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0% Present Value (Million dollar) NPV (Million dollar) 0 900 -900 1 -900.0 162.1 1 1 900 300 300 2 600 1.28 384 0 216 0.917431 198.2 2 1 900 600 600 2 1200 1.28 768 0 432 0.84168 363.6 3 1 900 900 900 2 1800 1.28 1152 0 0 648 0.772183 500.4 Don’t worry, we’ll go through this just now… ESD.70J Engineering Economy Module...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
dollar) Variable Cost (Million dollar) Investment (Million dollar) Salvage (Million dollar) Net value (Million dollar) Discount Factor @ 8.0% Present Value (Million dollar) NPV (Million dollar) 0 300 -300 1 -300.0 156.5 1 1 300 300 300 2 600 1.5 450 300 2 2 600 600 600 2 1200 1.5 900 300 -150 0.925926 -138.9 0 0.857339...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
Tables (1- and 2-way) – What Excel was presumably invented for… • Charts • Goal seek ESD.70J Engineering Economy Module - Session 1 27 Data Tables • Use Data Tables to see how different input values affect the output • Data Tables provide a shortcut for calculating, viewing and comparing multiple versions in one c...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
Cont) • Step 3: create new output values Select the range of cells containing the formulas and values (no labels!) – – Go to “Data” menu, select “Table” – – Data menu ⇒ ‘What-If Analysis’ in Excel 2007 • Reference “Column input cell” to the input variable whose value Excel varies as it iterates through the Data Tabl...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
1 36 Two-way Data Tables • Step 1: Create one column and one row varying input values for each of the inputs – Plan A variable cost varies from $1,200 to $1,450; for Plan B from $1,400 to $1,600 – Incremental step $100 Step 1 ESD.70J Engineering Economy Module - Session 1 37 Two-way Data Tables • Step 2: Enter the ...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
a try! Check with your neighbors… Check the solution sheet… Ask me questions… ESD.70J Engineering Economy Module - Session 1 41 Excel charts • With the sensitivity analysis, we’ve generated a lot data for ‘what-if’ scenarios • Would it be great to generate a visual aid to summarize this information? ESD.70J Engineer...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
Goal Seek • Vary (find) input value until the output equals a desired target value • Click “Goal Seek” under “Tools” menu on Mac, Data ⇒ ‘What-If Analysis’ on PC – “Set Cell”: cell whose value is changed to target value – “To Value”: desired target value – “By Changing Cell”: precedent cell value affecting desired ...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
than there are zeros in the format ESD.70J Engineering Economy Module - Session 1 52 Questions? Comments? Suggestions? ESD.70J Engineering Economy Module - Session 1 53 Summary • Excel is a powerful tool for decision-making • We’ve just scratched the surface • Good habits will make your life easier – Separate input ...	https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf
The heat and wave equations in 2D and 3D 18.303 Linear Partial Diﬀerential Equations Matthew J. Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath 2.3 – 2.5 § [Nov 2, 2006] Consider an arbitrary 3D subregion V of R3 (V deﬁned at all points x = (x, y, z) ﬁnd an equation governing u. The heat ener...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
�owing across the boundaries, we sum φ nˆ over the entire closed surface S, denoted by a double integral dS. Therefore, the conservation of energy principle becomes S · \| · · \| � � d dt V � � � cρu dV = nˆ dS + φ · − S � � V � � � Q dV (1) 1 1.1 Divergence Theorem (a.k.a. Gauss’s Theorem) ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
�z2 . Applying the Divergence theorem to (1) gives d dt V � � � cρu dV = − V ∇ · � � � φ dV + Q dV V � � � Since V is independent of time, the integrals can be combined as cρ V � � � � ∂u ∂t + φ − ∇ · Q dV = 0 � Since V is an arbitrary subregion of R3 and the integrand is assumed continuous, the in...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
2D and 3D Wave equation The 1D wave equation can be generalized to a 2D or 3D wave equation, in scaled coordinates, This models vibrations on a 2D membrane, reﬂection and refraction of electromagnetic (light) and acoustic (sound) waves in air, ﬂuid, or other medium. utt = 2 u ∇ (6) 3 Separation of variables i...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
