Assignment 1

Assignment 1-58
Let V be the set of variables and C the set of constraints.Suppose a soft constraint c applies to variable v and let (day(v), time(v)) be the value assigned to v in a solution S.

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With height of cylinder as two times its diameter, the ratio of solidification time for a sphere to that of cylinder (of the same diameter) will be(a) 36/25(b) 25/36(c) 64/25(d) 25/64Solution: V/A for sphere = 34π(D2 )3 = D 4π(D2 )2 6V/A for cylinder =2((π4π4DD22))× (22πDD)2 = D 5Solidification time for sphere / Solidification time for cylinder = (V/A)2sphere / (V/A)2cylinder = 25/36 (Ans)13.

The cylindrical mould have diameter-to-height ratio as 1.0.

Each soft constraint has a cost function giving the “penalty” for scheduling the meeting at a given time (lower costs are preferred).

The aim is to schedule all the required meetings so that the sum total of all the penalties is minimized, and all the constraints are satisfied.

The transitions in the state space implement domain splitting subject to arc consistency.

A goal state is an assignment of values to all variables that satisfies all the constraints.

The possible constraints are as follows: # binary constraints constraint, hm1i before hm2i constraint, hm1i same-day hm2i constraint, hm1i one-day-between hm2i # 1 full day between m1 and m2 constraint, hm1i one-hour-between hm2i # 1 hour between end m1 and start m2 # hard domain constraints domain, hmi, hdayi, hard domain, hmi, htimei, hard domain, hmi, hdayi htimei-hdayi htimei, hard # day-time range domain, hmi, morning, hard # finishes at or before 12pm domain, hmi, afternoon, hard # starts on or after 12pm domain, hmi, before hdayi, hard domain, hmi, before htimei, hard domain, hmi, before hdayi htimei, hard domain, hmi, after hdayi, hard domain, hmi, after htimei, hard domain, hmi, after hdayi htimei, hard # soft domain constraints domain, hmi, early-week, soft domain, hmi, late-week, soft domain, hmi, early-morning, soft domain, hmi, midday, soft domain, hmi, late-afternoon, soft Each soft constraint has a cost function, defining a “penalty” for only partially satisfying the constraint.

For example, a soft constraint that a meeting be early-week is satisfied if the meeting is on Tuesday, but with a cost (of 1, as defined below). The cost functions are defined as follows: early-week(d, t): the number of days from mon to d (0 if d = mon) late-week(d, t): the number of days from d to fri (0 if d = fri) early-morning(d, t): the number of hours from 9am to t midday(d, t): the number of hours from 12pm to t late-afternoon(d, t): the number of hours from t to 4pm Finally, to define the cost of a solution (that may only partially satisfy the soft constraints), add the costs associated with each soft constraint.

Considering total time of solidification as 2.0 min and mould constant as 1956000 sec/m2, the dimensions of the mould will be (a) D = H = 7.4 cm (b) D = H = 4.7 cm (c) D = H = 6.7 cm (d) D = H = 13.4 cm Solution: D/H =1, Time of solidification = Ts =2 min Cm = 1956000 sec/m2 = (1956000)/ (60 × 1002) =3.26 min/ cm2 V = πD2H /4 = π D3/ 4, though D/H =1 A = (2 ×πD2/4) (πDH) Ts = Cm (V/A)2 = 3.26 × (D2/36) = 2 D = 4.7 = H (Ans) 14.

The ratio of surface energy term to volume energy term for critical radius during solidification process is (a) -3/2 (b) -2/3 (c) -5/6 (d) -7/12 Solution: Surface energy term = 4 π r2 γ Volume energy term = (4/3) π r3 Δg Critical radius = (-2 γ)/Δg = r* Ratio of surface energy term to volume energy term = [4 π (r*)2 γ] / [(4/3) π (r*)3 Δg] = -3/2 (Ans) 15.


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