Game Theory and Strategy · End Term · May 2026 · AN

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** The seller retains the laptop, worth 2000 to him. Thus the requested coalition value is 2000. **Answer: 2000.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** Buyer 2 has no laptop and no seller in this coalition. Thus the requested coalition value is 0. **Answer: 0.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** The empty coalition has no players or asset. Thus the requested coalition value is 0. **Answer: 0.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** Buyer 3 cannot obtain the laptop within this singleton coalition. Thus the requested coalition value is 0. **Answer: 0.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** Assigning the seller’s laptop to buyer 2 gives total worth 2800, exceeding the seller’s 2000. Thus the requested coalition value is 2800. **Answer: 2800.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** Assigning the laptop to buyer 3 gives worth 3000. Thus the requested coalition value is 3000. **Answer: 3000.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** The buyers have no laptop without its owner, player 1. Thus the requested coalition value is 0. **Answer: 0.**

**1. Interpret coalition value:** Value is the maximum total worth obtainable using only assets and players in the coalition. Payments within it are transfers and cancel from total coalition value. **2. Identify the resource:** Player 1 alone owns the laptop. A coalition without player 1 cannot sell or allocate that laptop, so its value is zero. If player 1 is present, allocate the one laptop to the included player valuing it most: seller 2000, buyer 2 at 2800 or buyer 3 at 3000. **3. Apply to the requested coalition:** The one laptop goes to buyer 3, the highest valuer; buyer values are not added because there is only one laptop. Thus the requested coalition value is 3000. **Answer: 3000.**

**1. Construct the second-price payoff table:** Use thousands of INR. Mohan owns rows, Charan owns columns; each pair is (Mohan payoff, Charan payoff). A winner pays the other bid, while a tie gives half the win surplus in expectation. \[\begin{array}{c|ccc} &50&70&90\\ \hline50&(15,25)&(0,50)&(0,50)\\ 70&(30,0)&(5,15)&(0,30)\\ 90&(30,0)&(10,0)&(-5,5)\end{array}\] Mohan values the item at 80 and Charan at 100. For example, a tie at 90 gives Mohan \((80-90)/2=-5\) and Charan \((100-90)/2=5\). **2. Check Mohan’s row comparisons:** Row 70 weakly improves on row 50, but ties its zero payoff against Charan’s 90. Row 90 loses money against Charan’s 90. No pure row gives Mohan a strictly larger payoff against every column. Nor can a mixture strictly dominate row 70 or row 90 because each is best against some column; row 50 cannot be strictly dominated because its payoff against column 90 is already maximal at zero. **3. Check the whole claim:** Charan does have strict dominance of bid 50 by bid 90, but Mohan does not. Thus “for both” is false. **Answer: B — Selected: False.** A: **Incorrect:** Both players do not have a strictly dominated strategy. B: **Correct:** Charan has one, but Mohan has none, so the joint claim is false.

**1. Construct the second-price payoff table:** Use thousands of INR. Mohan owns rows, Charan owns columns; each pair is (Mohan payoff, Charan payoff). A winner pays the other bid, while a tie gives half the win surplus in expectation. \[\begin{array}{c|ccc} &50&70&90\\ \hline50&(15,25)&(0,50)&(0,50)\\ 70&(30,0)&(5,15)&(0,30)\\ 90&(30,0)&(10,0)&(-5,5)\end{array}\] Mohan values the item at 80 and Charan at 100. For example, a tie at 90 gives Mohan \((80-90)/2=-5\) and Charan \((100-90)/2=5\). **2. List best responses:** Against Charan’s bid 50, Mohan best responds with 70 or 90; against 70, he chooses 90; against 90, he chooses 50 or 70. Against Mohan’s 50, Charan chooses 70 or 90; against 70, he chooses 90; against 90, he chooses 90. **3. Find intersections:** The mutual best-response cells are (50,90) and (70,90), and no others. Therefore there are two pure Nash equilibria. Neither player gains by changing their bid alone in either cell. **Answer: B — Selected: 2.** A: **Incorrect:** There are two distinct mutual best-response cells. B: **Correct:** The equilibria are (50,90) and (70,90). C: **Incorrect:** Five overcounts cells where a player has a profitable deviation. D: **Incorrect:** Four overcounts the best-response intersections.