) The 3D Heat Equation implies ′T T 2X = ∇ = X λ = const − (10) where λ = const since the l.h.s. depends solely on t and the middle X ′′/X depends solely on x. The 3D wave equation becomes On the boundaries, ′′ T T = ∇ 2X X = λ = const − X (x) = 0, ∂D x ∈ The Sturm-Liouville Problem for X (x) is ∇ ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
t) = Xn (x) Tn (t). �� t + βn sin � λn t 4 (13) (14) (15) (16) 5 Uniqueness of the 3D Heat Problem Ref: Guenther & Lee 10.3 § We now prove that the solution of the 3D Heat Problem u (x, 0) = f (x) , x D ∈ is unique. Let u1, u2 be two solutions. Deﬁne v = u1 − D 2 v, vt = x u2. Then v satisﬁes ut...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
∂D � � But on ∂D, v = 0, so that the ﬁrst integral on the r.h.s. vanishes. Thus Also, at t = 0, dV dt (t) = 2 − dV v 2 \| 0 ≤ D \|∇ � � � V (0) = v 2 (x, 0) dV = 0 (17) (18) D � � � ≤ ≥ 0 and dV /dt Thus V (0) = 0, V (t) 0, i.e. V (t) is a non-negative, non-increasing function that starts at zero. Thus V...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
3D Wave Equation lead to the Sturm-Liouville problem ∇ 2X + λX = 0, X (x) = 0, D, ∂D. x x ∈ ∈ (19) 6.1 Green’s Formula and the Solvability Condition Ref: Guenther & Lee 8.3, Myint-U & Debnath 10.10 (Exercise 1) For the Type I BCs assumed here (u (x, t) = 0, for x ∂D), we now show that all eigenvalues ar...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
Result (22) applied to G = v2 and F = ∇ (v2∇ v1) · nˆdS = S � � V � � � 2 v1 + v2∇ v1 · ∇ v2 ∇ dV � � Subtracting (23) and (24) gives Green’s Formula (also known as Green’s second iden tity): (v1∇ v2 − v2∇ v1) nˆdS = · S � � V � � � � This is also known as a Solvability Condition, since the values of v1 an...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
v deﬁned on a volume V with closed smooth surface S. � � We now apply result (26) to a solution X (x) of the Sturm-Liouville problem (19). Letting v = X (x), S = ∂D and V = D, Eq. (26) becomes X X ∇ · nˆdS = 2XdV + X ∇ 2 dV X \| D \|∇ � � � � � � D ∂D � � Since X (x) = 0 for x ∂D, ∈ Also, from the PDE in (...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
integrating and imposing the BC X (x) = 0 X is nonzero ∂D gives X = 0 for all x ∇ dV > 0. Thus, If for x at some points in D, and hence by continuity of from (31), λ > 0. D , i.e. the trivial solution. Thus 2 X, D \|∇ ∇ X ∈ \| � � � 7 � 6.3 Orthogonality of eigen-solutions to ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
are orthogonal. � � � D 7 Heat and Wave problems on a 2D rectangle, ho mogeneous BCs Ref: Guenther & Lee [Nov 7, 2006] § 10.2, Myint-U & Debnath 9.4 – 9.6 § 7.1 Sturm-Liouville Problem on a 2D rectangle We now consider the special case where the subregion D is a rectangle D = (x, y) : 0 { ≤ x ≤ x0, 0 y ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
r.h.s. only depends on x, both sides must equal a constant, say µ, Y ′′ Y X ′′ − X + λ = = µ (36) The BCs (34) and (35) imply X (0) Y (y) = X (x0) Y (y) = 0, X (x) Y (0) = X (x) Y (y0) = 0, y x 0 0 ≤ ≤ ≤ ≤ y0, x0. To have a non-trivial solution, Y (y) must be nonzero for some y must be nonzero for so...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
Y + νY = 0, ≤ Y (0) = Y (y0) = 0 0 y0, y ≤ (40) µ. The solutions are the same as those for (38), with ν replacing µ: where ν = λ − Yn (y) = bn sin , νn = nπy y0 � � nπ y0 � � 2 , n = 1, 2, 3, ... (41) The eigen-solution of the 2D Sturm-Liouville problem (33) – (35) is vmn (x, y) = Xm (x) Yn (y) = cmn s...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
to the T (t) problem (13) and the Sturm-Liouville problem (33) – (35), respectively, to obtain (x, y) : 0 { x0, 0 y0} ≤ ≤ ≤ ≤ x y umn (x, y, t) = Amn sin = Amn sin mπx x0 mπx x0 � � � � sin sin nπy y0 nπy y0 � � � � e −λmnt −π2 e „ 2 2 m + n 2 2 y « x 0 0 t To satisfy the initial condition, we su...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
– (35). Multiplying both sides by v ˆ � ( ˆm, ˆn = 1, 2, 3, ...) and integrating over the rectangle D gives are the eigenfunctions of the 2D Sturm (x, y) mπx x0 sin mnˆ � � � f (x, y) v ˆ mnˆ (x, y) dA = Amn vmn (x, y) v ˆ mnˆ (x, y) dA (44) ∞∞ D � � m=1 n=1 � � D � � where dA = dxdy. Note that x0 vmn (x,...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
x, y) v ˆ y0 mnˆ D x0 � � f (x, y) sin 0 � 0 � (x, y) dA mπx x0 sin � � nπy y0 � � dydx (45) 7.3 Solution to wave equation on 2D rectangle, homogeneous BCs The solution to the wave equation on the 2D rectangle follows similarly. The general 3D wave problem (8) becomes (x, y) D, ∈ t > 0, utt = ∂2u ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
