The price at which there is neither a surplus nor a shortage.pdf

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The price at which there is neither a surplus nor a shortage is called: the adjustment price: the equal price. the equilibrium price. the fair prike. Question 40 1pts At a price of $40, the quantity demanded is 100 and the quantity supplied is 120 . This means that: there is a surplus of widgets at $40. hence the price of widgets will go down. a price floor is preventing the price of widgets from falling. Hoth A and B are possibly correct. None of the above.What would happen to the price of peanut butter if the price of peanuts went down, the price of jelly increased, more firms decided to produce peanut butter, and health officials announced that eating peanut butter was bad for you? The effect on both price and quantity is ambiguous. Price will rise and the effect on quantity is ambiguous. Quantity will fall and the effect on price is ambiguous. Price will fall and the effect on quantity is ambiguous.Consider the following Solow growth economy: 1. output y=k (where k is capital per person and y is output per person) 2. savings rate =1/2 3. depreciation rate of 50%. What is the steady-state value of k ? 5 1 2.5Consider the following Solow growth economy: 1. output y=k (where k is capital per person and y is output per person) 2. savings rate =1/2 3. depreciation rate of 50%. What is the steady-state value of y ? .5 1 2.5 5.

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The production function of a restaurant includes items such as labor (i.e., cooks, waiters, and a manager), eapital (i.e, ovens, counters, tables, chairs, and a building), and land (real estate). In the shoit nun, the owner of the restaurant will optimize production by employing a variable amount of given a fixed anount of land, labor and raw materials labor; capatal and raw materials land, capital and labor capital, labor and land Labor, capital and land.

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The parent Rubidium-87 has a half-life of 50 billion years. The daughter Strontium-87 increases as expected. Draw a graph to show the decrease in the parent isotope and the increase in the daughter product. Draw the graph on a blank piece of paper, take a picture of it and submit the picture here. Remember all graphs have both axis labeled and the points along the axis are reasonably proportional..

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The Rings of Saturn There are many planetary subsystems within our solar system. Most of the planets have moons orbiting them, but only theuj2 Jovian planets have rings. On no planet are the rings more enchanting to behold than those of Saturn. The sixth planet boasts 7 primary rings, A through G. From the planet outward they are: D,C,B,A,F,G, and E. The extent of the E ring is truly enormous. At 26 Saturn diameters, it is the largest planetary ring in the solar system. With a small telescope, however, it is the astonishing view of the A and B rings and the Cassini division between them that stands out so spectacularly. Figure 1 shows the orientation and relative positioning of the primary ring structures and moons of Saturn. rigure 1 - satum with Rings and Moons (Credit: NASA/IPL) The rings are composed of icy water and dust ranging in size from microscopic (Credit: William K. Hartman/NASA/IPL)Astronomy 1511 Laboratofy Mamal particulates to large meter sized boulders (Figure 2). Some of the ring material clumps together forming elongated aggregate debris. Some of these structures are loosely held together and may disaggregate according to the forces and momentum governing their motion. Data from the Cassini spacecraft suggests an interesting cascade of material from the inner rings onto Saturn. Complex inorganic grains embedded in water were discovered to be freefalling in Saturn's gravitational field onto the upper atmosphere. The material included particulates combined with gases such as: water, silicates, ammonia, methane, and carbon dioxide. Additionally, some of the gas had sufficient electric charge to be captured and accelerated by Saturn's planetary magnetic field causing those particles to spiral along trajectories determined by the field line orientation. Figare 3 - Influence of Saturn's Magnetic Field (Credit: NASA/IPL) Even though other forces generated from electric and magnetic fields have some influence on the ring material, it is gravity that keeps the rings bound in their orbits. The planet has help in this regard with a few moonlets such as Pan orbiting in the Encke Gap between the A and F rings to shepherd the material. The A, B, and C, main rings are less than 100 meters thick in most places and it is estimated that all the rings were developed in only the last few hundred million years. The shattering of a comet or a small moonlet torn apart by Saturn's gravity could have provided the material from which the rings are composed.1. If the average thickness of the rings is 100 meters, what is the ring volume in m3 ? Assume all the rings are combined into one ring structure of height 100m that extends from r= 6.7107m to R=1.4108m above Saturn's equator. The equation to calculate the volume is: VolumeofRing=(R2r2)h 2. If the density of the ring is similar to water at 1000kg/m3, what is the mass of the ring? Density=volumemass 3. Cassini measured the mass of Saturn's rings and it is now estimated to be about 40% of the mass of.

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The residents living in the community of "Joy" has a big problem to deal with this year. Due to a reasonably mild winter in 2021 , the mouse families enjoyed many social parties and had great times. There are now too many mice in this community. The municipal government has offered help to deal with this problem. Each household living in the community of "Joy" will receive a voucherworth $X dollar for each extermination service. When they hire mice extermination services, they can redeem this voucher to reduce the cost of dealing with this problem. (i) Use an appropriate economic analysis to show and explain the distribution of the benefits and costs of this municipal program to address the mice infestation problem in the community of "Joy". (ii) How would the "net impact" of this program change, if you incorporate the spill-over benefits of this program for the residents of "Moody", a community adjacent to the community of "Joy"? Explain your results. (Assume that the communities "Joy" and "Moody" belong to the same municipality)..

