Cost Allocations and Linear Programming

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1 Cost Allocations and Linear Programming by Richard A. Young 1 I. Introduction The heart of the accountant's job is allocation. Costs and revenues are allocated over time and across products. In this fashion, a product cost is constructed. This note uses linear programming to study cost allocation within a neo-classical, short-run, profit-maximization framework. 2 We learn that in this simple economic setting, cost allocations implicitly take place. These allocations are consistent with the economic setting, although they are not central to profit maximization. 3 We illustrate how cost allocation allows one to reframe the profit maximization problem so as to make it simpler to understand and solve. II. Short Run Profit Maximization Up to this point we have considered settings in which costs and revenues behave in a linear fashion. Of course linearity is only an approximation. And typically, it is valid only locally. If costs and revenues were globally linear in units sold, and selling price exceeded unit variable costs, an infinite amount of profit could be made by selling an infinite amount of product! So linearity is an assumption that is useful only in local analysis. Continuing, neo-classical economics tell us that increasing marginal (variable costs or decreasing marginal revenue (selling price imply there exists an optimal product mix. Yet, as accountants, we typically approximate the cost and revenue curves with straight lines. Can it make sense to talk of an optimal product mix under the accountant s linearity assumption? It can, if we introduce another frequently used managerial tool: short-run analysis. Short-run analysis assumes one or more inputs are fixed. Combining linearity with a short-run perspective, we can sensibly speak of an optimal product mix. The purpose of this note is to explore how the accountant s cost allocation procedure facilitates short-run profit planning. For concreteness, we begin with an example. Ralph is the owner of a manufacturing firm that is technologically able to manufacture Products 1 and 2. The manufacturing process requires that Product 2 pass through Machine A and Machine B. Product 1, however, requires no work on Machine B. The production standards are provided below. 1 The author wishes to thank Anil Arya and John Fellingham. Also, this note is heavily influenced by Joel Demski s text: Managerial Uses of Accounting Information, Kluwer Academic Publishers (1994. Finally, the author thanks Shannon Lilly and other former and present students for their comments and patience as he experimented with this note. 2 Linear programming is wonderful, whether it were useful (which it is or not. It is simple, yet elegant. And it is a parsimonious way to study equilibrium theory in economics. It has been said that any problem in the world can be solved using linear programming. Richard A. Young 3/2/00

2 Local linear approximations: Material cost = 5 (pounds (all variable Labor cost = 13 (labor hrs (all variable Overhead = 50 (all fixed Usage: Product 1 Product 2 Direct materials (in pounds 1 2 Direct labor (in labor hours 2 1 Machine A (in machine hours 1 1 Machine B (in machine hours 0 1 Machine A Machine B Machine hours available: As promised, we have chosen a local linear approximation for costs and revenue, and have assumed a constant-returns-to-scale technology (e.g., each unit of Product 1 requires two hours of labor. This allows us to formulate the optimal product mix problem as a linear program. Let and denote the quantities of Product 1 and 2, respectively. And let z 1 and z 2 denote the quantities of materials and labor. The linear program (P1 appears below. (P1: Max z 1-13 z 2,,z 1,z 2 subject to: z z 2 0,, z 1, z 2 The objective function is revenue minus variable costs, or total contribution margin. The fixed costs are omitted. This is not a problem, because the fixed costs are not affected by the levels of,, z 1 and z 2. Thus, the amount of fixed costs cannot affect the optimal amounts of the outputs and inputs. 4 The constraints tell us that the total amount of input used (demanded must not exceed the total supply of the resource. The first two constraints differ from the next two, because the supply of machine hours is regarded as fixed in the short run, while the supply of materials and labor is not. Note well, that the problem has been formulated without calculating any type of product cost. Cost is expressed as a function of inputs, and not as a function of outputs. No 3 These allocations are consistent with the economic setting, although they are not central to profit maximization. To bring the accountant s product costing activity to the center, one must add an ingredient or two to the neo-classical world. But that is a topic for advanced study. 4 The invariance of the solutions to the two linear programming models with and without fixed costs is an example of consistent framing. Richard A. Young 2 3/2/00

