P'' P' P ---------------------------------------------------------------- | | | | | |R'' |O'' |S'' | | |-------|-------| | | | | | | | |R' |Q'' | |S' | |---------------|----------------| | | |O' | | | | | | | | | | | | | | |R |Q' |O |S |--------------------------------|-------------------------------| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |Q | ---------------------------------------------------------------- FIGURE 1
TABLE 1: IF (the max value of AND (the max value of THEN (SR is) the entity is less the entity is than or equal to) greater than) 250,000 25,000 100,000 25,000 2,500 10,000 2,500 250 1,000 250 25 100 25 12.5 12.5 12.5 6.25 12.5 6.25 3.125 6.25 3.125 1.5625 3.125 1.5625 0.390625 1.5625 0.390625 0.0 0.1953125
The symbol: MML be defined as: Any Murrey Math Line
The symbol: MMML be defined as: Major Murrey Math Line
The symbol: mMML be defined as: Minor Murrey Math Line
The symbol: bMML be defined as: Baby Murrey Math Line
and, using the abbreviation MMI to mean "Murrey Math Interval", let;
The symbol: MMI be defined as: Any Murrey Math Interval
The symbol: MMMI be defined as: Major Murrey Math Interval = SR/8
The symbol: mMMI be defined as: Minor Murrey Math Interval = SR/8/8
The symbol: bMMI be defined as: Baby Murrey Math Interval = SR/8/8/8
where the symbol /8/8/8 means that SR is to be divided by 8 three times. For example, if SR = 100 then the Baby Murrey Math Interval bMMI is: 100/8/8/8 = 12.5/8/8 = 1.5625/8 = 0.1953125
100 -------------------------------------------- 8/8 MMML 87.5 -------------------------------------------- 7/8 MMML 75 -------------------------------------------- 6/8 MMML 62.5 -------------------------------------------- 5/8 MMML 50 -------------------------------------------- 4/8 MMML 37.5 -------------------------------------------- 3/8 MMML 25 -------------------------------------------- 2/8 MMML 12.5 -------------------------------------------- 1/8 MMML 0 -------------------------------------------- 0/8 MMML FIGURE 2
75 -------------------------------------------- 8/8 mMML 73.4375 -------------------------------------------- 7/8 mMML 71.875 -------------------------------------------- 6/8 mMML 70.3125 -------------------------------------------- 5/8 mMML 68.75 -------------------------------------------- 4/8 mMML 67.1875 -------------------------------------------- 3/8 mMML 65.625 -------------------------------------------- 2/8 mMML 64.0625 -------------------------------------------- 1/8 mMML 62.5 -------------------------------------------- 0/8 mMML FIGURE 3
For example, suppose a stock is trading within a range of 28 1/4 to 34 1/2. In this case the value of SR is 100. The MMMI is 12.5 (i.e. 100/8). The next smaller MMI is a mMMI = 12.5/8 = 1.5625. The MMML closest to 28 1/4 is the 2/8 th's (i.e. 2 x 12.5 = 25). The closest mMML (measured from 25) is also a 2/8 th's MML (i.e. 2 x 1.5625 = 3.125). So, the bottom of the square is 25 + 3.125 = 28.125 (i.e. 28 1/8).
The 28 1/8 MML is the base of the square in time. This MML satisfies rule 1 (it is an even numbered line, 2/8 th's) and it is close to 28 1/4 (28 1/4 - 28 1/8 = 1/8 = .125). The result of .125 is less than 4/8 th's of the next smaller octave which is a "baby" octave (bMMI = 1.5625/8 = .1953125). Specifically .125 is less than .78125 (4 x .1953125 = .781254).
NOTE: If the bottom MML of the square in time is an even MML, and the top MML of the square in time is 2, 4, or 8 MMI's above the bottom MML, then the top MML is also an even numbered MML.
For example, consider the same stock trading within the range 28 1/4 to 34 1/2. The base of the square in time was identified as the 2/8 th's mMML 28.125. In this case the top of the square is the mMML that is 4 mMMI's above the base: 28.125 + (4 x 1.5625) = 34.375. This MML can also be shown to be "close" to the high end of the trading range, since, 34.5 - 34.375 = .125 and .125 is less than .781254 (4 x .1953125 = .781254). Recall that .1953125 is the bMMI (i.e. the next smaller octave).
