Question
Three farms need exactly 1, 2 and 3 units of water tonight, respectively. Each farm must obtain all its water from one source. The shared pump charges each user $\(1+D\) per unit, where \(D\) is the total number of units demanded by all farms that choose the pump. A private tanker instead charges a fixed $5 per unit, independent of everyone else's choice. Everyone knows the demands and prices; each farm independently chooses pump or tanker, and nobody can revise their choice after seeing others' decisions.
Find every pure-strategy Nash equilibrium (a set of pump users where no single farm can strictly lower its own bill by switching). Then find the assignment minimising the sum of all three farms' bills. Quantify the extra total cost in the best and worst equilibrium. Ties count as no profitable deviation.
Get hint
Hint 1
For a proposed pump-user set \(S\), calculate \(D\) as the total water used by \(S\). A pump user pays \(d_i(1+D)\), while a tanker user pays \(5d_i\).
Check your answer
Solution
Let demands be \((d_1,d_2,d_3)=(1,2,3)\) and S the set of farms using the pump. Their combined pump demand is \(D=\sum_{i\in S}d_i\). A farm i in S pays \(d_i(1+D)\); otherwise it pays \(5d_i\). A unilateral move into the pump changes \(D\) to \(D+d_i\), a point that must be incorporated in the deviation calculation.
Enumerating all eight subsets gives (pump set; \(D\); total bill; Nash?): empty;0;$30;no. {1};1;$27;no, because farm 2 would join and pay $8 rather than $10. {2};2;$26;no, because farm 1 would join and pay $4 rather than $5. {3};3;$27;yes. {1,2};3;$27;yes. {1,3};4;$30;yes. {2,3};5;$35;no, because existing pump users prefer the tanker. {1,2,3};6;$42;no, because all would pay less using a tanker alone.
To verify the equilibria, at \(S=\{3\}\), farm 3 pays \(3(1+3)=\text{\$}12<\text{\$}15\) tanker; farm 1 joining would pay \(1(1+4)=\text{\$}5\), exactly its tanker price; farm 2 joining would pay \(2(1+5)=\text{\$}12>\text{\$}10\). At \(S=\{1,2\}\), the pump price per unit is $4, cheaper for both users than the $5 tanker; farm 3 joining would face $7 per unit, so stays on its tanker. At \(S=\{1,3\}\), the pump unit price is $5, making both users indifferent to leaving; farm 2 joining would face $7 per unit, so stays outside. Thus no unilateral strict improvements occur in exactly those three profiles.
Total cost as a function of \(D\) is \(C(D)=D(1+D)+(6-D)5=D^{2}-4D+30=(D-2)^{2}+26\). Feasible \(D\) values are 0 through 6 because subset sums of 1,2,3 cover every integer in that range. The unique minimum occurs at \(D=2\): only farm 2 uses the pump, with $6 pump cost plus $20 tanker cost = $26. It is not Nash-stable because farm 1 would join. The best equilibrium cost is $27, a $1 gap; the worst is $30, a $4 gap (price of anarchy \(\frac{30}{26}=\frac{15}{13}\)). Unequal, indivisible farm demands mean that setting two continuous-route costs equal is not a valid substitute for checking discrete deviations.
Assumptions
All farms have their stated fixed integer water demands and cannot split a demand across providers.
Pump unit price depends on the final total demand D, including the farm whose deviation is being considered.
Each farm minimises its own total dollar payment; the social objective sums those same payments.
Indifference does not count as a strictly profitable deviation.
Final Answer
Pure equilibria: pump users {farm 3}, {farms 1,2}, or {farms 1,3}; their total costs are $27,$27,$30. Social optimum: only farm 2 uses the pump, at total cost $26. Best equilibrium costs $1 extra; worst costs $4 extra.