3 cost allocation has been performed. Please resist the temptation to calculate a product cost, for now. In this form, the solution to the problem is a little hard to visualize (it's fourdimensional, but you can use a spreadsheet program to optimize it. The solution is as follows: * = 100 z 1 * = 200 TCM* = 1,200 * = 50 z 2 * = 250 Solution to (P1 Disaggregating the objective function, total revenue is 38( (50 = 5,450. Total material cost is 5(200 = 1,000 and total labor cost is 13(250 = 3,250. In addition, total profit is 1, = 1,150. If we substitute the optimal solution ( *, *, z 1 *, z 2 * into the constraints, we notice that the third and fourth constraints appear to be tight. For example, in the materials constraint z 1 is equal to, and not less than, zero. Why is this so? The logic goes like this. The variable z 1 affects only the third constraint, and increasing z 1 reduces the objective function. So if the materials needed for production (demand were less than those purchased (supply, one could decrease z 1 or increase or and make the objective function increase. The model tells us it is not a good idea to acquire more materials (z 1 than will be used in production ( + 2. Likewise, it tells us it is not a good idea to acquire more labor (z 2 than will be used in production (2. Interestingly, the above statement cannot necessarily be made about the first and second constraint, since the supply of machine hours (unlike the supply of materials and labor is fixed in the short run. The out-of-pocket cost of machine capacity is buried in the fixed cost number and hence not affected by and. To convince yourself that both the first and second constraints need not be tight, suppose the price of Product 1 were $1,000. Would any Product 2 be produced? Would the second constraint be tight? You see that optimality dictates which constraints are tight. Trusting our logic, let us assume the third and fourth constraints are tight: z 1 = + 2 and z 2 = 2. Substitute for z 1 and z 2 into the objective function to obtain: ( (2 = (38-5(1-13(2 + (33-5(2-13(1 = Now we can eliminate the first and second constraints. The re-framed linear program (R1 appears below. Richard A. Young 3 3/2/00

4 (R1: Max , subject to: , Before proceeding to solve (R1, observe that the unit contribution margins have appeared as coefficients in the new objective function. Unit contribution margins are found by allocating variable costs to each unit. A property of optimality (that the materials and labor constraints are tight has told us to allocate the costs, and how to do it. More will be said about this later. It is instructive to solve for * and * by graphical means. A graph of the set of points that satisfy both constraints appears below A B O C The set of feasible solutions is the region that is formed by drawing a polygon with nodes O (0,0, A (0, 50, B (100,50, and C (150, 0. A little intuition tells us straight away that the solution will be at points B or C. But, more generally, the fundamental theorem of linear programming tells us that an optimal solution must be at one of these four points. So we check the objective function (ΤCΜ at each of these points. 0 0 TCM , ,050 Verifying Solution to (R1 Richard A. Young 4 3/2/00

