IE341: Human Factors Engineering

Prof. Mohamed Zaki RamadanAhmed M. El-Sherbeeny, PhD

Information Theory• Information Processing is AKA:

– Cognitive Psychology– Cognitive Engineering– Engineering Psychology

• Objectives of Information Theory:– Finding an operational definition of information– Finding a method for measuring information– Note, most concepts of Info. Theory are descriptive (i.e.

qualitative vs. quantitative)

• Information (Defn):

– “Reduction of Uncertainty”– Emphasis is on “highly unlikely” events– Example (information in car):

• “Fasten seat belt”: likely event not imp. in Info. Th.⇒• “Temperature warning”: unlikely event imp.⇒

7

Unit of Measure of Information• Case 1: ≥ 1 equally likely alternative events:

– H : amount of information [Bits]– N: number of equally likely alternatives– e.g.: 2 equally likely alternatives ⇒

⇒ Bit (Defn): “amount of info. to decide between two equally likely (i.e. 50%-50%) alternatives”

– e.g.: 4 equally likely alternatives ⇒

– e.g.: equally likely digits (0-9)⇒

– e.g.: equally likely letters (a-z)⇒Note, for each of above, unit [bit] must be stated.

8

Cont. Unit of Measure of Information• Case 2: ≥ 1 non-equally likely alternatives:

– : amount of information [Bits] for single event, i

– : probability of occurrence of single event, i

– Note, this is not usually significant

(i.e. for individual event basis)

9

Cont. Unit of Measure of Information• Case 3: Average info. of non-equally likely

series of events:

– : average information [Bits] from all events– : probability of occurrence of single event, i– N : num. of non-equally likely alternatives/events– e.g.: 2 alternatives (N = 2)

• Enemy attacks by land, p1 = 0.9

• Enemy attacks by sea, p2 = 0.1

• ⇒

10

Cont. Unit of Measure of Information• Case 4: Redundancy:

– If 2 occurrences: equally likely ⇒• p1 = p2 = 0.5 (i.e. 50 % each)

• ⇒ H = Hmax = 1

– In e.g. in last slide, departure from max. info.• = 1 – 0.47 = 0.53 = 53%

– – Note, as departure from equal prob. ↑ %Red. ↑⇒– e.g.: not all English letters equally likely: “th”,“qu”

• ⇒ %Red. of English language = 68 %

• PS. How about Arabic language?

11

Choice Reaction Time Experiments• Experiments:

– Subjects: exposed to different stimuli– Response time is measured– e.g. 4 lights – 4 buttons

• Hick (1952):– Varied number of stimuli (eq. likely alternatives)– He found:

• As # of eq. likely alt. ↑ reaction time to stimulus ↑ ⇒

• Reaction time vs. Stimulus (in Bits): linear function

• Hyman (1953):– Kept number of stimuli (alternatives) fixed– Varied prob. of occurrence of events info. Varies⇒– He found: “Hick-Hyman Law”

• AGAIN: Reaction time vs. Stimulus (in Bits): linear function!

15

Solved Problems

1. The Hick-Hyman law provides one measure of information processing ability. Assume that an air traffic controller has a channel capacity bandwidth limit of 2.8 bits/second in decision making.

a. Assuming equally-likely alternatives, how many choices can this person make per second?

b. As the controller gains expertise he/she develops expectations of which routes different planes will fly. Explain how this will increase the controller's channel capacity on this task.

c. Describe at least three different general methods for improving the controller's information processing in this task. (Use methods we have studied in this course--don't just say "automate")

•1. a. H = 2.8/sec = log2(N); 10.84=N; N=6.92, or about 7.

1. b. Actually makes more knowledge, or less uncertainty. So, less potential knowledge gain. We can process less info. faster and more accurately than more info.

1. c. A few ideas:

· Allow more time and look-ahead in system · Earlier training on various target probabilities · Less targets per controller · Allow errors; may be redundance in system · Multiple channels or modalities in presentation · Better compatibility in the system.

2. 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, concisely state the meaning of the term "redundancy" in (b) above.

2. 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) = 2.01+.518 = 2.53 bits

2. b. Hmax = log2(6) = 2.59 bits

%redundancy = (1 - (2.53/2.59))*100% = 2.3% 2. c. Reduction in uncertainty due to unequally likely events.

