Choosing Among Discrete Affordable Bundles

Choosing from a Finite Menu

A consumer does not always choose from every mathematically possible combination of goods. A retailer may offer a short list of plans, a benefit program may provide a set of coverage designs, or a decision problem may specify several candidate bundles for comparison. In such cases, the task is to check which listed bundles the consumer can afford and then compare their utility. The highest-utility bundle in the whole menu is not necessarily the best available choice if it is over budget.

This is a finite-choice problem: the menu contains a limited number of bundles, and each can be evaluated directly. It is different from searching over every possible bundle. The methods here are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Definition: A bundle $(x,y)$ specifies quantities $x$ and $y$ of two goods. If their prices are $p_x$ and $p_y$, and the consumer has income $m$, the bundle is affordable when its expenditure satisfies $p_xx+p_yy\leq m$. For a finite menu $M$, the affordable menu is the subset of listed bundles that satisfy this inequality. A utility-maximizing choice is an affordable bundle whose utility is at least as high as the utility of every other affordable bundle.

The affordability condition uses a weak inequality: spending exactly the available income is allowed. The menu matters as well as the budget. If a bundle is affordable but is not listed, it is not a candidate in a problem that asks for a choice from that finite menu.

Utility represents the consumer's ranking of bundles. When a utility function assigns a higher value to one bundle than another, it represents the consumer as preferring the first. A utility number is not necessarily a dollar amount, a probability, or a measure of how many times more satisfied one bundle is. For this comparison, its role is to rank the choices.

A Reliable Comparison Method

For a finite menu, use the same sequence every time: calculate each listed bundle's expenditure, remove any bundle that fails the budget test, calculate utility for each remaining bundle, and compare the results. Keep every bundle tied for the greatest utility. An unaffordable bundle is excluded even if its utility exceeds the values of all affordable options.

Check the menu and budget.
Record the prices, income, and every listed bundle. Do not add unlisted bundles to the choice set.

Test affordability.
For each $(x,y)$, compute $p_xx+p_yy$ and compare it with $m$. Retain bundles with expenditure no greater than income.

Evaluate utility and compare.
Use the stated utility function on every retained bundle. The largest value identifies the maximizing bundle or bundles.

A short justification explains why this procedure works. The affordable menu is precisely the set of feasible candidates in the problem. A maximizing choice must be in that set, and its utility must be no lower than the utility of any other member. Therefore, after removing the infeasible options, comparing all remaining utility values directly identifies exactly the choices that meet the definition. If several have the same greatest value, each satisfies that comparison and is a maximizing choice.

M_{\mathrm{aff}}=\{(x,y)\in M:p_xx+p_yy\leq m\}, \qquad \text{maximizers}=\{(x,y)\in M_{\mathrm{aff}}:U(x,y)\geq U(x',y') \text{ for every }(x',y')\in M_{\mathrm{aff}}\}.

This also clarifies the meaning of a tie. A tie occurs when two or more affordable bundles have equal utility and no affordable bundle has greater utility. Equal utility among some bundles is not sufficient if another affordable bundle has a higher value.

Worked Example: Filtering the Menu Before Ranking

A consumer has income $m=\$18$. Good $x$ costs $p_x=\$3$ per unit and good $y$ costs $p_y=\$2$ per unit. The menu is $A=(2,4)$, $B=(4,2)$, $C=(1,6)$, $D=(3,3)$, and $E=(5,2)$. Preferences are represented by $U(x,y)=x+2y$. Which listed bundle maximizes utility subject to affordability?

Method. First apply the budget test, then evaluate utility only for bundles that pass. This avoids selecting a high-utility option that the consumer cannot buy.

For $A=(2,4)$, expenditure is $3(2)+2(4)=6+8=\$14$, so $A$ is affordable. Its utility is $2+2(4)=10$.

For $B=(4,2)$, expenditure is $3(4)+2(2)=12+4=\$16$, so $B$ is affordable. Its utility is $4+2(2)=8$.

For $C=(1,6)$, expenditure is $3(1)+2(6)=3+12=\$15$, so $C$ is affordable. Its utility is $1+2(6)=13$.