5 The solution to (R1 is * = 100 and * = 50. This is the same solution we obtained when we solved the original linear program (P1. Consistent framing has occurred. Nothing was lost by reducing (P1. 5 Even though Product 1 has the lower contribution margin, we make more Product 1 than 2. This can be attributed to the presence and form of the second constraint -- Product 1 uses no Machine B hours. In general, the incremental profit per unit of scarce resource is what matters. The trick is to figure out which are the scarce resources. We will soon return to this point. III. Shadow Prices - Informal Intuition So far we have learned that when choosing optimal product mix we are interested in the marginal profit associated with each product and its demand on resources. 6 A word of caution is in order: gross margins per unit misstate marginal contributions, because they include allocations of fixed costs (based on some arbitrary capacity level. More fundamentally, gross margins distort the relative marginal profits from each product. In managerial settings, we often find it a useful simplification to assume some resources could be changed in the short run (e.g., the supply of materials, but others could not (e.g., the supply of machine hours. Here, we used an LP model to tell us how to best utilize the capacity that is assumed fixed. A natural issue that arises in this context is when and where capacity should be expanded. We now discuss how the LP model also provides useful information regarding capacity expansion decisions. Interestingly, the same approach that is useful for providing information about where capacity should be expanded is also useful for allocating cost to products. Return to Ralph. Suppose he has the option of expanding capacity by renovating Machine A. What is the maximum amount Ralph should be willing to pay to expand capacity by one Machine A hour? Of course, it is the additional profit obtained by using this extra capacity optimally. Imagine that Ralph were handed one hour of capacity of Machine A. How would he optimally exploit the situation? For convenience, we think in terms of the re-framed linear program, the one where z 1 and z 2 are suppressed. Since both products use exactly one Machine A hour, we might feel that we should make one more of the product with the higher unit contribution margin. Under this premise we would like to make one more unit of Product 2, since the unit contribution margins of Products 1 and 2 are $7 and $10, respectively. If this were feasible, profit would go up by $10. However, we should reject 5 How do we retrieve z 1 * and z 2 *? Recall that the third and fourth constraints in the original program (P1 are tight, so we substitute * and * into them and solve. 6 The efficient organization properly integrates the accounting, marketing and manufacturing functions. Richard A. Young 5 3/2/00

6 this premise. Remember the Machine B constraint is also tight, since the optimal solution (before expansion is = 100, = 50. While we would like to make one more unit of Product 2, this is infeasible; making additional units of Product 2 would violate the Machine B constraint. Instead, we optimally use one more hour of Machine A capacity to make one more unit of Product 1, and profit increases by $7. This is the benefit from adding an hour, and we would choose to purchase this additional hour if and only if its cost were less than $7. Assume, instead, we add one more hour of Machine B time. That is, Machine A capacity is 150, but Machine B capacity is now 51. What should we do with the extra one hour of Machine B capacity? Since only Product 2 uses Machine B, and this constraint is tight, we would like to produce one more unit of Product 2 (it uses exactly one hour. If we could produce one more unit of Product 2 without affecting production of Product 1, profit would go up by $10. But remember that the Machine A constraint is also tight. In fact, since each unit of Product 2 uses one hour of Machine A time, there will be one less hour available for making Product 1. Further, since each unit of Product 1 also uses one hour of Machine A time, increasing production of Product 2 by one unit means we must decrease production of Product 1 by one unit. Therefore, the net benefit from expand Machine B capacity by one hour is: + 1($10-1($7 = $3. The maximum the company is willing to pay for one more hour of Machine B capacity is thus $3, since that would be the increase in profit if the extra hour were used optimally. We have used our intuition and the solution to (R1 to find the benefit obtained of adding one hour of capacity of Machine A and 2. A less thoughtful (but equally effective method is to: (1 set the right-hand side of the Machine A constraint to 151 and the right-hand side of the Machine B constraint at 50, (2 solve for the optimal solution, and (3 subtract the previous objective function value (1,200 from the new value (it will be 1,207. To obtain the benefit of a one unit increase in Machine B hours repeat this process, first setting the right-hand sides of the Machine A and B constraints at 150 and 51, respectively. IV. Shadow Prices as the Solution to the Dual Program If you already are familiar with linear programming, you recall that the marginal benefit of adding a unit of capacity to a right-hand side of a constraint is referred to as the shadow price of that resource. Another term for the shadow price is dual variable. Thus, in our original example the dual variable for the Machine A and B constraint are 7 and 3, respectively. We now take a more rigorous approach to studying shadow prices. It turns out that there is another, equivalent, way to frame our short-run, profitmaximization problem. Rather than thinking in terms of maximizing profit (contribution margin, we might think in terms of minimizing the opportunity cost of the scarce resources. The opportunity-cost-minimizing problem is also a linear program, called the dual Richard A. Young 6 3/2/00