3. How much Information (H) is contained in a fair roll of a 6-sided die? If an individual

realizes that the die is unfair, with 20%

chance of each of 4 sides appearing, and

10% on each of the other two, how does

H change? What does this mean?

3. A 6-sided die should land on any of its sides with equal probability. Thus, information (H) is log26 = log106 /.301 = 2.58 bits of information. If the die is unfair (and the gambler realizes it), these should be less potential information gain; or less information in the die. In other words, the gambler already has some pre-existing knowledge so his/her potential information gain from the die will be less. If a rapid decision were required based on the outcome of the die, it should be faster due to this preexisting bias. To quantify, Have = [(4).2log2(1/.2) + (2).1log2(1/.1)] = [4(.46) + 2(.33)] = 1.86 + .66 = 2.52 bits of information. This is, in fact, lower than the maximum information case above. This result means that the gambler already has the equivalent of .06 bit of pre-existing information. The computation of Redundancy is a well-accepted measure of this pre-existing information. Here, %Redundancy = [1-2.52/2.58]*100% = 2.3%.

4. A display can show one numerical digit (0-9) per second, with equally-likely digits, in a choice reaction time task. If an observer accurately processes the information from the display, what is the observer's channel capacity, in bits/second?

4. The observer must decide from among 10 choices/second.

This is log2(10) = 3.32 bits/sec.

This is known as the bandwidth, or channel capacity for rate of

processing.

5.What would your answer to Question 4 be if equally-likely double digits (0, 1, 2, ..., 99) could be presented? Why or why not does this make sense

to you?

5. This is now log2(100) = 6.62 bits/sec.

This makes sense if you consider that twice as many binary decisions were required--the first splits the numbers into groups of 50 and 50, the second splits one of these into two 25 groups, etc. Note that decision making ability is not linearly related to task complexity.

6. Consider Question 4 again, with unequally-likely digits. The probabilities of the digits appearing are shown below. Determine both the channel capacity and the redundancy.

 Digit 0 1 2 3 4 5 6 7 8 9

Prob 0 .08 .25 .12 .10 .08 .05 .10 .22 0

6. Hmax = log2(10)/sec = 3.32 bits/sec Hav = [2(.08log2(1/.08) + .25 log2(1/.25) + .12 log2(1/.12) + ... + .22 log2(1/.22)] = 2.81 bits/sec = new channel capacitySo, Redundancy = [1-(2.81/3.32)] *100% = 15.1%

7. Based upon his company experience, Ali knows that 50% of the chips are routed to Line 1, 30% to Line 2, and 20% to Line 3. Given a choice RT intercept of 250 msec, and a processing bandwidth of 7.5 bits/second, how much time does Ali require to make each routing decision? How much faster or slower is this, compared to the condition when all three routes are equally-likely?

7. The information (Hav) associated with each routing decision is: Hav = .5 log2 (1/.5) + .3 log2 (1/.3) + .2 log2 (1/.2) = 3.32 {.5 log10 (2) + .3 log10(3.33) + .2 log10 (5) } = 3.32(.15 + .157 + .14) = 1.48 bits A bandwidth of 7.5 bits/second equates to: 1000 ms/sec * 1/ 7.5 sec/bit = 133.3 msec/bit

The expected choice RT is then: 250 + 133.3 (Hav) = 250 + 133.3 (1.48) = 447 msec If all equally-likely, H=log2(3) = 1.59 bits RT = 250 + 133.3 (1.59) = 462 msec So, if Ali understands the stated probabilities, his expected decisions will be (462-447) = 15 msec faster than if all are equally-likely.

Solved Problems

Piano players must rapidly move their fingers between various keys on a piano keyboard. The keys are either black (0.5 inch wide) or white (1 inch wide). For a pianist, a Fitt's Law relationship was measured as:

Movement Time (msec) = -50 + 55(ID).Single movement time increase or decrease if the pianist moves to a black key instead of a white key?

1. How long will it take this pianist to move his finger from a 150 Hertz white key to a 2400 Hertz white key? Assume 7 keys/octave and negligible movement time for the actual key.

The pianist must make a single movement, over a 4 octave range.

Starting at 150 Hz, the 4 octaves are 300, 600, 1200, and 2400 Hz. At 7 keys/octave, this is 28 keys, with a center-to-center distance of 27 inches. The target key width is 1 inch. So, ID = log2[2(27)/1] = 5.7 bits. Using the given Fitt's Law equation, MT = -50 + 55(5.75) = 266 msec.

