Module MA3484: Methods of Mathematical Economics Hilary Term 2017 Part I (Sections 1 to 3)

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1 Module MA3484: Methods of Mathematical Economics Hilary Term 2017 Part I (Sections 1 to 3) D. R. Wilkins Copyright c David R. Wilkins Contents 1 Mathematical Programming Problems A Furniture Retailing Problem A Transportation Problem concerning Dairy Produce Finite-Dimensional Vector Spaces Real Vector Spaces Linear Dependence and Bases Dual Spaces The Transportation Problem The General Transportation Problem Transportation Problems where Supply equals Demand Bases for the Transportation Problem Basic Feasible Solutions of Transportation Problems The Northwest Corner Method The Minimum Cost Method for finding Basic Feasible Solutions Effectiveness of the Minimum Cost Method Formal Description of the Minimum Cost Method Formal Description of the Northwest Corner Method A Method for finding Basic Optimal Solutions Formal Analysis of the Solution of the Transportation Problem 55 i

2 1 Mathematical Programming Problems 1.1 A Furniture Retailing Problem A retail business is planning to devote a number of retail outlets to the sale of armchairs and sofas. The retail prices of armchairs and sofas are determined by fierce competition in the furniture retailing business. Armchairs sell for e700 and sofas sell for e1000. However the amount of floor space (and warehouse space) available for stocking the sofas and armchairs is limited; the amount of capital available for purchasing the initial stock of sofas and armchairs is limited; market research shows that the ratio of armchairs to sofas in stores should neither be too low nor too high. Specifically: there are 1000 square metres of floor space available for stocking the initial purchase of sofas and armchairs; each armchair takes up 1 square metre; each sofa takes up 2 square metres; the amount of capital available for purchasing the initial stock of armchairs and sofas is e351,000; the wholesale price of an armchair is e400; the wholesale price of a sofa is e600; market research shows that between 4 and 9 armchairs should be in stock for each 3 sofas in stock. We suppose that the retail outlets are stocked with x armchairs and y sofas. The armchairs (taking up 1 sq. metre each) and the sofas (taking up 2 sq. metres each) cannot altogether take up more than 1000 sq. metres of floor space. Therefore x + 2y 1000 (Floor space constraint). 1

3 The cost of stocking the retail outlets with armchairs (costing e400 each) and sofas (costing e600 each) cannot exceed the available capital of e Therefore 4x + 6y 3510 (Capital constraint). Consumer research indicates that x and y should satisfy 4y 3x 9y (Armchair/Sofa ratio). An ordered pair (x, y) of real numbers is said to specify a feasible solution to the linear programming problem if this pair of values meets all the relevant constraints. An ordered pair (x, y) constitutes a feasible solution to the the Furniture Retailing problem if and only if all the following constraints are satisfied: x 3y 0; 4y 3x 0; x + 2y 1000; 4x + 6y 3510; x 0; y 0; The feasible region for the Furniture Retailing problem is depicted below: 500 4x + 6y = 3510 x + 2y = /3 chairs per sofa 3 chairs per sofa 4y = 3x 3y = x Capital Floor space We identify the vertices (or corners) of the feasible region for the Furniture Retailing problem. There are four of these: there is a vertex at (0, 0); there is a vertex at (400, 300) where the line 4y = 3x intersects the line x + 2y = 1000; 2

4 there is a vertex at (510, 245) where the line x + 2y = 1000 intersects the line 4x + 6y = 3510; there is a vertex at (585, 195) where the line 3y = x intersects the line 4x + 6y = These vertices are identified by inspection of the graph that depicts the constraints that determine the feasible region. The furniture retail business obviously wants to confirm that the business will make a profit, and will wish to determine how many armchairs and sofas to purchase from the wholesaler to maximize expected profit. There are fixed costs for wages, rental etc., and we assume that these are independent of the number of armchairs and sofas sold. The gross margin on the sale of an armchair or sofa is the difference between the wholesale and retail prices of that item of furniture. Armchairs cost e400 wholesale and sell for e700, and thus provide a gross margin of e300. Sofas cost e600 wholesale and sell for e1000, and thus provide a gross margin of e400. In a typical linear programming problem, one wishes to determine not merely feasible solutions to the problem. One wishes to determine an optimal solution that maximizes some objective function amongst all feasible solutions to the problem. The objective function for the Furniture Retailing problem is the gross profit that would accrue from selling the furniture in stock. This gross profit is the difference between the cost of purchasing the furniture from the wholesaler and the return from selling that furniture. This objective function is thus f(x, y), where f(x, y) = 300x + 400y. We should determine the maximum value of this function on the feasible region. Because the objective function f(x, y) = 300x + 400y is linear in x and y, its maximum value on the feasable region must be achieved at one of the vertices of the region. Clearly this function is not maximized at the origin (0, 0)! Now the remaining vertices of the feasible region are at (400, 300), (510, 245) and (585, 195), and f(400, 300) = 240, 000, f(510, 245) = 251, 000, 3

5 f(585, 195) = 253, 500. It follows that the objective function is maximized at (585, 195). The furniture retail business should therefore use up the available capital, stocking 3 armchairs for every sofa, despite the fact that this will not utilize the full amount of floor space available. A linear programming problem may be presented as follows: given real numbers c i, A i,j and b j for i = 1, 2,..., m and j = 1, 2,..., n, find real numbers x 1, x 2,..., x n so as to maximize c 1 x 1 + c 2 x c n x n subject to constraints x j 0 for j = 1, 2,..., n, and A i,1 x 1 + A i,2 x A i,n x n b i for i = 1, 2,..., m. The furniture retailing problem may be presented in this form with n = 2, m = 4, (c 1, c 2 ) = (300, 400), A = , b 1 b 2 b 3 b 4 = Here A represents the m n whose coefficient in the ith row and jth column is A i,j. Linear programming problems may be presented in matrix form. We adopt the following notational conventions with regard to transposes, row and column vectors and vector inequalities:. vectors in R m and R n are represented as column vectors; we denote by M T the n m matrix that is the transpose of an m n matrix M; in particular, given b R m and c R n, where b and c are represented as column vectors, we denote by b T and c T the corresponding row vectors obtained on transposing the column vectors representing b and c; 4