πt � � × � m2 2 + x0 n2 2 + βnm sin πt y0 � � � m2 2 + x0 n2 2 y0 �� To satisfy the initial condition, we sum over all m, n to obtain the solution, in general form, ∞ ∞ u (x, y, t) = umn (x, y, t) (46) Setting t = 0 and imposing the initial conditions m=1 n=1 � � u (x, y, 0) = f (x, y) , ut (x, y, 0) =...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, (33) – (35). As above, multiplying both sides by v ˆ mnˆ � (x, y) ( ˆm, ˆn = 1, 2, 3, ...) and integrating over the rectangle D gives � � � αmn = βmn = 4 x0y0 � � 4 x0y0√λmn D f (x, y) vmn (x, y) dxdy g (x, y) vmn (x, y) dxdy D � � 8 Heat and Wave equations on a 2D circle, homo geneous BCs Ref: Guenther...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, t) = u (x, y, t) for 0 r ≤ ≤ 1, π − ≤ θ < π. 12 You can verify that The PDE becomes 2 v = ∇ ∂2v ∂x2 + ∂2v ∂y2 = 1 ∂ r ∂r r ∂w ∂r � � + 1 ∂2w r ∂θ2 2 1 ∂ r ∂r r ∂w ∂r � � + 1 ∂2w r2 ∂θ2 + λw = 0, 0 r ≤ ≤ 1, π − ≤ θ < π The BC (48) requires ≤ We use separation of vari...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
) which, in order to obtain non-trivial solutions (H (θ) = 0 for some θ), implies w (1, θ) = R (1) H (θ) = 0 R (1) = 0 (53) In the original (x, y) coordinates, it is assumed that v (x, y) is smooth (i.e. contin uously diﬀerentiable) over the circle. When we change to polar coordinates, we need to introduce an ex...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
m2 , Hm (θ) = am cos (mθ) + bm sin (mθ) Thus, in general, we may assume λ = m2, for m = 0, 1, 2, 3, ... The equation for R (r) in (52) becomes r d dr r � Rearranging gives 1 dR dr R (r) � + λr2 = µ = m 2 , m = 0, 1, 2, 3, ... 2 r d2Rm dr2 + r dRm dr + λr2 − 2 m Rm = 0; Rm (1) = 0, Rm (0) \| \| < ∞...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
the function Ym (s) is unbounded at s = 0. The general solution to the ODE is ¯Rm (s) = cm1Jm (s) + cm2Ym (s) where cmn are constants of integration. Our bound edness criterion at s = 0 implies c2m = 0. Thus R¯m (0) < � � � � ∞ ¯Rm (s) = cmJm (s) , � � � � Rm (r) = cmJm √λr . [Nov 9, 2006] The Bessel Function ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
� Note that the sum of numbers ( − 1)k s2k+m k! (k + m)!22k+m L2k+m k! (k + m)!22k+m ≤ � � � � � L2k+m k! (k + m)!22k+m k=0 � converges by the Ratio Test, since the ratio of successive terms in the sum is L2(k+1)+m (k+1)!(k+1+m)!22(k+1)+m L2k+m k!(k+m)!22k+m = L2 L2 (k + 1) (k + 1 + m) 4 ≤ (k + 1) 4 2 ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
Each Bessel function Jm (s) has an inﬁnite number of zeros (roots) for s > 0. Let jm,n be the n’th zero of the function Jm (s). Note that 1 j0,1 = 2.4048 j1,1 = 3.852 2 j0,2 = 5.5001 j1,2 = 7.016 3 j0,3 = 8.6537 j1,3 = 10.173 J0 (s) J1 (s) The second BC requires Rm (1) = Rm ¯ √λ = Jm √λ = 0 � This has an...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, y, t) = 0, ∂2u ∂y2 , (x, y) u (x, y, 0) = f (x, y) , (x, y) D, ∈ t > 0, ∂D, ∈ (x, y) D, ∈ where D is the circle D = . We reverse the separation of vari 1 } ables (9) and substitute solutions (14) and (42) to the T (t) problem (13) and the Sturm Liouville problem (47) – (48), respectively, to obtain (x, y) ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, 0) = vmn (x, y) ∞ ∞ m=1 n=1 � � We can use orthogonality relations to ﬁnd αmn and βmn. 9 The Heat Problem on a square with inhomoge neous BC We now consider the case of the heat problem on the 2D unit square D = (x, y) : 0 { ≤ x, y , 1 } ≤ (62) 16 where a hot spot exists on the left side, ut = u (x, y,...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, { u0/ε 0 � y y0\| < ε/2 } \| − otherwise on ∂D The PDE for uE is called Laplace’s equation. Laplace’s equation is an example of an elliptic PDE. The wave equation is an example of a hyperbolic PDE. The heat equation is a parabolic PDE. These are the three types of second order (i.e. involving double derivatives) ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
= c1e nπx + c2e −nπx An equivalent and more convenient way to write this is X (x) = c3 sinh nπ (1 x) + c4 cosh nπ (1 x) − − Imposing the BC at x = 1 gives X (1) = c4 = 0 and hence Thus the equilibrium solution to this point is X (x) = c3 sinh nπ (1 x) − ∞ uE (x, y) = An sinh (nπ (1 n=1 � x)) sin (nπy) − You ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
π) y0+ε/2 y0−ε/2 � sin (mπy) dy y0+ε/2 y0−ε/2 � 2u0 cos (mπy) = = = ε sinh (mπ) − mπ � 2u0 εmπ sinh (mπ) 4u0 sin (mπy0) sin εmπ sinh (mπ) � mπε 2 � 18 (cos (mπ (y0 − ε/2)) − cos (mπ (y0 + ε/2))) Thus ∞ sin (nπy0) sin nπε 2 uE (x, y) = 4u0 επ n sinh (nπ) � To solve the transient proble...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
