINCT *
                           FROM (SELECT DISTINCT * FROM names) -- use SELECT DISTINCT here to remove duplicatesFor the new id I have put a dummy value 888888, you may use a function or a sequence according to your needs.
    Regards.
    Al

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