6 given vectors u and v in R n for some positive integer n, we write u v (and v u) if and only if u j v j for j = 1, 2,..., n. Linear programming problems formulated as above may be presented in matrix notation as follows: Given an m n matrix A with real coefficients, and given column vectors b R m and c R n, find x R n so as to maximize c T x subject to constraints Ax b and x A Transportation Problem concerning Dairy Produce The Transportation Problem is a well-known problem and important example of a linear programming problem. Discussions of the general problem are to be found in textbooks in the following places: Chapter 8 of Linear Programming: 1 Introduction, by George B. Danzig and Mukund N. Thapa (Springer, 1997); Section 18 of Chapter I of Methods of Mathematical Economics by Joel N. Franklin (SIAM 2002). We discuss an example of the Transportation Problem of Linear Programming, as it might be applied to optimize transportation costs in the dairy industry. A food business has milk-processing plants located in various towns in a small country. We shall refer to these plants as dairies. Raw milk is supplied by numerous farmers with farms located throughout that country, and is transported by milk tanker from the farms to the dairies. The problem is to determine the catchment areas of the dairies so as to minimize transport costs. We suppose that there are m farms, labelled by integers from 1 to m that supply milk to n dairies, labelled by integers from 1 to n. Suppose that, in a given year, the ith farm has the capacity to produce and supply a s i litres of milk for i = 1, 2,..., n, and that the jth dairy needs to receive at least d j litres of milk for j = 1, 2,..., n to satisfy the business obligations. 5

7 The quantity m s i then represents that total supply of milk, and the quantity n d j represents the total demand for milk. We suppose that x i,j litres of milk are to be transported from the ith farm to the jth dairy, and that c i,j represents the cost per litre of transporting this milk. Then the total cost of transporting milk from the farms to the dairies is m n c i,j x i,j. The quantities x i,j of milk to be transported from the farms to the dairies should then be determined for i = 1, 2,..., m and j = 1, 2,..., n so as to minimize the total cost of transporting milk. However the ith farm can supply no more than s i litres of milk in a given year, and that jth dairy requires at least d j litres of milk in that year. It follows that the quantities x i,j of milk to be transported between farms and dairy are constrained by the requirements that n x i,j s i for i = 1, 2,..., m and m x i,j d j for j = 1, 2,..., n. Suppose that the requirements of supply and demand are satisfied. Then n m n m d j x i,j s i. Thus the total supply must equal or exceed the total demand. If it is the case that n x i,j < s i for at least one value of i then m n x i,j < m s i. Similarly if it is the case that m x i,j > d j for at least one value of j then m n x i,j > n d j. It follows that if total supply equals total demand, so that m n s i = d j, 6

8 then and n x i,j = s i for i = 1, 2,..., m m x i,j = d j for j = 1, 2,..., n. The following report, published in 2006, describes a study of milk transport costs in the Irish dairy industry: Quinlan C., Enright P., Keane M., O Connor D The Milk Transport Cost Implications of Alternative Dairy Factory Location. Agribusiness Discussion Paper No. 47. Dept of Food Business and Development. University College, Cork. The report is available at the following URL foodbusinessanddevelopment/paper47.pdf The problem was investigated using commercial software that implements standard linear programming algorithms for the solution of forms of the Transportation Problem. The description of the methodology used in the study begins as follows: A transportation model based on linear programming was developed and applied the Irish dairy industry to meet the study objectives. In such transportation models, transportation costs are treated as a direct linear function of the number of units shipped. The major assumptions are: 1. The items to be shipped are homogenous (i.e., they are the same regardless of their source or destination. 2. The shipping cost per unit is the same regardless of the number of units shipped. 3. There is only one route or mode of transportation being used between each source and each destination, Stevenson, (1993). Sources and Destinations 7

9 In 2004 there were about 25,000 dairy farmers in the Irish Republic. Hence identifying the location and size of each individual dairy farm as sources for the transportation model was beyond available resources. An alternative approach based on rural districts was adopted. There are 156 rural districts in the state and data for dairy cow numbers by rural district from the most recent livestock census was available from the Central Statistics Office (CSO). These data were converted to milk equivalent terms using average milk yield estimates. Typical seasonal milk supply patterns were also assumed. In this way an estimate of milk availability throughout the year by rural district was derived and this could then be further converted to milk tanker loads, depending on milk tanker size. The following is quoted from the conclusions of that report: A major report on the strategic development of the Irish dairy-processing sector proposed processing plant rationalization, Strategic Development Plan for the Irish Dairy Processing Sector Prospectus, (2003). It was recommended that in the long term the number of plants processing butter, milk powder, casein and whey products in Ireland should be reduced to create four major sites for these products, with a limited number of additional sites for cheese and other products. It was estimated that savings from processing plant economies of scale would amount to e20m per annum, Prospectus (2003). However, there is an inverse relationship between milk transport costs and plant size. Thus the optimum organisation of the industry involves a balancing of decreasing average plant costs against the increasing transport costs. In this analysis, the assumed current industry structure of 23 plants was reduced in a transportation modelling exercise firstly to 12 plants, then 9 plants and finally 6 plants and the increase in total annual milk transport costs for each alternative was calculated. Both a good location and a poor location 6 plant option were considered. The estimated milk transport costs for the different alternatives were; 4.60 cent per gallon for 23 plants; 4.85 cent per gallon for 12 plants; 5.04 cent per gallon for 9 plants; 5.24 cent per gallon for 6 plants (good location) and 5.75 cent per gallon (poor location) respectively. 8

10 In aggregate terms the results showed that milk transport costs would increase by e3, e5, e7 and e13 million per annum if processing plants were reduced from 23 to 12 to 9 to 6 (good location) and 6 (poor location) respectively. As the study of processing plant rationalization did not consider cheese plant rationalization in detail, it was inferred that the estimated saving from economies of scale of e20 million per annum was associated with between 6 and 12 processing sites. Excluding the 6 plant (poor location) option, the additional milk transport cost of moving to this reduced number of sites was estimated to be of the order of 5 million per annum. This represents about 25 per cent of the estimated benefits from economies of scale arising from processing plant rationalization. The transportation model also facilitated a comparison of current milk catchment areas of processing plants with optimal catchment areas, assuming no change in number of processing plants. It was estimated that if dairies were to collect milk on an optimal basis, there would be an 11% reduction from the current (2005) milk transport costs. In the benchmark model 23 plants were required to stay open at peak to accommodate milk supply and it was initially assumed that all 23 remained open throughout the year with the same catchment areas. However, due to seasonality in milk supply, it is not essential that all 23 plants remain open outside the peak. Two options were analysed. The first involved allowing the model to determine the least cost transport pattern outside the peak i.e. a relaxation of the constraint of fixed catchment areas throughout the year, with all plants available for milk intake. Further modest reductions in milk transport costs were realisable in this case. The second option involved keeping only the bigger plants open outside the peak period. A modest increase in milk transport costs was estimated for this option due to tankers having to travel longer distances outside the peak period. The analysis of milk transport costs in the Irish Dairy Industry is a significant topic in the Ph.D. thesis of the first author of the 2006 report from which the preceding quotation was taken: Quinlan, Carrie, Brigid, Optimisation of the food dairy coop supply chain. PhD Thesis, University College Cork. 9