An example of this can be seen on Chart #91 in Murrey's book. This is a chart of Chase Manhatten. In this case the bottom and top MML's of the square in time are the 5/8 th's and 7/8 th's MML's respectively. These are obviously odd MML's. Another example of an exception is Chart #83 in Murrey's book. In this case the bottom of the square in time is 37.5 (an odd 3/8 th's line) and the top of the square in time is 62.5 (an odd 5/8 th's line).
"Then you simply count up 2, 4, or 8 lines, and include the top of its trading range, as long as it's no higher than a) 19, b) 39, c) 78 cents above the 100% line. (there are exceptions where it will run up a full 12.5, or 25 or 50% line above the 100% line and come back down..."
PriceRange = 35.25 - 28.0 = 7.25
Murrey refers to this as "setting the rhythm" or identifying the "perfect square". Refer to TABLE 1 in this paper. Reading from TABLE 1 SR = 100 (This is because the high price for First American was 35.25. Since 35.25 is less than 250 but greater than 25, SR = 100).
Let's define two new parameters. The first parameter is RangeMMI. RangeMMI = PriceRange/MMI. RangeMMI measures the price range of First American (or any entity) in units of Murrey Math Intervals (MMI's).
The second parameter is OctaveCount. The purpose of OctaveCount will become evident shortly. The question to answer is, "What MMI should be used for creating the square in time?" This question will be answered by dividing the SR value by 8 until the "appropriate MMI" is found. So:
MMI = MMMI = SR/8 = 100/8 = 12.5
This is a MMMI. Is this the "appropriate MMI"? To answer that question divide PriceRange by this MMI.
RangeMMI = PriceRange/MMI = 7.25/12.5 = 0.58
Now compare RangeMMI to 1.25. If RangeMMI is less than 1.25 then a smaller MMI is needed. This is indeed the case because 0.58 is less than 1.25. Since the first MMI calculated was a MMMI, then the next MMI will be a mMMI. Simply divide the prior MMI by 8 to get the new MMI.
MMI = mMMI = MMMI/8 = 1.5625
This is a mMMI. Is this the "appropriate MMI"? To answer that question divide PriceRange by this latest MMI.
RangeMMI = PriceRange/MMI = 7.25/1.5625 = 4.64
Now compare RangeMMI to 1.25. If RangeMMI is less than 1.25 then a smaller MMI is needed. Since RangeMMI is 4.64 and 4.64 is greater than 1.25 we're done. The correct MMI to use is the mMMI which is 1.5625. (Naturally, in other cases, this process may be repeated further, continuing division by 8, until RangeMMI is greater than 1.25.)
Since we had to divide the perfect square (SR) by 8 two times to arrive at the appropriate MMI (SR/8/8 = 100/8/8 = 12.5/8 = 1.5625) we'll set the value of OctaveCount to be 2. The value of OctaveCount will act as a reminder as we proceed through this example.
Now the question of 1.25. Where did this number come from? Partly trial and error and partly reasoning. Remember that the parameter RangeMMI describes the trading range of First American in units of Murrey Math Intervals. Remember also that the rules for the square in time require that the square be at least 2 MMI's high, and that the square be close to the high and low values of the trading range.
If we used the MMMI to build the square in time for First American the result would have been a square with a height of (2 x 12.5) 25. Because First American has only traded within a range of 7.25 points, this square would not represent First American's' behavior very well. The trading range of First American should approximately fill the square. By choosing a smaller MMI (i.e. mMMI = 1.5625) the result is a square in time that will be 4 MMI's high (RangeMMI = 4.64 which is rounded to 4. The actual height selected for the square in time will be determined in STEP 4). Again, recall the rule that the square must be 2, 4, or 8 MMI's high. (Is the number 1.25 perfect? NO! But, tests conducted on the charts in the Murrey Math book indicate that 1.25 works in nearly all cases).