**1. Construct the second-price payoff table:** Use thousands of INR. Mohan owns rows, Charan owns columns; each pair is (Mohan payoff, Charan payoff). A winner pays the other bid, while a tie gives half the win surplus in expectation. \[\begin{array}{c|ccc} &50&70&90\\ \hline50&(15,25)&(0,50)&(0,50)\\ 70&(30,0)&(5,15)&(0,30)\\ 90&(30,0)&(10,0)&(-5,5)\end{array}\] Mohan values the item at 80 and Charan at 100. For example, a tie at 90 gives Mohan \((80-90)/2=-5\) and Charan \((100-90)/2=5\). **2. Compare Mohan’s bids 70 and 90:** Against Charan’s 50, both yield 30. Against Charan’s 70, bid 70 yields 5 but bid 90 yields 10. Against Charan’s 90, bid 70 yields zero and bid 90 yields -5. **3. Apply weak dominance:** A weakly dominating action must be at least as good against every opponent action and better against at least one. Bid 70 is worse against Charan’s 70, so it does not weakly dominate 90. Each of these bids is better in a different situation. **Answer: B — Selected: False.** A: **Incorrect:** Mohan earns 5 rather than 10 against Charan’s 70, violating the dominance condition. B: **Correct:** One counterexample column is enough to disprove the proposed weak dominance.

**1. Construct the second-price payoff table:** Use thousands of INR. Mohan owns rows, Charan owns columns; each pair is (Mohan payoff, Charan payoff). A winner pays the other bid, while a tie gives half the win surplus in expectation. \[\begin{array}{c|ccc} &50&70&90\\ \hline50&(15,25)&(0,50)&(0,50)\\ 70&(30,0)&(5,15)&(0,30)\\ 90&(30,0)&(10,0)&(-5,5)\end{array}\] Mohan values the item at 80 and Charan at 100. For example, a tie at 90 gives Mohan \((80-90)/2=-5\) and Charan \((100-90)/2=5\). **2. Compare Charan’s bids across Mohan’s rows:** At Mohan bids 50,70,90, Charan’s payoffs with bid 90 are (50,30,5). With bid 50 they are (25,0,0). **3. Check every comparison:** \(50>25\), \(30>0\) and \(5>0\). Bid 90 gives a strictly greater payoff against every available Mohan bid, so it strictly dominates 50. **Answer: A — Selected: True.** A: **Correct:** All three opponent-action comparisons are strict improvements. B: **Incorrect:** The proposed strict dominance is verified by every row.

**1. Read the preference convention:** Treat positive numbers as ranks with smaller numbers preferred; zero marks an unacceptable/unlisted partner rather than a first rank. In particular, c and d do not propose to A, and C does not accept a. This convention reproduces the intended incomplete preference lists. **2. Run men-proposing deferred acceptance:** a first proposes to C and is rejected as unacceptable, then proposes to A. b proposes to B; c and d propose to C. C retains c over d. Rejected d next proposes to B; B prefers d (rank 3) to b (rank 4), so b is rejected. b then proposes to A, which prefers b (rank 1) to a (rank 3). Rejected a next proposes to D, which accepts him. **3. Read the final matching:** \((A,b),(B,d),(C,c),(D,a)\). Every displaced man proposed next down his acceptable list. The retained partners satisfy stability under the stated convention; irrelevant equal-rank entries do not affect these decisions. **Convention note:** If zero were instead treated as a genuine best rank, the algorithm and matching would change. Here zero is interpreted as an unacceptable entry, as required by the source’s intended matching. **Requested pair:** A is paired with b. **Answer: D — Selected: b.** A: **Incorrect:** A ends with b rather than a. B: **Incorrect:** A ends with b rather than c. C: **Incorrect:** A ends with b rather than d. D: **Correct:** A retains b after rejecting a.