11 which is available at the following URL: /1197/QuinlanCB PhD2013.pdf The Transportation Problem, with equality of total supply and total demand, can be expressed generally in the following form. Some commodity is supplied by m suppliers and is transported from those suppliers to n recipients. The ith supplier can supply at most to s i units of the commodity, and the jth recipient requires at least d j units of the commodity. The cost of transporting a unit of the commodity from the ith supplier to the jth recipient is c i,j. The total transport cost is then m n c i,j x i,j. where x i,j denote the number of units of the commodity transported from the ith supplier to the jth recipient. The Transportation Problem can then be presented as follows: determine x i,j for i = 1, 2,..., m and j = 1, 2,..., n so as minimize i,j c i,j x i,j subject to the constraints x i,j 0 for all i and j, n x i,j s i and m x i,j d j, where m s i n d j. 10

12 2 Finite-Dimensional Vector Spaces 2.1 Real Vector Spaces Definition A real vector space consists of a set V on which there is defined an operation of vector addition, yielding an element v + w of V for each pair v, w of elements of V, and an operation of multiplication-by-scalars that yields an element λv of V for each v V and for each real number λ. The operation of vector addition is required to be commutative and associative. There must exist a zero element 0 V of V that satisfies v + 0 V = v for all v V, and, for each v V there must exist an element v of V for which v+( v) = 0 V. The following identities must also be satisfied for all v, w V and for all real numbers λ and µ: (λ + µ)v = λv + µv, λ(v + w) = λv + λw, λ(µv) = (λµ)v, 1v = v. Let n be a positive integer. The set R n consisting of all n-tuples of real numbers is then a real vector space, with addition and multiplication-byscalars defined such that and (x 1, x 2,..., x n ) + (y 1, y 2,..., y n ) = (x 1 + y 1, x 2 + y 2,... x n + y n ) λ(x 1, x 2,..., x n ) = (λx 1, λx 2,..., λx n ) for all (x 1, x 2,..., x n ), (y 1, y 2,..., y n ) R and for all real numbers λ. The set M m,n (R) of all m n matrices is a real vector space with respect to the usual operations of matrix addition and multiplication of matrices by real numbers. 2.2 Linear Dependence and Bases Elements u 1, u 2,..., u m of a real vector space V are said to be linearly dependent if there exist real numbers λ 1, λ 2,..., λ m, not all zero, such that λ 1 u 1 + λ 2 u λ m u m = 0 V. If elements u 1, u 2,..., u m of real vector space V are not linearly dependent, then they are said to be linearly independent. Elements u 1, u 2,..., u n of a real vector space V are said to span V if, given any element v of V, there exist real numbers λ 1, λ 2,..., λ n such that v = λ 1 u 1 + λ 2 u λ n u n. 11

13 A vector space is said to be finite-dimensional if there exists a finite subset of V whose members span V. Elements u 1, u 2,..., u n of a finite-dimensional real vector space V are said to constitute a basis of V if they are linearly independent and span V. Lemma 2.1 Elements u 1, u 2,..., u n of a real vector space V constitute a basis of V if and only if, given any element v of V, there exist uniquelydetermined real numbers λ 1, λ 2,..., λ n such that v = λ 1 u 1 + λ 2 u λ n u n. Proof Suppose that u 1, u 2,..., u n is a basis of V. Let v be an element V. The requirement that u 1, u 2,..., u n span V ensures that there exist real numbers λ 1, λ 2,..., λ n such that v = λ 1 u 1 + λ 2 u λ n u n. If µ 1, µ 2,..., µ n are real numbers for which then v = µ 1 u 1 + µ 2 u µ n u n, (µ 1 λ 1 )u 1 + (µ 2 λ 2 )u (µ n λ n )u n = 0 V. It then follows from the linear independence of u 1, u 2,..., u n that µ i λ i = 0 for i = 1, 2,..., n, and thus µ i = λ i for i = 1, 2,..., n. This proves that the coefficients λ 1, λ 2,..., λ n are uniquely-determined. Conversely suppose that u 1, u 2,..., u n is a list of elements of V with the property that, given any element v of V, there exist uniquely-determined real numbers λ 1, λ 2,..., λ n such that v = λ 1 u 1 + λ 2 u λ n u n. Then u 1, u 2,..., u n span V. Moreover we can apply this criterion when v = 0. The uniqueness of the coefficients λ 1, λ 2,..., λ n then ensures that if λ 1 u 1 + λ 2 u λ n u n = 0 V then λ i = 0 for i = 1, 2,..., n. Thus u 1, u 2,..., u n are linearly independent. This proves that u 1, u 2,..., u n is a basis of V, as required. Proposition 2.2 Let V be a finite-dimensional real vector space, let u 1, u 2,..., u n be elements of V that span V, and let K be a subset of {1, 2,..., n}. Suppose either that K = or else that those elements u i for which i K are linearly independent. Then there exists a basis of V whose members belong to the list u 1, u 2,..., u n which includes all the vectors u i for which i K. 12