7 program. The dual program can be obtained directly (and even mechanically from the original profit-maximizing program, which we shall call the primal program. This dual program's solution will produce as its solution the shadow prices. Below are written the primal and dual programs for our example. We continue to work with the re-framed version. Primal Program (R1: TCM Max , subject to: 150 (w 1 50 (w 2, Dual Program (D-R1: OC Min 150 w w 2 w 1, w 2 subject to: w 1 7 ( w 1 + w 2 10 ( w 1, w 2 What intuition can we use to guide us in writing down the dual program? First, w 1 and w 2 should be interpreted as the (shadow price, or per-unit opportunity cost, of a Machine A hour and a Machine B hour. The objective function is the price of Machine A (w 1 times the amount of Machine A hours plus the price of Machine B (w 2 times the amount of Machine B hours. Thus, the objective function is the total opportunity cost of all of the resources, which we seek to minimize. The easiest way to understand the dual program constraints is in terms of equilibrium pricing in competitive economies. The non-negativity constraints w i follow quite naturally from their interpretation as prices. Also, we are quite familiar with the notion that, under the usual assumptions of increasing marginal cost and decreasing marginal revenue, a firm producing a single output would produce up to the point where marginal revenue is equal to marginal cost. We apply this logic to a multi-product firm. Take the second constraint in the dual program, labeled for good reason as ( : w 1 + w The left-hand side is the opportunity cost of producing one unit of Product 2, equal to the amount of Machine A used to make one unit of Product 2 (1 times the price of Machine A (w 1 plus the amount of Machine B used to make one unit of Product 2 (1 times the price of Machine B (w 2. The right-hand side is the benefit obtained from one unit of Product 2. Similarly, the first constraint says that the cost of a unit of Product 1 is greater than or equal to the benefit. In light of the economic interpretation, you may ask why the constraints are "greater than or equal to" and not "equal to." The answer is that we do not require that a multi-product firm manufacture every product that is technologically feasible. If under the given Richard A. Young 7 3/2/00

8 technology, the cost of the resources is larger than the benefit, it is simply not produced. The product is too expensive to make. There are better uses of the resources. A vivid (but perhaps unpleasant example is silk diapers. It is technologically feasible to make diapers out of silk, but there are other, better uses of silk. So we observe no silk diapers produced. I earlier suggested that the dual program could be obtained mechanically. Inspection of the two programs reveals the following. We form the first constraint in the dual program by reading down the column of the primal program corresponding to the variable. The coefficient on in the (w 1 constraint of the primal is 1, and the coefficient on in the (w 2 constraint is zero, leading to 1 w w 2. The right-hand side of the ( dual constraint is the coefficient on in the primal objective function, 7. We form the second dual constraint by reading down the column of the primal program corresponding to the variable. The coefficient on in the (w 1 constraint of the primal is 1, and the coefficient on in the (w 2 constraint is one, leading to 1 w w 2. The right-hand side of the ( dual constraint is the coefficient on in the primal objective function, 10. The above dual program can be solved graphically, because there are only two dual variables. 7 The (shaded feasible region is bounded by the lines w 1 = 7 and w 1 + w 2 = 10 and w 2 = 0. One need not be concerned that the feasible region is only bounded from below because we are minimizing here. w A B 7 10 w 1 w 1 7 w 2 3 OC 1, ,500 Verifying the Solution to (D-R1 7 Of course, if the problem involved more dual variables (more constraints in the primal program one would try some intuition or seek out a spreadsheet optimizer. Richard A. Young 8 3/2/00

