not to be republished NCERT MATHEMATICAL MODELLING Appendix 2 A.2.1 Introduction A.2.2 Why Mathematical Modelling?

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1 256 MATHEMATICS A.2.1 Inroducion In class XI, we have learn abou mahemaical modelling as an aemp o sudy some par (or form) of some real-life problems in mahemaical erms, i.e., he conversion of a physical siuaion ino mahemaics using some suiable condiions. Roughly speaking mahemaical modelling is an aciviy in which we make models o describe he behaviour of various phenomenal aciviies of our ineres in many ways using words, drawings or skeches, compuer programs, mahemaical formulae ec. In earlier classes, we have observed ha soluions o many problems, involving applicaions of various mahemaical conceps, involve mahemaical modelling in one way or he oher. Therefore, i is imporan o sudy mahemaical modelling as a separae opic. In his chaper, we shall furher sudy mahemaical modelling of some real-life problems using echniques/resuls from marix, calculus and linear programming. A.2.2 Why Mahemaical Modelling? Sudens are aware of he soluion of word problems in arihmeic, algebra, rigonomery and linear programming ec. Someimes we solve he problems wihou going ino he physical insigh of he siuaional problems. Siuaional problems need physical insigh ha is inroducion of physical laws and some symbols o compare he mahemaical resuls obained wih pracical values. To solve many problems faced by us, we need a echnique and his is wha is known as mahemaical modelling. Le us consider he following problems: (i) (ii) (iii) (iv) To find he widh of a river (paricularly, when i is difficul o cross he river). To find he opimal angle in case of sho-pu (by considering he variables such as : he heigh of he hrower, resisance of he media, acceleraion due o graviy ec.). To find he heigh of a ower (paricularly, when i is no possible o reach he op of he ower). To find he emperaure a he surface of he Sun. Appendix 2 MATHEMATICAL MODELLING

2 (v) (vi) MATHEMATICAL MODELLING 257 Why hear paiens are no allowed o use lif? (wihou knowing he physiology of a human being). To find he mass of he Earh. (vii) Esimae he yield of pulses in India from he sanding crops (a person is no allowed o cu all of i). (viii) Find he volume of blood inside he body of a person (a person is no allowed o bleed compleely). (ix) Esimae he populaion of India in he year 2020 (a person is no allowed o wai ill hen). All of hese problems can be solved and infac have been solved wih he help of Mahemaics using mahemaical modelling. In fac, you migh have sudied he mehods for solving some of hem in he presen exbook iself. However, i will be insrucive if you firs ry o solve hem yourself and ha oo wihou he help of Mahemaics, if possible, you will hen appreciae he power of Mahemaics and he need for mahemaical modelling. A.2.3 Principles of Mahemaical Modelling Mahemaical modelling is a principled aciviy and so i has some principles behind i. These principles are almos philosophical in naure. Some of he basic principles of mahemaical modelling are lised below in erms of insrucions: (i) Idenify he need for he model. (for wha we are looking for) (ii) Lis he parameers/variables which are required for he model. (iii) Idenify he available releven daa. (wha is given?) (iv) Idenify he circumsances ha can be applied (assumpions) (v) Idenify he governing physical principles. (vi) Idenify (a) he equaions ha will be used. (b) he calculaions ha will be made. (c) he soluion which will follow. (vii) Idenify ess ha can check he (a) consisency of he model. (b) uiliy of he model. (viii) Idenify he parameer values ha can improve he model.

3 258 MATHEMATICS The above principles of mahemaical modelling lead o he following: seps for mahemaical modelling. Sep 1: Idenify he physical siuaion. Sep 2: Conver he physical siuaion ino a mahemaical model by inroducing parameers / variables and using various known physical laws and symbols. Sep 3: Find he soluion of he mahemaical problem. Sep 4: Inerpre he resul in erms of he original problem and compare he resul wih observaions or experimens. Sep 5: If he resul is in good agreemen, hen accep he model. Oherwise modify he hypoheses / assumpions according o he physical siuaion and go o Sep 2. The above seps can also be viewed hrough he following diagram: Fig A.2.1 Example 1 Find he heigh of a given ower using mahemaical modelling. Soluion Sep 1 Given physical siuaion is o find he heigh of a given ower. Sep 2 Le AB be he given ower (Fig A.2.2). Le PQ be an observer measuring he heigh of he ower wih his eye a P. Le PQ = h and le heigh of ower be H. Le α be he angle of elevaion from he eye of he observer o he op of he ower. Fig A.2.2