14 Proof We prove the result by induction on the number of elements in the list u 1, u 2,..., u n of vectors that span V. The result is clearly true when n = 1. Thus suppose, as the induction hypothesis, that the result is true for all lists of elements of V that span V and that have fewer than n members. If the elements u 1, u 2,..., u n are linearly independent, then they constitute the required basis. If not, then there exist real numbers λ 1, λ 2,..., λ n, not all zero, such that λ 1 u 1 + λ 2 u λ n u n = 0 V. Now there cannot exist real numbers λ 1, λ 2,..., λ n, not all zero, such that both n λ i u i = 0 V and also λ i = 0 whenever i K. Indeed, in the case where K =, this conclusion follows from the requirement that the real numbers λ i cannot all be zero, and, in the case where K, the conclusion follows from the linear independence of those u i for which i K. Therefore there must exist some integer i satisfying 1 i n for which λ i 0 and i K. Without loss of generality, we may suppose that u 1, u 2,..., u n are ordered so that n K and λ n 0. Then n 1 λ i u n = u i. λ n Let v be an element of V. Then there exist real numbers µ 1, µ 2,..., µ n such that v = n µ i u i, because u 1, u 2,..., u n span V. But then n 1 ( v = µ i µ ) nλ i u i. λ n We conclude that u 1, u 2,..., u n 1 span the vector space V. The induction hypothesis then ensures that there exists a basis of V consisting of members of this list that includes the linearly independent elements u 1, u 2,..., u m, as required. Corollary 2.3 Let V be a finite-dimensional real vector space, and let u 1, u 2,..., u n be elements of V that span the vector space V. Then there exists a basis of V whose elements are members of the list u 1, u 2,..., u n. Proof This result is a restatement of Proposition 2.2 in the special case where the set K in the statement of that proposition is the empty set. 13

15 2.3 Dual Spaces Definition Let V be a real vector space. A linear functional ϕ: V R on V is a linear transformation from the vector space V to the field R of real numbers. Given linear functionals ϕ: V R and ψ: V R on a real vector space V, and given any real number λ, we define ϕ + ψ and λϕ to be the linear functionals on V defined such that (ϕ + ψ)(v) = ϕ(v) + ψ(v) and (λϕ)(v) = λ ϕ(v) for all v V. The set V of linear functionals on a real vector space V is itself a real vector space with respect to the algebraic operations of addition and multiplication-by-scalars defined above. Definition Let V be a real vector space. The dual space V of V is the vector space whose elements are the linear functionals on the vector space V. Now suppose that the real vector space V is finite-dimensional. Let u 1, u 2,..., u n be a basis of V, where n = dim V. Given any v V there exist uniquely-determined real numbers λ 1, λ 2,..., λ n such that v = n λ j u j. It follows that there are well-defined functions ε 1, ε 2,..., ε n from V to the field R defined such that ε i ( n λ j u j ) = λ i for i = 1, 2,..., n and for all real numbers λ 1, λ 2,..., λ n. These functions are linear transformations, and are thus linear functionals on V. Lemma 2.4 Let V be a finite-dimensional real vector space, let u 1, u 2,..., u n be a basis of V, and let ε 1, ε 2,..., ε n be the linear functionals on V defined such that ε i ( n λ j u j ) = λ i for i = 1, 2,..., n and for all real numbers λ 1, λ 2,..., λ n. Then ε 1, ε 2,..., ε n constitute a basis of the dual space V of V. Moreover ϕ = n ϕ(u i )ε i for all ϕ V. 14

16 Proof Let µ 1, µ 2,..., µ n be real numbers with the property that n µ i ε i = 0 V. Then ( n ) 0 = µ i ε i (u j ) = n µ i ε i (u j ) = µ j for j = 1, 2,..., n. Thus the linear functionals ε 1, ε 2,..., ε n on V are linearly independent elements of the dual space V. Now let ϕ: V R be a linear functional on V, and let µ i = ϕ(u i ) for i = 1, 2,..., n. Now { 1 if i = j; ε i (u j ) = 0 if i j. It follows that ( n ) ( n µ i ε i λ j u j ) for all real numbers λ 1, λ 2,..., λ n. It follows that n ϕ = µ i ε i = = = n n µ i λ j ε i (u j ) = n µ j λ j ( n n ) λ j ϕ(u j ) = ϕ λ j u j n ϕ(u i )ε i. We conclude from this that every linear functional on V can be expressed as a linear combination of ε 1, ε 2,..., ε n. Thus these linear functionals span V. We have previously shown that they are linearly independent. It follows that they constitute a basis of V. Moreover we have verified that ϕ = n ϕ(u i )ε i for all ϕ V, as required. Definition Let V be a finite-dimensional real vector space, let u 1, u 2,..., u n be a basis of V. The corresponding dual basis of the dual space V of V consists of the linear functionals ε 1, ε 2,..., ε n on V, where ε i ( n λ j u j ) = λ i for i = 1, 2,..., n and for all real numbers λ 1, λ 2,..., λ n. Corollary 2.5 Let V be a finite-dimensional real vector space, and let V be the dual space of V. Then dim V = dim V. 15

17 Proof We have shown that any basis of V gives rise to a dual basis of V, where the dual basis of V has the same number of elements as the basis of V to which it corresponds. The result follows immediately from the fact that the dimension of a finite-dimensional real vector space is the number of elements in any basis of that vector space. Let V be a real-vector space, and let V be the dual space of V. Then V is itself a real vector space, and therefore has a dual space V. Now each element v of V determines a corresponding linear functional E v : V R on V, where E v (ϕ) = ϕ(v) for all ϕ V. It follows that there exists a function ι: V V defined so that ι(v) = E v for all v V. Then ι(v)(ϕ) = ϕ(v) for all v V and ϕ V. Now and ι(v + w)(ϕ) = ϕ(v + w) = ϕ(v) + ϕ(w) = (ι(v) + ι(w))(ϕ) ι(λv)(ϕ) = ϕ(λv) = λϕ(v) = (λι(v))(ϕ) for all v, w V and ϕ V and for all real numbers λ. It follows that ι(v + w) = ι(v) + ι(w) and ι(λv) = λι(v) for all v, w V and for all real numbers λ. Thus ι: V V is a linear transformation. Proposition 2.6 Let V be a finite-dimensional real vector space, and let ι: V V be the linear transformation defined such that ι(v)(ϕ) = ϕ(v) for all v V and ϕ V. Then ι: V V is an isomorphism of real vector spaces. Proof Let u 1, u 2,..., u n be a basis of V, let ε 1, ε 2,..., ε n be the dual basis of V, where { 1 if i = j, ε i (u j ) = 0 if i j, and let v V. Then there exist real numbers λ 1, λ 2,..., λ n such that v = n λ i u i. Suppose that ι(v) = 0 V. Then ϕ(v) = E v (ϕ) = 0 for all ϕ V. In particular λ i = ε i (v) = 0 for i = 1, 2,..., n, and therefore v = 0 V. We conclude that ι: V V is injective. Now let F : V R be a linear functional on V, let λ i = F (ε i ) for i = 1, 2,..., n, let v = n λ i u i, and let ϕ V. Then ϕ = n ϕ(u i )ε i (see 16

