AMS 303 Second Test Solutions

AMS 303 SECOND TEST REVIEW Solutions to Past Tests

2011 AMS 303 Second Test
1. (x sub 3)^4.
2. a (-,oo), b(g-,1), c(a+,1), d(i-,1), e(h-,1), f(j-,1), g(c+,1), h(b+,1), i(f+,1), j(c+,1), k(e+,1 or d+,1), z(k+,1). One new solution b-h, c-g, d-i, e-k, f-j.
3. a) digital sum of initial pile is 111, remove 1 from pile 2, b) remove 4 from pile 3, c) g(0)=0, g(1)=1, g(2)=2, g(3)=0, g(4)=1, g(5)=2, g(6)=3, g(7)=0, sum is 010, remove 2 from pile pile 3.
4. (1/6)[(b+w)^9 + 2(b^3 + w^3)^3 + (b+w)(b^2 + w^2)^4].
5. Not possible, only 4 wins available for 5 games to be played.
6. Initial solution: x11=20, x12=30, x22=20, x32=20, x33=40. Prices u1=$10, u2=$8, u3=$9, v1=v2=v3=$14. Select x13. Increase x13 and x32 by 30, reduce x12 and x33 by 30.

2010 AMS 303 Second Test
1. x sub 10.
2. a (-,oo), [all remaining second labels are 1, f(a+), g(f+), j(f+), d(g-), c(j-), k(d+), i(c+), e(k-), b(i-), h(e+ or b+); new solution with e-h, b-i, c-j,d-k,f-g.
3. a) Digital sum of initial pile is 011; several moves of kernel, e.g., remove 3 from first pile; b) remove 1 from first pile; c) g(0)=0, g(1)=1, g(2)=0, g(3)=1, g(4)=2, g(5)=3, g(6)=2-- dig. sum is 001, remove 1 from first pile.
4. (1/12)[(x1)^7+ 2x1x6+ 2x1(x3)^2 + 4x1(x2)^3 + 3(x1)^3(x2)^2], set xi = (b^i + w^i).
5. Not possible because two leading teams play each other 3 times but each can win only one more game-- must build network model and show flows.
6. a) x11=40, x21=30, x22=20, x23=10, x33=40, b) u1=10, u2=9,u3=9, v1= 14,v2=13,v3=15, c) pick (1,3) or (3,1), d) for (1,3): x11=30, x13=10, x21=40, x22=20, x33=40.

2009 AMS 303 Second Test
1. xsub9.
2. a (-,oo), b (a+,1) c (j-,1), d (h-,1), e (i-,1), f (k-,1), g (c+,1)or(f+,1), h (b+,1), i (b+,1), j (d+1), k (e+,1), z (g+,1), New match: b-h, c-g, d-j, e-i, f-k or b-i, c-j, d-h, e-k, f-g.
3. a) Gr = 2, b) remove 2 from pile 3 or 4, c) already at Gr = 2, d) remove 2 from first pile.
4. (1/4)[(b+w)^8 + (b^2+w^2)^4 + 2(b+w)^2(b^2+w^2)^3]
5. Not possible, T + T can win only one more game but they play each other 3 times.
6a) Initial Northwest corner solution: x11=40, x12=30, x22=10, x32=20, x33=40; b) u1=$10, u2=$9, u3=$11, v1=$15, v2=$14, v3=$15, c) x13 saves $2 a unit; d) x11=40, x12=0, x13=30, x22=10, x32=50, x33=10.