**1. Read the preference convention:** Treat positive numbers as ranks with smaller numbers preferred; zero marks an unacceptable/unlisted partner rather than a first rank. In particular, c and d do not propose to A, and C does not accept a. This convention reproduces the intended incomplete preference lists. **2. Run men-proposing deferred acceptance:** a first proposes to C and is rejected as unacceptable, then proposes to A. b proposes to B; c and d propose to C. C retains c over d. Rejected d next proposes to B; B prefers d (rank 3) to b (rank 4), so b is rejected. b then proposes to A, which prefers b (rank 1) to a (rank 3). Rejected a next proposes to D, which accepts him. **3. Read the final matching:** \((A,b),(B,d),(C,c),(D,a)\). Every displaced man proposed next down his acceptable list. The retained partners satisfy stability under the stated convention; irrelevant equal-rank entries do not affect these decisions. **Convention note:** If zero were instead treated as a genuine best rank, the algorithm and matching would change. Here zero is interpreted as an unacceptable entry, as required by the source’s intended matching. **Requested pair:** B is paired with d. **Answer: D — Selected: d.** A: **Incorrect:** B rejects b after d proposes. B: **Incorrect:** C retains c; B retains d. C: **Incorrect:** a ends at D, not B. D: **Correct:** B prefers d to its earlier tentative partner b.

**1. Read the preference convention:** Treat positive numbers as ranks with smaller numbers preferred; zero marks an unacceptable/unlisted partner rather than a first rank. In particular, c and d do not propose to A, and C does not accept a. This convention reproduces the intended incomplete preference lists. **2. Run men-proposing deferred acceptance:** a first proposes to C and is rejected as unacceptable, then proposes to A. b proposes to B; c and d propose to C. C retains c over d. Rejected d next proposes to B; B prefers d (rank 3) to b (rank 4), so b is rejected. b then proposes to A, which prefers b (rank 1) to a (rank 3). Rejected a next proposes to D, which accepts him. **3. Read the final matching:** \((A,b),(B,d),(C,c),(D,a)\). Every displaced man proposed next down his acceptable list. The retained partners satisfy stability under the stated convention; irrelevant equal-rank entries do not affect these decisions. **Convention note:** If zero were instead treated as a genuine best rank, the algorithm and matching would change. Here zero is interpreted as an unacceptable entry, as required by the source’s intended matching. **Requested pair:** C is paired with c. **Answer: C — Selected: c.** A: **Incorrect:** b ends at A. B: **Incorrect:** C rejects d in favor of c. C: **Correct:** C retains c in the proposal process. D: **Incorrect:** C treats a as unacceptable under the intended zero-entry convention.

**1. Read the preference convention:** Treat positive numbers as ranks with smaller numbers preferred; zero marks an unacceptable/unlisted partner rather than a first rank. In particular, c and d do not propose to A, and C does not accept a. This convention reproduces the intended incomplete preference lists. **2. Run men-proposing deferred acceptance:** a first proposes to C and is rejected as unacceptable, then proposes to A. b proposes to B; c and d propose to C. C retains c over d. Rejected d next proposes to B; B prefers d (rank 3) to b (rank 4), so b is rejected. b then proposes to A, which prefers b (rank 1) to a (rank 3). Rejected a next proposes to D, which accepts him. **3. Read the final matching:** \((A,b),(B,d),(C,c),(D,a)\). Every displaced man proposed next down his acceptable list. The retained partners satisfy stability under the stated convention; irrelevant equal-rank entries do not affect these decisions. **Convention note:** If zero were instead treated as a genuine best rank, the algorithm and matching would change. Here zero is interpreted as an unacceptable entry, as required by the source’s intended matching. **Requested pair:** D is paired with a. **Answer: A — Selected: a.** A: **Correct:** a reaches D after rejection at A and is accepted. B: **Incorrect:** d ends at B. C: **Incorrect:** b ends at A. D: **Incorrect:** c is retained by C.

**1. Write a firm’s profit:** With zero costs and rival price Pj, \(\pi_i=P_i(18-2P_i+P_j)\), for nonnegative prices in the supplied demand model. **2. Optimize own price:** \(\partial\pi_i/\partial P_i=18-4P_i+P_j\). The second derivative is -4, so the best response is \[BR_i(P_j)=\frac{18+P_j}{4}.\] **3. Solve mutual best responses:** \(4P_1-P_2=18\) and \(4P_2-P_1=18\) imply P1=P2=6. Each firm then has demand 12 and profit 72. The linear equations have a unique solution. **Requested function:** The best response is (18+Pj)/4. **Answer: B.** A: **Incorrect:** Dividing by two misses the factor four from differentiating the quadratic revenue term. B: **Correct:** Setting the own-price profit derivative to zero gives this expression. C: **Incorrect:** 9+Pj/2 equals the incorrect division-by-two expression. D: **Incorrect:** The denominator four is required by the derivative.