In STEP 3 above, we selected the appropriate value for the MMI and calculated the final value of RangeMMI. Given the value of RangeMMI, TABLE 2 may be used to select the actual height of the square in time.
TABLE 2 ALLOWED SQUARES IN TIME: RangeMMI Square in Time is Bounded by These MML's 1.25 < RangeMMI < 3.0 (0,2) (1,3) (2,4) (3,5) (4,6) (5,7) (6,8) (7,1) 3.0 <= RangeMMI < 5.0 (0,4) (2,6) (4,8) (6,2) 5.0 <= RangeMMI < ... (0,8) (4,4)
TABLE 2 was arrived at using trial and error. The results of the C program I had written were compared to the charts in the back of the Murrey Math book. Is TABLE 2 perfect? NO! But it works fairly well. TABLE 2 specifies the allowed upper and lower MML numbers that may be used to create the square in time. Note that once the upper and lower MML's are specified so is the height of the square. TABLE 2 attempts to accomodate Murrey's rules for creating the square in time as well as the exceptions to those rules.
The first row of TABLE 2 addresses squares that are two MMI's high. Note that the exception of having squares in time with odd top and bottom MML's is included.
The second row of TABLE 2 addresses squares that are four MMI's high. Note that these squares are required to lie on even MML's only.
The third row of TABLE 2 addresses squares that are eight MMI's high. Note that these squares are required to lie on (0,8) or (4,4) MML's only. The notation (0,8) means that the bottom of the square will be a 0/8 th's MML and the top of the square will be an 8/8 th's MML.
Continuing with First American, recall that RangeMMI = 4.64. Reading from TABLE 2 we see that the square in time will be 4 MMI's high and will lie on one of the MML combinations (0,4), (2,6), (4,8), or (6,2).
The objective of this step is to find the MML that is closest to the low value of First American's trading range (i.e. 28.0). This MML must be a mMML since the MMI we are using is a mMMI (i.e. 1.5625). Actually, the MML we will find in this step is the mMML that is closest to but is less than or equal to First American's low value.
This is fairly simple. To repeat, the MML type must correspond to the MMI type that was selected. We chose an MMI that is a mMMI (i.e. 1.5625), hence, the MML must be a mMML. We now make use of the parameter OctaveCount. In this example, OctaveCount = 2. Since OctaveCount = 2 we will perform 2 divisions by 8 to arrive at the desired MML.
MMI = MMMI = SR/8 = 100/8 = 12.5
The base of the perfect square is 0.0, so subtract the base from the low value of First American's trading range (28.0 - 0.0 = 28.0). Now we find the MMML that is less than or equal to 28.0. In other words, how many MMMI's could we stack up from the base (i.e. 0.0) to get close to (but less than 28.0).
28.0/MMMI = 28.0/12.5 = 2.24 ==> 2 (Since there are no partial MMI's)
0.0 + (2 x 12.5) = 25.0
25.0 is the 2/8 th's MMML that is closest to but less than 28.0
Since OctaveCount = 2, this process will be repeated a second time for the mMMI. The only difference is that the base line is the MMML from the prior step. So, once again, subtract the base (i.e. 25) from the low value of First American's trading range (28 - 25 = 3.0). Now find the mMML that is less than or equal to 28.0. In other words, how many mMMI's could we stack up from the base (i.e. 25) to get close to (but less than 28.0).
3.0/mMMI = 3.0/1.5625 = 1.92 ==> 1 (Since there are no partial MMI's)
25 + (1 x 1.5625) = 26.5625
26.5625 is the 1/8 th mMML that is closest to but less than 28.0
So, mMML = 26.5625
This mMML is the "best first guess" for the bottom of the square in time. But there is a problem...
By the end of STEP 5, a square in time has been defined that will be 4 mMMI's in height and have a base on the 1/8 th mMML = 26.5625. Recall, however, that the rules in TABLE 2 state that a square that is 4 MMI's in height must lie on an even numbered MML. A 1/8 th line is odd. So, two choices are available. Referring to TABLE 2 we can choose either a (0,4) square or a (2,6) square. Which do we choose?
