I think that there is a problem in the question, or that there is a part of the question you omit to consider or post. With one equation and two unknowns, it's not possible to get those answers, because there are infinity solutions.
The question says
The interferer
is ON with probability .25 and OFF with probability .75.
The average transmit power is 10 mW,
And solution says
0.75P1 + 0.25P2 = 10
and then he gives the answer
P1= 12.25 mW
P2=3.25mW
How he calculated P1 and P2
This is complete question
The channel has a combination of AWGN n[k] and interference I[k]. We model I[k] as AWGN. The interferer
is on (i.e. the switch is down) with probability .25 and off (i.e. the switch is up) with probability .75.
The average transmit power is 10 mW, the noise spectral density is 10−8 W/Hz, the channel bandwidth B is
10 KHz (receiver noise power is NoB), and the interference power (when on) is 9 mW.
(a) What is the Shannon capacity of the channel if neither transmitter nor receiver know when the interferer
is on?
(b) What is the capacity of the channel if both transmitter and receiver know when the interferer is on?
and this is solution
The solution
(a) If neither transmitter nor receiver knows when the interferer is on, they must transmit assuming
worst case, i.e. as if the interferer was on all the time,
C = B log
µ
1 + S
N0B + I
¶
= 10:7Kbps:
(b) Suppose we transmit at power S1 when jammer is o® and S2 when jammer is o®,
C = B max
·
log
µ
1 + S1
NoB
¶
0:75 + log
µ
1 + S2
NoB + I
¶
0:25
¸
subject to
0:75S1 + 0:25S2 = S:
This gives S1 = 12:25mW, S2 = 3:25mW and C = 53:21Kbps.