**1. Write a firm’s profit:** With zero costs and rival price Pj, \(\pi_i=P_i(18-2P_i+P_j)\), for nonnegative prices in the supplied demand model. **2. Optimize own price:** \(\partial\pi_i/\partial P_i=18-4P_i+P_j\). The second derivative is -4, so the best response is \[BR_i(P_j)=\frac{18+P_j}{4}.\] **3. Solve mutual best responses:** \(4P_1-P_2=18\) and \(4P_2-P_1=18\) imply P1=P2=6. Each firm then has demand 12 and profit 72. The linear equations have a unique solution. **4. Judge the requested incorrect statement:** The game is symmetric and its unique equilibrium is the symmetric price pair (6,6). A claimed unique equilibrium at price 9 is false: against 9 the best response is (18+9)/4=6.75, not 9. **Answer: D — Selected: This game has a unique Nash equilibrium in which P=9.** A: **Incorrect:** The payoff functions have the same form after swapping the players, so this is true. B: **Incorrect:** The equilibrium (6,6) is symmetric, so this statement is true. C: **Incorrect:** The equilibrium calculation gives price 6 for each firm, so this is true. D: **Correct:** Selected as incorrect: price 9 is not a mutual best response.

**1. Build quarterly payoffs:** Rows belong to firm 1, columns to firm 2, and each pair is (firm 1 payoff, firm 2 payoff). E is immediate exit, T is exit after this quarter and N is exit after next quarter. \[\begin{array}{c|ccc} &E&T&N\\ \hline E&(0,0)&(0,3)&(0,6)\\ T&(3,0)&(-1,-1)&(-1,2)\\ N&(6,0)&(2,-1)&(-2,-2)\end{array}\] For example, T against N loses 1 in the shared first quarter, while the N firm then gains 3 alone, producing (-1,2). **2. Compare candidate dominance:** E earns zero against all columns. T and N beat it against E but lose to it against N. N beats T against E and T, but T beats N against N. **3. Check strict mixed domination as well:** E is a best response to N and N is a best response to E, so neither can be strictly dominated. To beat T against E a mixture needs more than half weight on N, but to beat T against N it needs less than half on N. Those requirements conflict. Thus no strategy is strictly dominated. **Answer: A — Selected: True.** A: **Correct:** Neither pure nor mixed alternatives strictly dominate any of the three actions. B: **Incorrect:** There is no action uniformly strictly improved upon.

**1. Build quarterly payoffs:** Rows belong to firm 1, columns to firm 2, and each pair is (firm 1 payoff, firm 2 payoff). E is immediate exit, T is exit after this quarter and N is exit after next quarter. \[\begin{array}{c|ccc} &E&T&N\\ \hline E&(0,0)&(0,3)&(0,6)\\ T&(3,0)&(-1,-1)&(-1,2)\\ N&(6,0)&(2,-1)&(-2,-2)\end{array}\] For example, T against N loses 1 in the shared first quarter, while the N firm then gains 3 alone, producing (-1,2). **2. Compare pure alternatives:** N is better than T against E and T, but worse against N. E is better than T against T and N, but worse against E. The comparisons with E and N also cross. **3. State the intended scope:** No action is weakly dominated by another pure action, so the intended answer is None. **Convention note:** If mixed-strategy domination is allowed, T is weakly dominated by the half-E, half-N mixture: its payoffs are (3,1,-1), versus T’s (3,-1,-1). The source’s None answer therefore uses pure-strategy comparisons. This distinction matters. **Answer: D — Selected: None.** A: **Incorrect:** No single other pure action weakly dominates E. B: **Incorrect:** T has no weakly dominating pure action; it does have the explicitly noted dominating mixture. C: **Incorrect:** No other pure action weakly dominates N. D: **Correct:** None is correct under the source’s pure-strategy dominance convention.