For $D=(3,3)$, expenditure is $3(3)+2(3)=9+6=\$15$, so $D$ is affordable. Its utility is $3+2(3)=9$.

For $E=(5,2)$, expenditure is $3(5)+2(2)=15+4=\$19$, which exceeds the $\$18$ budget. It is not a feasible choice, even though its utility would be $5+2(2)=9$.

Among the affordable bundles, the utility values are $U(A)=10$, $U(B)=8$, $U(C)=13$, and $U(D)=9$. The greatest is $13$, so $C$ is the unique affordable maximizer. As a check, its expenditure of $\$15$ is within the budget, and none of the other affordable options reaches utility $13$.

Ties and Utility Representation

A finite menu can have multiple maximizing choices. For example, the utility function may assign the same highest value to several bundles. In that case, report all of them and state their common utility. Do not break a tie by choosing the bundle that costs more, uses more of one good, or spends the most income unless the problem supplies an additional preference or rule that ranks those bundles.

The numerical scale of utility is not unique. If $U$ represents a preference ranking and $g$ is strictly increasing, then $g(U)$ gives the same ranking: whenever $U(A)>U(B)$, strict increase implies $g(U(A))>g(U(B))$; and equal utility values remain equal after applying $g$. Thus changing from $U$ to a strictly increasing transformation cannot change which affordable bundles maximize utility. This conclusion assumes the transformation is strictly increasing over the utility values being compared.

Worked Example: Identifying Every Tied Maximizer

A consumer has income $m=\$20$. Good $x$ costs $p_x=\$4$ and good $y$ costs $p_y=\$2$. The menu is $A=(2,4)$, $B=(3,2)$, $C=(1,6)$, and $D=(4,3)$. Utility is $U(x,y)=2x+y$. Identify all affordable utility-maximizing choices.

Method. Calculate both expenditure and utility for every menu item. A tie is a maximizing tie only if no other affordable bundle has a larger utility.

For $A$, expenditure is $4(2)+2(4)=8+8=\$16$, so it is affordable. Its utility is $2(2)+4=8$.

For $B$, expenditure is $4(3)+2(2)=12+4=\$16$, so it is affordable. Its utility is $2(3)+2=8$.

For $C$, expenditure is $4(1)+2(6)=4+12=\$16$, so it is affordable. Its utility is $2(1)+6=8$.

For $D$, expenditure is $4(4)+2(3)=16+6=\$22$, which exceeds income. Its utility would be $2(4)+3=11$, but it cannot be chosen from the affordable menu.

The affordable options $A$, $B$, and $C$ all have utility $8$, and there is no other affordable option. Therefore, all three are maximizing choices, tied at utility $8$. The unaffordable $D$ does not defeat the tie. The method avoids treating the highest utility anywhere in the menu as the answer without first enforcing the budget.

Nonlinear Utility and a Common Exam Trap

The comparison procedure is unchanged when utility is nonlinear. Evaluate the stated function at each affordable bundle, taking care to substitute both quantities correctly. If a calculation uses square roots or another non-integer result, keep enough precision during the comparison and round only after identifying the maximum. Rounding early can make close utility values appear tied when they are not.

Worked Example: A Nonlinear Ranking with an Unaffordable Temptation

A consumer has income $m=\$12$, with $p_x=\$2$ and $p_y=\$1$. The menu consists of $A=(4,4)$, $B=(1,8)$, $C=(2,7)$, $D=(3,6)$, and $E=(4,5)$. Utility is $U(x,y)=\sqrt{x}+\sqrt{y}$. Find the affordable maximizer. Report utility values to four decimal places.

Method. Apply the expenditure test first; then evaluate the square-root utility for each affordable option. This prevents a tempting but infeasible bundle from being included in the ranking.

For $A$, expenditure is $2(4)+1(4)=\$12$, so $A$ is affordable. Its utility is $\sqrt{4}+\sqrt{4}=2+2=4.0000$.

For $B$, expenditure is $2(1)+1(8)=\$10$, so $B$ is affordable. Its utility is $\sqrt{1}+\sqrt{8}=1+2.828427\ldots=3.8284$.