x) ≈ enπ(1−x) enπ = e −nπx Thus the terms decrease in magnitude (x > 0) and hence uE (x, y) can be approxi mated the ﬁrst term in the series, uE (x, y) 4u0 sin (πy0) sin sinh (π) � ≈ επ πε 2 � sinh (π (1 − x)) sin (πy) A plot of sinh (π (1 the square is approximately − x)) sin (πy) is given below. The temp...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
u = u0 on one side, and u = 0 on the other 3 sides. Let α = uE . Rotating the plate by 90o will not alter uE , since this is the center of the plate. � Let uEsum be the sums of the solutions corresponding to the BC u = u0 on each of the � four diﬀerent sides. Then by linearity, uEsum = u0 on all sides and hence uE...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
when the hot spot is placed in the center of the side, i.e. y0 = 1/2. 9.5 Solution to inhomogeneous heat problem on square We now use the standard trick and solve the inhomogeneous Heat Problem (63), ut = u (x, y, t) = 2 u, u0/ε, 0, ∇ � u (x, y, 0) = f (x, y) D (x, y) ∈ x = 0, { y y0\| < ε/2 } \| − otherwise...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
sin (mπx) sin (nπy) e −π2(m +n2)t 2 where m=1 n=1 � � 1 1 Amn = 4 (f (x, y) 0 � Thus we have found the full solution u (x, y, t). � 0 − uE (x, y)) sin (mπx) sin (nπy) dydx 10 Heat problem on a circle with inhomogeneous BC Consider the heat problem ut = 2 u, ∇ (x, y) D ∈ (x, y) ∂D ∈ (65) u (x, y, t) =...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
for The problem for uE becomes 0 r ≤ ≤ 1, π − ≤ θ < π. 1 ∂ r ∂r r ∂wE ∂r + 1 ∂2wE r2 ∂θ2 = 0, � w (r, � π) = w (r, π) , − 0 r ≤ ≤ 1, π − ≤ θ < π wθ (r, π) = wθ (r, π) , − w (0, θ) \| \| < w (1, θ) = gˆ (θ) , ∞ π − ≤ θ < π where ˆg (θ) = g (x, y) for (x, y) We separate variables ∈ ∂D and θ = arct...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
π) , 22 with eigen-solutions Hm (θ) = am cos (mθ) + bm sin (mθ) , m = 0, 1, 2, ... The problem for R (r) is 0 = r d dr dR dr − � r � 2R m = r 2 d2R dr2 + r dR dr − m 2R Try R (r) = rα to obtain the auxiliary equation α (α − 1) + α − m 2 = 0 whose solutions are α = For m > 0, r−m blows up as...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
(mθ) + Bm sin (mθ)) = gˆ (θ) (71) The orthogonality relations are m=0 � π cos (mθ) sin (nθ) dθ = 0 −π � cos (mθ) cos (nθ) sin (mθ) sin (nθ) π −π � � dθ = � � π, m = n 0, m = n , (m > 0) . Multiplying (71) by sin nθ or cos nθ and applying these orthogonality relations gives π gˆ (θ) cos (mθ) dθ (72) g...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
uE = u0 + 2π ∞ r m m=1 � � u0 sin (mθ0) mπ (θ0 + π) cos (mθ) − 10.1.2 Interpretation u0 (cos (mθ0) ( − mπ (θ0 + π) − 1)m) sin (mθ) � The convergence of the inﬁnite series is rapid if r required for accuracy. ≪ 1. If r ≈ 1, many terms are The center temperature (r = 0) at equilibrium (steady-state) is uE (...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
. We call these lines the “heat ﬂow lines” or the “orthogonal trajectories”, and draw these as dashed lines in the ﬁgure below. ∇ ∇ ∇ Note that lines of symmetry correspond to (heat) ﬂow lines. To see this, let nl u nl. be the normal to a line of symmetry. Then the ﬂux at a point on the line is · Rotate the image...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
2 ∞ r m=1 � 2n−1 sin ((2n 2n 1) θ) − 1 − Use the BCs for the boundary. Note that the solution is symmetric with respect to the y-axis (i.e. even in x). The solution is discontinuous at , or r = 1, θ = 0, π { π/2. The sum for uE is messy, so we use intuition. We start with the boundary conditions and use cont...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
ﬂux lines must leave the hot spot and extend to various θ = points on the boundary. − 10.2 Solution to inhomogeneous heat problem on circle We now use the standard trick and solve the inhomogeneous Heat Problem (65), ut = 2 u, ∇ u (x, y, t) = g (x, y) , u (x, y, 0) = f (x, y) (x, y) D ∈ (x, y) ∂D ∈ (73) us...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
t). − uE (x, y). 26 11 Mean Value Property for Laplace’s Eq. Ref: Guenther & Lee 8.4, Myint-U & Debnath 8.4 § § Theorem [Mean Value Property] Suppose v (x, y) satisﬁes Laplace’s equation in a 2D domain D, 2 v = 0, ∇ (x, y) D. ∈ (74) Then at any point (x0, y0) in D, v equals the mean value of the tempera...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