18 Lemma 2.4), and therefore ι(v)(ϕ) = ϕ(v) = n λ i ϕ(u i ) = n F (ε i )ϕ(u i ) ( n ) = F ϕ(u i )ε i = F (ϕ). Thus ι(v) = F. We conclude that the linear transformation ι: V V is surjective. We have previously shown that this linear transformation is injective. There ι: V V is an isomorphism between the real vector spaces V and V as required. The following corollary is an immediate consequence of Proposition 2.6. Corollary 2.7 Let V be a finite-dimensional real vector space, and let V be the dual space of V. Then, given any linear functional F : V R, there exists some v V such that F (ϕ) = ϕ(v) for all ϕ V. 17

19 3 The Transportation Problem 3.1 The General Transportation Problem The Transportation Problem can be expressed in the following form. Some commodity is supplied by m suppliers and is transported from those suppliers to n recipients. The ith supplier can supply at most s i units of the commodity, and the jth recipient requires at least d j units of the commodity. The cost of transporting a unit of the commodity from the ith supplier to the jth recipient is c i,j. The total transport cost is then m n c i,j x i,j. where x i,j denote the number of units of the commodity transported from the ith supplier to the jth recipient. The Transportation Problem can then be presented as follows: determine x i,j for i = 1, 2,..., m and j = 1, 2,..., n so as minimize c i,j x i,j subject to the constraints x i,j 0 for all i i,j n and j, x i,j s i and m x i,j d j, where s i 0 for all i, d j 0 for all i, and m s i n d j. The quantities (s 1, s 2,..., s m ) representing the quantities of the transported commodity supplied by the suppliers are the components of an m- dimensional vector (s 1, s 2,..., s m ). We refer to this vector as the supply vector for the transportation problem. The quantities (d 1, d 2,..., d m ) representing the quantities of the transported commodity demanded by the recipients are the components of an n-dimensional vector (d 1, d 2,..., d n ). We refer to this vector as the demand vector for the transportation problem. The quantities c i,j that represent the cost of transporting the commodity from the ith supplier to the jth recipient are the components of an m n matrix. We refer to this matrix as the cost matrix for the transportation problem. 18

20 3.2 Transportation Problems where Supply equals Demand Consider a transportation problem with m suppliers and n recipients. The following proposition shows that a solution to the transportation problem can only exist if total supply of the relevant commodity exceeds total demand for that commodity. Proposition 3.1 Let s 1, s 2,..., s m and d 1, d 2,..., d n be non-negative real numbers. Suppose that there exist non-negative real numbers x i,j for i = 1, 2,..., m and j = 1, 2,..., n that satisfy the inequalities n x i,j s i and m x i,j d j. Then Moreover if it is the case that n d j n d j = m s i. m s i. then and n x i,j = s i for i = 1, 2,..., m m x i,j = d j for j = 1, 2,..., n. Proof The inequalities satisfied by the non-negative real numbers x i,j ensure that n m n m d j x i,j s i. Thus the total supply must equal or exceed the total demand. Now s i n x i,j 0 for i = 1, 2,..., m. It follows that if s i > n x i,j for at least one value of i then m s i > m n x i,j. Similarly m x i,j d j 0 19

21 for j = 1, 2,..., n. It follows that if it is the case that m x i,j > d j for at least one value of j then m n x i,j > n d j. It follows that if total supply equals total demand, so that m n s i = d j, then and as required. n x i,j = s i for i = 1, 2,..., m m x i,j = d j for j = 1, 2,..., n, We analyse the Transportation Problem in the case where total supply equals total demand. The optimization problem in this case can then be stated as follows: determine x i,j for i = 1, 2,..., m and j = 1, 2,..., n so as minimize c i,j x i,j subject to the constraints x i,j 0 for all i i,j n and j, x i,j = s i and m x i,j = d j, where s i 0 and d j 0 for all i and j, and m s i = n d j. Definition A feasible solution to a transportation problem (with equality of total supply and total demand) is represented by real numbers x i,j, where x i,j 0 for i = 1, 2,..., m and j = 1, 2,..., n; n x i,j = s i for = 1, 2,..., m; m x i,j = d j for j = 1, 2,..., n. Definition A feasible solution (x i,j ) of a transportation problem is said to be optimal if it minimizes cost amongst all feasible solutions of that transportation problem. 20

22 3.3 Bases for the Transportation Problem Definition Let I = {1, 2,..., m} and J = {1, 2,..., n}, where m and n are positive integers. Then a subset B of I J is said to be a basis for the transportation problem with m suppliers and n recipients if, given any vectors y R m and z R n satisfying m (y) i = n (z) j, there exists a unique m n matrix X with real coefficients satisfying the following properties: (i) n (X) i,j = (y) i for i = 1, 2,..., m; (ii) m (X) i,j = (z) j for j = 1, 2,..., n; (iii) (X) i,j = 0 unless (i, j) B. Lemma 3.2 Let I = {1, 2,..., m} and J = {1, 2,..., n}, where m and n are positive integers. and let B = {(i, j) I J : i = m or j = n}. Then B is a basis for a transportation problem with m suppliers and n recipients. Proof The result can readily be verified when m = 1 or n = 1. We therefore restrict attention to cases where m > 1 and n > 1. Let B = {(i, j) I J : i = m or j = n}, where m > 1 and n > 1. Then, given any vectors y R m and z R n that satisfy m y i = n z j, there exists a unique m n matrix X with real coefficients with all the following properties: (i) n (X) i,j = y i for i = 1, 2,..., m; (ii) m (X) i,j = z j for j = 1, 2,..., n; (iii) (X) i,j = 0 unless (i, j) B. 21