9 Thus, the solution is a familiar one: the dual variables (shadow prices are 7 and 3. It may be a bit intriguing that the dual program's objective function when evaluated at its optimal solution is equal to the primal program's objective function when evaluated at its optimal solution (see Appendix. We can say more about the relationship between the primal and dual programs. For constraint ( in the dual program, w 1 + w 2 is the total opportunity cost of the resources used to make a unit of Product 2, and 10 is the benefit of making a unit of Product 2. What would it mean if the left-hand side were strictly greater than the right? It would mean the opportunity cost exceeded the benefit, implying one should not make that particular output. In other words, if a dual constraint is not tight, the value of the corresponding output should be zero. Conversely, if the optimal amount of the output is positive, the constraint will be tight. That is, if * > 0 (which it is in our original example, then w 1 + w 2 = 10 at the optimal solution to the dual program. Doesn't this sound like the shadow prices on the constraints in (D-R1 are the optimal solution to (R1? In fact, this is true; the dual of the dual program is the primal program! 8 In our example, all primal variables were positive at the optimal solution, so both of the dual constraints are tight. Of course, one sees immediately that, if the dual constraints are tight, the only solution is w 1 = 7 and w 2 = 3. By the way, suppose that at the optimal solution one of the machine hour constraints were not tight. What would be the benefit of its relaxation? The answer is, of course, zero. To see this, consider a different setting and its associated LP model, (R1', where the selling price of Product 1 is moved from $38 to $45. The optimal solution is * = 150, * = 0, and w 1 * = 14, w 2 * = 0. The optimal solution moves from (100, 50 to (150, 0, and only the first constraint is tight. 0 0 TCM , ,100 Verifying Solution to (R1' The relationship between a constraint and its shadow price can be summarized as follows. If a constraint is not tight, the marginal benefit of its relaxation is zero. Conversely, 8 For fun, try this: begin with the above dual program, turn it into a maximization program subject to less than or equal to constraints, apply the procedure used above, and see if you obtain a program which is identical to the primal. You may have to do some multiplying by - 1. Richard A. Young 9 3/2/00

10 if the marginal benefit obtained by relaxing a constraint is non-zero, it must be tight. This important idea is commonly referred to as complementary slackness. V. Shadow Prices and Cost Allocations - the Simple Case Let us return to the original version of Ralph s LP model (P1. We might as well write down its dual as well. (P1: Max z 1-13 z 2,,z 1,z 2 subject to: 150 (w 1 50 (w z 1 0 (w z 2 0 (w 4,, z 1, z 2 (D-P1: Min 150 w w 2 w i subject to: w w w 4 38 ( w 1 + w w 3 + w 4 33 ( - w 3-5 (z 1 - w 4-13 (z 2 w 1, w 2, w 3, w 4 The optimal solution is as follows. Primal Dual TCM = OC = 100 w 1 = 7 1,200 = 50 w 2 = 3 z 1 = 200 w 3 = 5 z 2 = 250 w 4 = 13 Optimal Solution to Primal (P1 and Dual (D-P1 We notice something quite intuitive about the solution. The shadow price on the material and labor constraints are 5 and 13, which is equal to their variable cost. The intuition is as follows. If someone handed Ralph one pound of materials, he would purchase one less pound and so save the cost per pound of $5. Note well, the amount of labor used has no effect on the amount of material used. Since $5 is the opportunity costs of the input, this is the rate at which its cost should be allocated to products. Notice that the left-hand side of the ( and ( constraints in the dual mimic unit contribution margin calculations. We see that it is okay in our example (and quite easy to let our intuition guide the calculation of the marginal (variable cost of Product 1 and 2 before optimizing over their quantities. Hence, we can calculate contribution margins and explicitly consider only the supply-demand constraints on the fixed resources (machine hours, here. But things are not Richard A. Young 10 3/2/00