−π � gˆ (θ) dθ (76) On the boundary of the circle (of radius 1), (x, y) = (cos θ, sin θ) and the BC implies that uE = gˆ (θ) on that boundary. Thus, ˆg (θ) = uE (cos θ, sin θ) and (76) becomes uE (0, 0) = A0 = We now consider the region π 1 2π −π � uE (cos θ, sin θ) dθ. (77) B(x0,y0) (R) = (x, y) : (x 2 x0...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
�) (ˆx, yˆ) : ˆx 2 + ˆy 2 1 ≤ � ∈ � 27 ∇ where ˆ 2 = (∂2/∂xˆ2, ∂2/∂yˆ2). The solution is given by Eqs. (70) and (72), and we found the center value above in Eq. (77). Reversing the change of variable (79) in Eq. (77) gives v (x0, y0) = π 1 2π −π � v (x0 + R cos θ, y0 + R sin θ) dθ as required. � For th...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
any disc B(x0,y0)(R) of radius R > 0 centered at (x0, y0), v (x0, y0) = v (x0 + R cos θ, y0 + R sin θ) dθ = Average of v on boundary of circle π 1 2π −π � But the average is always between the minimum and maximum. Thus v (x0, y0) must be between the minimum and maximum value of v (x, y) on the boundary of the ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
∂D, ∇ ∇ 0 = ∂D � � v v ∇ · nˆdS = � � � D Thus, by continuity, D. v \| \|∇ = 0 and ∇ 2 v + v ∇ v 2 \| \|∇ dV = v 2 dV \| D \|∇ � � � � � v = 0 everywhere in D and hence v = const in 13 Eigenvalues on diﬀerent domains In this section we revisit the Sturm-Liouville problem ∇ 2 v + λv = 0, v = 0, D ∂D x ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
��ned on a volume V with closed smooth surface S. v v ∇ · nˆdS = ∂D � � � � � D � 2 v + v ∇ v ∇ · ∇ v dV � In the statement of the theorem, we assumed that v = 0 on ∂D, and hence D ∇ � � � vdV = v · ∇ − � � � D 2vdV v ∇ (81) φn} Let { which satisfy be an orthonormal basis of eigenfunctions on D, i....	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
n m � � � � � D 2vdV = v ∇ − � � � D ∞∞ m=1 n=1 � � Substituting (84) into (81) gives λnAnAm φnφmdV = � � � D D φn 2 dV = 1) (83) ∞ � � � A2 n =1 n � ∞ − n=1 � λn A2 n (84) D ∇ � � � vdV = v · ∇ − � � � D 2vdV = v ∇ 2 λnAn ∞ n=1 � (85) Substituting (83) and (85) into (80) gives R (v) = � � � D ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, so that R (v) > λ1. � 30 � � Theorem If two domains D and D in R2 satisfy D D � i.e., D D but D = D, ⊂ then the smallest eigenvalues of the Sturm-Liouville problems on D and D λˆ1, respectively, satisfy � � � , λ1 and λ 1 < λ1 � In other words, the domain D smaller eigenvalue....	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
is continuous, since v1 is zero on the boundary of D. Applying the and function ˆv1 (which satisﬁes all the requirements previous theorem to the region D of the theorem) gives � Equality happens only if ˆv1 is the eigenfunction corresponding to λˆ1. � λˆ1 ≤ R (ˆv1) . smallest eigenvalue λˆ1 on D Useful fact [state...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
+ 1 2 y0 � � 2 The smallest eigenvalue on the circle of radius 1 is λ01 = J0 ,1 where the ﬁrst zero of the Bessel function J0 (s) of the ﬁrst kind is J0,1 = 2.4048. Since √λ multiplied r in the Bessel function, then for a circle of radius R, we’d rescale by the change of rˆ where ˆr goes from 0 to 1. Thus variab...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
π − π π � − π � R � 0 = 2 rdrdθ dv dr v2rdrdθ � = 6 R2 > λ01 You can show that and R (v) = � � � � This conﬁrms the ﬁrst theorem, since v (r) is smooth on D2, v (R) = 0 (zero on the � boundary of D2), and v is nonzero in the interior. � 32 13.1 Faber-Kahn inequality Thinking about th...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
1CIRC, as the Faber-Kahn inequality states. 14 Nodal lines Consider the Sturm-Liouville problem ∇ 2 v + λv = 0, v = 0, D ∂D x x ∈ ∈ (89) Nodal lines are the curves where the eigenfunctions of the Sturm-Liouville problem are zero. For the solution to the vibrating membrane problem, the normal modes unm (x,...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
boundary ∂D. Since λmn = λnm, then the function fnm = Avmn + Bvnm is also an ≥ 33 eigenfunction with eigenvalue λmn, for any constants A, B. The nodal lines for fnm can be quite interesting. Examples: we draw the nodal lines on the interior and also the lines around the boundary, where vnm = 0. (i) m = ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
Since λ31 = λ13 = 10π2/a2, this is a solution to SL problem (89) on D with λ = λ13. To ﬁnd the nodal lines, we use the identity sin 3θ = (sin θ) 4 sin2 θ to write − 3 − � � v13 = sin v31 = sin v13 − v31 = 4 sin sin sin sin � � � � πy a πy a πy a πx a πx a πx a � � � 3 4 sin2 − πy a � � 3 � � − sin2 ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