2. By how much will a single movement time increase or decrease if the pianist moves to a black key instead of a white key?

Now the pianist must move to a black key, which decreases the distance by 0.25 inch, and decreases the target width to 0.5 inch. So, ID = log2[2(26.75)/0.5] = 6.7 bits, which is a harder task. Now, MT = -50 + 55(6.7) = 320.82 msec, or 55 msec slower for the movement.

3. By how much will a single movement time increase or decrease if the pianist starts at a black key instead of a white key?

Two answers were accepted to this--since the target width doesn't change, you could say there will be no difference. However, if the starting key is black instead of white, the overall distance decreases by 0.25 inch, which decreases the movement time by under 1 msec.

5. How else could time be predicted for the pianists finger movements between keys?

As explained both in class and in a handout, the MTM-2 system could be used instead of Fitt's Law, using Tolerance Puts. Explanations involving choice reaction time were not correct, as these don't try to predict movement, they only try to model decision making. (Did your fingers ever leave the keys in the choice reaction time lab?--they shouldn't have.)

Solved Problems

1. Compare the inspection capability of Inspector A and Inspector B, determing d' and decision criterion. Inspector A located 26 of 28 defective parts, but also called 2 of 15 good parts defective. Inspector B found 29 of 30 defects, but called 6 of

20 good parts defective. In this case, the Value of a 'hit' was greater than the Cost of a 'false alarm'.

Inspector A: HR = 26/28 = .93; Z = 1.48 FAR = 2/15 = .13; Z = 1.13 We draw a picture of the model, and determine that the decision criterion is located between the means of the two distributions; therefore, we add the Z-scores from each distribution to compute d'. Note that this inspector is fairly neutral (neither liberal or conservative). d' = 1.48 + 1.13 = 2.61

Inspector B: HR = 29/30 = .97; Z = 1.88 FAR = 6/20 = .3; Z = .52 Similarly, we add the two Z-scores to compute d'. Note that, if FAR had been greater than .5, we would have to subtract one of the Z-values. This inspector was somewhat more liberal than inspector A. d' = 1.88 + .52 = 2.40.

Thus, Inspector A was a bit better discriminator. If the value of a hit is greater than the cost of a FA, then we seek a very liberal inspector. We might decide to go with inspector B, if the cost of a FA is small enough and the value of a Hit is great enough.

Ali is a rain forecaster at 'Predicta-Weather'. Over a 3-month period, he forecast that no rain would fall on 60 of the 67 days on which no rain actually fell. He also (incorrectly) forecast that no rain would fall on 2 of the 22 days in which rain actually fell. For the following analyses, assume that a 'signal' is a rainy day.a. Determine Ali's d', stating whether he is a liberal

or conservative forecaster. b. Management is very concerned that Ali is making too many False Alarms, and would like to see these reduced to a probability of 0.001. Determine Ali's resultant Hit Rate with this reduced False Alarm Rate, assuming the same

d' as found in (a).

Betty Said....  Actual Rainy Days

Actual Non-Rainy Days

"Rain"  20 7

 "No Rain" 2 60

 Totals 22 67

a. We (arbitrarily) define a signal as a rainy day. A simple response matrix follows:

From the table, the Hit Rate (HR) = 20/22 = 0.91Likewise, the False Alarm Rate (FAR) = 7/67 = 0.10Drawing a picture, we see the following:

Looking up the Z values for tail areas of 0.1 and 0.09, we find Zn = 1.28 and Zsn = 1.34. The d' = 1.28 + 1.34 = 2.62.

Since Ali's decision criterion is just towards the left of the crossing point between the distributions, he is pretty much a neutrally-biased responder, with possibly a very slight tendency towards being a liberal responder (making more FARs in order to increase his HR).b. Looking up the Zn for a tail area of 0.001, we find that Zn = 3.1. Redrawing the criterion in the above figure (shown in yellow), we see that the criterion has moved towards the right, or conservative side (fewer FAs, but also fewer HITS). Since the FAR is so small, the criterion is actually to the right of the mode on the 'Rain' distribution. We know that 3.1 - Zsn = 2.62, so Zsn = 0.48. Looking up the area associated with a Z = 0.48 gives us an area of 0.31. Thus, the area to the right of the criterion = Hit Rate = 0.31.

Try to solve this

Try to solve this