Let's define an error function and choose the square that minimizes this error. The error function is:
Error = abs(HighPrice - TopMML) + abs(LowPrice - BottomMML)
Where:
Having now defined an error function it can now be applied to the problem at hand. The square in time that was determined in STEP 5 has a bottom MML of 26.5625 and a height of 4 mMMI's. The top MML is therefore 26.5625 + (4 x 1.5625) = (26.5625 + 6.25) = 32.8125. Recall, however, this is still the square lying upon the 1/8 mMML (a (1,5) square on odd MML's). We want to use the error function to distinguish between the (0,4) square and the (2,6) square.
The (0,4) square is simply the (1,5) square shifted down by one mMMI and the (2,6) square is the (1,5) square shifted up by one mMMI.
0/8 th mMML = 26.5625 - 1.5625 = 25.0
4/8 th's mMML = 32.8125 - 1.5625 = 31.25
So, the bottom of the (0,4) square is 25.0 and the top of the (0,4) square is 31.25.
Likewise for the (2,6) square:
2/8 th's mMML = 26.5625 + 1.5625 = 28.125
6/8 th's mMML = 32.8125 + 1.5625 = 34.375
So, the bottom of the (2,6) square is 28.125 and the top of the (2,6) square is 34.375.
Now apply the error function to each square to determine "the best square in time".
Error(0,4) = abs(35.25 - 31.25) + abs(28.0 - 25.0) = 7.0
Error(2,6) = abs(35.25 - 34.375) + abs(28.0 - 28.125) = 1.0
Clearly the (2,6) square is the better fit (has less error). Finally, we have arrived at a square in time that satisfies all of the rules. We can now divide the height of the square by 8 to arrive at the 1/8 lines for the square in time.
(34.375 - 28.125)/8 = 6.25/8 = .78125
So the final square is:
100.0% 34.375 87.5% 33.59375 75.0% 32.8125 62.5% 32.03125 50.0% 31.25 37.5% 30.46875 25.0% 29.6875 12.5% 28.90625 0.0% 28.125
Exactly as seen on Chart #85B of the Murrey Math book.
PriceRange = 437.5 - 433.5 = 4.0
Refer to TABLE 1: SR = 1000
Octave 1:
Octave 2:
Octave 3:
Since the scale factor SR was divided by 8 three times, OctaveCount = 3.
Refer to TABLE 2: RangeMMI = 2.048 so the height of the square is 2.
First Octave:
Second Octave:
Third Octave:
This results in a square with a height of 2 bMMI's and a base on the 5/8 th's bMML 431.64.
The result of STEP 5 is a square with a height of 2 bMMI's and a base on the 5/8 th's bMML 431.64. Refer to TABLE 2: The likely "best square" is either the (5,7) or the (6,8).
The bottom and top of the (5,7) square are:
Bottom: 431.64
Top: 431.64 + (2 x 1.953125) = 435.55
The bottom and top of the (6,8) square are:
Bottom: 431.64 + 1.953125 = 433.59
Top: 435.55 + 1.953125 = 437.50
Calculate the fit errors:
The "best square" is the (6,8) square since the (6,8) square has the smallest error.
So the final square is:
100.0% 437.5 87.5% 437.01 75.0% 436.52 62.5% 436.03 50.0% 435.54 37.5% 435.05 25.0% 434.57 12.5% 434.08 0.0% 433.59
PriceRange = 7170 - 7110 = 60.0
Refer to TABLE 1: SR = 10000
Octave 1:
Octave 2:
Octave 3:
Since the scale factor SR was divided by 8 three times, OctaveCount = 3.
Refer to TABLE 2: RangeMMI = 3.072 so the height of the square is 4.
First Octave:
Second Octave:
Third Octave:
This results in a square with a height of 4 bMMI's and a base on the 4/8 th's bMML 7109.375.
The result of STEP 5 is a square with a height of 4 bMMI's and a base on the 4/8 th's bMML 7109.375. Refer to TABLE 2: The likely "best square" is the (4,8). One could, of course, perform a test using the error function and check other squares as was done in the prior examples. A quick visual check of Chart #300, however, shows that the (2,6) or (6,2) squares will result in errors that are greater than the error associated with the (4,8) square.