the square plate: y = x + ka y = x, y = x + a − v31 are the sides and diagonals of the square. Thus the nodal lines of v13 − Let DT be the isosceles right triangle whose hypotenuse lies on the bottom hor v31 is zero on the boundary ∂DT , izontal side of the square. The function v13 − positive on the interior o...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
λ13 of the Sturm- Liouville problem (89) on Dc. πy a sin2 3 2 − = � � � � . (vii) Find the ﬁrst eigenvalue on the right triangle DT 2 = 0 y ≤ ≤ √3x, x 0 ≤ ≤ � 1 . � Note that separation of variables is ugly, because you’d have to impose the BC X (x) Y √3x = 0 � � We proceed by placing the triangle i...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
n2 and look for repeated values: � � n, m 1 2 3 4 5 1 4 7 12 19 28 2 13 16 21 28 37 4 3 28 31 36 43 The smallest repeated value of 3m2 + n2 is 28: λ31 = λ24 = λ15 = 28π2/3 The linear combination of the corresponding eigenfunctions is itself an eigenfunction of the SL problem (89) on the rect...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
another method of ﬁnding the smallest eigenvalue. The Rayleigh quotient can help us identify an upper bound. 14.2 Nodal lines for the disc (circle) For the disc of radius 1, we found the eigenfunctions and eigenvalues to be vmnS = Jm (rjm,n) sin mθ, vmnC = Jm (rjm,n) cos mθ 36 with λmn = π2j2 m,n , n, m = 1...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
, z, t). We assume the axis of the cylinder is on the z-axis and (r, θ, z) are cylindrical coordinates. Initially, the temperature is u (r, θ, z, 0). The ends are kept at a temperature of u = 0 and sides kept at u (a, θ, z, t) = g (θ, z). The steady-state temperature uE (r, θ, z) in the 3D cylinder is given by 2 ...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
z) d2Z (z) dz2 = 0. Rearranging gives 1 d rR (r) dr dR (r) dr r � � + 1 r2H(θ) d2H(θ) dθ2 = 1 −Z (z) d2Z (z) dz2 = λ (92) where λ is constant since the l.h.s. depends only on r, θ while the middle depends only on z. The function Z (z) satisﬁes The BCs at z = 0, L imply d2Z dz2 + λZ = 0 0 = u (r...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
θ. We assume g (θ, z) is smooth and 2π-periodic in θ. Hence H ′ ( π) = H ′ (π), − H ′ ( π) = H ′ (π) − We solved this problem above and found that µ = m2 for m = 0, 1, 2, ..., H0(θ) = const and Hm(θ) = Am cos mθ + Bm sin mθ, m = 1, 2, 3, ... Multiplying (93) by R (r) gives r d dr � r dR (r) dr − � λnr 2 + m...	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
(r, θ, z) = Im nπr L sin nπz L [Amn cos (mθ) + Bm sin (mθ)] m=0 n=1 � � � In theory, we can now impose the condition u (a, θ, z) = g (θ, z) and ﬁnd Amn, Bmn using orthogonality of sin, cos. � � � 39	https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf
(R\|Ψ) = cos Ψ + sin Ψ = 2 1/2 cos(Ψ - π/4) 2. Identify Sample Space 0 π/2 ψ 3. Probability Law over Sample Space: Invoke isotropy implying uniformity of angle fΨ(ψ) 2/ π 0 π/2 ψ 4. Find CDF (cid:139) FR(r) = P{R < r} = P{21/2 cos(Ψ-π/4) < r} (cid:139) FR(r) = P{R < r} = P{cos(Ψ-π/4) < r/ 21/2 } g(Ψ) 1 r/21/2 1/21/2...	https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf
139) σR/E[R] = 0.098, implies very robust A Quantization Problem NYC Marine Transfer Station Tug Delivers LIGHTS Tug Picks Up HEAVIES LIGHT and HEAVY Barges Stored Loading Loading Barge Barge Loading Loading Barge Barge Barges Shifted By Hand Or Tug Refuse Inflow λi (t) Fresh Kills Landfill Digger HEAVY BARGES UNLOA...	https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf
Θare independent. b) Θis uniformly distributed over [0, 1] fD,Θ(d, θ) = fD(d) fΘ(θ) = fD(d)(1) = fD(d), d > 0, 0 <θ <1 x θ 1 K = 0 K = 1 K = 2 1-x 0 1 x 1-x 2 K = 3 3 d 4. Working in the Joint Sample Space Look at E [K \|D = d ] Let d = i + x 0 < x <1 E [K \|D = i + x ] = i (1 - x) + (i + 1) x = i + x = d Implies E [ K ...	https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf
18.445 Introduction to Stochastic Processes Lecture 17: Martingle: a.s convergence and Lp-convergence Hao Wu MIT 15 April 2015 Hao Wu (MIT) 18.445 15 April 2015 1 / 10 Recall Martingale : E[Xn \| Fm] = Xm for n ≥ m. Optional Stopping Theorem : E[XT ] = E[X0] ? Today’s goal a.s.martingale convergence Doob’s maximal ineq...	https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf
limit. Hao Wu (MIT) 18.445 15 April 2015 4 / 10 Examples Example 1 Let (ξj )j≥1 be independent random variables with mean zero such that ∞ E[\|ξj \|] . Set < ∞ (cid:80) j= 1 X0 = 0, Xn = ξj . n (cid:88) j=1 1 (Xn)n≥0 is a martingale bounded in L . Xn converges a.s. to X∞ = ∞ In fact, Xn also converges to X∞ in L1. j= ξ1...	https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf
