A 32 – bit wide main memory unit with a capacity of 1 GB is built using 256M X 4-bit DRAM chips

A 32 – bit wide main memory unit with a capacity of 1 GB is built using 256M X 4-bit DRAM chips. The number of rows of memory cells in the DRAM chip is 2^14. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closet integer) of the time available for performing the memory read/write operations in the main memory unit is _______ .

I calculated that the no of DRAM chips needed is 32. Now each DRAM have rows = 2^14
and columns 2^16 also as we can refresh the rows in parallel and since for one memory cell the time is 50 nanoseconds so for 2^16 columns we will need 2^16 * 50 nano sec ?…Is my approach right

or if i consider that 50 nanoseconds is the time for refresh of a complete row then also it would need a total of 50 nanosec to refresh all in parallel

Do multiple sources of counting as one size larger for carrying capacity stack?

I am currently mocking up a Goliath Barbarian, and was wondering if there is a limit to the amount of sources of “You count as if you were one size larger for the purpose of determining your carrying capacity” that stack.

From prior browsing, I’ve found that Goliath’s Powerful Build & Totem Barbarian’s Bear Aspect feature stack in that regard, but could you stack, for example:

  • Powerful Build (innate Goliath feature)
  • 6th level Totem Barbarian Bear Aspect
  • Brawny feat

Essentially, could you double your carrying capacity and lift/pull capacity thrice?

Capacity of broadcast channels

It is given in the book by El Gamal that

The capacity region of the DM-BC depends on the channel conditional pmf $ p(y_1 , y _2 |x)$ only through the conditional marginal pmfs $ p(y_1 |x)$ and $ p(y_ 2 |x)$

The statement made is about the entire capacity region. Every point in the capacity region means that there exists a sequence of $ (2^{nR_1},2^{nR_2},n)$ (considering only private messages) codes with 0 error. I am wondering about the converse.

For a given sequence of codes, would broadcast channels with identical marginals have the same achievability, ie, if it is achievable in the first channel, would it be achievable in the second channel and vice versa?

Does a sprite familiar’s equipment count against its carrying capacity?

Under Lifting and Carrying (PHB 176) it says:

Your carrying capacity is your Strength score multiplied by 15. This is the weight (in pounds) that you can carry… You can push, drag, or lift a weight in pounds up to twice your carrying capacity (or 30 times your Strength score).

and also specifies that a tiny creature can carry half as much

A sprite familiar summoned through a warlock’s Pact of Chain feature has a strength score of 3 and so has a carrying capacity of 22.5 pounds and a push, drag, lift limit of 45 pounds. The sprite stat block also specifies, however, that they wear leather armor and carry a longsword and shortbow.

Supposedly these weigh 10 pounds, 3 pounds and 2 pounds respectively but considering they are smaller than those worn by a medium humanoid it is unlikely that they weigh as much.

Is there any official guidance as to how much the sprite familiar’s equipment counts against its carrying capacity?

Shortest sequence of jobs, with dependencies, subject to capacity constraints

Suppose I have $ n$ courses, some with some prerequisites, and I can take up to $ k$ courses in a semester. I want to compute the least number of semesters needed to complete all courses, while respecting prerequisites.

Equivalently: suppose I have a dag $ G=(V,E)$ , and a positive integer $ k$ . The desired output is a sequence $ S_1,\dots,S_m$ of vertices that minimizes $ m$ , subject to the constraint that $ S_1 \cup \dots \cup S_m = V$ , $ |S_i| \le k$ for all $ k$ , and every edge goes from some set $ S_i$ to $ S_j$ where $ i<j$ (i.e., there is no edge $ v \to w$ where $ v\in S_i$ , $ w\in S_j$ , and $ i \ge j$ ).

Is there a polynomial-time algorithm for this problem?

Approaches I’ve considered:

The obvious greedy strategy is a variant of Kahn’s algorithm is: in each semester, arbitrarily pick $ k$ courses whose prerequisites have all been previously taken, and take those $ k$ courses. Unfortunately, this algorithm is not guaranteed to generate an optimal schedule. For example, in the graph with vertices $ v_1,v_2,v_3,v_4$ and the single edge $ v_1 \to v_2$ , with $ k=2$ this algorithm might generate the schedule $ \{v_3,v_4\},\{v_1\},\{v_2\}$ , which is longer than the optimal schedule $ \{v_1,v_3\},\{v_2,v_4\}$ .

The next natural idea is to modify the above approach by breaking ties in favor of vertices that are part of longer dependency chains. I’m not sure whether this works or not.

Inspired by taking school courses efficiently.

How does the help action affect lifting capacity?

Let’s say there are two PCs, both with 15 Strength. We’re using the Variant: Encumbrance rules and there is a body that they want to try carry between them which weights 250 lb.

How would I calculate their combined lift? Would I just add their strength together? In that case they would be heavily encumbered which would slow their movement by 20 ft.

Capacity of a discrete memoryless channel

For an integer $ I$ , the input-output relationship of a discrete memoryless channel is given by:

$ Y = X + Z$ (mod I, i.e. sum indicates a modular addition)

where $ I ≥ 2$ , and

• X is an integer chosen from the alphabet Ax = {1,…,2I},

• Z is noise which is a uniform Bernoulli random variable. This means that Az = {0,1}, and

Pr{Z = 0} = Pr{Z = 1} = 0.5.

How can we calculate the capacity of this channel?

0-1 knapsack problem with minimum and maximum weight capacity

In classical 0-1 knapsack problem we have maximum allowed value for the weight – weight capacity.

Let’s restrict total knapsack weight by min and max values

$ $ M \leq \sum_{i=1}^{n}{w_i x_i} \leq W $ $

Is there any known algorithm for this problem? Is there anything known which is better than brute force?

I failed to find anything about the given knapsack problem variation.