**1. Build quarterly payoffs:** Rows belong to firm 1, columns to firm 2, and each pair is (firm 1 payoff, firm 2 payoff). E is immediate exit, T is exit after this quarter and N is exit after next quarter. \[\begin{array}{c|ccc} &E&T&N\\ \hline E&(0,0)&(0,3)&(0,6)\\ T&(3,0)&(-1,-1)&(-1,2)\\ N&(6,0)&(2,-1)&(-2,-2)\end{array}\] For example, T against N loses 1 in the shared first quarter, while the N firm then gains 3 alone, producing (-1,2). **2. Find best responses:** Against E, a firm prefers N (6 over 3 and 0). Against T, it prefers N (2 over 0 and -1). Against N, it prefers E (0 over -1 and -2). **3. Find mutual best responses:** They are (E,N) and (N,E). Of the listed alternatives only (E,N) appears, so B is selected. **Answer: B — Selected: (E,N).** A: **Incorrect:** Against E, the second firm would improve from T to N. B: **Correct:** The first firm best responds to N with E, and the second best responds to E with N. C: **Incorrect:** The first firm can improve from T to N against E. D: **Incorrect:** The first firm can improve from T to E against N.

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The left pair coalition has values a=20+10=30. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 30. **Answer: 30.**

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The grand-coalition value is b=30+40=70. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 70. **Answer: 70.**

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The top pair coalition gives 40=c+15, so c=40-15=25. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 25. **Answer: 25.**

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The bottom-left pair coalition gives 15=d+10, so d=15-10=5. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 5. **Answer: 5.**

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The right pair coalition gives 35=10+e, so e=35-10=25. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 25. **Answer: 25.**

**1. Interpret game addition:** The diagram states that the left cooperative game is the sum of the two right games. For every identical coalition S, values add pointwise: \(v(S)=v_1(S)+v_2(S)\). **2. Match the same diagram position:** The right singleton gives 20=20+f, so f=0. **3. Check the operation:** Add or subtract values belonging to the same coalition, not values at different player subsets. The requested missing value is 0. **Answer: 0.**

**1. Define the Shapley value:** Average a player’s marginal contribution as the players enter in each possible order. **2. Simplify the supplied game:** v1=v2=5, v3=0, v12=10, v13=v23=5 and v123=10. Adding player 1 raises any predecessor coalition’s value by 5. The same holds for player 2. Adding player 3 never changes value, so player 3 is a dummy player. **3. Average:** The marginal-contribution vector is (5,5,0) in every ordering, so the Shapley vector is \((\phi_1,\phi_2,\phi_3)=(5,5,0)\). It sums to grand-coalition value 10. **Requested component:** Player 1 receives 5. **Answer: 5.**

**1. Define the Shapley value:** Average a player’s marginal contribution as the players enter in each possible order. **2. Simplify the supplied game:** v1=v2=5, v3=0, v12=10, v13=v23=5 and v123=10. Adding player 1 raises any predecessor coalition’s value by 5. The same holds for player 2. Adding player 3 never changes value, so player 3 is a dummy player. **3. Average:** The marginal-contribution vector is (5,5,0) in every ordering, so the Shapley vector is \((\phi_1,\phi_2,\phi_3)=(5,5,0)\). It sums to grand-coalition value 10. **Requested component:** Player 2 receives 5. **Answer: 5.**

**1. Define the Shapley value:** Average a player’s marginal contribution as the players enter in each possible order. **2. Simplify the supplied game:** v1=v2=5, v3=0, v12=10, v13=v23=5 and v123=10. Adding player 1 raises any predecessor coalition’s value by 5. The same holds for player 2. Adding player 3 never changes value, so player 3 is a dummy player. **3. Average:** The marginal-contribution vector is (5,5,0) in every ordering, so the Shapley vector is \((\phi_1,\phi_2,\phi_3)=(5,5,0)\). It sums to grand-coalition value 10. **Requested component:** Player 3 receives 0. **Answer: 0.**

**1. Find uncontested claims:** The estate is 250. Creditor 1 claims 200 and leaves 50 unclaimed for creditor 2. Creditor 2 claims 150 and leaves 100 unclaimed for creditor 1. **2. Divide only the disputed remainder:** Give the uncontested amounts first: 100 to creditor 1 and 50 to creditor 2. The remaining \(250-100-50=100\) is contested by both and is split equally, 50 each. **3. Add awards:** Creditor 1 gets 150 and creditor 2 gets 100. The awards total 250 and remain below their respective claims. **Requested award:** Creditor 1 receives 150. **Answer: C — Selected: 150.** A: **Incorrect:** 100 is only creditor 1’s uncontested portion. B: **Incorrect:** 170 does not follow from an equal split of the contested remainder. C: **Correct:** 100 uncontested plus 50 contested gives 150. D: **Incorrect:** 125 would split the whole estate equally and ignore uncontested claims.

