IE442 FIRST TUTORIAL (SPRING 2011) 1. A mail-sorter must place customers' mail into one of 256 bins. If the mail-sorter assumes that all bins are equally-likely to receive mail, Calculate the information associated with each sort decision. Here we have 256 alternatives, so N=256. Since all bins are equally-likely to receive mail, then the formula to use is:H=log2 N Solve for H, H=log2 256= log10 256/.301=2.408/0.301=8 bits

2. Another, very experienced mail sorter also must place mail into one of 15 customers' bins. If he knows and understands that certain bins receive more mail than others, calculate the information associated with each sort decision. Here, no probabilities were given for the events. We cannot give the exact amount of information. But at least we can give a value for which the information must be lower because the maximum information happens when all the alternatives have equal probabilities. If the alternatives were equally-likely, we would expect H = log2(15) = 3.9 bits. Since they are not equally-likely, the decision maker has additional preexisting information, which lessens the information present within the task. This further reduces H below 3.9 bits. So the answer is H<3.9 bits.

3. After watching a dice-rolling game, you notice that a one side of a die appears twice as often as it should. All other sides of the die appear with equal probability. a. Compute the information that is present in the unfair die. b. Determine the redundancy present in the unfair die. c. In your own words, state the meaning (in brief) of the term "redundancy" in (b) above. a. A die has 6 sides; since one side is twice as likely as any other, we have: (2x) + 5x =1; So, x=1/7 [note: 2/6 vs. 5/6 does not add to one!]Hav = 5/7log2(1/1/7) + 2/7log2(1/2/7) = = 2.53 bits. b. Hmax = log2(6) = 2.59 bits %redundancy = (1 - (2.53/2.59))*100% = 2.3% c. Reduction in uncertainty due to unequally likely events.