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The Protocol We present the protocol sequentially, though if we are concerned with minimizing the number of back-and-forth rounds we can send some of the messages in parallel with each other (and use some additional shortcuts) to achieve a 4 round protocol. - Input: A has input (a0,a1){0,1} 2 and B has input b{0,1}. - Desired Output: A should receive no output while B should receive ab{0 ,1}. 1. BA:B chooses u{0,1}n, draws (k,z)Com(u) and sends z to A. 7 2. AB: A chooses u{0,1}n and sends u to B. 3. AB:A draws (N,e,d)RSAGen() and sends (N,e) to B. 4. BA:B sends (y0,y1)ZN2 to A where y0 and y1 are prepared as follows (recall b is B's input bit): - yb is set to yb=xe for a random xZN; . y1b is set to y1b=uu where u and u are the strings used in rounds 1 and 2( indicates the bit-wise XOR of two strings; we are using that since n=log(N), any n-bit binary string can be converted to an integer modN ). 5. BA : B and A use the zero knowledge proof system with B playing as the prover and A playing as the receiver, where: Statement: The statement which is used as common input to both players is (z,u,y0,y1)L5. BA:B and A use the zero knowledge proof system with B playing as the prover and A playing as the receiver, where: - Statement: The statement which is used as common input to both players is (z,u,y0,y1)L where membership in L holds if there exists (k,u,i) such that Decom(z,k)=u and yi=uu -Witness: B uses (k,u,1b) as his witness, where k is the decommitment string from the first round, u is the committed value in the first round and b{0,1} is B's input bit. If A rejects B's proof then A aborts the protocol; otherwise A continues. 6. AB:A computes (x0,x1)ZN by setting xi=yid for i=0,1 where d is the secret RSA exponent generated in round 3. Additionally, A draws r0,r1{0,1}n and sends (r0,r1,w0,w1) to B where wi=xi,riai{0,1} for i=0,1. - Output: B outputs the bit wbx,rb{0,1}. Intuition. In order to understand the protocol, first imagine that the protocol consists only of rounds 3,4,6 and in round 4 , B chooses yb as stated above, but draws y1bZN. This simpler protocol will satisfy correctness (since B will output ab ), and it will be secure against a corrupt A (since the only information sent by B is (y0,y1) and both yi are simply random elements of ZN ). However, security against B is problematic since B could choose x0,x1ZN and set yi=xie for i=0,1 and then would be able to learn both of A's bits (a0,a1). So the role of the extra rounds is essentially to boost security against B. So to summarize, the simpler scheme is secure as long as B generates (y0,y1) as he is supposed to (i.e., if yb=xe for random xZN and y1bZN ), but fails if B is able to deviate. At a high level, the function of the extra rounds is to ensure that one of the two yi is random in ZN. This works by running a type of "coin flipping" procedure in rounds 1 and 2 , and then using the ZK proof in round 5 to prove that one of the yi is equal to the output of this procedure (B proves th.

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The protocol will use the following building blocks. The Inner Product Prediction Challenge for RSA. Recall that the RSA based cryptographic constructions involve a composite integer NN where N=pq for large primes p,qN. Additionally, a public exponent e is chosen along with N, whereas the secret exponent d satisfies de =1mod(p1)(q1) and is known by only one of the two players. Recall that de=1mod(p1)(q1) ensures xed=xmodN, whereas given (N,e), recovering d is as hard as factoring N. In the protocol below we will write (N,e,d)RSA.Gen() to indicate that (N,e,d) are generated as above. Finally, recall that the RSA assumption assumes that when (N,e,d)RSA .Gen () and xZN is drawn, one cannot output x if given (N,e,y) where y=xe (of course if one is given (N,e,d,y) then one can simply compute x=yd ). In homework 6 1, problem 3 we saw the following "inner product prediction" version of the RSA game: given (N,e,y,r) where (N,e,d)RSA. Gen(),xZN,y=xe and r{0,1}n where n=log(N), output x,r. We showed that any adversary who can win the inner product prediction game for RSA with probability 1/2+ for non-negligible >0 can be used to build (via the techniques of Goldreich-Levin) another adversary who can break the RSA assumption with non-negligible probability. Blum's Commitment Scheme. The protocol below will use Blum's non-interactive commitment scheme as a building block. While Blum's scheme allows Alice to commit to a single bit, Alice can commit to an n-bit string by committing independently to each bit. For an n-bit string u{0,1}n, we write (z,k)Com(u) to indicate that z and k are respectively the collections of commitment and decommitment strings when Blum's scheme is used to commit to each bit of x. Recall binding says that for any z, there is at most one k such that the decommitment procedure succeeds when run with input (z,k). Recall the hiding game for this commitment scheme is played between a challenger C and adversary A as follows. - A sends (u0, u1) to C where u0,u1{0,1}n; - C draws b{0,1} and (z,k)Com(ub) and sends z to A; - A sends b{0,1} to C signaling the end of the game; A wins if b=b. Hiding security demands that no efficient adversary can win this hiding game except with probability 1/2+ for negligible >0.Zero-Knowledge Proofs. Our OT protocol will use a ZK proof system as a key building block for one of the parties to prove to the other that a particular quantity in the protocol has been computed correctly. As a simple example to help understand the role that the ZK proof plays, imagine that early on in the protocol, A sends z to B where (z,k)Com(0). Then while z is supposed to be a commitment to 0 , hiding says that B cannot actually tell what value is hidden inside, so it is possible that an adversarial A might have actually drawn (z,k)Com(1). In this case, it might be useful to have A prove to B using a ZK proof that in fact z is a commitment to 0 . For this to work, the statement is " zL " where membership in L holds whenever ther.

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