For $C$, expenditure is $2(2)+1(7)=\$11$, so $C$ is affordable. Its utility is $\sqrt{2}+\sqrt{7}=1.414213\ldots+2.645751\ldots=4.0600$.

For $D$, expenditure is $2(3)+1(6)=\$12$, so $D$ is affordable. Its utility is $\sqrt{3}+\sqrt{6}=1.732050\ldots+2.449489\ldots=4.1815$.

For $E$, expenditure is $2(4)+1(5)=\$13$, which exceeds the budget. Its utility would be $\sqrt{4}+\sqrt{5}=4.2361$, but it is not an available choice.

Among affordable bundles, $D$ has the greatest utility, $4.1815$, so it is the unique maximizer. As a final check, $D$ costs exactly $\$12$, and its utility exceeds those of $A$, $B$, and $C$. A common incorrect calculation is to select $E$ because $4.2361$ is greater than $4.1815$. The correction is to exclude $E$ before comparing utility: $2(4)+5=13>12$.

Interpreting the Result Carefully

The answer should identify the bundle or bundles, not merely give the largest utility number. When useful, state both the maximum utility and the affordability calculation. This makes clear that the result is a feasible choice rather than simply the most attractive item on the full menu.

A few checks help prevent errors:

Use the stated prices and quantities. Expenditure is price times quantity for each good, added across goods.
Use the budget as a constraint. A bundle with expenditure above income is excluded, regardless of its utility.
Compare every affordable option. Finding one high utility value is not enough to establish a maximum.
Preserve ties. If several affordable bundles share the greatest utility, report all of them.
Respect the menu. A feasible bundle outside a specified finite menu is not an option in that comparison.

These steps apply whether utility is linear, multiplicative, logarithmic, or otherwise specified. They do not require assumptions about how the consumer would choose among all possible bundles; they answer the narrower question of which listed, affordable bundle or bundles rank highest.

Key takeaway: For a finite choice problem, filter the menu by the budget inequality before ranking utility. The affordable bundle or bundles with the greatest utility are the maximizing choices; an unaffordable option cannot win, and a genuine highest-value tie must be reported in full.

Check Your Understanding

For each problem, first identify the affordable options and then compare their utility. Give all maximizing choices when there is a tie.

A consumer has income $\$10$, with $p_x=\$2$ and $p_y=\$1$. The menu is $A=(3,2)$, $B=(2,3)$, and $C=(4,3)$. If $U(x,y)=x+2y$, which affordable bundle maximizes utility? (Answer: $B$, with utility $8$; $C$ costs $\$11$ and is unaffordable.)
A consumer has income $\$12$, with $p_x=\$1$ and $p_y=\$3$. The menu is $A=(3,2)$, $B=(2,3)$, $C=(6,2)$, and $D=(1,4)$. If $U(x,y)=3x+y$, identify the affordable maximizer. (Answer: $C$, with utility $20$; $D$ is unaffordable.)
A consumer has income $\$10$, with $p_x=\$2$ and $p_y=\$1$. The menu is $A=(2,4)$, $B=(3,3)$, $C=(4,2)$, and $D=(1,5)$. If $U(x,y)=xy$, identify the maximizing choice and its utility. (Answer: $B$, with utility $9$.)
A consumer has income $\$8$, with $p_x=\$1$ and $p_y=\$2$. The menu is $A=(4,2)$, $B=(1,3)$, $C=(0,4)$, and $D=(5,2)$. If $U(x,y)=\sqrt{x}+\sqrt{y}$, identify the affordable maximizer. (Answer: $A$, with utility $2+\sqrt{2}$, approximately $3.4142$; $D$ is unaffordable.)
A consumer has income $\$12$, with $p_x=\$3$ and $p_y=\$1$. The menu is $A=(2,4)$, $B=(3,3)$, $C=(1,5)$, and $D=(4,2)$. If $U(x,y)=x+3y$, identify all affordable maximizers and verify that no excluded option changes the result. (Answer: $C$, with utility $16$; $D$ is unaffordable and has utility $10$.)

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