23 This matrix X has coefficients as follows: X i,j = 0 if i < m and j < n; X i,n = y i for i < m; X m,j = z j for j < n; X m,n = w, where n 1 w = y m z j = z n m 1 y i. This matrix X is thus of the form y y 2 X = y m 1 z 1 z 2... z n 1 w, where n 1 w = y m z j = z n m 1 It follows from the definition of bases for transportation problems that the subset B of I J is a basis for a transportation problem with m suppliers and n recipients. This completes the proof. We now introduce some notation for use in discussion of the theory of transportation problems. For each integer i between 1 and m, let e (i) denote the m-dimensional vector whose ith component is equal to 1 and whose other components are zero. For each integer j between 1 and n, let ê (j) denote the n-dimensional vector whose jth component is equal to 1 and whose other components are zero. Thus y i. (e (i) ) k = { 1 if i = k, 0 if i k, and (ê (j) ) l = { 1 if j = l; 0 if j l. Moreover y = m (y) i e (i) for all y R m and z = n (y) i e (i) for all z R n. Also, for each ordered pair (i, j) of integers with 1 i m and 1 j n, let E (i,j) denote the m n matrix that has a single non-zero coefficient equal to 1 located in the ith row and jth column of the matrix. Thus { 1 if k = i and j = l; (E (i,j) ) k,l = 0 if k i or j l. 22

24 Moreover X = m n (X) i,j E (i,j) for all m n matrices X with real coefficients. We let ρ: M m,n (R) R m and σ: M m,n (R) R n be the linear transformations defined such that (ρ(x)) i = n (X) i,j for i = 1, 2,..., m and (σ(x)) j = m (X) i,j for j = 1, 2,..., n. Then ρ(e (i,j) ) = e (i) for i = 1, 2,..., m and σ(e (i,j) ) = ê (j) for j = 1, 2,..., n. A feasible solution of the transportation problem with given supply vector s, demand vector d and cost matrix C is represented by an m n matrix X satisfying the following three conditions: The coefficients of X are all non-negative; ρ(x) = s; σ(x) = d. The cost functional f: M m,n (R) R is defined so that f(x) = m n c i,j (X) i,j = trace(c T X) i=0 j=0 for all X M m,n (R), where C is the cost matrix and c i,j = (C) i,j for i = 1, 2,..., m and j = 1, 2,..., n. A feasible solution ˆX of the Transportation problem is optimal if and only if f( ˆX) f(x) for all feasible solutions X of that problem. Lemma 3.3 Let X be an m n matrix, let ρ(x) R m and σ(x) R n be defined so that (ρ(x)) i = n (X) i,j for i = 1, 2,..., m and (σ(x)) j = m (X) i,j for j = 1, 2,..., n, and let W = { Then (ρ(x), σ(x)) W. } (y, z) R m R n : m (y) i = n (z) j. 23

25 Proof Summing the components of the vectors ρ(x) and σ(x), we find that m (ρ(x)) i = m n (X) i,j = n (σ(x)) j. Thus (ρ(x), σ(x)) W, as required. Given a subset K of I J, where I = {1, 2,..., m} and J = {1, 2,..., n}, we denote by M K the vector subspace of the space M m,n (R) of m n matrices with real coefficients defined such that M K = {X M m,n (R) : (X) i,j = 0 unless (i, j) K}. The definition of bases for transportation problems then ensures that a subset B of I J is a basis for a transportation problem with m suppliers and n-recipients if and only if the linear transformation θ B : M B W is an isomorphism of vector spaces, where { } W = (y, z) R m R n : m (y) i = n (z) j, and θ B (X) = (ρ(x), σ(x)) for all X M B, where (ρ(x)) i = i = 1, 2,..., m and (σ(x)) j = m (X) i,j for j = 1, 2,..., n. n (X) i,j for Proposition 3.4 A basis for a transportation problem with m suppliers and n recipients has m + n 1 elements. Proof Let I = {1, 2,..., m} and J = {1, 2,..., n} and, for all (i, j) I J, let E (i,j) denote the m n matrix defined so that { 1 if k = i and j = l; (E (i,j) ) k,l = 0 if k i or j l. Let B be a basis for the transportation problem with m suppliers and n recipients. Then the m n matrices E (i,j) for which (i, j) B constitute a basis of the vector space M B where M B = {X M m,n (R) : (X) i,j = 0 unless (i, j) B}. It follows that the dimension of the vector space M B is equal to the number of elements in the basis B. 24

26 Let W = { } (y, z) R m R n : m (y) i = n (z) j, and let θ B : M B W be defined so that θ B (X) = (ρ(x), σ(x)) for all X M B, where ρ(x) i = n (X) i,j for i = 1, 2,..., m, and σ(x) j = m (X) i,j for j = 1, 2,..., n. Now the definition of bases for transportation problems ensures that θ: M B W is an isomorphism of vector spaces. Therefore dim M B = dim W. It follows that any two bases for a transportation problem with m suppliers and n recipients have the same number of elements. Lemma 3.2 showed that {(i, j) I J : i = m or j = n} is a basis for a transportation problem with m suppliers and n recipients. This basis has m + n 1 elements. It follows that dim W = m + n 1, and therefore every basis for a transportation problem with m suppliers and n recipients has m + n 1 elements, as required. Proposition 3.5 Let I = {1, 2,..., m} and J = {1, 2,..., n}, where m and n are positive integers, and let K be a subset of I J. Suppose that, given any vectors y R m and z R n satisfying m (y) i = n (z) j, there exists an m n matrix X with real coefficients belonging to M K with the following properties: (i) n (X) i,j = y i for i = 1, 2,..., m; (ii) m (X) i,j = z j for j = 1, 2,..., n; (iii) (X) i,j = 0 unless (i, j) K. Then there exists a basis B for the transportation problem satisfying B K. Proof First we define bases for the vector spaces involved in the proof. For each integer i between 1 and m, let e (i) R m be defined such that { 1 if i = k; (e (i) ) k = 0 if i k. 25