4 MATHEMATICAL MODELLING 259 Le l = PC = QB Now an α = AC H h = PC l or H = h + l an α... (1) Sep 3 Noe ha he values of he parameers h, l and α (using sexan) are known o he observer and so (1) gives he soluion of he problem. Sep 4 In case, if he foo of he ower is no accessible, i.e., when l is no known o he observer, le β be he angle of depression from P o he foo B of he ower. So from ΔPQB, we have PQ h anβ = = or l = h co β QB l Sep 5 is no required in his siuaion as exac values of he parameers h, l, α and β are known. Example 2 Le a business firm produces hree ypes of producs P 1, P 2 and P 3 ha uses hree ypes of raw maerials R 1 and R 3. Le he firm has purchase orders from wo cliens F 1 and F 2. Considering he siuaion ha he firm has a limied quaniy of R 1 and R 3, respecively, prepare a model o deermine he quaniies of he raw maerial R 1 and R 3 required o mee he purchase orders. Soluion Sep 1 The physical siuaion is well idenified in he problem. Sep 2 Le A be a marix ha represens purchase orders from he wo cliens F 1 and F 2. Then, A is of he form P1 P2 P3 F = 1 A F 2 Le B be he marix ha represens he amoun of raw maerials R 1 and R 3, required o manufacure each uni of he producs P 1, P 2 and P 3. Then, B is of he form R1R2R3 P1 B= P 2 P 3

5 260 MATHEMATICS Sep 3 Noe ha he produc (which in his case is well defined) of marices A and B is given by he following marix R1R2R3 F = 1 AB F 2 which in fac gives he desired quaniies of he raw maerials R 1 and R 3 o fulfill he purchase orders of he wo cliens F 1 and F 2. Example 3 Inerpre he model in Example 2, in case A=, B = and he available raw maerials are 330 unis of R 1, 455 unis of R 2 and 140 unis of R 3. Soluion Noe ha AB = R1 R 2 R3 F = F This clearly shows ha o mee he purchase order of F 1 and F 2, he raw maerial required is 335 unis of R 1, 467 unis of R 2 and 147 unis of R 3 which is much more han he available raw maerial. Since he amoun of raw maerial required o manufacure each uni of he hree producs is fixed, we can eiher ask for an increase in he available raw maerial or we may ask he cliens o reduce heir orders. Remark If we replace A in Example 3 by A 1 given by A 1 = i.e., if he cliens agree o reduce heir purchase orders, hen A 1 B =

6 MATHEMATICAL MODELLING 261 This requires 311 unis of R 1, 436 unis of R 2 and 138 unis of R 3 which are well below he available raw maerials, i.e., 330 unis of R 1, 455 unis of R 2 and 140 unis of R 3. Thus, if he revised purchase orders of he cliens are given by A 1, hen he firm can easily supply he purchase orders of he wo cliens. Noe One may furher modify A so as o make full use of he available raw maerial. Query Can we make a mahemaical model wih a given B and wih fixed quaniies of he available raw maerial ha can help he firm owner o ask he cliens o modify heir orders in such a way ha he firm makes he full use of is available raw maerial? The answer o his query is given in he following example: Example 4 Suppose P 1, P 2, P 3 and R 1, R 3 are as in Example 2. Le he firm has 330 unis of R 1, 455 unis of R 2 and 140 unis of R 3 available wih i and le he amoun of raw maerials R 1 and R 3 required o manufacure each uni of he hree producs is given by R1 R2 R3 P B= P P How many unis of each produc is o be made so as o uilise he full available raw maerial? Soluion Sep 1 The siuaion is easily idenifiable. Sep 2 Suppose he firm produces x unis of P 1, y unis of P 2 and z unis of P 3. Since produc P 1 requires 3 unis of R 1, P 2 requires 7 unis of R 1 and P 3 requires 5 unis of R 1 (observe marix B) and he oal number of unis, of R 1, available is 330, we have Similarly, we have 3x + 7y + 5z = 330 (for raw maerial R 1 ) 4x + 9y + 12z = 455 (for raw maerial R 2 ) and 3y + 7z = 140 (for raw maerial R 3 ) This sysem of equaions can be expressed in marix form as x y z 140