n Xk . Then, for all p > 1, we have \|\|X ∗\|\|n p ≤ p p − 1 \|\|X n\|\|p. Recall Hölder inequality : p > 1, q > 1 and 1/p + 1/q = 1, then E[\|XY \|] ≤ E[\|X p 1\| ] /p × E[\|Y q 1\| ] /q. Hao Wu (MIT) 18.445 15 April 2015 7 / 10 Lp Convergence for p > 1 Theorem Let X = (Xn)n≥0 be a martingale and p > 1, then the following statemen...	https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf
= 1, X n n = Πj=1ξj . 1 (Xn)n≥0 is a non-negative martingale. 2 Xn converges a.s. to some limit X∞ ∈ L1. Question : 1 Do we have E[X∞] = 1 ? Answer : Set aj = E[ (cid:112) ξj ] ∈ (0, 1]. 1 2 If Πj aj > 0, then X converges in L1 and E[X∞] = 1. (Next lecture) If Πj aj = 0, then X∞ = 0 a.s. Hao Wu (MIT) 18.445 15 April 20...	https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf
MODULE ConcurrentTransactions [ V, % Value S, % State of database T % Transaction ID ] EXPORT Begin, Do, Commit = CONST s0:S := init() % initial state A = S -> [v,s] E = [a: A, v: V] H = SEQ E Y = T -> H % Action % Event % History % histories of transactions TS = SET T XC = (T, T)-> Bool % eXternal Cons...	https://ocw.mit.edu/courses/6-826-principles-of-computer-systems-spring-2002/5ffe1e8aa354be6f61ceb68a116f8392_20slides.pdf
=ts /\ Consistent(to, xc) /\ Valid(y,to)) FUNC Invariant(com: TS, act: TS, xc, y) -> Bool = Serializable(com, xc, y) APROC Begin() -> T = << VAR t: T \| ~ t IN (active \/ committed) => y(t) := {}; active := active \/ {t}; xc(t,t) := true; DO VAR t’ :IN committed \| ~xc.closure(t’,t) => xc(t’,t):=true OD; >> ...	https://ocw.mit.edu/courses/6-826-principles-of-computer-systems-spring-2002/5ffe1e8aa354be6f61ceb68a116f8392_20slides.pdf
current) ==> Serializable(ts, xc0, y0)) CC: Serializable(current, xc0, y0) EO: (ALL t :IN act \| EXISTS ts \| com <= ts <= current /\ Serializable(ts + {t}, xc0, y0)) OD: (ALL t :IN act \| EXISTS ts \| AtBegin(t) <= ts <= current /\ Serializable(ts + {t}, xc0, y0)) OC1: (ALL t :IN act, h :IN Prefixes(y0(t)) \| EXIST...	https://ocw.mit.edu/courses/6-826-principles-of-computer-systems-spring-2002/5ffe1e8aa354be6f61ceb68a116f8392_20slides.pdf
= h.last.v)) NC: true FUNC Prefixes(h: T) -> SET H = RET {h’ \| h’ M= h /\ h’ # {}} FUNC AtBegin(t: T) -> TS = RET {t’ \| xc.closure(t’,t)} FUNC IsInterleaving(h: H, s: SET H) -> Bool = ... sequence h is interleaving of sequences from the set s ... TYPE Lk Lks A = String = SET Lk = S -> [v: V, s: S] CONST p...	https://ocw.mit.edu/courses/6-826-principles-of-computer-systems-spring-2002/5ffe1e8aa354be6f61ceb68a116f8392_20slides.pdf
New Bedford Steel Coking Coal Supply Problem 1 New Bedford Steel (NBS) is a small steel manufacturing company. Coking coal is a necessary raw material in the production of steel, and NBS procures 1.0 - 1.5 million tons of coking coal per year. It is now time to plan for the 1997 production, and Stephen Coggins, coa...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
Finally, Steve Coggins needs to keep in mind that capacity for bringing in coal by rail is limited to roughly 650 mtons per year, and capacity for bring in coal by truck is limited to 720 mtons per year. Questions: 1. How much coal should Coggins contract for from each supplier? 2. What will be NBS's total cost ...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
.50 80.00 15.066J 3 Summer 2003 Excel Spreadsheet for New Bedford Steel Optimal Solution • The third row gives the optimal ...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
Original Value Final Value $D$8 $E$8 $F$8 $G$8 $H$8 $I$8 $J$8 $K$8 Amount from Ashley Amount from Bedford Amount from Consol Amount from Dunby Amount from Earlam Amount from Florence Amount from Gaston Amount from Hopt 0 0 0 0 0 0 0 0 55 600 0 20 100 0 450 0 Constraint Cell Name Ce...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
100 $H$8<=$H$9 0 $I$8<=$I$9 450 $J$8<=$J$9 0 $K$8<=$K$9 Binding Binding Binding Not Binding Not Binding Not Binding Not Binding Binding Not Binding Not Binding Binding Not Binding Binding Not Binding Binding Not Binding Not Binding Not Binding Not Binding Binding Not Binding 0 0 0 145 125 55 60...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
$C$16 Rail $C$14 Union 1225 0 720 505 125 61.5 3 -1 0 0 1225 0 720 650 0 5 20 20 1E+30 125 25 100 3.33 145 1E+30 15.066J 6 Summer 2003 Sensitivity Report : Notes for NBS Case Discussion Shadow Price for a constraint is how much the objective function for the ...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
