Russia and Qatar made the first serious moves in October 2008 toward forming an OPEC-style cartel for natural gas. The two strat
egies these countries face are to comply with the cartel agreement or to cheat on the cartel agreement. If both countries comply, the economic profit for each will be $140 million. If one country cheats, that country earns $200 million in economic profit and the other country will have an economic loss of $10 million. If all countries cheat, they break even. What is the outcome of this game if it is only played once
1 answer:
Answer:
The outcome of this game is that both countries will cheat
Explanation:
Solution
Recall that
For Russia:
If Qatar cheats or found cheating is = zero cheat
If Qatar co-operates = 140 million (co-operates or comply)
For Qatar:
If Russia cheats or found cheating is = it is considered as a zero cheat
If Russia cooperates = 140 million (co-operates or comply)
So,
The outcome of equilibrium = (0,0)
Therefore the outcome of this game is that both countries will cheat
Note: Kindly find an attached copy of part of the solution to the question given.
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Here is the correct question.
A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.
Answer:
the minimum design factor that will guarantee that the part will not fail. = 1.434
Explanation:
Looking at the uncertainty; loss of strength must be raised to due to the stress that causes the failure (strength) is known within ±15% uncertainty.
Looking at the uncertainty; the maximum allowable load must be reduced to because the load is known with an uncertainty of ±10.
Looking at the uncertainty; the diameter must be raised to because the diameter is known within an uncertainty of ±5.
The decrease in the maximum allowable stress can be estimated as:
where,
stress
P = load
A = cross-sectional area of the cylinder
∴
replacing P' with and d' with
, we have:
Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045
Now, the minimum design factor that will ascertain that the part will not fail can be computed as:
where;
the design factor =
the design factor = 1.434.
Thus, the minimum design factor that will guarantee that the part will not fail. = 1.434