The bottom and top of the (4,8) square are:
Bottom: 7109.375
Top: 7109.375 + (4 x 19.53125) = 7187.5
Since the original price values were multiplied by 10000, the reverse operation is performed to arrive at MML values that match the quoted prices of the Deutsche Mark.
The "corrected" bottom and top of the (4,8) square are:
Bottom: .7109
Top: .7187
So the final square is:
100.0% .7187 87.5% .7177 75.0% .7168 62.5% .7158 50.0% .7148 37.5% .7138 25.0% .7129 12.5% .7119 0.0% .7109
PriceRange = 10275 - 10205 = 70.0
Refer to TABLE 1: SR = 10000
Octave 1:
Octave 2:
Octave 3:
Since the scale factor SR was divided by 8 three times, OctaveCount = 3.
Refer to TABLE 2: RangeMMI = 3.584 so the height of the square is 4.
First Octave:
Second Octave:
Third Octave:
This results in a square with a height of 4 bMMI's and a base on the 2/8 th's bMML 10195.3125.
The result of STEP 5 is a square with a height of 4 bMMI's and a base on the 2/8 th's bMML 10195.3125. Refer to TABLE 2: The likely "best square" is the (2,6). One could, of course, perform a test using the error function and check other squares as was done in the prior examples. A quick visual check of Chart #298, however, shows that the (0,4) or (4,8) squares will result in errors that are greater than the error associated with the (2,6) square.
The bottom and top of the (4,8) square are:
Bottom: 10195.3125
Top: 10195.3125 + (4 x 19.53125) = 10273.4375
Since the original price values were multiplied by 100, the reverse operation is performed to arrive at MML values that match the quoted prices of the 30 Yr Bond.
The "corrected" bottom and top of the (4,8) square are:
Bottom: 101.95
Top: 102.73
So the final square is:
100.0% 102.73 87.5% 102.63 75.0% 102.54 62.5% 102.44 50.0% 102.34 37.5% 102.24 25.0% 102.15 12.5% 102.05 0.0% 101.95
PriceRange = 32.75 - 26.75 = 6.0
Refer to TABLE 1: SR = 100
Octave 1:
Octave 2:
Since the scale factor SR was divided by 8 two times, OctaveCount = 2.
Refer to TABLE 2: RangeMMI = 3.84 so the height of the square is 4.
First Octave:
Second Octave:
This results in a square with a height of 4 mMMI's and a base on the 1/8 th's mMML 26.5625
The result of STEP 5 is a square with a height of 4 mMMI's and a base on the 1/8 th's mMML 26.5625. Refer to TABLE 2: Two squares are candidates for the "best square", the (0,4) square and the (2,6) square.
The bottom and top of the (0,4) square are:
Bottom: 26.5625 - 1.5625 = 25.0
Top: 25.0 + (4 x 1.5625) = 31.25
The bottom and top of the (2,6) square are:
Bottom: 26.5625 + 1.5625 = 28.125
Top: 28.125 + (4 x 1.5625) = 34.375
Now apply the error function to each square to determine "the best square in time".
Error(0,4) = abs(32.75 - 31.25) + abs(26.75 - 25.0) = 3.25
Error(2,6) = abs(32.75 - 34.375) + abs(26.75 - 28.125) = 3.0
The (2,6) square has the smallest error and one would expect it to be the square of choice. Refer to Chart #85 in the Murrey Math book. The square selected in the book was the (0,4) square.