11 always quite this easy. Notice here that w 4 does not appear in the third constraint in the dual program. Nor does w 3 appear in the fourth. However, we shall see in the next section that there are circumstances where inputs are interdependent and, hence, so are their costs. This makes calculation of the marginal costs and contribution margins a little more delicate. VI. Shadow Prices and Cost Allocations - Interdependencies Textbooks state that the reciprocal method of service cost allocation has a distinct advantage over the direct or step methods. The texts assert that this method allocates costs in a better way because it takes into account the reciprocal relationship among the service departments. What does this mean? Further, it asserts that the reciprocal method can be used to identify the company's marginal cost, leading to better planning. But recall that the reciprocal method in the typical textbook begins with a budget specifying the level of activity and total costs. Apparently, someone has decided on the product (output mix. Shouldn't we be using the marginal cost information to decide the best product mix? And if the product mix is already chosen, why does it matter how we allocate the costs? We next demonstrate that, just as with materials and labor (which are not interdependent, there is a way to allocate the service costs first, so they can be factored into the solution to the optimal product mix problem. We continue by appending to our ongoing example. Assume that in addition to using labor and materials in production, Ralph's company also uses power and repair. To plan production, we need to know how many units of each service are used to produce one unit of each output ( and. The difference between inputs such as power and repair and those such as labor and materials is one generally must use power to increase repair activities and repair hours to increase the supply of power. This is what is meant by reciprocation. In the table below, a column represents the consumer of the input (service department or product; a row represents the provider of the input. Note, we have self-service in the power department, as evidenced by the 0.1 entry in the matrix. Power is used to supply power. OUTPUTS Product 1 Product 2 Power Repair Materials Labor INPUTS: Materials Labor Power Repair Again, an important feature that permits the use of linear programming is we assume that the above technological relationships (as indicated in the input-output matrix hold for any arbitrary combination of production levels of and. Local linear approximations for costs are as follows: Richard A. Young 11 3/2/00

12 The program to optimize is: Material cost = 5 (pounds Labor cost = 13 (labor hrs Power cost = (kw-hrs Repair cost = (repair-hrs (P2: Max z 1-13 z z z 4 subject to: 150 (w 1 50 (w z 1 (w 3 2 z 2 (w z z 4 z 3 (w z 3 z 4 (w 6,, z 1, z 2, z 3, z 4 The objective function is the revenue from sale of the outputs minus the costs of materials, labor, power and repair. The unusual constraints in this program are (w 5 and (w 6. These constraints reflect that the total power or repair supplied must not be less than the total demanded. We are going to write down and solve the dual program corresponding to (P2, so it is useful to write the decision variables on the left-hand side of the inequalities: (P2': Max z 1-13 z z z 4 subject to: 150 (w 1 50 (w z 1 0 (w z 2 0 (w z z 4 0 (w z 3 - z 4 0 (w 6,, z 1, z 2, z 3, z 4 The program above uses six variables. If one had a computer routine available, the solution could be attained immediately. However, it will prove instructive to solve this problem by hand. For one, it makes us confront the effect of reciprocation on the shadow prices for the service department constraints. Before we begin to solve for these dual variables, we address their interpretation. We examine the power constraint: z z 4 z 3. As explained above for the inputs labor and materials, we look at the effect on the objective function of being handed one Richard A. Young 12 3/2/00

13 unit of power for free. If this were to happen, the amount demanded can exceed z 3 by one. That is, we will be able to make the same amount of and as before we added one to the right-hand side, but be able to do so supplying one less unit of power internally. Thus, the dual variable w 5 is the marginal cost of supplying power. Likewise, w 6 is the marginal cost of supplying repair. One might be tempted to guess that w 5 is equal to $.90. Here s where reciprocation enters. If we reduce the amount of power we supply, we are able to produce less repair. And further, if we use less repair services, we can use even less power. 9 Thus, $.90 would understate the cost of power. And $2.60 would understate the cost of repair. Let us now find the marginal cost of power and repair by solving the dual program, which is written below: (D2: Min 150 w w 2 subject to: w 1 + w w 4 + w w 6 38 ( w 1 + w 2 +2 w 3 + w w 5 + w 6 33 ( - w 3-5 (z 1 - w 4-13 (z w w (z 3.2 w 5 - w (z 4 w 1, w 2, w 3, w 4, w 5, w 6 For now, we can guess that some of Product 1 or 2 will be made at optimality. (We can always check later. From the production requirement matrix, we know that both Products 1 and 2 use some of both power and repair; thus, z 3 * > 0 and z 4 * > 0. Since z 3 * and z 4 * are the shadow prices on the last two constraints in (D2, we know that they are both tight: w w 6 = w 5 - w 6 = Because these constraints will be satisfied as equalities, we need not worry about the objective function. There is one and only one value of w 5 and w 6 which makes both constraints tight: w 5 * = 2 and w 6 * = 3. Note, we have obtained the marginal costs of these services without first choosing a production plan. Contrast this with the textbook approach, where we use the reciprocal method to derive the marginal cost after-the-fact. Of course, we can use the same logic to obtain w 3 * = 5 and w 4 * = 13. Next, we can use the shadow prices to calculate the contribution margins on Product 1 and 2, which are as follows. Richard A. Young 13 3/2/00

