We have given the timings of the gold medalists in the 400-metre race from the time the event was included in the Olympics, in the table below. Construct a mathematical model relating the years and timings. Use it to estimate the timing in the next Olympics.
Year Timing (in second)
1964 52.01
1968 52.03
1972 51.08
1976 49.28
1980 48.88
1984 48.83
1988 48.65
1992 48.83
1996 48.25
2000 49.11
2004 49.41
Mathematical Modelling
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Question

We have given the timings of the gold medalists in the 400-metre race from the time the event was included in the Olympics, in the table below. Construct a mathematical model relating the years and timings. Use it to estimate the timing in the next Olympics.
Year Timing (in second)
1964 52.01
1968 52.03
1972 51.08
1976 49.28
1980 48.88
1984 48.83
1988 48.65
1992 48.83
1996 48.25
2000 49.11
2004 49.41

Mathematical Modelling

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KnowledgeBoat · Accepted Answer

Let us first convert the problem into a mathematical problem. Step 1 : Formulation : Let us take 1964 as 0th year, and write 1 for 1968, 2 for 1972 and so on. So table will be : Year | Timing (in second) --------|------------------ 0 | 52.01 1 | 52.03 2 | 51.08 3 | 49.28 4 | 48.88 5 | 48.83 6 | 48.65 7 | 48.83 8 | 48.25 9 | 49.11 10 | 49.41 The difference in timings of gold medalist in 400 meters race in Olympics is given in the following table : Year | Timing (in second) | Difference -------- | --------------------- | -------- 0 | 52.01 | 0 1 | 52.03 | 0.02 2 | 51.08 | -0.95 3 | 49.28 | - 1.8 4 | 48.88 | -0.4 5 | 48.83 | -0.05 6 | 48.65 | -0.18 7 | 48.83 | 0.18 8 | 48.25 | -0.58 9 | 49.11 | 0.86 10 | 49.41 | 0.3 Total | | ΣDifference = -2.6 At the end of 4 years period from 1964 - 1968 the timing has increased by 0.02 second from 52.01 to 52.03 second. At the end of second Olympic the reduction in timing is 0.95 second from 52.03 to 51.08. from the table above we cannot find a definite relationship between the number of years and change in timing. By formula, Mean difference = Mean of differences = = -0.26 We have assumed that the timing in 400 m race of Olympic reduced at the rate of 0.26 per year. So, timing in the first year = 52.01 - 0.26 = 51.75 Timing in the second year = 52.01 - 2 x 0.26 = 52.01 - 5.2 = 51.49 Similarly in the n^th year ⇒ t = 52.01 - 0.26 n for n ≥ 1 ......(1) Now we have to find t for n = 11. Step 2 : Finding a solution : Substituting n = 11 in equation (1) we get : ⇒ t = 52.01 - 0.26 x 11 ⇒ t = 52.01 - 2.86 ⇒ t = 49.15 seconds. Step 3 : Interpretation : Since, we are dealing with a real life situation, we have to see to what extent this value matches with the real situation. Step 4 : Validation : Let us find the values for the years we already know, using formula/equation (1) and compare it with known values by finding the difference. the values are given in the table : Year | Timing (in second) | Timing using equation (1) | Deviation --------|----------------------------|--------------|---------- 0 | 52.01 | 52.01 | 0 1 | 52.03 | 51.75 | 0.28 2 | 51.08 | 51.49 | -0.41 3 | 49.28 | 51.23 | -1.95 4 | 48.88 | 50.97 | -2.09 5 | 48.83 | 50.71 | -1.88 6 | 48.65 | 50.45 | -1.8 7 | 48.83 | 50.19 | -1.36 8 | 48.25 | 49.93 | -1.68 9 | 49.11 | 49.67 | -0.56 10 | 49.41 | 49.41 | 0.00 Total | | | Σ Deviation = -11.45 Suppose we decide that this error is negligible. In this case formula/equation (1) is our mathematical model and if we decide that this error is not acceptable. Then we have to go back to step 1, the formulation and change equation (1). Step 1 : Reformulation : We still assume that the values decrease steadily by 0.26, but we will now introduce a correction factor to reduce the error. For this, we find the mean of all the deviations. This is : We take the mean of the errors and correct our formula by this value. Step 2 : Revised Mathematical Description : Let us now add the mean of the deviations to our formula/equation (1). So, our corrected formula is : ⇒ t = 52.01 - 0.26 n - 1.145 ⇒ t = 50.865 - 0.26 n ......(2) for n ≥ 1 Step 3 : Altered solution : By solving equation (2) for t, Putting n = 1 in equation (2) we get : ⇒ t = 50.865 - 0.26(1) ⇒ t = 50.865 - 0.26 = 50.605 Putting n = 2 in equation (2) we get : ⇒ t = 50.865 - 0.26(2) ⇒ t = 50.865 - 0.52 = 50.345 Putting n = 11 in equation (2) we get : ⇒ t = 50.865 - 0.26(11) ⇒ t = 50.865 - 2.86 ⇒ t = 48.005 seconds. Step 3 : Interpretation : So in 11^th year(in next Olympics) t = 48.005 seconds Step 4 : Validation : Once again, let us compare the values got by using formula/equation (2) with the actual values. Table is given below for comparison. Year | Timing (in second) | Timing using equation (2) | Deviation -------- | --------------------- | ---------------- | ------------ 0 | 52.01 | 52.01 | 0 1 | 52.03 | 50.605 | 1.425 2 | 51.08 | 50.345 | 0.735 3 | 49.28 | 50.085 | -0.805 4 | 48.88 | 49.825 | -0.945 5 | 48.83 | 49.565 | -0.735 6 | 48.65 | 49.305 | -0.655 7 | 48.83 | 49.045 | -0.215 8 | 48.25 | 48.785 | -0.535 9 | 49.11 | 48.525 | 0.585 10 | 49.41 | 48.265 | 1.145 Total | | |Σ Deviation = 0 As we can see many of the values that formula/equation (2) gives are closer to the actual value than the values that formula/equation (1) gives the mean of the deviations is 0. So, formula/equation (2) is our mathematical description that gives a mathematical relationship between years and timing for race. Hence, timing in next olympics is 48.005 seconds.

Mathematics

Mathematical Modelling

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Year	Timing (in second)
1964	52.01
1968	52.03
1972	51.08
1976	49.28
1980	48.88
1984	48.83
1988	48.65
1992	48.83
1996	48.25
2000	49.11
2004	49.41