CS 145
Sorry, the textbook is not clear. Let's collect all these bugs to make our lives easier.Tuesday, October 26, 2004
Hw 4.3
Most people are doing well.
One problem:
"For a causal system (A,E,D), A = A*E ..."
There is no assumption here whether the system is causal or not. And you don't really need this condition at all. The problem statment says "input process A has an envelope of E". You can only have A < = A*E.
One problem:
"For a causal system (A,E,D), A = A*E ..."
There is no assumption here whether the system is causal or not. And you don't really need this condition at all. The problem statment says "input process A has an envelope of E". You can only have A < = A*E.
Tuesday, October 19, 2004
hw 4.2
most popular mistake:
A(t) < = b + sigma+ rho*t (1)
D(t) < = sigma+ rho*t (2)
So: X(t) = A(t) - D(t) < = b (3)
(or other equivalent bugs)
You cannot get (3) from (1) and (2) since it is inequality in (1) or (2).
Some of you write down A(t) = b + sigma + rho * t or D(t) = sigma + rho*t. These are regarded as the same kind of error.
Lesson:
1. when you see "A(t) is regulated by (sigma, rho)", it only means A(t) <= sigma + rho*t, the equality holds only in some special case (sigma packets are sent at time zero and the flow is constant-rate of rho packets per second).
2. When there is inequality, be careful.
A(t) < = b + sigma+ rho*t (1)
D(t) < = sigma+ rho*t (2)
So: X(t) = A(t) - D(t) < = b (3)
(or other equivalent bugs)
You cannot get (3) from (1) and (2) since it is inequality in (1) or (2).
Some of you write down A(t) = b + sigma + rho * t or D(t) = sigma + rho*t. These are regarded as the same kind of error.
Lesson:
1. when you see "A(t) is regulated by (sigma, rho)", it only means A(t) <= sigma + rho*t, the equality holds only in some special case (sigma packets are sent at time zero and the flow is constant-rate of rho packets per second).
2. When there is inequality, be careful.
hw 4.1
The most "popular" mistake:
Step 1: try to make a scenario (usually setting s=0) and claim X(t)={A(t)-c*t} for any t > 0.
Step 2: try to prove that this is the worst case scenario.
I do not think step 1 is correct. You may not find a scenario that X(t)={A(t)-c*t}. (e.g. A(t) < c(t) all the time. Then X(t) is always equal to zero).
Lesson:
1. You cannot "prove a general theorem" by a single scenario or some (uncomplete) set of scenarios
2. When you write "Let q= sup S" or "q = inf S", you have to be sure that S is not an empty set.
Step 1: try to make a scenario (usually setting s=0) and claim X(t)={A(t)-c*t} for any t > 0.
Step 2: try to prove that this is the worst case scenario.
I do not think step 1 is correct. You may not find a scenario that X(t)={A(t)-c*t}. (e.g. A(t) < c(t) all the time. Then X(t) is always equal to zero).
Lesson:
1. You cannot "prove a general theorem" by a single scenario or some (uncomplete) set of scenarios
2. When you write "Let q= sup S" or "q = inf S", you have to be sure that S is not an empty set.
Tuesday, October 12, 2004
Some equivalent notations in the textbook
What is "A(t) is (sigma, rou) constrained"
"A(t) is (sigma, rou) constrained" means the shape of A is like the output of a (sigma, rou) regulator. That is:
A(t) <= inf { A(s) + sigma + rou (t-s) }
A(t) <= inf { A(s) + sigma + rou (t-s) }
What is a (sigma, rou, R) regulator?
A (sigma, rou, R) regulator is a regulator with (sigma,rou) control, with a maximum output rate of R.
Without the constraint R, at most sigma packets can be sent in a burst instantly. With the constraint R, the sigma packets is sent in a rate no faster than R.
Without the constraint R, at most sigma packets can be sent in a burst instantly. With the constraint R, the sigma packets is sent in a rate no faster than R.
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