7 262 MATHEMATICS Sep 3 Using elemenary row operaions, we obain x y This gives x = 20, y = 35 and z = 5. Thus, he firm can produce 20 unis of P 1, 35 unis of P 2 and 5 unis of P 3 o make full use of is available raw maerial. Remark One may observe ha if he manufacurer decides o manufacure according o he available raw maerial and no according o he purchase orders of he wo cliens F 1 and F 2 (as in Example 3), he/she is unable o mee hese purchase orders as F 1 demanded 6 unis of P 3 where as he manufacurer can make only 5 unis of P 3. Example 5 A manufacurer of medicines is preparing a producion plan of medicines M 1 and M 2. There are sufficien raw maerials available o make boles of M 1 and boles of M 2, bu here are only boles ino which eiher of he medicines can be pu. Furher, i akes 3 hours o prepare enough maerial o fill 1000 boles of M 1, i akes 1 hour o prepare enough maerial o fill 1000 boles of M 2 and here are 66 hours available for his operaion. The profi is Rs 8 per bole for M 1 and Rs 7 per bole for M 2. How should he manufacurer schedule his/her producion in order o maximise profi? Soluion Sep 1 To find he number of boles of M 1 and M 2 in order o maximise he profi under he given hypoheses. Sep 2 Le x be he number of boles of ype M 1 medicine and y be he number of boles of ype M 2 medicine. Since profi is Rs 8 per bole for M 1 and Rs 7 per bole for M 2, herefore he objecive funcion (which is o be maximised) is given by Z Z (x, y) = 8x + 7y The objecive funcion is o be maximised subjec o he consrains (Refer Chaper 12 on Linear Programming) x y x y z x y x 0, y (1) Sep 3 The shaded region OPQRST is he feasible region for he consrains (1) (Fig A.2.3). The co-ordinaes of verices O, P, Q, R, S and T are (0, 0), (20000, 0), (20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respecively.

8 MATHEMATICAL MODELLING 263 Noe ha Z a P (0, 0) = 0 Fig A.2.3 Z a P (20000, 0) = = Z a Q (20000, 6000) = = Z a R (10500, 34500) = = Z a S = (5000, 40000) = = Z a T = (0, 40000) = = Now observe ha he profi is maximum a x = and y = and he maximum profi is Rs Hence, he manufacurer should produce boles of M 1 medicine and boles of M 2 medicine in order o ge maximum profi of Rs Example 6 Suppose a company plans o produce a new produc ha incur some coss (fixed and variable) and le he company plans o sell he produc a a fixed price. Prepare a mahemaical model o examine he profiabiliy. Soluion Sep 1 Siuaion is clearly idenifiable.

9 264 MATHEMATICS Sep 2 Formulaion: We are given ha he coss are of wo ypes: fixed and variable. The fixed coss are independen of he number of unis produced (e.g., ren and raes), while he variable coss increase wih he number of unis produced (e.g., maerial). Iniially, we assume ha he variable coss are direcly proporional o he number of unis produced his should simplify our model. The company earn a cerain amoun of money by selling is producs and wans o ensure ha i is maximum. For convenience, we assume ha all unis produced are sold immediaely. The mahemaical model Le x = number of unis produced and sold C = oal cos of producion (in rupees) I = income from sales (in rupees) P = profi (in rupees) Our assumpions above sae ha C consiss of wo pars: (i) fixed cos = a (in rupees), (ii) variable cos = b (rupees/uni produced). Then C = a + bx... (1) Also, income I depends on selling price s (rupees/uni) Thus I = sx... (2) The profi P is hen he difference beween income and coss. So P = I C = sx (a + bx) =(s b) x a... (3) We now have a mahemaical model of he relaionships (1) o (3) beween he variables x, C, I, P, a, b, s. These variables may be classified as: independen x dependen C, I, P parameers a, b, s The manufacurer, knowing x, a, b, s can deermine P. Sep 3 From (3), we can observe ha for he break even poin (i.e., make neiher profi a nor loss), he mus have P = 0, i.e., x unis. s b Seps 4 and 5 In view of he break even poin, one may conclude ha if he company a produces few unis, i.e., less han x unis, hen he company will suffer loss s b