27 For each integer j between 1 and n, let ê (j) R n be defined such that { 1 if j = l; (ê (j) ) l = 0 if j l. For each ordered pair (i, j) of integers with 1 i m and 1 j n, let E (i,j) M n (R) be defined such that { 1 if k = i and j = l; (E (i,j) ) k,l = 0 if k i or j l. Let M K denote the vector subspace of the space M m,n (R) of m n matrices with real coefficients defined such that let M K = {X M m.n (R) : (X) i,j = 0 unless (i, j) K}, W = { } (y, z) R m R n : m (y) i = n (z) j, and let θ K : M K W be the linear transformation defined so that θ K (X) = (ρ(x), σ(x)) for all X M m,n (R), where ρ(x) i = n (X) i,j for i = 1, 2,..., m and σ(x) j = m (X) i,j for j = 1, 2,..., n. Then X = (i,j) K (X) i,j E (i,j) for all X M K, and therefore θ K (X) = (X) i,j θ(e (i,j) ) = (i,j) K (i,j) K (X) i,j (e (i), ê (j) ) for all X M K. The conditions of the proposition ensure that that the ordered pairs (e (i), ê (j) ) of basis vectors for which (i, j) belongs to K span the vector space W. It then follows from standard linear algebra that there exists a subset B of K such that those ordered pairs (e (i), ê (j) ) for which (i, j) belongs to B constitute a basis for the vector space W (see Corollary 2.3). Thus, given any ordered pair (y, z) of vectors belonging to W, there exist uniquely determined real numbers x i,j for all (i, j) B such that (y, z) = x i,j (e (i), ê (j) ). (i,j) B 26

28 Let X M B be the m n matrix defined such that (X) i,j = x i,j for all (i, j) B and (X) i,j = 0 for all (i, j) (I J) \ B. Then X is the unique m n matrix with the properties that ρ(x) = y, σ(x) = z and X (i,j) = 0 unless (i, j) B. It follows that the subset B of K is the required basis for the transportation problem. Proposition 3.6 Let m and n be positive integers, let I = {1, 2,..., m} and J = {1, 2,..., n}, and let K be a subset of I J. Suppose that there is no basis B of the transportation problem for which K B. Then there exists a non-zero m n matrix Y with real coefficients which satisfies the following conditions: n (Y ) i,j = 0 for i = 1, 2,..., m; m (Y ) i,j = 0 for j = 1, 2,..., n; (Y ) i,j = 0 when (i, j) K. Proof For each integer i between 1 and m, let e (i) R m be defined such that { 1 if i = k; (e (i) ) k = 0 if i k. For each integer j between 1 and n, let ê (j) R n be defined such that (ê (j) ) l = { 1 if j = l; 0 if j l., and let W = { (y, z) R m R n : m (y) i = } n (z) j. Now follows from Proposition 2.2 that if the elements (e (i), ê (j) ) for which (i, j) K were linearly independent then there would exist a subset B of I J satisfying K B such that the elements (e (i), ê (j) ) for which (i, j) B would constitute a basis of W. It would then follow that, given any ordered pair (y, z) of vectors belonging to W, there would exist a unique m n matrix X with real coefficients with the properties that m (X) i,j = (y) i for i = 1, 2,..., m, n (X) i,j = (z) i for j = 1, 2,..., n, and (X) i,j = 0 unless (i, j) B. The subset B of I J would thus be a basis for the transportation problem. But the subset K is not contained in any basis for the Transportation Problem. It follows that the elements (e (i), ê (j) ) for which 27

29 (i, j) K must be linearly dependent. Therefore there exists a non-zero m n matrix Y with real coefficients such that (Y ) i,j = 0 when (i, j) K and m n (Y ) i,j (e (i), ê (j) ) = (0, 0). But then m n (Y ) i,j e (i) = 0 and m n (Y ) i,j ê (j) = 0, and therefore and n (Y ) i,j = 0 for i = 1, 2,..., m m (Y ) i,j = 0 for j = 1, 2,..., n. Also (Y ) i,j = 0 unless (i, j) K. The result follows. 3.4 Basic Feasible Solutions of Transportation Problems Consider the transportation problem with m suppliers and n recipients, where the ith supplier can provide at most s i units of some given commodity, where s i 0, and the jth recipient requires at least d j units of that commodity, where d j 0. We suppose also that total supply equals total demand, so that m n s i = d j, The cost of transporting the commodity from the ith supplier to the jth recipient is c i,j. Definition A feasible solution (x i,j ) of a transportation problem is said to be basic if there exists a basis B for that transportation problem such that x i,j = 0 whenever (i, j) B. Example Consider a transportation problem where m = n = 2, s 1 = 8, s 2 = 3, d 1 = 2, d 2 = 9, c 1,1 = 2, c 1,2 = 3, c 2,1 = 4 and c 2,2 = 1. 28

30 A feasible solution takes the form of a 2 2 matrix ( ) x1,1 x 1,2 x 2,1 x 2,2 with non-negative components which satisfies the two matrix equations ( ) ( ) ( ) x1,1 x 1,2 1 8 = x 2,1 x 2,2 1 3 and ( 1 1 ) ( x 1,1 x 1,2 x 2,1 x 2,2 ) = ( 2 9 ). A basic feasible solution will have at least one component equal to zero. There are four matrices with at least one zero component which satisfy the required equations. They are the following: ( ), ( ), ( ), ( The first and third of these matrices have non-negative components. These two matrices represent basic feasible solutions to the problem, and moreover they are the only basic feasible solutions. The costs associated with the components of the matrices are c 1,1 = 2, c 1,2 = 3, c 2,1 = 4 and c 2,2 = 1. ( ) 0 8 The cost of the basic feasible solution is 2 1 8c 1,2 + 2c 2,1 + c 2,2 = = 33. ( ) 2 6 The cost of the basic feasible solution is 0 3 2c 1,1 + 6c 1,2 + 3c 2,2 = = 25. ( ) x1,1 x Now any 2 2 matrix 1,2 satisfying the two matrix equations x 2,1 x 2,2 ( x1,1 x 1,2 x 2,1 x 2,2 ) ( 1 1 ( 1 1 ) ( x 1,1 x 1,2 x 2,1 x 2,2 ) = ( 8 3 ), ) = ( 2 9 ) ). 29