14 Price $ 38 $ 33 Materials $ 5 (1 = 5 $ 5 (2 = 10 Labor $13 (2 = 26 $13 (1 = 13 Power $ 2 (1 = 2 $ 2 (2 = 4 Repair $ 3 (1 = 6 $ 3 (1 = 3 Unit CM ($ 1 $ 3 Cost Allocations Using Shadow Prices We can now write down a re-framed version of (P2': (R2 Max subject to: 150 (w 1 50 (w 2, The solution to (R2 is easy to see: * = 0 and * = 50. All that remains is to calculate the amount of power and repair needed to meet this production plan. We already argued that w 1 * through w 6 * are all positive Thus, the (w 3 through (w 6 constraints in the original primal (P1' are all tight. Solving the equations below provides the optimal input quantities. 1(0 + 2(50 - z 1 = 0 2(0 + 1(50 - z 2 = 0 1(0 + 2( z z 4 = 0 2(0 + 1( z 3 - z 4 = 0. Referring to our two LP s, we see that the ( constraint in (D2 is not tight (since * = 0 but the ( constraint is (since * > 0. Also, substitution of * = 0 and * = 50, reveals the (w 1 constraint in (P2' is not tight and, hence, w 1 = 0. Combining these two facts, [w 1 + w w 3 + w w 5 + w 6 ] = [0 + w 2 + 2( (2 + 3] = 33 implies w 2 * = 3. If one relaxes the second constraint by one hour one can make one more unit of, increasing profit by 1. w 2 * = 3. The solution is summarized below Have no fear; this cycling converges. 10 Precisely, z 3 * = 2,750/21 and z 4 * = 625/7. Richard A. Young 14 3/2/00

15 Primal Dual TCM = OC = 0 w 1 = = 50 w 2 = 3 z 1 = 100 w 3 = 5 z 2 = 50 w 4 = 13 z 3 = w 5 = 2 z 4 = w 6 = 3 Primal and Dual Program Solutions One more thing -- we can check our cost allocations. Recall all costs have been allocated to, since Product 1 is not being produced. Per-unit, the allocated costs are $30 = ( , so the total allocated is $30 (50 = $1,500. Below we verify that the total costs allocated are equal to the total costs incurred. Materials $5 (100 = $ Labor $13 (50 = Power $0.90 ( = Repair $2.60 (89.29 = Total costs $1, VI. Conclusions In this note, we have demonstrated that the short-run profit planning problem can be reframed in a consistent manner through cost allocations. Consistent framing means the same decision results from either framing. In the example, the problem was reduced from one of six decision variables and six constraints to one of two decision variables and two constraints. Intuition for the invariance of the solutions was provided and it was demonstrated that extra care has to be taken when the inputs are interdependent. This note concludes by presenting some self-study exercises and a technical Appendix. Richard A. Young 15 3/2/00