your economic intuition to assure that you are making the right interpretation. Also, different solvers may refer to dual prices, rather than shadow prices: they are the same. 15.066J 7 Summer 2003 Now, let's consider the volatility constraint from the NBS case. Suppose that we re- write the constr...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
for 1 ton of volatile matter). We can reinterpret this in terms of the volatility content of coal. Each percentage point of volatility will supply 0.01 tons of volatile matter per ton of coal. Since the cost to supply 1 ton of volatile matter is $300, each percentage point of volatility has a value of $3 per ton of...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
goes down, how much of a penalty should NBS insist on in their contract? 15.066J 9 Summer 2003 New Bedford Steel Coking Coal Supply Problem - Addendum From the optimal solution we can use the shadow prices on the resource and quality constraints to infer an ex-post imputed cost...	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
, F and H are at their lower bound of 0. The difference between their imputed costs and $61.50 (shadow price for the supply constraint) is their reduced costs. 15.066J 10 Summer 2003	https://ocw.mit.edu/courses/15-066j-system-optimization-and-analysis-for-manufacturing-summer-2003/60367b40dc46191eb2ff5a1d44b8b66e_ses1_freund.pdf
Lecture # 22 Date: 4/30/09 8.592J–HST.452J: Statistical Physics in Biology 1 Kinetics of protein–DNA interaction 1.1 Reaction Kinetics 1 The rate of change with time of the concentration of a protein–DNA complex is the sum of two terms. A positive contribution due to complex formation between a previously free speciﬁc ...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
, (2) DNA]) as the characteristic time scale for a free where we have identiﬁed τ = 1/(ka[P repressor to locate the operator sequence. For the measured value of ka, this search time is of order τ 0.1s. \| ∼ 1.2 Debye-Smoluchowski theory In this section we will compute the on–rate ka. The classical theory of the on–rate ...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
and C(b) = 0 (because diﬀusing particles disappear at r = b). Spherical symmetric solution is 2C = 0 ∇ (4) C(r) = C0 1 (cid:20) b r , (cid:21) − (5) where C0 = CR/(1 the radial inward direction is ~J = equals: b/R) − ≈ CR for b ≪ D3∇ − I = J(r)4πr2 = − R. The diﬀusion current density of proteins along D3bC0/r2 ˆer, so ...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
combination of 1D (sliding) and 3D (jumps) diﬀusion to quickly ﬁnd the target site on the DNA (Figure 1). Proteins are able TF Figure 1: Schematics of 1D/3D search for target site on the DNA. Dashed lines represent 3D diﬀusion trajectories and thick lines are 1D sliding footprints. to bind to any site on the DNA and th...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
∞ NR = NRq(1 NR=1 X − 3 q)NR−1 = 1 q = M n (8) The average search time for protein to ﬁnd the target site is: ts = NR(τ1 + τ3) = M n (τ1 + τ3), (9) where τ3 is average time of 3D jump. In reality sliding events don’t take ﬁxed amount of time. Protein detach from the DNA with rate k(ns) d = 1/τ1 and time of each slidin...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
h ts h i ts h i ts ts h h i i = = = = ∞ NR NR=1 X NR NR NR h h h i i i τ3 + τ3 + h (cid:0) R NR [τ1,i + τ3] ! * i=1 X ∞ p NR−1 q(τ1,NR) [1 q(τ1,i)] + − i=1 Y )NR−2 [ h i (NR 1) q h i (1 − q − h NR=1 (cid:26) X NR h (cid:26) + τ3 τ1i [ h i = τ1i − h Mb 2√D1τ1 qτ1i ] + NR qτ1i ih h (cid:27) (cid:1) It is interesting to ...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
binding energy Ens since this interaction is predominantly electrostatic. One would expect that at standard physiological conditions τ1 is close to the optimal value τ3, but it turns out that protein spends more time sliding than jumping τ1 > τ3. Typical measured 10−3s, while we can estimate the typical jumping time sl...	https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/6039e032dce5df28031d2100fe2fe33f_MIT8_592JS11_lec19.pdf
Chapter 11B Design of Seals Design of Face Seals Wear of Face Seals (From Ayala, et al. 1998) Diagram removed for copyright reasons. See Ayala, H.M., Hart, D.P., Yeh, O.C., Boyce, M.C. "Wear of Elastomeric Seals in Abrasive Slurries", Wear, 220, 9-21, 1998. Percentage of lip worn as a function of the number...	https://ocw.mit.edu/courses/2-800-tribology-fall-2004/605d69102d68f79501ee511dae851cf7_ch11b_seal_des.pdf

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