**1. Find uncontested claims:** The estate is 250. Creditor 1 claims 200 and leaves 50 unclaimed for creditor 2. Creditor 2 claims 150 and leaves 100 unclaimed for creditor 1. **2. Divide only the disputed remainder:** Give the uncontested amounts first: 100 to creditor 1 and 50 to creditor 2. The remaining \(250-100-50=100\) is contested by both and is split equally, 50 each. **3. Add awards:** Creditor 1 gets 150 and creditor 2 gets 100. The awards total 250 and remain below their respective claims. **Requested award:** Creditor 2 receives 100. **Answer: B — Selected: 100.** A: **Incorrect:** 150 is the full claim rather than the contested-garment award. B: **Correct:** 50 uncontested plus 50 contested gives 100. C: **Incorrect:** 125 incorrectly divides the whole estate equally. D: **Incorrect:** 75 understates the computed award.

**1. Consider one A-speaker deviating from no learning:** Without learning, this person cannot communicate with the 700 B-speakers. Learning B creates 700 additional communication benefits. **2. Compare cost:** Net individual gain is \(700-C_A\). **3. Prevent a profitable deviation:** Require \(700-C_A\le0\), or \(C_A\ge700\). Equality allows indifference. For the full no-learning equilibrium, B-speakers must also have CB≥300. **Answer: B — Selected: CA ≥ 700.** A: **Incorrect:** Costs below 700 encourage an A-speaker to learn. B: **Correct:** CA≥700 makes a unilateral A-side learning deviation unprofitable. C: **Incorrect:** 500 is not the size of the other language group. D: **Incorrect:** CA≥500 alone permits profitable deviations when CA is below 700.

**1. Consider a B-speaker:** Learning A when nobody else learns permits communication with all 300 A-speakers, creating benefit 300. **2. Form the net gain:** \(300-C_B\). **3. Apply Nash’s no-profitable-deviation condition:** \(C_B\ge300\). The companion A-side condition is CA≥700. Weak inequalities include indifference at equality. **Answer: D — Selected: CB ≥ 300.** A: **Incorrect:** The relevant group size is 300, not 700, and a low-cost bound does not deter learning. B: **Incorrect:** CB≥700 is sufficient but unnecessarily restrictive; it is not the exact threshold. C: **Incorrect:** CB≤300 encourages or permits a gain rather than ruling out all profitable learning. D: **Correct:** CB≥300 is the exact B-side no-learning threshold.

**1. Count the social communication gain:** If all 300 A-speakers learn B, every A–B pair can communicate. There are 300(700) cross-group pairs, and both people gain one per link, yielding \(2(300)(700)=420{,}000\). **2. Compare against no learning:** The A-side learning cost is \(300C_A\). Nonnegative total net gain requires \(C_A\le1400\). **3. Compare which group should learn:** Having all B-speakers learn A delivers the same cross-group benefit but costs \(700C_B\). A-side learning is at least as cheap when \(300C_A\le700C_B\), or \(C_B\ge(3/7)C_A\). **4. Combine:** CA≤1400 and CB≥(3/7)CA. Under this complete two-group setup and uniform within-group costs, partial learning does not improve on the relevant full-group/no-learning alternatives. **Answer: A.** A: **Correct:** Both the nonnegative social-gain condition and the cheaper-group condition are correct. B: **Incorrect:** It reverses both necessary comparison directions. C: **Incorrect:** 700 is the private A-speaker benefit, not the social benefit including gains to B-speakers. D: **Incorrect:** This combines the wrong benefit threshold/direction for social efficiency.

**1. Consider an A-speaker’s incentive:** When all other A-speakers learn B and no B-speaker learns A, an A-speaker who also learns can communicate with the 700 B-speakers. Without learning, that person loses those 700 links. **2. Compare benefit with cost:** Learning is a best response when \(700-C_A\ge0\), so CA≤700. **3. Check B-speakers:** Since all A-speakers now know B, a B-speaker already communicates with them and gains no new links by learning A. With nonnegative CB, not learning is a best response. Thus the stated CA condition supports this equilibrium. **Answer: A — Selected: CA ≤ 700.** A: **Correct:** CA≤700 makes the benefit cover the A-speaker’s cost. B: **Incorrect:** CA above 700 would make an A-speaker prefer not to learn. C: **Incorrect:** 300 counts the learner’s own group rather than new B-speaking contacts. D: **Incorrect:** A lower bound does not characterize the required private learning incentive.