31 must be of the form ( x1,1 x 1,2 x 2,1 x 2,2 ) = ( λ 8 λ 2 λ 1 + λ for some real number ( λ. ) λ 8 λ But the matrix has non-negative components if and 2 λ 1 + λ only if 0 λ 2. It follows that the set of feasible solutions of this instance of the transportation problem is {( ) } λ 8 λ : λ R and 0 λ 2. 2 λ 1 + λ The costs associated with the components of the matrices are c 1,1 = 2, c 1,2 = 3, c 2,1 = 4 and c 2,2 = 1. Therefore, for each ( real number λ ) satisfying λ 8 λ 0 λ 2, the cost f(λ) of the feasible solution is given 2 λ 1 + λ by f(λ) = 2λ + 3(8 λ) + 4(2 λ) + (1 + λ) = 33 4λ. ( ) 2 6 Cost is minimized when λ = 2, and thus is the optimal solution of 0 3 this transportation problem. The cost of this optimal solution is 25. Proposition 3.7 Given any feasible solution of a transportation problem, there exists a basic feasible solution with whose cost does not exceed that of the given solution. Proof Let m and n be positive integers, and let let the m n matrix X represent a feasible solution of a transportation problem with supply vector s, demand vector d and cost matrix C, where C is an m n matrix with real coefficients. Then s i 0 for i = 1, 2,..., m and d j 0 for j = 1, 2,..., n, where s = (s 1, s 2,..., s m ), d = (d 1, d 2,..., d n ). n Also x i,j 0 for all i and j, x i,j = s i for i = 1, 2,..., m and m x i,j = d j for j = 1, 2,..., n. The cost of the feasible solution X is then m ) n c i,j x i,j, where c i,j is the coefficient in the ith row and jth column of the cost matrix C. If the feasible solution X is itself basic then there is nothing to prove. Suppose therefore that X is not a basic solution. We show that there then 30

32 exists a feasible solution X with fewer non-zero components than the given feasible solution. Let I = {1, 2,..., m} and J = {1, 2,..., n}, and let K = {(i, j) I J : x i,j > 0}. Because X is not a basic solution to the Transportation Problem, there does not exist any basis B for the transportation problem satisfying K B. It therefore follows from Proposition 3.6 that there exists a non-zero m n matrix Y whose coefficients y i,j satisfy the following conditions: n y i,j = 0 for i = 1, 2,..., m; m y i,j = 0 for j = 1, 2,..., n; y i,j = 0 when (i, j) K. We can assume without loss of generality that m n c i,j y i,j 0, where the quantities c i,j are the coefficients of the cost matrix C, because otherwise we can replace Y with Y. Let Z λ = X λy for all real numbers λ, and let z i,j (λ) denote the coefficient (Z λ ) i,j in the ith row and jth column of the matrix Z λ. Then z i,j (λ) = x i,j λy i,j for i = 1, 2,..., m and j = 1, 2,..., n. Moreover n z i,j (λ) = s i ; m z i,j (λ) = d j ; z i,j (λ) = 0 whenever (i, j) K; m n c i,j z i,j (λ) m n c i,j x i,j whenever λ 0. Now the matrix Y is a non-zero matrix whose rows and columns all sum to zero. It follows that at least one of its coefficients must be strictly positive. Thus there exists at least one ordered pair (i, j) belonging to the set K for which y i,j > 0. Let { } xi,j λ 0 = minimum : (i, j) K and y i,j > 0. y i,j 31

33 Then λ 0 > 0. Moreover if 0 λ < λ 0 then x i,j λy i,j > 0 for all (i, j) K, and if λ > λ 0 then there exists at least one element (i 0, j 0 ) of K for which x i0,j 0 λy i0,j 0 < 0. It follows that x i,j λ 0 y i,j 0 for all (i, j) K, and x i0,j 0 λ 0 y i0,j 0 = 0. Thus Z λ0 is a feasible solution of the given transportation problem whose cost does not exceed that of the given feasible solution X. Moreover Z λ0 has fewer non-zero components than the given feasible solution X. If Z λ0 is itself a basic feasible solution, then we have found the required basic feasible solution whose cost does not exceed that of the given feasible solution. Otherwise we can iterate the process until we arrive at the required basic feasible solution whose cost does not exceed that of the given feasible solution. A transportation problem has only finitely many basic feasible solutions. Indeed there are only finitely many bases for the problem, and any basis is associated with at most one basic feasible solution. Therefore there exists a basic feasible solution whose cost does not exceed the cost of any other basic feasible solution. It then follows from Proposition 3.7 that the cost of this basic feasible solution cannot exceed the cost of any other feasible solution of the given transportation problem. This basic feasible solution is thus a basic optimal solution of the Transportation Problem. The transportation problem determined by the supply vector, demand vector and cost matrix has only finitely many basic feasible solutions, because there are only finitely many bases for the problem, and each basis can determine at most one basic feasible solution. Nevertheless the number of basic feasible solutions may be quite large. But it can be shown that a transportation problem always has a basic optimal solution. It can be found using an algorithm that implements the Simplex Method devised by George B. Dantzig in the 1940s. This algorithm involves passing from one basis to another, lowering the cost at each stage, until one eventually finds a basis that can be shown to determine a basic optimal solution of the transportation problem. 3.5 The Northwest Corner Method Example We discuss in detail how to find an initial basic feasible solution of a transportation problem with 4 suppliers and 5 recipients, using a method known as the Northwest Corner Method. This method does not make use of cost information. The course of the calculation is determined by the supply vector s and 32

34 the demand vector d, where s = (9, 11, 4, 5), d = (6, 7, 5, 3, 8). We need to fill in the entries in a tableau of the form x i,j s i d j In the tableau just presented the labels on the left hand side identify the suppliers, the labels at the top identify the recipients, the numbers on the right hand side list the number of units that the relevant supplier must provide, and the numbers at the bottom identify the number of units that the relevant recipient must obtain. Number in the bottom right hand corner gives the common value of the total supply and the total demand. The values in the individual cells must be non-zero, the rows must sum to the value on the left, and the columns must sum to the value on the bottom. The Northwest Corner Method is applied recursively. At each stage the undetermined cell in at the top left (the northwest corner) is given the maximum possible value allowable with the constraints. The remainder of either the first row or the first column must then be completed with zeros. This leads to a reduced tableau to be determined with either one fewer row or else one fewer column. One continues in this fashion, as exemplified in the solution of this particular problem, until the entire tableau has been completed. The method will also determine a basis associated with the basic feasible solution determined by the Northwest Corner Method. This basis lists the cells that play the role of northwest corner at each stage of the method. At the first stage, the northwest corner cell is associated with supplier 1 and recipient 1. This cell is assigned a value equal to the mimimum of the corresponding column and row sums. Thus, this example, the northwest corner cell, is given the value 6, which is the desired column sum. The remaining cells in that row are given the value 0. The tableau then takes the following form: 33