16 Self-study Exercises Consider each item independently. In each case use the same material and labor requirements. 1. Modify the problem above only be changing the price of Product 1 from $38 to $40. Repeat. 2. Modify the problem above as follows. Change the prices of Product 1 and 2 from $38 to $47 and from $33 to $40. Also replace the production requirements for services by the following: Product 1 Product 2 Power Repair Power Repair Modify the problem above as follows. Replace the (a capacity constraints, (b production requirements for services, (c variable service costs, and (d output prices as follows. (a + 3 < 500 (b Product 1 Product 2 Power Repair 2 < 500 Power Repair (c Unit Variable Costs for services: Power $2 Repair $1 (d Output prices: Product 1 $41 Product 2 $31 Check Figures: 1. w 1 = 1 = 100 w 2 = 2 = 50 w 5 = 2 w 6 = 3 2. w 1 = 6.09 = 100 w 2 = 2.55 = 50 w 5 = 2.27 w 6 = w 1 = = 0 w 2 = 0 = 166 2/ 3 w 5 = 2.71 w 6 = 1.71 Richard A. Young 16 3/2/00

17 Appendix This appendix describes a remarkable property of linear programming problems. Recall that in the example above the primal program objective function at the optimal solution was 7 ( (50 = 1,200. Now let's try substituting the optimal solution to the dual program into the dual objective function: 150 ( (2 = 1,200. The value of the primal objective function at the optimal solution to the primal program and the value of the dual objective function at the optimal solution to the dual objective function are equal. Is this a coincidence, or is this a property of linear programming that holds in general? Remarkably, the answer is the latter. Formally: OBSERVATION: Suppose q = (,..., q n is a feasible solution to the primal program with m constraints, and w = (w 1,..., w m is a feasible solution to the corresponding dual program. Then if the primal objective function evaluated at q is equal to the dual objective function evaluated at w, then both problems are optimized by q and w. What a powerful result! Suppose we "guessed" that the solutions might be q = (100, 50 and w = (7, 3. q is feasible in the primal and w is feasible in the dual, and each solution when substituted into the corresponding objective function was equal to 1,200. By the above observation, we immediately know that q and w are optimal solutions. (Of course, in large problems one would have trouble guessing the optimal solution. Another important point is the result also "goes the other way." That is, if the primal and dual objective functions are not equal at a particular feasible solution, then we know those solutions are not both optimal. Therefore, from now on we can use this result to check whether we have correctly found the optimal solution to the primal and the dual. If we solve (say, graphically both problems, but do not get the same objective function value, we know we did something wrong. The Observation is not difficult to prove. We prove it for a special case: two constraints and two variables. The primal program in the two-variable, two-constraint situation is: Richard A. Young 17 3/2/00

18 Max c 1 + c 2 subject to A + B b 1 The dual program is: C + D b 2, Min b 1 w 1 +b 2 w 2 subject to A w 1 + C w 2 c 1 B w 1 + D w 2 c 2 w 1, w 2 First, using dual feasibility, multiply the first dual constraint by (which will not change the direction of the inequality. Multiply the second dual constraint by as well. This produces: [A w 1 + C w 2 ] c 1 [B w 1 + D w 2 ] c 2. These constraints together imply: c 1 + c 2 [A w 1 + C w 2 ] + [B w 1 + D w 2 ]. Rearrange the right-hand side of this inequality: [A w 1 + C w 2 ] + [B w 1 + D w 2 ] = [A + B ] w 1 + [C + D ] w 2. Therefore: c 1 + c 2 [A + B ] w 1 + [C + D ] w 2. Now, by the constraints in the primal program: c 1 + c 2 [A + B ] w 1 + [C + D ] w 2 b 1 w 1 + b 2 w 2 But this states that the primal objective function can never be larger than the dual objective function, if the proposed solutions are feasible in their respective programs. So, if c 1 + c 2 = b 1 w 1 + b 2 w 2, we know we have made the primal objective as large as it can possibly be -- it is maximized! In parallel fashion, if c 1 + c 2 = b 1 w 1 + b 2 w 2, we know we have made the dual objective function as small as it can be -- it is minimized! Thus, if feasible dual and primal solutions lead to the same objective function values, then both solutions are optimal. Richard A. Young 18 3/2/00

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