**1. Reverse the learning arrangement:** If B-speakers learn A while A-speakers do not learn B, each B-speaker gains communication with 300 A-speakers. **2. Check individual learning incentive:** \(300-C_B\ge0\), so CB≤300. **3. Check the nonlearners:** A-speakers already communicate with the B-speakers who learned A and gain nothing additional by learning B. Nonnegative CA makes not learning a best response. Therefore the exact B-side condition is CB≤300. **Answer: D — Selected: CB ≤ 300.** A: **Incorrect:** 700 is not the number of new A-speaking contacts for a B-speaker. B: **Incorrect:** A large cost does not support the proposed B-side learning behavior. C: **Incorrect:** The inequality should favor learning costs no greater than the 300-person benefit. D: **Correct:** CB≤300 is the correct private-benefit threshold.

**1. Identify the largest unilateral gain:** With nobody learning, learning A creates 500 contacts for a B- or C-speaker. Learning B creates at most 300; learning C at most 200. **2. Prevent every profitable one-language deviation:** A common per-language cost C must be at least 500. At that level, other single-language gains are also no greater than cost. **3. Check multiple-language deviations:** Learning two languages costs 2C; their combined group sizes are at most 800, below 2C when C≥500. Thus C≥500 supports no learning for every group. **Answer: C — Selected: C ≥ 500.** A: **Incorrect:** At cost 200, learning A can still give 500 new contacts. B: **Incorrect:** At cost 300, B- and C-speakers can still gain by learning A. C: **Correct:** C≥500 deters all unilateral learning choices. D: **Incorrect:** The exact common-cost threshold is supplied in C.

**1. Require a no-learning equilibrium:** Nobody learning is sustainable when C≥500, the largest other-language group size. **2. Check the alternative A-language coordination:** If B and C both learn A, a B-speaker gains links with 500 native A-speakers and 200 C-speakers, benefit 700. A C-speaker gains links with 500 A-speakers and 300 B-speakers, benefit 800. Their learning incentives require C≤700 and C≤800, respectively. **3. Check native A-speakers and intersect:** They already communicate with all groups and gain nothing by buying another language. Thus the alternative equilibrium needs C≤700. Combining with no-learning’s C≥500 gives \(500\le C\le700\), including indifference endpoints. **Answer: A — Selected: 500 ≤ C ≤ 700.** A: **Correct:** The intersection of the two equilibrium conditions is [500,700]. B: **Incorrect:** Below 500 the no-learning outcome has profitable deviations. C: **Incorrect:** Above 700 B-speakers would drop A, so the upper endpoint 800 is too high. D: **Incorrect:** This excludes valid costs below 700 and includes costs where B-speakers stop learning.

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 1 pays 10+6=16. **Answer: 16.**

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 2 pays 10. **Answer: 10.**

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 3 pays 10+6+12=28. **Answer: 28.**

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 4 pays 10+6+7=23. **Answer: 23.**

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 5 pays 10+8+5=23. **Answer: 23.**

**1. Use the tree cost-sharing rule:** An edge is needed whenever at least one downstream customer connects. Its Shapley cost is divided equally among the customers whose unique station path uses it; each is equally likely to be the first to require that edge in a random ordering. **2. Allocate shared edges:** The station–2 edge costs 60 and serves all six customers, so everyone pays 10. Edge 2–1 costs 18 and serves customers 1,3,4, giving each 6. Edge 2–6 costs 16 and serves 5,6, giving each 8. **3. Allocate leaf edges:** Edge 1–3 costs 12 and is paid by 3; edge 1–4 costs 7 is paid by 4; edge 6–5 costs 5 is paid by 5. Thus the full shares are (16,10,28,23,23,18). **4. Check budget balance:** Shares sum to 118, exactly \(60+18+16+12+7+5\), the total tree cost. **Requested customer:** Player 6 pays 10+8=18. **Answer: 18.**