Error(0,4) = abs(32.75 - 31.25) + abs(26.75 - 25.0) = 3.25
Error(2,6) = abs(32.75 - 34.375) + abs(26.75 - 28.125) = 3.0
TABLE 3 # of MMI's in Square 0.0% 12.5% 25% 37.5% 50% 62.5% 75% 87.5% 100% 2 0/8 2/8s 4/8s 6/8s 1/8 2/8s 4/8s 6/8s 2/8 2 1/8 2/8s 4/8s 6/8s 2/8 2/8s 4/8s 6/8s 3/8 2 2/8 2/8s 4/8s 6/8s 3/8 2/8s 4/8s 6/8s 4/8 2 3/8 2/8s 4/8s 6/8s 4/8 2/8s 4/8s 6/8s 5/8 ** 2 4/8 2/8s 4/8s 6/8s 5/8 2/8s 4/8s 6/8s 6/8 2 5/8 2/8s 4/8s 6/8s 6/8 2/8s 4/8s 6/8s 7/8 2 6/8 2/8s 4/8s 6/8s 7/8 2/8s 4/8s 6/8s 8/8 2 7/8 2/8s 4/8s 6/8s 8/8 2/8s 4/8s 6/8s 1/8 4 0/8 4/8s 1/8 4/8s 2/8 4/8s 3/8 4/8s 4/8 4 2/8 4/8s 3/8 4/8s 4/8 4/8s 5/8 4/8s 6/8 4 4/8 4/8s 5/8 4/8s 6/8 4/8s 7/8 4/8s 8/8 4 6/8 4/8s 7/8 4/8s 8/8 4/8s 1/8 4/8s 2/8 8 0/8 1/8 2/8 3/8 4/8 5/8 6/8 7/8 8/8 8 2/8 3/8 4/8 5/8 6/8 7/8 8/8 1/8 2/8 8 4/8 5/8 6/8 7/8 8/8 1/8 2/8 3/8 4/8 8 6/8 7/8 8/8 1/8 2/8 3/8 4/8 5/8 6/8
TABLE 4: TRENDLINES Line Line Points Forming the Line: Trend Slope Point 1 Point 2 + 8:8 O X + 8:7 O G' + 8:6 O F' + 8:5 O E' + 8:4 O D' + 8:3 O C' + 8:2 O B' + 8:1 O A' + 1:8 O Q + 2:8 O R + 3:8 O S + 4:8 O T + 5:8 O U + 6:8 O V + 7:8 O W - 8:8 O' P - 8:7 O' G - 8:6 O' F - 8:5 O' E - 8:4 O' D - 8:3 O' C - 8:2 O' B - 8:1 O' A - 1:8 O' W - 2:8 O' V - 3:8 O' U - 4:8 O' T - 5:8 O' S - 6:8 O' R - 7:8 O' Q TABLE 5: MOMENTUM LINES Line Line Points Forming the Line: Trend Slope Point 1 Point 2 + 1:1 G Q + 2:2 F R + 3:3 E S + 4:4 D T + 5:5 C U + 6:6 B V + 7:7 A W + 8:8 O X + 7:7 G' Q' + 6:6 F' R' + 5:5 E' S' + 4:4 D' T' + 3:3 C' U' + 2:2 B' V' + 1:1 A' W' - 1:1 G' W - 2:2 F' V - 3:3 E' U - 4:4 D' T - 5:5 C' S - 6:6 B' R - 7:7 A' Q - 8:8 O' P - 7:7 G W' - 6:6 F V' - 5:5 E U' - 4:4 D T' - 3:3 C S' - 2:2 B R' - 1:1 A Q' O' A' B' C' D' E' F' G' ------|------|------|------|------|------|------|------- X 8/8 | | | W' | W 7/8 | | | V' | V 6/8 | | | U' | U 5/8 | | | T' | T 4/8 | | | S' | S 3/8 | | | R' | R 2/8 | | | Q' | Q 1/8 | | ------|------|------|------|------|------|------|------- P 0/8 O A B C D E F G 0/8 1/8 2/8 3/8 4/8 5/8 6/8 7/8 8/8 TIME ===> FIGURE 4
1/8 th + .78 cents 50% of the time = 2.34 2/8 ths (3.125) 75% of the time = 3.12 3/8 ths (4.68) 85% of the time = 4.68 4/8 ths (6.25) 90% of the time = 6.25 5/8 ths (7.81) 95% of the time = 7.81
If Price Moves By: The probability of reversal is: (1 x mMMI) + (4 x bMMI) 50% (2 x mMMI) 75% (3 x mMMI) 85% (4 x mMMI) 90% (5 x mMMI) 95%
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