10 MATHEMATICAL MODELLING 265 and if i produces large number of unis, i.e., much more han a s b unis, hen i can make huge profi. Furher, if he break even poin proves o be unrealisic, hen anoher model could be ried or he assumpions regarding cash flow may be modified. Remark From (3), we also have d P s b dx = This means ha rae of change of P wih respec o x depends on he quaniy s b, which is he difference of selling price and he variable cos of each produc. Thus, in order o gain profi, his should be posiive and o ge large gains, we need o produce large quaniy of he produc and a he same ime ry o reduce he variable cos. Example 7 Le a ank conains 1000 lires of brine which conains 250 g of sal per lire. Brine conaining 200 g of sal per lire flows ino he ank a he rae of 25 lires per minue and he mixure flows ou a he same rae. Assume ha he mixure is kep uniform all he ime by sirring. Wha would be he amoun of sal in he ank a any ime? Soluion Sep 1 The siuaion is easily idenifiable. Sep 2 Le y = y () denoe he amoun of sal (in kg) in he ank a ime (in minues) afer he inflow, ouflow sars. Furher assume ha y is a differeniable funcion. When = 0, i.e., before he inflow ouflow of he brine sars, y = 250 g 1000 = 250 kg Noe ha he change in y occurs due o he inflow, ouflow of he mixure. Now he inflow of brine brings sal ino he ank a he rae of 5 kg per minue (as g = 5 kg) and he ouflow of brine akes sal ou of he ank a he rae of y y y 25 kg per minue (as a ime, he sal in he ank is kg) Thus, he rae of change of sal wih respec o is given by or dy y = 5 d 40 (Why?) dy 1 y d + = 5... (1) 40

11 266 MATHEMATICS This gives a mahemaical model for he given problem. Sep 3 Equaion (1) is a linear equaion and can be easily solved. The soluion of (1) is given by ye where, c is he consan of inegraion. Noe ha when = 0, y = 250. Therefore, 250 = C or C = 50 Then (2) reduces o or or Therefore = 40 = 200e + C or y () = C e... (2) 40 y = e... (3) y 200 = e = 40log 40 e 50 y 200 e 50 y 200 Here, he equaion (4) gives he ime a which he sal in ank is y kg. e... (4) Sep 4 Since 40 is always posiive, from (3), we conclude ha y > 200 a all imes Thus, he minimum amoun of sal conen in he ank is 200 kg. Also, from (4), we conclude ha > 0 if and only if 0 < y 200 < 50 i.e., if and only if 200 < y < 250 i.e., he amoun of sal conen in he ank afer he sar of inflow and ouflow of he brine is beween 200 kg and 250 kg. Limiaions of Mahemaical Modelling Till oday many mahemaical models have been developed and applied successfully o undersand and ge an insigh ino housands of siuaions. Some of he subjecs like mahemaical physics, mahemaical economics, operaions research, bio-mahemaics ec. are almos synonymous wih mahemaical modelling. Bu here are sill a large number of siuaions which are ye o be modelled. The reason behind his is ha eiher he siuaion are found o be very complex or he mahemaical models formed are mahemaically inracable.

12 MATHEMATICAL MODELLING 267 The developmen of he powerful compuers and super compuers has enabled us o mahemaically model a large number of siuaions (even complex siuaions). Due o hese fas and advanced compuers, i has been possible o prepare more realisic models which can obain beer agreemens wih observaions. However, we do no have good guidelines for choosing various parameers / variables and also for esimaing he values of hese parameers / variables used in a mahemaical model. Infac, we can prepare reasonably accurae models o fi any daa by choosing five or six parameers / variables. We require a minimal number of parameers / variables o be able o esimae hem accuraely. Mahemaical modelling of large or complex siuaions has is own special problems. These ype of siuaions usually occur in he sudy of world models of environmen, oceanography, polluion conrol ec. Mahemaical modellers from all disciplines mahemaics, compuer science, physics, engineering, social sciences, ec., are involved in meeing hese challenges wih courage.